# Proof in Serre/Fulton's rep. theory of Artin-Wedderburn for $\mathbb C[G]$

I have figured out a proof myself for the following theorem, but in both Serre's "Linear Representation of Finite Groups" and Fulton's "Representation Theory" books, I don't understand their comments on the proof. Here it is :

Let $\rho^i : G \to \mathrm{GL}(W_i)$ for $1 \le i \le h$ be the distinct irreducible representations of $G$. This gives us algebra homomorphisms $\widetilde{\rho}^i : \mathbb C[G] \to \mathrm{End}_{\mathbb C}(W_i)$ and we can take their products to get a map $$\widetilde{\rho} : \mathbb C[G] \to \prod_{i=1}^h \mathrm{End}_{\mathbb C}(W_i), \quad \widetilde{\rho}(s) = (\rho_s^1,\cdots, \rho_s^h).$$ The claim is that this map is an isomorphism (which is essentially a special case of Artin-Wedderburn's theorem).

We show that $\widetilde{\rho}$ is injective. Suppose that $\widetilde{\rho} \left( \sum_{s \in G} a_s s \right) = 0$, so that for $1 \le i \le h$, we have $\sum_{s \in G} a_s \rho_s^i = 0$. It suffices to show that $a_1 = 0$, for then we can define $b_s = a_{st}$ and $$\widetilde{\rho}\left( \sum_{s \in G} b_s s \right) = \widetilde{\rho} \left( \sum_{s \in G} a_{st} s \right) = \widetilde{\rho} \left( \sum_{s \in G} a_s st^{-1} \right) = \widetilde{\rho} \left( \sum_{s \in G} a_s s \right) \widetilde{\rho}(t) = 0,$$ from which it will follow that $a_t = b_1 = 0$.

Now $\sum_{s \in G} a_s \rho_s^i = 0$ implies that $\sum_{s \in G} a_s \chi_i(s) = 0$, and since $\chi_i$ is a class function this means that $$\sum_{s \in G} \left( \sum_{t \in G} a_{tst^{-1}} \right) \chi_i(s) = \sum_{t \in G} \sum_{s \in G} a_s \chi_i(t^{-1}st) = 0.$$ Now $\psi : G \to \mathbb C$ defined by $\psi(s) = \sum_{t \in G} a_{t(s^{-1})t^{-1}}$ is clearly a class function satisfying $(\psi, \chi_i) = 0$ for $1 \le i \le h$, hence since the irreducible characters form a basis of the class function space, $\psi = 0$ ; in particular, $|G| a_1 = \psi(1) = 0$, so $a_1 = 0$.

My problem is that Serre says "$\widetilde{\rho}$ is surjective, since otherwise there is a non-zero linear form on $\prod \mathrm{End}_{\mathbb C}(W_i)$ which vanishes on $\mathrm{im}(\widetilde{\rho})$, a contradiction because this gives a non-trivial relation between the coefficients of the representations $\rho_i$, which is impossible by the orthogonality formulas (c.f. Serre's book, section 2.2)."

Fulton is even less obvious, he only states "$\widetilde{\rho}$ is injective because the action of $G$ is faithful".

My question : If anyone understand any of those two comments, any help would be appreciated. I did give them a big and long thought, it just never pushed completely through.

• If r is not faithful that means there is some nonzero f \in k[G] which acts by zero on all irreducibles, which means it acts by zero on all representations. This is untrue because it doesn't act by zero on k[G] (f * 1 = f) – user148177 Jun 3 '14 at 5:02
• @user148177 : What is $r$? You mean $\widetilde{\rho}$? But I understand your comment I guess. Let me think it through. – Patrick Da Silva Jun 3 '14 at 5:05
• r is rho. Dont know how to do latex in comments – user148177 Jun 3 '14 at 5:06
• @user148177 : Put your latex stuff between cash symbols. – Patrick Da Silva Jun 3 '14 at 5:06
• @user148177 : Okay so you explained to me Fulton's proof, that's wonderful! Feel free to answer your comment, I'll upvote. What about Serre's proof? – Patrick Da Silva Jun 3 '14 at 5:09

As for what Serre wants to say:

First of all, for Serre's argument it is important that you fix a matrix representation for every irreducible representation. I.e. we have a map $$\tilde \rho \colon \mathbb C[G] \to \prod_{k=0}^h M_{n_k} \mathbb C=:M$$ as in Serre (just before Prop 10). Let us denote by $c^k_{i,j}\colon M \to \mathbb C$ the $(i,j)$-th entry of the $k$-th matrix. Note that these functions give a basis of the dual space $M'$.

Now, Serre proves in Section 2.2 (Remark (1)) that the functions $r^k_{i,j}= c^k_{i,j} \circ \tilde \rho \colon G \to \mathbb C$ are linear independent in the vector space $\mathbb C^G$.

Assume $\ell\in M'$ and write $\ell = \sum \lambda^k_{i,j} c^k_{i,j}$. Assuming $\ell \circ \tilde \rho = 0$ we get $$0 = \sum \lambda^k_{i,j} c^k_{i,j} \circ \tilde \rho = \sum \lambda^k_{i,j} r^k_{i,j} \in \mathbb C^G,$$ thus all $\lambda$ are zero and thus $\ell = 0$.

(Note that: $$|G| \cdot \sum_{s\in G} a_s \chi_i(s) = \sum_{t\in G}\sum_{s\in G} a_s \chi_i(s) = \sum_{t\in G}\sum_{s\in G} a_s \chi_i(t^{-1}st) =\sum_{t\in G}\sum_{s\in G} a_{tst^{-1}} \chi_i(t(t^{-1}st)t^{-1}) \qquad= \sum_{s\in G}\sum_{t\in G} a_{tst^{-1}} \chi_i(s)$$ because $s\mapsto tst^{-1}$ is a group automorphism.)
• So if I understood Serre's proof properly, it only works in the complex case? (since they use $r_{ij}^k(t^{-1}) = \overline{r_{ji}^k(t)}$ at some point) Thanks by the way, it took me a few months to actually take the time to dig the details, but that was exactly what I needed. – Patrick Da Silva Oct 8 '14 at 2:06