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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).

Now here is my proof, so this is not found in any book, I am quite traumatized about this :

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.

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  • $\begingroup$ 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) $\endgroup$ – user148177 Jun 3 '14 at 5:02
  • $\begingroup$ @user148177 : What is $r$? You mean $\widetilde{\rho}$? But I understand your comment I guess. Let me think it through. $\endgroup$ – Patrick Da Silva Jun 3 '14 at 5:05
  • $\begingroup$ r is rho. Dont know how to do latex in comments $\endgroup$ – user148177 Jun 3 '14 at 5:06
  • $\begingroup$ @user148177 : Put your latex stuff between cash symbols. $\endgroup$ – Patrick Da Silva Jun 3 '14 at 5:06
  • $\begingroup$ @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? $\endgroup$ – Patrick Da Silva Jun 3 '14 at 5:09
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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 $\ell = 0$.

As for your proof:

Serre might have complained about how you write up your proof also. Or probably it would have taken him much less time than me to find remedies for your explanations. Never the less, I come to the conclusion that your proof of injectivity is also correct. =)

(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.)

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    $\begingroup$ 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. $\endgroup$ – Patrick Da Silva Oct 8 '14 at 2:06

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