In Serre's book "Linear Representations of Finite Groups", very early on we are presented with a theorem:

Theorem 2: Every representation is a direct sum of irreducible representations

This proceeds in an easy to understand manner: since we are dealing with finite groups, we can always restrict to the case where $V$, the representation, is finite dimensional, then there is an induction.

This is all fine and I understand it. But for most examples, the group we are dealing with is so small that you already kind of know what the answer is. For instance, there is an example in section 2.6 where the group is the group with two elements. Here, there are obviously two degree 1 irreducible representations corresponding to sending the nontrivial element to 1 or -1. In a similar way, the canonical decomposition of any representation of this group will have two summands, the symmetric and antisymmetric elements.

But for a more complicated group and representation, what is the general algorithm for doing this? It's one thing to say you can just reduce a representation, but how do you do it generally?

I ask because there is this line in section 2.7:

Thus the decomposition of a representation $V$ can be done in two stages. First the canonical decomposition $V_1 \oplus \ldots \oplus V_n$ is determined; this can be done easily using the formulas giving the projections $p_i$.

For reference, the formula is:

$$p_i = \frac{n_i}{g} \sum_{t \in G} \chi_i(t)^* \rho_t$$

But this depends on the degree $n_i$ and the character $\chi_i$ of each irreducible summand $W_i$ in the decomposition of $V$ into irreducibles. Since there is no easy algorithm given for working out the irreducible decomposition of a representation, how is this "done easily"?

I could have missed something very obvious, I am only a beginner to representation theory.

  • 4
    $\begingroup$ You can apply the formula starting with any irreducible character whatsoever. If it doesn't appear in $V$, the corresponding projection will be zero. So the problem reduces to computing the character table of $G$. $\endgroup$ – Qiaochu Yuan Aug 4 '18 at 23:21
  • $\begingroup$ Oh right, that makes sense. I guess I was confused since I thought the $i$ was indexing over the $V_i$, but I was only thinking about the nonzero $V_i$, of course the $W_i$ that don't appear in $V$ are included, the corresponding $V_i$ is just zero. Thanks a lot. $\endgroup$ – pizzaroll Aug 4 '18 at 23:43

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