A quick lemma on group representations

So, since I’ll be talking about the symmetric group a bit, and since I still don’t have enough time for a deep post on it, I’ll take the opportunity to cover a quick and relevant lemma in group representation theory (referring as usual to the past blog post as background).

A faithful representation of a finite group ${G}$ is one where different elements of ${G}$ induce different linear transformations, i.e. ${G \rightarrow Aut(V)}$ is injective. The result is

Lemma 1 If ${V}$ is a faithful representation of ${G}$ , then every simple representation of ${G}$ occurs as a direct summand in some tensor power ${V^{\otimes p}}$ .

To prove this, let ${\chi_V}$ be the character of ${V}$ and ${\psi}$ the character of ${W}$ , for ${W}$ some irreducible in ${Rep(G)}$ . Then we need to show for some ${n}$ , in view of the orthonormality relations,

$\displaystyle \sum_{g \in G} \chi_V^n(g) \overline{\psi(g)} \neq 0 \ \ \ \ \ (1)$

Now, let ${\{a_i, 1 \leq i \leq m \}}$ be the set of values assumed by ${\chi_V}$ and let ${A_i \subset G}$ be the set where ${\chi_V}$ takes the value ${a_i}$ . If ${b_i := \sum_{A_i} \overline{\psi(g)}}$ , then (1) implies

$\displaystyle \sum_i a_i^n b_i = 0$

for all ${n}$ . But this implies that each ${b_i=0}$ by taking a van der Monde determinant. If, say, ${A_j = \{ 1\}}$ —by faithfulness ${\chi_V(g) = \dim V}$ iff ${g=1}$ —then ${b_j = 0}$ , which implies ${\dim W = 0}$ .

Note that the proof (due to Brauer) actually gives an effective bound: we can take the tensor power to be at most ${m-1}$ , where ${m}$ is as in the proof of the result. This follows again from van der Monde determinants.

The case that interests us is the symmetric group ${S_n}$ , where we have a canonical regular representation ${\mathfrak{h}}$ spanned by basis vectors ${e_1, \dots, e_n}$ with ${\sigma(e_i) = e_{\sigma(i)}}$ for ${\sigma \in S_n}$ . This is faithful, so we find that every simple representation of ${S_n}$ is a summand of some ${\mathfrak{h}^{\otimes p}}$ .