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A quick lemma on group representations September 23, 2009

Posted by Akhil Mathew in algebra, representation theory.
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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}}.

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