Abstract Nonsense

Crushing one theorem at a time

A Characterization of Irreducibility


Point of post: In this post we give an easy way to check whether or not a given representation of a finite group is irreducible in terms of the representations character.

Motivation

Often given a representation of a finite group it is difficult to look at it and decide whether or not it is irreducible. In this post we prove a theorem which enable us to decide whether or not a given representation is irreducible by checking whether or not the inner product of its induced character with itself is unity.

Numerical Check for Irreducibility

Let’s get right to it.

Theorem: Let G be a finite group and \rho:G\to\mathcal{U}\left(\mathscr{V}\right) a representation of G with character \chi_\rho. Then, \rho is irreducible if and only if \left\langle\chi_\rho,\chi_\rho\right\rangle=1

Proof: Evidently if \rho is irreducible then \rho\in\alpha for some \alpha\in\widehat{G} and thus \chi_\rho=\chi^{(\alpha)} and the rest follows from the orthonormality relations for the irreducible characters.

Conversely, suppose that \left\langle \chi_\rho,\chi_\rho\right\rangle=1. We know that for any chosen set of representations \rho^{(\alpha)} for each \alpha\in G there exists m^{(\alpha)}\in\mathbb{N}\cup\{0\} and a unitary W such that

\displaystyle \rho(g)=W\bigoplus_{\alpha\in\widehat{G}}m^{(\alpha)}\rho^{(\alpha)}W^{-1}

 

it follows then that \displaystyle \chi_\rho=\sum_{\alpha\in\widehat{G}}m^{(\alpha)}\chi^{(\alpha)} and thus

\displaystyle \begin{aligned}\left\langle \chi_\rho,\chi_\rho\right\rangle &= \sum_{\alpha\in\widehat{G}}\sum_{\beta\in\widehat{G}}m^{(\alpha)}m^{(\beta)}\left\langle \chi^{(\alpha)},\chi^{(\beta)}\right\rangle\\ &=\sum_{\alpha\in\widehat{G}}\sum_{\beta\in\widehat{G}}m^{(\alpha)}m^{(\beta)}\delta_{\alpha,\beta}\\ &= \sum_{\alpha\in\widehat{G}}\left(m^{(\alpha)}\right)^2\end{aligned}

 

and thus by assumption that \left\langle \chi_\rho,\chi_\rho\right\rangle=1 we may conclude that there exists some \alpha_0\in\widehat{G} such that

 

m^{(\alpha)}=\begin{cases}1 & \mbox{if}\quad \alpha=\alpha_0\\ 0 & \mbox{if}\quad \alpha\ne\alpha_0\end{cases}

 

from where it follows that \rho\cong\rho^{(\alpha_0)} and thus \rho is irreducible as desired. \blacksquare

References:

1. Isaacs, I. Martin. Character Theory of Finite Groups. New York: Academic, 1976. Print.

2. Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Mathematical Society, 1996. Print.

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March 7, 2011 - Posted by | Algebra, Representation Theory | , , ,

6 Comments »

  1. […] so by previous theorem we have that is irreducible. Let . One clearly has that and thus by Schur’s lemma one has […]

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  2. […] . Now, the only question is whether is irreducible. To do this, it suffices to check our alternative characterization of irreducibility. […]

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  3. […] generator for . A quick check shows then that , , , and . It’s clear then that and so by our useful characterization of irreducibility that is not irreducible in the classic sense. Even simpler, we know that can’t be […]

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  5. […] for any linear representation one has that for every ) from where the conclusion follows from our previous characterization of irreducibility. To see that we merely note […]

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  6. […] To be direct, we have yet to decide when is actually an irrep of . The idea is simple, namely we know that being an irrep is equivalent to having but from Frobenius Reciprocity theorem we know that […]

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