Abstract Nonsense

Crushing one theorem at a time

Mackey Irreducibility Criterion

Point of Post: In this post we prove the Mackey Irreducibility Criterion.

\text{ }


In this post we put to rest probably one of the most niggling questions about induced representations. Namely, we have so far figured out how to take representations on H\leqslant G and create a representation on G but have yet to give a condition for which the a property we are so fond of us is true. To be direct, we have yet to decide when \text{ind}^G_H(\psi) is actually an irrep of G. The idea is simple, namely we know that being an irrep is equivalent to having \left\langle \text{ind}^G_H(\chi_\psi),\text{ind}^G_H(\chi_\psi)\right\rangle=1 but from Frobenius Reciprocity theorem we know that this is equivalent to showing the slightly more scary looking   \left\langle \chi_\psi,\text{Res}^H_G\left(\text{ind}^G_H(\chi_\psi)\right)\right\rangle=1. But, thanks to Mackey we have information on this right hand term.

\text{ }

Mackey’s Irreducibility Criterion

\text{ }

Before we get to the theorem we need a few definitions. Namely, let G be any group and \rho,\psi to representations on G. We call \rho and \psi disjoint if they have no equivalent subrepresentations and we denote this property by \rho\perp \psi. This notation is supposed to be provocative since it’s evident that \rho\perp\psi if and only if \left\langle \chi_\rho,\chi_\psi\right\rangle=0, in other words if the characters of \rho and \psi are orthogonal. Also, as was mentioned in our last post given a subgroup H of G and a representation \rho:H\to\mathcal{U}\left(\mathscr{V}\right) we can define a representation on sHs^{-1} for any s\in G given by \rho^s:sHs^{-1}\to\mathcal{U}\left(\mathscr{V}\right) by \rho^s:shs^{-1}\mapsto \rho_h. We denote the character of \rho_s by \chi^s. We note that if s\in H then \rho^s:H\to\mathcal{U}\left(\mathscr{V}\right) and for any h\in H we have that \text{tr}(\rho^s(h))=\text{tr}(\rho(s)^{-1}\rho(h)\rho(s))=\text{tr}(h)=\chi(h) and so \chi^s=\chi. With these the irreducibility criterion becomes simple. Namely:

\text{ }

Theorem (Mackey Irreducibility): Let G be a finite group, H\leqslant G, and \psi:H\to\mathcal{U}\left(\mathscr{V}\right) a representation of H with character \chi. Then, the induced representation \text{ind}^G_H(\psi) is an irreducible G-representation if and only if \psi is irreducible and for every s\in G-H one has that \text{Res}^{H\cap sHs^{-1}}_H(\psi)\perp\text{Res}^{H\cap sHs^{-1}}_{sHs^{-1}}(\psi^s).

Proof: As stated in the motivation the proof is simple, namely we know that \text{ind}^G_H(\psi) is irreducible if and only if \left\langle \text{Ind}^G_H(\chi),\text{Ind}^G_H(\chi)\right\rangle=1. But, using Frobenius Reciprocity and the our knowledge of \text{Res}^H_G\circ\text{Ind}^G_H we know that, picking some transversal \mathscr{S} for the set of double cosets H\backslash G/H we see that

\text{ }

\displaystyle \begin{aligned}\left\langle\text{Ind}^G_H(\chi),\text{Ind}^G_H(\chi)\right\rangle &= \left\langle \text{Res}^H_G\left(\text{Ind}^G_H(\chi)\right),\chi\right\rangle\\ &= \sum_{s\in\mathscr{S}}\left\langle \text{Ind}^H_{H\cap sHs^{-1}}\left(\text{Res}^{H\cap sHs^{-1}}_H(\chi^s)\right),\chi\right\rangle\\ &= \sum_{s\in\mathscr{S}}\left\langle \text{Res}^{H\cap sHs^{-1}}_{sHs^{-1}}(\chi^s),\text{Res}^{H\cap sHs^{-1}}_H(\chi)\right\rangle\end{aligned}

\text{ }

Then, making our previous observation that for the s\in HeH=H we have by the above observation that \text{Res}^{H\cap sHs^{-1}}_H(\chi^s)=\chi we see that the above is equal to

\text{ }

\left\langle \text{Ind}^G_H(\chi),\text{Ind}^G_H(\chi)\right\rangle=\displaystyle \left\langle\chi,\chi\right\rangle+\sum_{\substack{s\in \mathscr{S}\\ s\notin H}}\left\langle \text{Res}^{H\cap sHs^{-1}}_{sHs^{-1}}\left(\chi^s\right),\text{Res}^{H\cap sHs^{-1}}_H(\chi)\right\rangle

\text{ }

And since \left\langle\chi,\chi\right\rangle\geqslant 1 we see from the above that \left\langle \text{Ind}^G_H(\chi),\text{Ind}^G_H(\chi)\right\rangle=1 if and only if \left\langle\chi,\chi\right\rangle=1 and

\text{ }

\left\langle \text{Res}^{H\cap sHs^{-1}}_{sHs^{-1}}(\chi^s),\text{Res}^{H\cap sHs^{-1}}_H(\chi)\right\rangle=0

\text{ }

for s\in\mathscr{S}-H. But, unraveling this it says precisely that \text{Ind}^G_H(\rho) is irreducible if and only if \rho is irreducible and using our disjoint notation  \text{Res}^{H\cap sHs^{-1}}_{sHs^{-1}}(\psi^s)\perp\text{Res}^{H\cap sHs^{-1}}_H(\psi) as desired. But, since the possible choices for a transversal \mathscr{S} with s\notin H is precisely G-H the conclusion follows. \blacksquare

\text{ }

Of course, if H\unlhd G the above greatly simplifies greatly since H\cap sHs^{-1}=H and evidently H\backslash G/H=G/H.

\text{ }

\text{ }


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


May 6, 2011 - Posted by | Algebra, Representation Theory | , , , , , ,

No comments yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: