# Abstract Nonsense

## Mackey Irreducibility Criterion

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

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Motivation

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.

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Mackey’s Irreducibility Criterion

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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:

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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

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\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}

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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

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

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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

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

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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. $\blacksquare$

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

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References:

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