Abstract Nonsense

Induced Class Functions and the Space of Integral Class Functions (Pt. I)

Point of Post: In this post we set up the frame work to say the Frobenius reciprocity theorem intelligently. Namely, we discuss the space of all linear characters $\mathbb{Z}\left(\text{irr}(G)\right)$ of a group and describe the natural maps $\bigtriangleup:\mathbb{Z}\left(\text{irr}(H)\right)\to\mathbb{Z}\left(\text{irr}(G)\right)$ and $\bigtriangledown:\mathbb{Z}\left(\text{irr}(G)\right)\to\mathbb{Z}\left(\text{irr}(H)\right)$.

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Motivation

The beginnings of the theory in this post can be traced back to our first derivation for the induced character of an induced representation when one noticed that by extending every (irreducible) character on $H$ to a character (thus a class function) we’ve automatically defined a map $\text{Ind}^G_H:\text{Cl}\left(H\right)\to\text{Cl}\left(G\right)$ by extending the map $\chi\mapsto \text{Ind}^G_H(\chi)$ by linearity (recalling that the irreducible characters form a basis for $\text{Cl}(H)$). But, what we’ll note that is if we restrict $\text{Cl}(H)$ to the space $\text{span}_{\mathbb{Z}}(\text{irr}(H))\overset{\text{def.}}{=}\mathbb{Z}\left(\text{irr}(H)\right)$ we get that $\text{Ind}^G_H\left(\mathbb{Z}\left(\text{irr}(H)\right)\right)\subseteq\mathbb{Z}\left(\text{irr}(G)\right)$. Thus when we restrict $\text{Ind}^G_H$ to $\mathbb{Z}\left(\text{irr}(H)\right)$ we get a map $\bigtriangleup:\mathbb{Z}\left(\text{irr}(H)\right)\to\mathbb{Z}\left(\text{irr}(G)\right)$. The subject of this post will be to explore theses topics and their dual concepts–namely the obviously defined map $\text{Res}^H_G:\text{Cl}(G)\to\text{Cl}(H)$ and $\bigtriangledown:\mathbb{Z}\left(\text{irr}(G)\right)\to\mathbb{Z}\left(\text{irr}(H)\right)$.

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Induced Class Functions

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Let $G$ be a finite group, $H\leqslant G$, and $\chi\in\text{irr}(H)$. Recall that we could then define the induced character $\text{Ind}^G_H(\chi)$ on $G$ which has explicit formula

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$\displaystyle \text{Ind}^G_H(\chi)(g)=\sum_{t\in\mathscr{T}}\chi^\circ\left(t^{-1}gt\right)$

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where $\mathscr{T}$ is any transveral of $G/H$. Since $\text{Ind}^G_H(\chi)$ is a character and thus trivially a class function we now have a mapping $F:\text{irr}(H)\to\text{Cl}(G)$ thus we definitively have a map $D:\text{span}_\mathbb{C}\left(\text{irr}(H)\right)\to\text{span}_\mathbb{C}\left(\text{Cl}(G)\right)$ by extending $F$ by linearity. But this is very pedantic since we know that $\text{span}_\mathbb{C}\left(\text{irr}(H)\right)=\text{Cl}(H)$ since  we know $\text{irr}(H)$ is an (orthonormal) basis for $\text{Cl}(H)$. Moreover, we know that $\text{span}_\mathbb{C}\left(\text{Cl}(G)\right)$ since we have proven that $\text{Cl}(G)$ is a subalgebra of the group algebra $\mathcal{A}(G)$. So, explicitly we have a map $D:\text{Cl}(H)\to\text{Cl}(G)$ given by

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$\displaystyle D:\sum_{\alpha\in\widehat{H}}z^{(\alpha)}\chi^{(\alpha)}\mapsto \sum_{\alpha\in\widehat{H}}z^{(\alpha)}\text{Ind}^G_H\left(\chi^{(\alpha)}\right)$

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We denote the map more formally as $\text{Ind}^G_H$. While fairly obvious we prove the following fact: namely, if $f\in\text{Cl}(H)$ define $f^\circ\in\mathcal{A}(G)$ by $f^\circ(h)=h$ if $h\in H$ and $0$ otherwise. Then,

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Theorem: Let the map $\text{Ind}^G_H:\text{Cl}(H)\to\text{Cl}(G)$ be as above, then for any transveral $\mathscr{T}$ of $G/H$ and any $f\in\text{Cl}(H)$ one has

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$\displaystyle \text{Ind}^G_H(f)(g)=\sum_{t\in\mathscr{T}}f^\circ\left(t^{-1}gt\right)$

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Proof: We merely note that by definition

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$\displaystyle f=\sum_{\alpha\in\mathcal{H}}z^{(\alpha)}\chi^{(\alpha)}$

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for some $z^{(\alpha)}\in\mathbb{C}$. Thus, by definition

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\displaystyle \begin{aligned}\text{Ind}^G_H(f)(g) &=\sum_{\alpha\in\widehat{G}}z^{(\alpha)}\text{Ind}^G_H\left(\chi^{(\alpha)}\right)\\ &=\sum_{\alpha\in\widehat{H}}z^{(\alpha)}\sum_{t\in\mathscr{T}}\left(\chi^{(\alpha)}\right)^\circ\left(t^{-1}gt\right)\\ &=\sum_{t\in\mathscr{T}}\sum_{\alpha\in\widehat{H}}z^{(\alpha)}\left(\chi^{(\alpha)}\right)^\circ\left(t^{-1}gt\right)\end{aligned}\quad\quad\mathbf{(1)}

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Note though that if $k\in H$ that

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$\displaystyle \sum_{\alpha\in\widehat{H}}z^{(\alpha)}\left(\chi^{(\alpha)}\right)^\circ(k)=\sum_{\alpha\in\widehat{H}}z^{(\alpha)}\chi^{(\alpha)}(k)=f(k)$

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and if $k\notin H$ then

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$\displaystyle \sum_{\alpha\in\widehat{H}}z^{(\alpha)}\left(\chi^{(\alpha)}\right)^{\circ} (k)=0=f(k)$

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and thus

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$\displaystyle f^\circ=\sum_{\alpha\in\widehat{H}}z^{(\alpha)}\left(\chi^{(\alpha)}\right)^\circ$

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the result then follows from $\mathbf{(1)}$. $\blacksquare$

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References:

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

April 27, 2011 -