# Abstract Nonsense

## The Character of an Induced Representation

Point of Post: In this post we discuss the notion of the induced character of an induced representation and provide  a formulaic relationship between the induced character and the original character.

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Motivation

As we have already seen given a finite group $G$ a subgroup $H\leqslant G$ and a representation $\rho:H\to\mathcal{U}\left(\mathscr{V}\right)$ we can create a representation $\text{Ind}^G_H(\rho):G\to\mathcal{U}\left(\mathscr{X}\right)$. Of course though, representations aren’t just the main focus in representation theory, we are also heavily interested in the representations characters since they holds so much information about the representations they come from as well as the underlying group as well. Thus, in this post we derive a relationship between the character of a representation of $H$ and the character of the associated induced representation for $G$.

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Induced Characters

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Let $G$ be a finite group and $H\leqslant G$. Furthermore, suppose that $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$ is a representation and consider the associated induced representation $\displaystyle \text{ind}^G_H(\rho):G\to\mathcal{U}\left(\bigoplus_{t\in\mathscr{T}}\mathscr{V}_t\right)$. Let then $\chi$ be the character of $\rho$. We say then that the character of $\text{ind}^G_H(\rho)$ is induced by the character $\chi$ \$ and denote it by $\text{Ind}^G_H(\chi)$ when convenient. We now find a formulaic relationship between $\chi$ and $\text{ind}^G_H(\chi)$. Indeed:

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Theorem: Let $G$ be a finite group, $H\leqslant G$, and $\rho:H\to\mathcal{U}\left(\mathscr{V}\right)$ be a representation with character $\chi$ . Let $\displaystyle \text{ind}^G_H(\rho):G\to\mathcal{U}\left(\bigoplus_{t\in\mathscr{T}}\mathscr{V}_t\right)$ denote the associated induced representation for some choice of transveral $\mathscr{T}$ of $G/H$ and $\text{ind}^G_H(\chi)$ the associated character. Then, for any $g\in G$ one has

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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, for every $k\in K$, one has that $\chi^\circ(k)=\chi(k)$ if $k\in H$ and $0$ otherwise.

Proof: We begin by fixing an ordered basis $(v_1,\cdots,v_n)$ for $\mathscr{V}$ and note that it’s obvious then that if we order the transveral $\mathscr{T}$ as $\{t_1,\cdots,t_m\}$ then $\mathcal{B}=(x_{t_1}v_1,\cdots,x_{t_1}v_m,\cdots,x_{t_n}v_1,\cdots,x_{t_n}v_m)$ is an ordered basis for $\displaystyle \bigoplus_{t\in\mathscr{T}}\mathscr{V}_t$. We next define then, for every $h\in H$, the matrix elements of $[\rho_h]_{(v_1,\cdots,v_n)}$ as $M_{i,j,h}\;\; i,j[n]$. Then, we note that by definition

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$\displaystyle \chi(h)=\text{tr}(\rho_h)=\sum_{j=1}^{n}M_{j,j,h}$

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We next define $M_g$ to be the matrix $\left[\text{ind}^G_H(\rho)_g\right]_\mathcal{B}$ and we seek to find the diagonal entries of $M_g$ since their sum will be equal to $\text{ind}^G_H(\chi)(g)$. To do this we note that by definition

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$\displaystyle \text{ind}^G_H(\rho)_g(x_{t_i}v_j)=x_{P_1(g,t_i)}\rho_{P_2(g,t_i)}(v_j)$

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clearly then the diagonal entry associated to $x_{t_i}v_j$ will be zero unless $t_i=P(g,t_i)$. What we claim is that this is true precisely when $t_i^{-1}gt_i\in H$. Indeed, suppose first that $t_i=P(g,t_i)$ then by definition there exists some $h\in H$ such that $gt_i=t_i h$ so that $t_i^{-1}gt_i=h\in H$. Conversely, if $t_i^{-1}gt_i\in H$ then since $gt_i=t_i t_i^{-1}gt_i$ one has that $P_1(g,t_i)=t_i$. Thus, the diagonal entry related to $x_{t_i}v_j$ will be zero unless $t_i^{-1}gt_i\in H$. Note then that if this is true we have by the above discussion that $P_2(g,t_i)=t_i^{-1}gt_i$. It’s clear to see then that if $D_{i,j}$ denotes the diagonal entry of $M_g$ associated to $x_{t_i}v_j$ then

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$\displaystyle D_{i,j}=\begin{cases}M_{j,j,t_i^{-1}gt_i} & \mbox{if}\quad t_i^{-1}gt_i\in H\\ 0 & \mbox{if}\quad t_i^{-1}gt_i\notin H\end{cases}$

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Thus, it clearly follows that

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$\displaystyle \sum_{j=1}^{n}D_{i,j}=\begin{cases}\chi\left(t_i^{-1}gt_i\right) & \mbox{if}\quad t_i^{-1}gt_i\in H\\ 0 & \mbox{if}\quad t_i^{-1}gt_i\notin H\end{cases}=\chi^\circ\left(t_i^{-1}gt_i\right)$

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Thus,

\displaystyle \begin{aligned}\text{ind}^G_H(\chi)(g) &= \text{tr}(M_g)\\ &= \sum_{i=1}^{m}\sum_{j=1}^{n}D_{i,j}\\ &= \sum_{i=1}^{m}\chi^\circ\left(t_i^{-1}gt_i\right)\\ &= \sum_{t\in\mathscr{T}}\chi^\circ\left(t^{-1}gt\right)\end{aligned}

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$\blacksquare$

April 25, 2011 -