Abstract Nonsense

Crushing one theorem at a time

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

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April 25, 2011 - Posted by | Algebra, Representation Theory | , , , , ,

3 Comments »

  1. […] In our last post we derived a formula for the induced character of a representation. The interesting thing is that […]

    Pingback by Different Formula For the Character of an Induced Representation « Abstract Nonsense | April 26, 2011 | Reply

  2. […] 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 […]

    Pingback by Induced Class Functions and the Space of Integral Class Functions (Pt. I) « Abstract Nonsense | April 27, 2011 | Reply

  3. […] group and some there is a natural map by extending linearly the map for every where is the induced character. We saw dually there was a map which just took a class function on and restricted it to . In […]

    Pingback by Frobenius Reciprocity « Abstract Nonsense | May 1, 2011 | Reply


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