# Abstract Nonsense

## Another Way of Looking at Induced Representations (Pt. I)

Point of Post: In this post we discuss another way to view induced representations and and show that it is actually equivalent to our previous definition since the two different constructed representations of the ambient group will be equivalent.

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Motivation

Of course, as is true for most things in mathematics, the notion of induced representation has several equivalent formulations. In this post we describe another one which may seem, at first, to be a simpler construction than our previous one. In fact, both are equally useful and come up in different aspects of the coming theory. Thus, it is wise to have both of these (among many more) equivalent formulations sitting in our back pocket.

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Induced Representations Again

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Let $G$ be a finite group and $H\leqslant G$. Assume that there is a representation $\psi:H\to\mathcal{U}\left(\mathscr{V}\right)$. We begin our construction by fixing a transversal $\mathscr{T}$ for the set of cosets $G/H$. We then consider then the direct sum of $(G:H)$ copies of $\mathscr{V}$ indexed by the elements of $\mathscr{T}$ where each copy is distinguished. Said more concretely for each $t\in\mathscr{T}$ we define $\mathscr{V}_t$ to be equal to $\mathscr{V}$ and consider the space $\displaystyle \bigoplus_{t\in\mathscr{T}}\mathscr{V}_t$ given the usual inner product structure on direct sum spaces with a normalization factor of $|G|^{-1}$–we will think of the elements of this space as $\displaystyle \sum_{t\in\mathscr{T}}x_tv_t$ where $v_t\in\mathscr{V}_t$ and $x_t$ is a formal symbol ‘tagging’ $v_t$. We then note that since the set $\left\{Ht\right\}_{t\in\mathscr{T}}$ forms a partition of $G$ for each $g\in G$ and $t\in\mathscr{T}$ there exists unique $h\in H$ and $t'\in\mathscr{T}$ such that $gt=t'h$, we denote this $t'$ by $P_1(g,t)$ and this $h$ as $P_2(g,t)$. We then define, for each $g\in G$ the map

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$\displaystyle \rho_g:\bigoplus_{t\in\mathscr{T}}\mathscr{V}_t\to\bigoplus_{t\in\mathscr{T}}\mathscr{V}_t:\sum_{t\in\mathscr{T}}x_t v_t\mapsto \sum_{t\in\mathscr{T}}x_{P_1(g,t)}\psi_{P_2(g,t)}(v_t)$

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We claim that in fact $\rho_g$ is in $\displaystyle \mathcal{U}\left(\bigoplus_{t\in\mathscr{T}}\mathscr{V}_t\right)$. Indeed, to see that $\rho_g$ is a linear transformation we note that for any $z,z'\in\mathbb{C}$ and $\displaystyle \sum_{t\in\mathscr{T}}x_t v_t,\sum_{t\in\mathscr{t}}x_t v'_t$

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\displaystyle \begin{aligned}\rho_g\left(z\sum_{t\in\mathscr{T}}x_t v_t+z'\sum_{t\in\mathscr{T}}x_t v'_t\right) &= \rho_g\left(\sum_{t\in\mathscr{T}}x_t\left(xv_t+z'v'_t\right)\right)\\ &= \sum_{t\in\mathscr{T}}x_{P_1(g,t)}\psi_{P_2(g,t)}\left(zv_t+z'v'_t\right)\\ &= z\sum_{t\in\mathscr{T}}x_{P_1(g,t)}\psi_{P_1(g,t)}(v_t)+z'\sum_{t\in\in\mathscr{T}}x_{P_1(g,t)}\psi_{P_2(g,t)}(v'_t)\\ &= z\rho_g\left(\sum_{t\in\mathscr{T}}x_t v_t\right)+z'\rho_g\left(\sum_{t\in\mathscr{T}}x_t v'_t\right)\end{aligned}

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To see that it’s unitary we merely note that

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\displaystyle \begin{aligned}\left\langle\rho_g\left(\sum_{t\in\mathscr{T}}x_tv_t\right),\rho_g\left(\sum_{t\in\mathscr{T}}x_v w_t\right)\right\rangle &= \left\langle \sum_{t\in\mathscr{T}}x_{P_1(g,t)}\psi_{P_2(g,t)}(v_t),\sum_{t\in\mathscr{T}}x_{P_1(g,t)}\psi_{P_2(g,t)}(w_t)\right\rangle\\ &= \frac{1}{|G|}\sum_{t\in\mathscr{T}}\left\langle \psi_{P_2(g,t)}(v_t),\psi_{P_2(g,t)}(w_t)\right\rangle_\mathscr{V}\\ &= \frac{1}{|G|}\sum_{t\in\mathscr{V}}\left\langle v_t,w_t\right\rangle_\mathscr{V}\\ &= \left\langle \sum_{t\in\mathscr{T}}x_t v_t,\sum_{t\in\mathscr{T}}x_t w_t\right\rangle\end{aligned}

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and thus $\rho_g$ is in $\displaystyle \mathcal{U}\left(\bigoplus_{t\in\mathscr{T}}\mathscr{V}_t\right)$ as claimed. Our last claim is that the map $G\ni g\mapsto \rho_g$ is a homomorphism. To do this we first claim that for any $g,g'\in G$ and any $t\in\mathscr{T}$ one has that $P_1(gg',t)=P_1\left(g,P_1(g',t)\right)$ and $P_2(gg',t)=P_2\left(g',P_1(g',t)\right)P_2(g',t)$. To see this we merely note that if $g'=P_1(g',t)P_2(g',t)$ then $gg't=gP_1(g',t)P_2(g',t)=P_1\left(g,P_1(g',t)\right)P_2\left(g,P_1(g',t)\right)P_2(g',t)$. Then, for any $g,g'\in G$ and any $\displaystyle \sum_{t\in\mathscr{T}}x_tv_t\in\bigoplus_{t\in\mathscr{T}}\mathscr{V}_t$ one has that

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\displaystyle \begin{aligned}\rho_{gg'}\left(\sum_{t\in\mathscr{T}}x_t v_t\right) &= \sum_{t\in\mathscr{T}}x_{P_1(gg',t)}\psi_{P_2(gg',t)}(v_t)\\ &= \sum_{t\in\mathscr{T}}x_{P_1(g,P_1(g',t))}\psi_{P_2(g,P_1(g',t))P_2(g',t)}(v_t)\\ &= \sum_{t\in\mathscr{T}}x_{P_1(g,P_1(g',t))}\psi_{P_2(g,P_1(g',t))}\left(\psi_{P_2(g',t)}(v_t)\right)\\ &= \rho_g\left(\sum_{t\in\mathscr{T}}x_{P_1(g',t)}\psi_{P_2(g',t)}(v_t)\right)\\ &= \rho_g\left(\rho_{g'}\left(\sum_{t\in\mathscr{T}}x_tv_t\right)\right)\end{aligned}

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and since $\displaystyle \sum_{t\in\mathscr{T}}x_t v_t$ was arbitrary we may conclude that $\rho_{gg'}=\rho_g\circ\rho_{g'}$ and thus $\rho$ is a homomorphism as claimed. Summing this all up we have that:

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Theorem: Let $G$ be a finite group, $H\leqslant G$,$\mathscr{T}$ a transversal of $G/h$, and $\psi:H\to\mathcal{U}\left(\mathscr{V}\right)$ a representation of $H$. Then, if the space $\displaystyle \bigoplus_{t\in\mathscr{T}}\mathscr{V}_t$ (as described above) is given the usual inner product structure on direct sums with a normalization factor $|G|^{-1}$ then the map

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$\displaystyle \rho:G\to\mathcal{U}\left(\bigoplus_{t\in\mathscr{T}}\mathscr{V}_t\right),\;\; \rho_g:\sum_{t\in\mathscr{T}}x_t v_t\mapsto \sum_{t\in\mathscr{T}}x_{P_1(g,t)}\psi_{P_2(g,t)}(v_t)$

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(where $P_1(g,t),P_2(g,t)$ are the unique elements of $\mathscr{T},H$ respectively such that $gt=P_1(g,t)P_2(g,t)$) is a representation.

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We shall call this representation $\text{ind}^G_H(\psi)$.

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References:

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

April 24, 2011 -

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