# Abstract Nonsense

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

Point of Post: This post is a continuation of this one.

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Our Two Formulations Are Equivalent

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What we’d now like to show is that $\text{Ind}^G_H(\rho)$  and $\text{ind}^G_H(\rho)$ are actually equivalent. Indeed:

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Theorem: Let $G$ be a finite group, $H\leqslant G$, and $\rho:H\to\mathcal{U}\left(\mathscr{V}\right)$. Then, if $\text{Ind}^G_H(\rho):G\to\mathcal{U}\left(\mathscr{X}\right)$ is our first definition of induced representation and $\displaystyle \text{ind}^G_H(\rho):G\to\mathcal{U}\left(\bigoplus_{t\in\mathscr{T}}\mathscr{V}_t\right)$ the one developed in this post, then $\text{Ind}^G_H(\rho)\cong\text{ind}^G_H(\rho)$.

Proof: We define

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$\displaystyle T:\bigoplus_{t\in\mathscr{T}}\mathscr{V}_t\to\mathscr{X}$

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by the rule

$\displaystyle \left(T\left(\sum_{t\in\mathscr{T}}x_t v_t\right)\right)(sh)=\rho_h^{-1}(v_s)$

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where $sh$ is the unique representation of an element of $G$ as the product of something in $\mathscr{T}$ and something in $H$. This is well-defined, in the sense that $\text{im}(T)\subseteq\mathscr{X}$, since if $g=sh\in G$ and $h'\in H$ then

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$\displaystyle \left(T\left(\sum_{t\in\mathscr{T}}x_t v_t\right)\right)(shh')=\rho_{hh'}^{-1}(v_s)=\rho_{h'}^{-1}\left(\rho_h^{-1}(v_s)\right)=\rho_{h'}^{-1}\left(\left(T\left(\sum_{t\in\mathscr{T}}x_t v_t\right)\right)(sh)\right)$

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In fact, we’ve already seen this map in  a sense before when finding $\deg\text{Ind}^G_H(\rho)$ and have seen that it is a bijection and has inverse

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$\displaystyle \mathscr{X}\ni f\overset{T^{-1}}{\longmapsto}\sum_{t\in\mathscr{T}}x_t f(t)$

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What we next claim is that for every $g\in G$ one has $T^{-1}\text{Ind}^G_H(\rho)_g T=\text{ind}^G_H(\rho)_g$. To do this we first note that

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\displaystyle \begin{aligned}\left(\text{Ind}^G_H(\rho)_g T\left(\sum_{t\in\mathscr{T}}x_t v_t\right)\right)(t_0) &=\left(T\left(\sum_{t\in\mathscr{T}}x_t v_t\right)\right)\left(g^{-1}t_0\right)\\ &=\left(T\left(\sum_{t\in\mathscr{T}}x_t v_t\right)\right)\left(P_1(g^{-1},t_0)P_2(g^{-1},t_0)\right)\\ &= \rho_{P_2(g^{-1},t)}^{-1}\left(v_{P_1(g^{-1},t)}\right)\end{aligned}

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So that by definition (using the obvious fact that for fixed $g$, $P_1(g,t)$ is a bijection $\mathscr{T}\to\mathscr{T}$)

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$\displaystyle \left(T^{-1}\text{Ind}^G_H(\rho)_g T\right)\left(\sum_{t\in\mathscr{T}}x_t v_t\right)=\sum_{t\in\mathscr{T}}x_t \rho^{-1}_{P_2(g^{-1},t)}\left(v_{P_1(g^{-1},t)}\right)=\sum_{t\in\mathscr{T}}x_{P_1(g,t)}\rho^{-1}_{P_2\left(g^{-1},P_1(g,t)\right)}\left(v_{P_1\left(g^{-1},P_1(g,t)\right)}\right)\quad\mathbf{(1)}$

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But, from previous work with $P_1,P_2$ we know that

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$P_1\left(g^{-1},P_1(g,t)\right)=P_1\left(g^{-1}g,t\right)=P_1(e,t)=t$

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and

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$P_2\left(g^{-1},P_1(g,t)\right)=P_2(g^{-1}g,t)P_2(g,t)^{-1}=P_2(g,t)^{-1}$

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so that $\mathbf{(1)}$ may be restated as

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$\displaystyle \left(T^{-1}\text{Ind}^G_H\left(\rho\right)_g T\right)\left(\sum_{t\in\mathscr{T}}x_t v_t\right)=\sum_{t\in\mathscr{T}}x_{P_1(g,t)}\rho_{P_2(g,t)}(v_t)=\left(\text{ind}^G_H(\rho)_g\right)\left(\sum_{t\in\mathscr{T}}x_t v_t\right)$

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And, since $g\in G$ was arbitrary we may appeal to a previous theorem. $\blacksquare$

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Remark: Note that this proves as well the implied fact that, up to equivalence, the construction of $\text{ind}^G_H(\rho)$ is independent of choice of transveral $\mathscr{T}$.

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References:

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