Abstract Nonsense

Crushing one theorem at a time

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.

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April 24, 2011 - Posted by | Algebra, Representation Theory | , , , , , ,

1 Comment »

  1. [...] that since if we define in the obvious way (the character of ) we may conclude that [...]

    Pingback by The Character of an Induced Representation « Abstract Nonsense | April 25, 2011 | Reply


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