Abstract Nonsense

Crushing one theorem at a time

Induced Representation is ‘Inductive’

Point of Post: In this post we use the Frobenius Reciprocity theorem to show that the construction of Induced representation is inductive, in the sense that \text{Ind}^H_K\circ\text{Ind}^G_H=\text{Ind}^G_K for any K\leqslant H\leqslant G.

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An obvious question is the following: given groups K\leqslant H\leqslant G there are two ways we can create a representation on G from a representation on K. Namely, we take a representation of \rho of K we can then obviously consider the induced representation \text{Ind}^G_K(\rho), but we could also consider creating a H-representation by considering \text{Ind}^H_K(\rho) and then successively considering the G-representation \text{Ind}^G_H\left(\text{Ind}^H_K(\rho)\right). The obvious question though is that if we do this, do we get two different representation. The answer turns out to be no, in other words it’s true that given any K-representation \rho the representations \text{Ind}^G_K(\rho) and \text{Ind}^G_H\left(\text{Ind}^H_K(\rho)\right) are the same representation. In general this may be a very messy task, but thanks to the Frobenius Reciprocity theorem this becomes almost obvious in the sense that since \text{Ind}^G_K:\text{Cl}(K)\to\text{Cl}(G) is the adjoint of \text{Res}^K_G:\text{Cl}(G)\to\text{Cl}(K) it suffices to show that the map \text{Ind}^G_H\circ\text{Ind}^H_K:\text{Cl}(K)\to\text{Cl}(G) is the adjoint of \text{Res}^K_G and we will be done by the uniqueness of the adjoint. The fact that this is true is often stated that inducing representations is inductive.

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Induced Representation is Inductive

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We begin by noting the following fact. Namely, let G be finite and K\leqslant H\leqslant G. Let then \text{Res}^K_H:\text{Cl}(H)\to\text{Cl}(K), \text{Res}^H_G:\text{Cl}(G)\to\text{Cl}(H), and \text{Res}^K_G:\text{Cl}(G)\to\text{Cl}(K) be the restriction maps from H to K, G to H, and G to K respectively. We note then the obvious fact that \text{Res}^K_G=\text{Res}^K_H\circ\text{Res}^H_G. The reason this is ‘obvious’ is that the map is just restriction. From this we get that the analogous result for the maps \text{Ind}_K^H,\text{Ind}_H^G, and \text{Ind}^G_K. Namely:

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Theorem: Let G be a finite group and K\leqslant H\leqslant G, then \text{Ind}^G_K=\text{Ind}_H^G\circ\text{Ind}^H_K.

Proof: We note that for any f_1\in\text{Cl}(K) and f_2\in\text{Cl}(G) we have by (double) application of the Frobenius Reciprocity theorem

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\displaystyle \begin{aligned}\left\langle \text{Ind}^G_H\left(\text{Ind}^H_K(\left(f_1\right)\right),f_2\right\rangle_{\text{Cl}(G)} &= \left\langle \text{Ind}^H_K\left(f_1\right),\text{Res}^H_G(f_2)\right\rangle_{\text{Cl}(H)}\\ &= \left\langle f_1,\text{Res}^K_H\left(\text{Res}^H_K(f_2)\right)\right\rangle_{\text{Cl}(K)}\\ &= \left\langle f_1,\text{Res}^K_G\left(f_2\right)\right\rangle_{\text{Cl}(K)}\end{aligned}

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Thus, \text{Ind}^G_H\circ\text{Ind}^K_H is adjoint to \text{Res}^K_G. But, by Frobenius Reciprocity we know that \text{Ind}^G_K is an adjoint to \text{Res}^K_G and since the adjoint is unique we may conclude that \text{Ind}^G_H\circ\text{Ind}^H_K=\text{Ind}^G_K. \blacksquare

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A corollary of this which is mostly of novel interest is that:

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Theorem: Let f\in\text{Cl}(K) then

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\displaystyle \frac{1}{|K|}\sum_{g\in G}f^{\circ_K}\left(g^{-1}ag\right)=\frac{1}{|H||K|}\sum_{g\in G}\left(\sum_{h\in H}f^{\circ_K}\left(h^{-1}g^{-1}agh\right)\right)^{\circ_H}=\frac{1}{|H||K|}\sum_{g\in G}\sum_{h\in H}\left(f^{\circ_K}\left(\left(gh\right)^{-1}agh\right)\right)^{\circ_H}

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where ^{\circ_H} (resp. ^{\circ_K}) are obviously the notation to indicate the the function is itself when in H (resp. K) and 0 when not in H (resp. K).

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


May 1, 2011 - Posted by | Algebra, Representation Theory | , , , , , ,

1 Comment »

  1. […] has its usual meaning. But, since is a transversal fro the above may be […]

    Pingback by Composition of the Restriction Map and the Induction Map « Abstract Nonsense | May 4, 2011 | Reply

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