Abstract Nonsense

Crushing one theorem at a time

Composition of the Restriction Map and the Induction Map


Point of Post: In this post we discuss what happens when we compose the maps \text{Ind}^G_H and \text{Res}^H_G.

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Motivation

So in our recent discussion of the induction map \text{Ind}^G_H and \text{Res}^H_G a very obvious question is…what happens if we compose the two maps? Namely, we have that \text{Ind}^G_H:\text{Cl}(H)\to\text{Cl}(G) and \text{Res}^H_G:\text{Cl}(G)\to\text{Cl}(H) so what does the map \text{Res}^H_G\circ\text{Ind}^G_H:\text{Cl}(H)\to\text{Cl}(H) look like? This is a question which was first solved by Mackey. In this post we prove a more general result, which in particular finds \text{Res}^H_G\circ\text{Ind}^G_K where H,K\leqslant G.

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What is \text{Res}^H_G\circ\text{Ind}^G_K?

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We begin by define a way to correspond class functions on subgroups of G and class functions on their conjugates. Namely, if H\leqslant G and s\in G we can define a map ^s:\text{Cl}(H)\to\text{Cl}\left(sHs^{-1}\right) given by f^s\left(shs^{-1}\right)=f(h), in other words f^s(x)=f\left(s^{-1}xs\right). This is evidently well defined map, in the sense that f^s\in\text{Cl}\left(sHs^{-1}\right) since given sh_1s^{-1},sh_2s^{-1}\in sHs^{-1} we have that

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f^s\left(sh_1s^{-1}sh_2s^{-1}\right)=f^s\left(sh_1h_2s^{-1}\right)=f\left(h_1h_2\right)=f\left(h_2h_1\right)=f^s\left(sh_1s^{-1}sh_2s^{-1}\right)

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and thus f^s\in\text{Cl}\left(sHs^{-1}\right) by a previous theorem. With this in mind we can state the following theorem:

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Theorem: Let H,K\leqslant G and let \mathscr{S} be a transveral of the set of double cosets H\backslash G/K then for any f\in\text{Cl}(K) one has that

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\displaystyle \left(\text{Res}^H_G\circ\text{Ind}^G_K\right)\left(f\right)=\sum_{s\in\mathscr{S}}\left(\text{Ind}^H_{H\cap sKs^{-1}}\circ\text{Res}^{sKs^{-1}}_{H\cap sKs^{-1}}\right)\left(f^s\right)

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Proof: We use the orbit-stabilizer theorem we write G as

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\displaystyle G=\bigsqcup_{s\in\mathscr{S}}HsK

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But, by a previous theorem we may rewrite this as

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\displaystyle G=\bigsqcup_{s\in\mathscr{S}}\bigsqcup_{xK\in\mathcal{O}_{sK}}xK\quad\mathbf{(1)}

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but it’s a common fact that that \mathcal{O}_{sK}=\left\{h\cdot sK:h\in\mathscr{T}_s\right\}=\left\{hsK:h\in \mathscr{R}_s\right\} where \mathscr{R}_s is a transversal for H/\text{stab}(sK) which by a previous observation is equal to H/\left(H\cap sKs^{-1}\right). Thus, if we let \mathscr{T} be the union over \mathscr{S} of \mathscr{R}_s then by \mathbf{(1)} we have that

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\displaystyle G=\bigsqcup_{t\in\mathscr{T}}tK

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and thus \mathscr{T} forms a transversal for G/K. Thus, by definition for any f\in\text{Cl}(K) and h\in H we have that

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\displaystyle \begin{aligned}\text{Ind}^G_K(f)(h) &= \sum_{t\in\mathscr{T}}f^\circ\left(t^{-1}ht\right)\\ &= \sum_{s\in\mathscr{S}}\sum_{t\in\mathscr{R}_s}f^\circ\left((rs)^{-1}rs\right)\\ &= \sum_{s\in\mathscr{S}}\sum_{r\in\mathscr{R}_s}f^\circ\left(s^{-1}r^{-1}hrs\right)\\ &= \sum_{s\in\mathscr{S}}\sum_{\substack{r\in\mathscr{R}_s\\ r^{-1}hr\in sKs^{-1}}}f^s\left(r^{-1}hr\right)\\ &= \sum_{s\in\mathscr{S}}\sum_{\substack{r\in\mathscr{R}_s\\ r^{-1}hr\in sKs^{-1}}}\text{Res}^{H\cap sKs^{-1}}_{sKs^{-1}}\left(f^s\left(r^{-1}hr\right)\right)\\ &= \sum_{s\in\mathscr{S}}\sum_{r\in\mathscr{R}_s}\left(\text{Res}^{H\cap sKs^{-1}}_{sKs^{-1}}\left(f^s\left(r^{-1}hr\right)\right)\right)^{\circ_{H\cap sKs^{-1}}}\end{aligned}

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where ^{\circ}_{H\cap sKs^{-1}} has its usual meaning. But, since \mathscr{R}_s is a transversal fro H/\left(H\cap sKs^{-1}\right) the above may be rewritten

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\displaystyle \text{Ind}^G_K\left(f\right)(h)=\sum_{s\in\mathscr{S}}\text{Ind}^{H}_{H\cap sKs^{-1}}\left(\text{Res}^{H\cap sKs^{-1}}_{sKs^{-1}}\left(f^s(h)\right)\right)

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and since h\in H was arbitrary the conclusion follows. \blacksquare

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From this we get the obvious corollary:

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Theorem: Let G be a finite group, H,K\leqslant G and \psi:K\to\mathcal{U}\left(\mathscr{V}\right) a representation. Then, if \text{ind}^G_H(\psi) is the usual induced representation then for any 

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\displaystyle \text{Res}^H_G\left(\text{ind}^G_K\left(\psi\right)\right)\cong\bigoplus_{s\in\mathscr{S}}\text{Ind}^H_{H\cap sKs^{-1}}\left(\psi^s\right)

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where \psi^s:H\cap sKs^{-1}\to\mathcal{U}\left(\mathscr{V}\right) is given by \psi^s\left(shs^{-1}\right)=\psi(h).

Proof: This is obvious since the above theorem proves that these two representations of H admit the same character. \blacksquare

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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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May 4, 2011 - Posted by | Algebra, Representation Theory | , , , , ,

1 Comment »

  1. […] this is equivalent to showing the slightly more scary looking   . But, thanks to Mackey we have information on this right hand […]

    Pingback by Mackey Irreducibility Criterion « Abstract Nonsense | May 6, 2011 | Reply


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