# Abstract Nonsense

## 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