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$\newcommand{\Ind}{{\text{Ind}}}$ $\newcommand{\Res}{{\text{Res}}}$ $\newcommand{\ds}{{\displaystyle}}$ $\newcommand{\inv}{{^{-1}}}$

I am doing Exercise 3.16 from Fulton Harris. http://bit.ly/JeTz1J

I have already read http://bit.ly/HYm9Fx, but it's much too advanced an answer.

Goals: a) $U \otimes \Ind(W) =\Ind (\Res (U) \otimes W) $ and b)$\Ind^G_H(W) =\Ind^G_K (\Ind^K_H(W))$ where $H<K<G$

My attempts:

a) $U \otimes \Ind W = U \otimes \ds\bigoplus_{\sigma \in G/H} \sigma W = \ds\bigoplus_{\sigma \in G/H} U \otimes \sigma W = \ds\bigoplus_{\sigma \in G/H} \sigma( \sigma\inv U \otimes W)$. Now $\Ind (\Res (U) \otimes W) = \ds\bigoplus_{\sigma \in G/H} \sigma (\Res U \otimes W) = \ds\bigoplus_{\sigma \in G/H} \sigma \Res U \otimes \sigma W$.

In particular if you let $W$ be the trivial representation then $\Res U$ can be expressed as $\Res( U) \otimes W$ and thus $\Ind(\Res (U))= \Ind(\Res (U) \otimes W) = U \otimes \Ind W= U \otimes P$ Using example 3.13.

b) By defintion we can expand $\Ind^G_H(W) = \bigoplus _{\sigma \in G/H} \sigma W$ and

$\Ind^G_K (\Ind^K_H(W)) = \Ind^G_K (\bigoplus _{\tau \in K/H} \tau W) =\bigoplus _{\gamma\in G/K} \gamma (\bigoplus _{\tau \in K/H} \tau W) = \bigoplus _{\gamma\in G/K} \bigoplus _{\tau\in K/H} \gamma \tau W$.

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1 Answer 1

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For (b), note that tensor product is associative:

$$\begin{eqnarray}\text{Ind}_K^G(\text{Ind}_H^KW) &&= k[G]\otimes_{k[K]}\text{Ind}_H^KW \\ &&=k[G]\otimes_{k[K]} (k[K]\otimes_{k[H]}W) \\ &&=(k[G]\otimes_{k[K]} k[K])\otimes_{k[H]}W\\&&=k[G]\otimes_{k[H]} W \\&&=\text{Ind}_H^GW.\end{eqnarray}$$

For (a), use the universal property of the induced representation.

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I don't know what $k[G]\otimes_{k[K]}$ means. Is there a way to do it the way I sort of was? –  Steven-Owen Apr 20 '12 at 20:43
    
@Bruno: I'm working through this same problem, and I'd like to prove (a) by Frobenius reciprocity (which I assume is the universal property you refer to), but I don't see how it follows. Could you elaborate on that? –  Bey Nov 13 '12 at 4:36

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