Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am reading a book of Fulton and Harris "Representation theory, a first course". Now it's all about representation theory of finite groups, and there is one exercise, which I can't solve:

If $U$ is a representation of $G$ and $W$ a representation of $H$ (here $H$ is a subgroup of $G$; further induced and restricted representations are all with respect to $H$), show that $U \otimes \operatorname{Ind} W = Ind(\operatorname{Res}U \otimes W)$. In particular, $Ind(\operatorname{Res}U) = U \otimes P$, where $P$ is the permutation representation of $G$ on $G/H$.

Actually, it's seems that it only requires to use definitions, but I stuck with it! The thing I don't really understand is : as I see, restriction doesn't serve the whole information about the original representation, but the formula says other things.

If it's possible, give a detailed explanation, please. Thank you very much.

P.S. Sorry for duplicating this question. There were 2 aswers given before: 1 and 2. but the first was really hard to comprehend, and there was no full answer in the second. (The hint was to use Frobenius reciprocity, but the thing is in the book Frobenius reciprocity goes only after this exercise and I am not capable to go on until I completely understand this problem)

share|cite|improve this question
The answer to the second question you link tells you what the isomorphisms are explicitly. Could you say what more you are looking for? – m_t_ Jan 30 '13 at 15:51
There are 2 issues: first is dealt with module map $\alpha$ : why it sends elements to $kG \otimes H(W \otimes U_{H})$, but not to $kG \otimes H(W \otimes U)$?And the second is a bit psychological: it doesn't seems that restriction serves the whole information about the original representation. (apart from the $U_{H}$ I do understand everything from the proof, but psychologically I can't admit it, because there are many representations, restrictions of which collides (there are trivial and alternating representations of $Sym(\[n\])$, restrictions to $A_n$ of which equal),despite the formula. – Sergey Finsky Jan 30 '13 at 16:40
$U$ and $U_H$ are the same as vector spaces, so there's not really an issue there. You're certainly right that information is lost when you restrict to a subgroup, in the sense that two different modules can have the same restriction. I suppose the moral is that information is lost when tensoring too: you can't hope to recover $U$ from $U$ tensored with something. You could think of this result as telling you how much is lost when you apply $-\otimes \operatorname{Ind}(W)$: if $U_H$ and $V_H$ are the same, so are $U \otimes \operatorname{Ind}(W)$ and $V\otimes (\operatorname{Ind}(W)$ – m_t_ Jan 30 '13 at 16:46
For the 1st problem: (at first, just to be sure that I understood everything correctly, we need $H$-modules $\alpha$ and $\beta$, don't we?) and the problem is: what would happen if we wanted to send some element of the form $(x \otimes_H w) \otimes yu$, where $x^{-1}y \ne 1 mod H$, we would send it to $x \otimes (w \otimes_H x^{-1}yu$, but $H$ acts differently on $U$ and $x^{-1}yU$ in general case... That's seems to be a problem here. For the second part: let we consider $U \otimes (\sum_{g \in G} gx)$ for some $x \in W$, it seems that we can recover action of $G$ on $U$. – Sergey Finsky Jan 30 '13 at 17:23
$\alpha$ and $\beta$ must be maps of $kG$-modules, not just $kH$-modules. I don't understand your next objection, indeed $U = x^{-1}yU$ since both $x$ and $y$ act as bijections. As for the last bit, $W$ is an $H$-module not a $G$-module so you are not able to form that sum. – m_t_ Jan 30 '13 at 17:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.