Take the 2-minute tour ×
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.

Let $G$ be a (finite) group and $N$ a normal subgroup. Given an irreducible representation $\pi$, how can I decompose $Ind_N^G \pi$?

I'd be happy also about a good reference for this.

share|improve this question
    
What kind of answer are you looking for? Would Frobenius reciprocity be what you want? –  Soarer Jul 13 '11 at 7:18
    
What Geoff Robinson points out in his answer, that I needed Frob.rec. + Clifford's theorem=) –  plusepsilon.de Jul 13 '11 at 10:56
add comment

2 Answers

up vote 5 down vote accepted

If you read about Clifford's theorem and its consequences in a text on representation theory (eg Curtis and Reiner: Representation Theory of finite groups and associative algebras Wiley,1962, or Isaacs Character Theory), that should help to answer your question. I presume you are thinking of complex representations, though Clifford theory can be done in a more general context.

The problem breaks into two (or possibly three, depending on your point of view), steps. If the isomorphism type of $\pi$ is $G$-stable, then it is possible (in the complex case) to replace $G$ by a central extension $H$ by a cyclic central subgroup so that $\pi$ extends to an irreducible representation of $H$. So if we consider the case that $\pi$ extends to a representation of $G$, the induced representation breaks up as the tensor product of (the extension of) $\pi$ and the regular representation of $G/N$. If $\pi$ is $G$-stable, but does not extend, we need to look at a twisted group algebra for $G/N$ instead, (this is explained in the texts referred to above).

In general, the elements of $G$ such that $\pi^{g} \cong \pi$ as a representation of $N$ form a subgroup of $G$ called the inertial subgroup of $\pi$, which I denote by $I$. If $I = G$ we are in the case above. If not, then the irreducible components of $Ind_{N}^{G}(\pi)$ are in bijection with the irreducible components of ${\rm Ind}_{N}^{I}(\pi)$ and an irreducible component for $G$ is just induced from the corresponding component for $I$.

share|improve this answer
    
I actually had both of the books you mention on my desk and was also aware of some of the facts. But your presentation helped me a lot in understanding how I should read the theorems in these books. Thanks a lot. –  plusepsilon.de Jul 13 '11 at 10:52
    
You are welcome. Glad you found it helpful. –  Geoff Robinson Jul 13 '11 at 16:22
add comment

The easiest way I know is by modified group projectors technique [Damnjanovic, Milosevic, J.Phys. A 28 (1995) 669-79]:

To shorten notation, $D$ is your induced rep of $\pi$ of $N$, $d$ is arbitrary irreducible representation of $G$. You want to find frequency numbers $f_{d}$ in the decomposition (direct sum over irreducible reps of $G$):

$$D=\sum_{d} f_{d} d.$$

So, find modified projector ($\otimes$ is tenzor product of matrices, $d^*$ conjugated representation $d$):

$$P_d=1/|N| \sum_n \pi(n) \otimes d^*(n).$$

Then: $f_{d}=\mathrm{Tr}\, P_{d}$ (Tr is trace).

You should do this for all irreducible representations of $G$ (of course, you can stop when the dimension of $D$, $|D|=|d| |G|/|N|$, is exhausted).

Note that here only subduced (restricted) to $N$ irreducible reps of $G$ comes. Further, if you have infinite discrete group $N$, the problem can be reduced to generators only, and if it is Lie group, again only Lie generators are used.

share|improve this answer
add comment

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.