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.

  • $\begingroup$ What kind of answer are you looking for? Would Frobenius reciprocity be what you want? $\endgroup$ – Soarer Jul 13 '11 at 7:18
  • $\begingroup$ What Geoff Robinson points out in his answer, that I needed Frob.rec. + Clifford's theorem=) $\endgroup$ – Marc Palm Jul 13 '11 at 10:56

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 $I$.

  • $\begingroup$ 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. $\endgroup$ – Marc Palm Jul 13 '11 at 10:52
  • $\begingroup$ You are welcome. Glad you found it helpful. $\endgroup$ – Geoff Robinson Jul 13 '11 at 16:22

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.


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