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There are already some questions with almost the same title, but they are more restrictive.

Let $G$ be a finite group, $H$ a subgroup, $V$ an irreducible representation of $H$, and $W=Ind_H^G V$ the induced representation.

Is there a simple way to compute the number of irreducible components of $W = Ind_H^G V$?

Let me precise my question by saying what kind of answers I would hope for. There is a well-known description of $Res_G^H Ind_H^G V = Res_G^H W$ as follows. Let $S$ be a set of representatives of the double classes of $H \backslash G / H$. For $s \in S$, let $H^s$ be the group $sHs^{-1} \cap H$ and $V^s$ the representation of $H^s$ on the space $V$ where $x \in H^s$ acts by $s^{-1}xs$ on $V$. Then $$Res_G^H W = \oplus_{s \in S} Ind_{H_s}^H V^s.$$ Using this description, one gets using Frobenius reciprocity twice that $$Hom_G (W,W) = Hom_H(V,Res_G^H W) = \oplus_{s \in S} Hom_H(V, Ind_{H^s}^H V^s) = \oplus_{s \in S} Hom_{H^s} (V,V^s)$$ which implies the well-known criterion of Mackey: $W$ is irreducible if and only if for all $s \in S$, $V$ and $V^s$ have no $H^s$-irreducible subrepresentations in common. So assuming we can compute $S$, the groups $H^s$, the representations $V^s$, and how $V$ and $V^s$ decomposes as representations of $H^s$ (this may be hard in practice, but let's assume we can do that), then we know when $W$ is irreducible, and in general we know how to determine the dimension of $Hom_G(W,W)$. (Of course all of this is standard cf. Serre Linear Representations for instance).

Yet this falls short of answering the question, because if for example you know that the dimension of $Hom_G(W,W)$ is $4$, that doesn't tell you if $W$ is the sum of 4 non-isomorphic irreducible representations, or of 2 copies of the same irreducible representation. What I'd like would be a way to tell us, by looking at the $H^s$ and the $V^s$ etc, of determine which one it is. Or more generally, how to "read" the decomposition of $W$ into irreducible reps. in terms of the $V^s$, $H^s$ etc. If it is not possible, I would also like an explanation why.

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2 Answers 2

The answer to the highlighted question is almost certainly no, depending on what "simple" means in this context. While it's reasonable to look for refinements of Mackey's methods (as I believe people have done for a long time), it's necessary to look at well-studied classes of finite groups in order to get perspective on what is possible.

First, I'm assuming you are working in the setting of a splitting field of characteristic 0 for $G$ and its subgroups (with $\mathbb{C}$ being a convenient though very large choice). I also assume you are looking for the number of distinct irreducible constituents of an induced character. Here you run into the immediate problem of dealing with induction from the trivial character of the trivial subgroup: this gives all irreducible characters, each with the multiplicity of its own degree. But it takes some subtle reasoning to see that the total number of distinct irreducible characters equals the number of conjugacy classes in $G$. (And what is that number for symmetric groups?) In the case of a finite simple group of Lie type, the computation of the number of classes is delicate and requires case-by-case study as well.

In the case of groups of Lie type, there are many possibilities for induction from nontrivial subgroups (including Gelfand-Graev induction), but it's usually quite difficult to work out the number of constituents much less their degrees and characters. For instance, this is already subtle when $H$ is a Borel subgroup and you start with its trivial character. By now a lot is known about groups of Lie type, but the actual results of induction from a variety of subgroups don't suggest any easy general pattern and often rely on external notions such as Iwahori's "Hecke algebras".

The point is not to discourage further study of induction methods, but rather to emphasize how diverse the actual results are as the groups and subgroups vary. Even Mackey's original methods are not usually easy to apply in practice even though they provide some theoretical unity.

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Thanks for this thoughtful answer. Yes, I was thinking of complex representations, I should have said it. So we could say that decomposing induced representations is an art, not a technics, couldn't we ? –  Joël Feb 25 '13 at 2:06
    
@Joel: Speaking as a nonexpert in finite group representations, I'd certainly view it as an art form. –  Jim Humphreys Feb 25 '13 at 18:11

The proof of Mackey's theorem on intertwiners actually tells you how to construct the endomorphism algebra of an induced representation, not just its dimension. So, if you work a little harder, you may be able to get the additional information that you want.

To see how this algebra can be found, you may refer to my notes http://www.imsc.res.in/~amri/html_notes/notesch1.html#x4-70001.4 A convolution product can be defined on $\Delta$'s in the notes, which will correspond to multiplication in the endomorphism algebra.

Added:

An interesting special case is decomposing parabolically induced representations of general linear groups over finite fields, which is beautifully explained in the notes of Howe and Moy Harish-Chandra homomorphisms for $p$-adic Groups.

Such algebras are often called Hecke algebras, and there is a vast literature on them.

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Dear Amritanshu, thanks for your answer. But Jim Humphreys' answer convinced me that you need to work more than "a little harder" to get that kind of information - as he mentions, if $H=1$, $W$ is the regular representation, and the method of Mackey gives us Dim $Hom_G(W,W)=|G|^2$, but one need extra (and somewhat harder) arguments to get the number of irreducible components in $W$, which certainly cannot be read on the decomposition of $Res_G^H W$, which is just a sum of $|G|$ trivial representation. Could you please explain, in a special case, of you envision this "work a little harder"? thx –  Joël Feb 25 '13 at 14:09
    
Dear Joël, thanks for your comment; I have edited my answer. The main thing to note is that the idea behind Mackey does not just give the dimension of the endomorphism algebra, but also the structure of the algebra; so if $H=1$, you don't just get the dimension of the regular representation, but actually its endomorphism algebra, from which you can (in principle) deduce the representation theory. –  Amritanshu Prasad Feb 26 '13 at 5:52
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Just thought I'd add the following note: As an example of just how complicated this can become you might want to look at the article "Induced Cuspidal Representations and Generalised Hecke Rings" by Howlett and Lehrer. Even in this situation (which is reasonably optimal in the theory of finite reductive groups) they are left with an unknown 2-cocycle which one still has to compute to get an explicit description of the endomorphism algebra. –  Jay Taylor Feb 26 '13 at 7:36

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