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If we consider an induced representation $Ind_{H}^{G}1_{H}$ of finite groups, where $1_{H}$ is the trivial caracter of $H$, how we decompose this representation into irreducible ? There is a decomposition $Ind_{H}^{G}1_{H}=V\oplus W$, where $V$ is formed of canstant functions on $G$ and $W$ is formed of functions $f:G\longrightarrow\mathbb{C}$, $f\in Ind_{H}^{G}1_{H}$, such taht $$\sum_{x\in G/H}f(x)=0$$ but I don't know if $ W $ is irreducible and if it is not irreducible how decomposed there into irreducible ?

The question arises also for profinite groups : If $G$ is totaly disconnected compact group and $H$ is an open subgroup, how to decompose the representation $Ind_{H}^{G}1_{H}$ into irreducibles ?

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Take for $H$ the trivial subgroup, you see that $W$ is reducible in general. Now, if you are happy with an abstract decomposition $Ind_H^G 1_H=\oplus_{\pi\in\hat{G}} m_\pi.\pi$, the answer is given by Frobenius reciprocity. If you want an explicit decomposition of $\mathbb{C}(G/H)$, you probably need a description of the minimal central idempotents in the group algebra of $G$, which is a much more difficult problem. – Alain Valette Feb 19 '12 at 6:17
One should mention the dimension of the centralizer algebra for a permutation module is the number of orbits of the group on ordered pairs. Indeed if f is the permutation character then $\langle f,f\rangle=\langle f^2,1\rangle$. The former is the dimension of the centralizer and the latter the number of orbits on pairs. From this one sees two transitive means the centralizer of W is C. – Benjamin Steinberg Feb 19 '12 at 15:49
up vote 7 down vote accepted

Since it has not really been made completely explicit in the answers above, I point out that the specific question of whether your invariant subspace $W$ is irreducible or not is easier to check, at least in the case of finite groups. It rarely is. As is implicit in the earler answers, there is a formula due to Mackey for the norm (squared) of the induced character (and Mackey's formula works in more general contexts. In the case of finite groups, it is a double application of Frobenius reciprocity). We have $\langle {\rm Ind}_{H}^{G}(1), {\rm Ind}_{H}^{G}(1) \rangle = |T|$ , where $T$ is a full set of representatives for the distinct $(H,H)$-double cosets (so $G$ is the disjoint union $\cup_{t \in T} HtH ).$ The space $W$ is irreducible if and only if this norm is $2,$ which happens if and only if there are just two $(H,H)$-double cosets. This in turn happens precisely when $G$ is doubly transitive in the usual action on the ( say right) cosets of $H.$

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You will have to compute the algebra of intertwiner $$A = Endo_G( Ind_H^G 1) \cong \bigoplus\limits_{\pi} M_{m(\pi) \times m(\pi)}(\mathbb{C}),$$ where $Ind_H^G 1 = \bigoplus_\pi m(\pi) (V_\pi,\pi)$.

There are two thing, which I usually use:

  1. There is a $*$ algebra isomorphism $$ A \cong C[ H \backslash G /H]$$

  2. Use Frobenius reciprocity and the induction restriction formula $$ A \cong Hom_H( 1, Res_H Ind_H^G 1) = \bigoplus_{\gamma \in H \backslash G /H} Hom_H ( 1, Ind_{H^g \cap H}^H 1).$$

For locally profinite groups, you do the same thing and everything what I said, remains true by replacing group rings by $C_c^\infty$ and isomorphism by dense embeddings.

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I might have confused, left and right adjoints in the Frobenius reciprocity, but note that, if the induced representation is admissible representation, compact induction and induction are isomorphic, do it does not matter here. – Marc Palm Feb 19 '12 at 6:51

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