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Let $G$ be a group, usually infinite. I am interested in finite-dimensional complex unitary representations of $G$, i.e. group homomorphisms $G \rightarrow U_n(\mathbb{C})$. The category of these representations is an exact category, and so we can form the Grothendieck group $K_0(G)$ of this category. If $H$ is a finite-index subgroup of $G$, we have an induction map $K_0(H) \rightarrow K_0(G)$. Given a collection of subgroups $H_i$ of $G$, we can ask whether the map obtained by induction

$\oplus_i K_0(H_i) \rightarrow K_0(G)$

is onto, or onto after tensoring with $\mathbb{Q}$. Let us say that the collection is good if this map is onto. If $G$ is finite, for example the Brauer induction theorem or the Artin induction theorem give answers: The collection of elementary subgroups of $G$ is good. If $G$ is infinite and maps to a finite group $F$, we can pull back the elementary subgroups of $F$ to $G$ which yields a good collection of subgroups of $G$.

My question is: Are there other known cases of infinite groups $G$ with a good collection of subgroups? Also, I would be interested in any references where finite-dimensional representations of infinite groups are studied.

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up vote 2 down vote accepted

I believe that all examples must come from finite quotients:

  1. A collection of subgroups is necessarily good if the element $1\in K_0(G)$ is in the subgroup generated by the images of the induction maps $K_0(H)\to K_0(G)$. That's because these images are ideals, using the formula $V\otimes Ind(W)=Ind(Res(V)\otimes W)$.

  2. Therefore every good collection contains a finite good collection. And of course for any collection of finite index subgroups $H_i$ there is a finite index normal subgroup $N$ of $G$ that is contained in them all.

  3. And then the collection ${H_i}$ will be good in $G$ if and only if the collection $H_i/N$ is good in the finite group $G/N$. (This last requires a little argument splitting representations of $G$, or $H_i$, into the part fixed by $N$ and its orthogonal complement and noting that this splitting is compatible with induction.)

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Yes you are right. Thank you! –  Fabian Lenhardt Apr 26 '12 at 14:19
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Hi, its seems that your question is rather soft, so I will give some arguments, which come close to a description you seem to search for.

I'll start with Clifford theory, Kirillov orbit method or the Mackey Machine in the case of compact groups. I will not use the notation $K_0$, since you can argue with irreducible representations directly in this context, so with the generators of $K_0$. I learnt this theory from this paper: http://arxiv.org/abs/0807.4684. It's theorem 2.1.

Let $G$ be a compact group and $N$ a closed, normal subgroup, then $G$ acts by conjugaction on $N$, hence on the "set" of its irreducible representations. Let $\pi$ be an irreducible representation of $G$.

Facts:

  1. Cliffords theorem: $Res_N \pi$ contains exactly one $G$-orbit $\{ \sigma \}$ of irreducible representations of $N$.
  2. Let $G_\sigma$ be the stabilizer of $\sigma$, then we have a one-to-one correspondance of irred. reps. $\pi$ of $G$, which contain $\sigma$ in $Res_N \pi$, and irred. reps $\pi' $ of $G_\sigma$, which contain $\sigma$ in $Res_N \pi$, by the induction $$ \pi' \mapsto Ind_{G^\sigma}^{G} \pi.$$

This theory is in particular helpful, when $N$ is abelian. In fact, it will work equally well for finite index or cocompact normal groups, which are type 1. In general you should expect a direct integral.

To sum up, we have an isomorphism

$$ \bigoplus_{\{ \sigma \} } K_0( G^\sigma) \cong K_0(G)$$

Further comments mostly for reductive groups over local fields and Lie groups (don't take them to serious though):

  • The Mackey machine is available in the case of unitary representation of locally compact separabel groups, where $N$ is a type I group. It is Mackey's paper on group extensions. I think that it is hard to single out the finite dimensional representations.

  • The finite dimensional representation of a reductive Lie group are glued together from finite dimensional representations of the maximal compact subgroup, but most of them are not unitary (Weyl's unitary trick). For $SL_2(\mathbb{R})$, the only unitary, finite dimensional representation is trivial. For the general linear group they factor through the determinant. So I guess, you might want to drop unitarizability here. All finite dimensional representation of reductive Lie groups appear as sub-module, or sub-quotient of parabolic induced representations, and are almost never unitary.

  • I recall that there is a statement that the unitary representation theory of a general Lie group depends essentially on the representation theory of reductive group and nilpotent groups. I know that this is a rigorous statement for algebraic group, since there every group is an extension of a reductive one by a unipotent one.

  • I'll give one concrete example here: Consider the upper triangular matrices of $\mathbb{R}$. All finite dimensional unitary representations will be unitary representations of the diagonal matrices. In general, it seems that you only have to answer the question for reductive groups.

  • Also index theory has been used to construct discrete series by Atiyah-Schmid, but I do not know much about that. For $SL_2(\mathbb{R})$, knowing in which parabolic induction to find the finite-dimensional reps is equivalent to finding the discrete series reps. Discrete series are the "prime objects".

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Regarding comment 3, take a free group, or a surface group, or a lamplighter ... Is it possible that all these statements you allude to are only for certain Type I groups? –  Yemon Choi Apr 20 '12 at 17:53
    
Also, I maybe being slow here, but what is the relevance of the Baum-Connes conjecture here? –  Yemon Choi Apr 20 '12 at 17:53
    
better now?.... –  plusepsilon.de Apr 20 '12 at 19:00
    
much better, thanks –  Yemon Choi Apr 20 '12 at 19:46
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