MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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.)

share|cite|improve this answer
Yes you are right. Thank you! – Fabian Lenhardt Apr 26 '12 at 14:19

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: 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 $G$.


  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".

share|cite|improve this answer
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?.... – Marc Palm Apr 20 '12 at 19:00
much better, thanks – Yemon Choi Apr 20 '12 at 19:46

Your Answer


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.