Let $G$ be a discrete group and consider the reduced group C* algebra $C_r^\ast(G)$, viewed as an algebra of bounded operators on $\ell^2(G)$ by the regular representation. The canonical trace on $C_r^\ast(G)$ is simply defined to be: $\tau(T) = (T \delta_e, \delta_e)$ where $\delta_e$ is the basis vector of $\ell^2(G)$ corresponding to the identity. This trace gives rise to a homomorphism $\tau_\ast: K_0(C_r^\ast(G)) \to \mathbb{R}$ on K-theory.

It is a very deep topological fact that the image of $\tau_\ast$ consists of integers for a huge class of groups, specifically those groups for which the Baum-Connes conjecture is true. To see this, let $M$ be a closed manifold with fundamental group $G$, let $\mu: K_0(M) \to K_0(C_r^\ast(G))$ be the assembly map on K-homology, and note that $\tau_\ast \circ \mu$ agrees with the (integer valued) index map $K_0(M) \to \mathbb{R}$. Thus $\tau_\ast$ is integer-valued so long as the assembly map is surjective, and the Baum-Connes conjecture asserts that it is an isomorphism.

Here are my questions:

Are there easy proofs of the integrality of $\tau_\ast$ in (nontrivial) special cases, or at least proofs that don't resort to heavy-duty topological machinery?

Even better,

Are there groups for which the Baum-Connes conjecture is not known but $\tau_\ast$ is known to be integer-valued?

These two questions stem from my intuition that this should be a problem in analysis or possibly representation theory rather than topology. But in the likely event that this intuition is wrong...

Is the integrality of $\tau_\ast$ related (or equivalent) to any more transparently topological phenomena?

  • $\begingroup$ I'm sure you've already thought of this, but does Joel Cohen's argument in ams.org/mathscinet-getitem?mr=546507 give a proof if integrality for nonabelian free groups? I phrase this as a question because it's been a long while since I read the paper and my current uni library subscription doesn't give online access to "old" JFA articles $\endgroup$ – Yemon Choi Mar 15 '11 at 0:38
  • $\begingroup$ As for your second question, I think integrality of the trace would be enough to prove the Kadison-Kaplansky conjecture (for a given group), so you are now into the territory of groups which are known to satisfy Kad-Kap but not known to satisfy BC. Lattices in higher-rank Lie groups might be candidates, from my dim recollections of hearing people talk about these things. $\endgroup$ – Yemon Choi Mar 15 '11 at 0:42
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    $\begingroup$ One would conjecture only for torsionfree groups that the trace is $\mathbb Z$-valued. $\endgroup$ – Andreas Thom Apr 10 '11 at 16:27

It is shown that all torsion-free cocompact lattices in any SL$(n,\mathbb{C})$ produce the desired integrality of the trace at least on the group ring, under a very weak Spin$^c$-assumption, in my thesis. (More precisely, the trace on the algebraic K-theory of the group ring with coefficients in the ring of rapidly decaying matrices is shown to be integer-valued, and this implies Kadison-Kaplansky for the group ring.) Few, if any, of these groups are known to satisfy Baum-Connes for $n\ge 3$. I think none of these are known for $n\ge 4$, and for $n=3$ they are known by the work of V. Lafforgue. For SL$(2, \mathbb{C})$, any discrete subgroup satisfies Baum-Connes shown by Julg and Kasparov, as it acts on hyperbolic space.

The Kadison-Kaplansky-conjecture, in turn, was known much earlier for these groups; this goes back to Hyman Bass. Moreover, I think that the integrality on $\ell^1 \Gamma$ for many lattices follows from the proof of the Bost conjecture, due to V. Lafforgue.

Therefore, the answer depends strongly on the type of algebra you require the trace to be defined on. For the $C^\ast$-algebra, the question is completely open in the cases not covered by Baum-Connes, at least to my knowledge (and I would be excited to hear to the contrary!)

The idea of the integrality proof in my thesis is related to the standard Baum-Connes proof: it is a careful version of the Dirac-dual Dirac construction, smooth enough for (a variant of) cyclic homology, and then resort to the integrality of an elliptic operator's index, as you said. Maybe this answers part of your question.

Perhaps one can say, therefore, that since all known proofs of trace integrality are geometric-analytic in nature, the problem can be viewed as a geometrical one as opposed to a representation theoretical one (at least insofar as the group ring is concerned).

By the way, finitely generated free groups satisfy Baum-Connes; for instance because they have the rapid decay property (or are the archtypical Gromov-hyperbolic groups), however, one can, as far as I remember, directly understand the right hand-side and write down generators of the K-theory, see Bruce Blackadar's book.

You can read my thesis here: http://arxiv.org/abs/math/0612023 .


For abelian groups $G$ (maybe a trivial case?), the canonical trace is given by integration over the Pontryagin dual $\hat{G}$ (with respect to normalized Haar measure), so we have equivalences:

(1) $\tau_*$ is integer-valued on $K_0(C(\hat{G}))$; (2) the Kaplansky-Kadison conjecture holds for $C^*_r(G)$; (3) $\hat{G}$ is connected; (4) $G$ is torsion-free.

In general, for non-abelian $G$, proving the integrality of the trace is done by means of an index theorem, so that it is deeply related to surjectivity of the Baum-Connes map. This is to say that, to my knowledge, there are no examples of groups where integrality of the trace is known, but Baum-Connes is not. There are a few cases however, where the Kaplansky-Kadison conjecture has been established independently of the Baum-Connes conjecture. This is the case for torsion-free hyperbolic groups, see Michael Puschnigg, The Kadison-Kaplansky conjecture for word-hyperbolic groups. Invent. Math. 149, no. 1, 153--194, 2002.


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