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There are many versions of the Baum-Connes conjecture (the original, coarse, with coefficients, etc.). I would like to know what group theory results are needed in order to prove or disprove one of these conjectures.

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    I'm sure a lot has happened in the last 6 years, but Wolfgang Lück has a rather extensive survey in the Handbook of -theory (containing a chapter about the status of the conjectures). It focuses more on the positive aspects, though. You can find it at http://arxiv.org/abs/math/0402405 – Theo Buehler Jan 31 '11 at 22:49
  • Vincent Lafforgue has (also) written some expository articles on Baum--Connes, one of them fairly recent. (Available on his website http://www.math.jussieu.fr/~vlafforg/)

    I appologize in advance, if this remark is to general to be useful; I just happended to remember this fact, but am not well-informed on this subject.

     –  Jan 31 '11 at 22:55
  • Paul Baum gave a series of lectures on BC at Vanderbilt in Spring 2008 (I believe you were on sabbatical, Mark). I recall him saying that he expected the original BC conjecture to be true for all exact groups, but that he didn't know of a counterexample to original conjecture. But I guess his point was that one should study non-exact groups. – Dan Ramras Feb 06 '11 at 21:34
  • @Dan: thanks! But existence of a compact aspherical manifold with non-exact fundamental group would not automatically disprove the conjecture? –  Feb 06 '11 at 23:55
  • I don't think Baum was suggesting that every non-exact group with a nice classifying space would give a counterexample. He just indicated that this is a good place to look for counterexamples. – Dan Ramras Feb 07 '11 at 23:33

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I want to second David Ben-Zvi's recommendation of Nigel Higson's ICM lecture, though there is a bit of progress that has been published since it was written. Fortunately, you needn't venture further than Nigel's webpage to learn more: his paper "Counterexamples to the Baum-Connes Conjecture" with Lafforgue and Skandalis gives some more recent counter-examples to BC with coefficients and coarse BC.

A few other comments...

As far as I know, the ordinary Baum-Connes conjecture is known to be true for more or less every group that anyone can name. The experts (of which I am not one) seem to suspect that the conjecture is not true in general but that finding a counter-example requires new techniques for constructing discrete groups.

There are a few fairly general techniques for proving the conjecture. The first and most conceptually satisfying is the "Dirac / dual Dirac" method which involves using KK-theory to imitate Atiyah's operator theoretic proof of Bott periodicity. As far as I know there is no evidence that this technique is not powerful enough to prove injectivity in general, but there are examples of groups for which BC is true but the Dirac / dual Dirac method cannot prove surjectivity. I suggest Higson and Guentner's "Group C* algebra's and K-theory" for a discussion of these issues - the paper starts more or less from scratch and reads like a short textbook.

The second method is more of a philosophy: BC tends true for groups which act on nice spaces in a minimally nice way. For example, it is known to be true for a-T-menable groups (see "E-theory and KK-theory for groups which act properly and isometrically on Hilbert space" by Higson and Kasparov). One of the frontiers of the conjecture is in the realm of p-adic groups - I think Paul Baum is still quite active here - and one of the basic techniques seems to be to get p-adic groups to act on buildings.

I hope that helps! For more advice, I would really suggest talking directly to Nigel - I don't think BC is quite as central to his interests these days, but I would guess he's as up on the subject as anyone.

Paul Siegel
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  • @Paul: I know about the paper by Higson, Lafforgue and Skandalis. In particular, they are using Gromov's f.g. group containing expander to construct a counterexample to one of the versions of Baum-Connes conjecture. I wonder what kind of groups do they need to construct a counterexample to the main (original) Baum-Connes conjecture. As far as I know the conjecture is not proved even for uniform lattices in $SL_3(\mathbb{R})$. But perhaps those groups are just not wild enough for this. So what kind of wild groups are needed? –  Feb 01 '11 at 14:07
  • It's a good question, and I can only be of limited use in this regard. I have asked a few people (including Nigel, I think) outright if there are any known nontrivial geometric sufficient conditions on a group which would make it a counterexample, and the answer has consistently been a qualified no. It seems that for the conjecture to be false for a group there has to be some sort of mismatch between its large scale geometry and its harmonic analysis. To try to speculate any further would be straying too far from my (already limited) comfort zone. – Paul Siegel Feb 01 '11 at 17:32
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Nigel Higson's 1998 ICM address, "The Baum-Connes Conjecture" (available eg. on his website or on the IMU page) is a great survey of the subject (pre-Lafforgue). In general I highly recommend Higson's writings on the subject (relevant papers, books and talk slides are available at the above link).

David Ben-Zvi
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This January in New Orleans Paul Baum gave a rather extensive survey of the history and status of Baum/Connes, so I am guessing that he has a historical survey written up, or quasi-written up, since I am not seeing it on his web page), so I would strongly suggest just asking him. Sadly, I don't believe he is a MO participant.

Igor Rivin
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    I am not sure he is the main "player" (it is like asking S. P. Novikov about the current status of Novikov's conjecture). Perhaps some Penn State students know the answer. There are several of them on MO. –  Jan 31 '11 at 22:05
  • I don't know about Novikov, but Baum gave a rather detailed 1 hour talk, so it sounded like he was very much in the loop. – Igor Rivin Feb 01 '11 at 01:08
  • @Igor: OK, then this lecture should have been recorded and saved. It is not difficult. Strange that AMS does not do it. –  Feb 01 '11 at 01:20
  • @Mark: actually, I am not certain AMS does not do it, but I think that if we started to discuss all the ways in which the AMS is behind the times, that would take us a very long time... – Igor Rivin Feb 01 '11 at 01:36
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    @Igor: OK, I just paid my AMS membership fee, and feel like complaining. Similar thing happens on April 15 when I pay my taxes.:) I don't think AMS records invited lectures at AMS meetings. Anyway I will ask Braun and other people who are mentioned here. –  Feb 01 '11 at 04:21