Do torsion-free groups give projectionless group (\$C^\ast\$) algebras? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T07:50:16Z http://mathoverflow.net/feeds/question/3468 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/3468/do-torsion-free-groups-give-projectionless-group-c-ast-algebras Do torsion-free groups give projectionless group (\$C^\ast\$) algebras? Dave Penneys 2009-10-30T20:07:07Z 2010-01-21T19:14:49Z <p>One of the reasons I study von Neumann algebras is that they always have plenty of projections. There are many projectionless \$C^\ast\$-algebras (\$0\$ and possibly \$1\$ are the only projections), but the von Neumann algebras they generate must have nontrivial projections (unless it's just the complex numbers, of course). A good example of this is the reduced group \$C^\ast\$-algebra of any free group \$F_n\$. If \$n=1\$, then \$C_r^\ast(Z)\cong C(S^1)\$ via the Gelfand transform, which is clearly projectionless. If \$n\geq 2\$, the proof is fairly complicated. See <a href="http://www.amazon.com/Algebras-Example-Fields-Institute-Monographs/dp/0821805991" rel="nofollow">Davidson's book</a> for a proof when \$n=2\$.</p> <p>If \$G\$ is a torsion-free group, is the reduced group \$C^\ast\$-algebra of \$G\$ projectionless? This \$C^\ast\$-algebra always contains the group algebra \$C[G]\$, so a simpler question is whether \$C[G]\$ is projectionless if \$G\$ is torsion-free.</p> <p>Note that torsion-free is a necessary condition as one gets a projection from summing up the elements in the cyclic group generated by a torsion element and dividing by the order of the element.</p> <p>EDIT: changed typestting. still some bugs... help please?</p> http://mathoverflow.net/questions/3468/do-torsion-free-groups-give-projectionless-group-c-ast-algebras/3478#3478 Answer by Yemon Choi for Do torsion-free groups give projectionless group (\$C^\ast\$) algebras? Yemon Choi 2009-10-30T22:12:10Z 2009-10-31T23:57:01Z <p>Heh, you've picked an open problem: this is the <em>Kadison-Kaplansky conjecture</em>... I would answer it, but first I have to find a sufficiently big margin in which to write the proof.</p> <p>To be less flippant, it is known to follow (but I don't understand <em>exactly</em> how) from the <a href="http://www.math.psu.edu/higson/GFA/bc/special%5Fcases.html" rel="nofollow">Baum-Connes conjecture</a>: thus, if a torsion-free discrete group satisfies BC, then its reduced group C*-algebra contains no non-trivial projections.</p> <p>Trying to answer this question was, I think, one of the original motivations of Connes and others in some of the older work on cyclic cohomology and souped-up versions thereof. See e.g.</p> <p>M. Puschnigg, The Kadison-Kaplansky conjecture for word-hyperbolic groups. Invent. Math. 149 (2002), no. 1, 153--194.</p> <p>for some relatively recent work on those lines. Since I'm not an expert, I'd suggest Googling some combination of Kadison-Kaplansky and Baum-Connes and going from there.</p> http://mathoverflow.net/questions/3468/do-torsion-free-groups-give-projectionless-group-c-ast-algebras/3628#3628 Answer by MTS for Do torsion-free groups give projectionless group (\$C^\ast\$) algebras? MTS 2009-11-01T01:04:42Z 2009-11-01T01:04:42Z <p>The book <em>Introduction to the Baum-Connes Conjecture</em>, by Alain Valette, begins with a discussion of this problem.</p>