Do torsion-free groups give projectionless group ($C^\ast$) algebras? - MathOverflow

Do torsion-free groups give projectionless group ($C^\ast$) algebras?

2009-10-30T20:07:07Z

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 Davidson's book for a proof when $n=2$.

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.

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.

EDIT: changed typestting. still some bugs... help please?

Answer by Yemon Choi for Do torsion-free groups give projectionless group ($C^\ast$) algebras?

2009-10-30T22:12:10Z

Heh, you've picked an open problem: this is the Kadison-Kaplansky conjecture... I would answer it, but first I have to find a sufficiently big margin in which to write the proof.

To be less flippant, it is known to follow (but I don't understand exactly how) from the Baum-Connes conjecture: thus, if a torsion-free discrete group satisfies BC, then its reduced group C*-algebra contains no non-trivial projections.

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.

M. Puschnigg, The Kadison-Kaplansky conjecture for word-hyperbolic groups. Invent. Math. 149 (2002), no. 1, 153--194.

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.

Answer by MTS for Do torsion-free groups give projectionless group ($C^\ast$) algebras?

2009-11-01T01:04:42Z

The book Introduction to the Baum-Connes Conjecture, by Alain Valette, begins with a discussion of this problem.