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](http://rads.stackoverflow.com/amzn/click/0821805991) 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?



Do torsion-free groups give projectionless group ($C^\ast$) algebras?
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](http://www.math.psu.edu/higson/GFA/bc/special_cases.html): 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.


