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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?

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    $\begingroup$ Note that this would be false for the full $C^*$-algebra $C^*(G)$: if $G$ is an infinite group with Kazhdan's property (T), then $C^*(G)$ has non-trivial projections, even if $G$ is torsion-free... $\endgroup$
    – Alain Valette
    Jul 2, 2011 at 20:05
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    $\begingroup$ even though this is an older question: here arxiv.org/pdf/1801.06974.pdf theorem 4.1 is written down a proof that for a torsion-free nilpotent discrete group, $C^*(G)$ has no non-trivial projections $\endgroup$
    – Sabrina Gemsa
    Jun 5, 2018 at 12:53

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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.

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  • $\begingroup$ thanks! i had no idea that this was a conjecture already. i just thought it was nontrivial... $\endgroup$
    – Dave Penneys
    Oct 31, 2009 at 21:50
  • $\begingroup$ The link seems to be dead - and I wasn't able to find it in the Wayback Machine. (The new website of the author seems to be here - but I wasn't able to guess what exactly was linked there.) $\endgroup$
    – Martin Sleziak
    Jul 24, 2022 at 11:37
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The book Introduction to the Baum-Connes Conjecture, by Alain Valette, begins with a discussion of this problem.

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