Kadison–Kaplansky conjecture

Remarks

The hypothesis that $G$ is torsion-free is needed. If $g \in G$ has finite order $n > 1$, then $p = \frac{1}{n}(1 + g + \cdots + g^{n-1})$ is a projection different from $0$ and $1$.

The conjecture implies the idempotent part of Kaplansky's conjectures for the field $\C$. Indeed $\C[G] \subseteq C^*_r(G)$, and every idempotent of a C*-algebra is similar to a projection.

The canonical trace $\tau$ on $C^*_r(G)$, given by $\tau(x) = \ip{x \delta_e}{\delta_e}$, is faithful, and $0 \le \tau(p) \le 1$ for every projection $p$. So it suffices to show that $\tau(p)$ is an integer. For torsion-free $G$, this integrality follows from surjectivity of the assembly map of the Baum–Connes conjecture. In this way the conjecture is known for free groups, by Pimsner and Voiculescu, for torsion-free groups with the Haagerup property, and for torsion-free hyperbolic groups. It is open in general.