Counterexamples to Baum–Connes and Kadison–Kaplansky

The explanation

Start with a group that has two elements, 1 and g, where g times g is 1. Build formal sums a·1 + b·g with complex numbers a and b. Now take e = (1 + g)/2. Squaring gives (1 + 2g + g²)/4 = (2 + 2g)/4, which is e again. An element equal to its own square is a projection. The trace of a sum is its coefficient on 1, so e has trace 1/2. This projection exists because g has finite order. An element of finite order other than 1 is called torsion.

Torsion-free groups seem to behave differently. Take the integers under addition. Their operator algebra turns out to be the continuous functions on a circle. A projection is then a function taking only the values 0 and 1. On a connected circle, such a function must be constant. So the only projections are 0 and 1.

The Kadison–Kaplansky conjecture says this holds for every torsion-free group. The algebra is the reduced group C*-algebra, the closure of the group's sums acting on square-summable sequences. Pimsner and Voiculescu proved the conjecture for free groupsSource 1. A main route to it is the Baum–Connes conjecture. That conjecture says an assembly map, from the group's topology to the algebra's K-theory, is an isomorphismSource 2. K-theory sorts projections into classes. Baum and Connes circulated it in 1982 and published it in 2000Source 3. It is known for large classes of groups, including all countable amenable groupsSource 4. Earlier counterexamples of Higson, Lafforgue and Skandalis used extra coefficients. They do not break the coefficient-free formSource 5. The obstacle was to find a group, with no coefficients, whose algebra escapes the topology.

The three papers claim to do that. The first builds a finitely generated torsion-free group whose assembly map sends a class of infinite order to zeroSource 6. The group is finitely generated, but finite presentation is not part of the conclusionSource 7. It disproves neither maximal rational injectivity nor the classical Novikov conjectureSource 8. The second builds a finitely generated torsion-free group with a projection other than 0 and the identitySource 9. Its trace lies strictly between 0 and 1/2. That projection is in the completed algebra, not in the group ring of finite sumsSource 10. The third uses Lück's theorem: every assembly class has a rational traceSource 11. It then builds a finitely generated group with a projection of irrational traceSource 12. That projection cannot come from assembly. The constructions use expander graphs, sparse networks that are still very well connected.

How sure is this? A computer proof checker called Lean has not checked any part of it. Each claim rests on a long chain of graph, spectral and operator estimates. No specialist review of the papers has been published.

What it doesn't show. This does not disprove the classical Novikov conjecture or maximal rational injectivity, and it gives no finitely presented example.