The ℓ¹-Bass conjecture for all discrete groups

The explanation

Take a group with two elements: 1 and a flip s, where flipping twice gives 1. Form sums like a + b·s with complex numbers a and b, and multiply them out using s·s = 1. This is the group ring. Now try e = ½ + ½s. Squaring gives ¼ + ½s + ¼s·s, which is ¼ + ½s + ¼, or ½ + ½s again. So e·e = e. An element with this property is called an idempotent.

The trick used that s has finite order: some power of s is 1. Try the same with a group where nothing but 1 has finite order, such as the whole numbers under addition, written as powers of t. Then (½ + ½t)² is ¼ + ½t + ¼t², which is not ½ + ½t. Kaplansky's idempotent problem asks whether this is always so. In a group with no elements of finite order other than 1, are 0 and 1 the only idempotents? Such a group is called torsion-free.

Bass asked a sharper question. Take an idempotent square matrix over the group ring. Add up its diagonal entries and sort the coefficients by conjugacy class. A conjugacy class is the set of elements that become each other after a change of viewpoint, g turned into h·g·h⁻¹. This sorted sum is the Hattori–Stallings trace. The Bass trace conjecture says its coefficients are zero on every class of infinite orderSource 1. Finite-order classes may be nonzero. In the flip example, e has coefficient ½ on the class of sSource 2.

Many cases were known. Bass proved the vanishing for linear groups, groups made of invertible matricesSource 3. Farrell and Linnell proved the complex statement for elementary amenable groupsSource 4. Berrick, Chatterji and Mislin covered amenable groups through a stronger statement on a larger algebraSource 5. For idempotents, Formanek handled torsion-free groups with a chain condition on subgroupsSource 6. Each method needed a hypothesis on the group.

One paper proves the complex Bass trace conjecture for every group, with no hypothesisSource 7. For torsion-free groups it deduces Kaplansky's conjecture in characteristic zeroSource 8. That means coefficients like whole numbers, rationals or complex numbers, and any domain built on them. A second paper claims the ℓ¹ versionSource 9. There the sums may have infinitely many terms, as long as the sizes of their coefficients add up to a finite number. It claims the trace lives on finitely many finite-order classesSource 10.

The single obstacle is that the group is arbitrary. Earlier methods asked the group to be amenable, hyperbolic or of finite dimension in some sense. The group-ring paper turns one trace coefficient into a number computed on a high-dimensional geometric objectSource 11. It then shows that number must be zero in high enough dimension, by cutting the object into small piecesSource 12. For the ℓ¹ paper, products of coefficients can involve elements far from the identitySource 13. Its estimates have to stay under control as the number of factors grows.

How sure is this? A computer proof checker called Lean has a formal proof of the complex Bass trace conjecture for every group. It also covers the torsion-free trace identity and Kaplansky's conjecture over every characteristic-zero domain. The ℓ¹ result lies outside that check, so for it the paper only claims a proof. No specialist review of either paper has been published.

What it doesn't show. This does not settle Kaplansky's idempotent problem for coefficients of positive characteristic, such as the integers mod 2.