The blockwise Alperin weight conjecture

The explanation

Take the six symmetries of a triangle, the group S₃, and the prime 3. Count its simple representations over a field where 3 = 0. They match the classes of elements whose order is not divisible by 3: the identity and the three flips. So the count is 2.

Now count something local. Look at subgroups whose size is a power of 3: the trivial one, and the three rotations, of size 3. For each, take the larger group that keeps it in place, divided by the subgroup itself. For the trivial subgroup this is all of S₃. Its ordinary characters have degrees 1, 1 and 2, and none is divisible by 3. So it contributes nothing. For the rotations, the quotient has size 2, and both of its two characters count. The local count is 2, matching the first.

Try the prime 2 instead. Here S₃ splits into two blocks, pieces that can be studied separately. One block contains just the two-dimensional representation. The other contains a single simple representation, because the trivial and sign representations become the same when 2 = 0. Each block has count 1. On the local side, the trivial subgroup gives the degree-2 character, matched to the first block. A flip subgroup of size 2 gives one more, matched to the second. The counts agree block by block.

This is Alperin's weight conjecture. It predicts an exact local-to-global count, block by blockSource 1. A weight is a pair: a subgroup of prime-power size, and a character of its local quotient. That character must have defect zero, meaning its degree carries the full power of the prime. Alperin stated it at the 1986 Arcata conferenceSource 2. Early cases included symmetric groups and general linear groups, by Alperin and FongSource 3. Later work reduced it to questions about finite simple groupsSource 4. Feng, Li and Zhang then proved it for several familiesSource 5.

This paper claims the full blockwise conjecture, for every prime and every finite groupSource 6. It proves equality of the two counts, not a natural matching between themSource 7. The proof does not use the classification of finite simple groups or an inductive conditionSource 8. Established consequences, such as inequalities between local and global counts, follow at onceSource 9.

The route is unusual. First, algebra turns the equality into a statement about alternating sums over chains of subgroupsSource 10. This leaves a counting problem about tuples of group elements that satisfy one equationSource 11. The equation is the one that describes loops on a doughnut-shaped surface with punctures. The paper must show these counts are divisible by ever higher powers of the prime.

The one obstacle is uniformity. The divisibility must hold however many punctures there are, with one threshold for all of themSource 12. Only then do the error terms in the trace sums shrink to zeroSource 13. The paper gets this from geometry. It builds special genus-one curves with marked points, over arithmetic where p equals zero. Their covers obey a genus bound that becomes exact for large powers of p. That forces the needed divisibility.

How sure is this? No Lean formalization is listed for this paper. The claim rests on the paper's own long argument, which combines group algebra with the geometry of curves. No specialist review of the paper has been published.

What it doesn't show. This does not give a natural one-to-one matching between simple modules and weights, only equal counts.