A perfect fingerprint, proved at last?
Two AI preprints claim that von Neumann algebras of lattices like SL(3,ℤ) remember the group
Covers result 286 in OpenAI’s catalog: all 2 papers. See its connections to other results and the work it cites.
3:31. English captions available. The video loads from YouTube when you press play.
Chapters
The claim, in plain words
Every group has a von Neumann algebra, an algebra of operators that records how the group acts on itself. When every element other than the identity has infinitely many conjugates, this algebra is a factor, with a trace that measures the size of its projections continuously. In SL₃(ℤ), the conjugates of a basic matrix keep multiplying (1, 7, 40, 204, 925 as our search radius grows), while −I in SL₂(ℤ) has no conjugate but itself. Around 1980, Connes asked whether such a factor remembers the group, for rigid groups with Kazhdan’s property (T), like SL₃(ℤ). Amenable groups of this kind give no hope: they all produce the same factor. In August 2026, counterexamples among property (T) groups showed that the answer is no in general, one from OpenAI and one by Zhou with AI help. Lattices like SL₃(ℤ) remained open.
Result 286 is two OpenAI preprints, 190 pages, dated September 23, 2026. For such groups, commensurable with lattices in products of simple groups with property (T), over the reals, the complex numbers or the p-adic numbers, the first paper classifies every finite-index bimodule into a twisted group factor of any group with infinite conjugacy classes: each one is built from finite-index subgroup isomorphisms and finite-dimensional projective representations. The second paper deduces that these factors remember everything. If a stretched copy of one twisted group factor equals another, the stretch is one, the groups are isomorphic and the twists agree. So SL₃(ℤ) and its relatives would be W*-superrigid, and their factors would have trivial fundamental group.
What it doesn’t claim
Lattices in groups without property (T), like SL₂(ℤ); fields of positive characteristic; or property (T) groups in general, where the conjecture is already known to fail.
Where it stands
- Posted as preprints
- OpenAI, two papers, 190 pages, dated September 23, 2026
- Peer reviewed
- Not yet
- Checked in Lean
- No (about 2,300 unlinked lines cover some algebra of the obvious bimodules)
Last checked October 19, 2026. The preprints have not been revised. In Lean, only some algebra of the obvious bimodules is formalized, about 2,300 lines not linked to the item. OSVerify has not reviewed them. Cyril Houdayer, who had worked on this problem, wrote of being “genuinely thrilled by the announced solution” and wanting to understand the proof. We found no published check.
How the proof goes
- Bridges between algebras. Think of bimodules as bridges, each with a left and a right size. The obvious ones come from matching finite-index subgroups, such as the congruence subgroup of level 3, whose index in SL₃(ℤ) is 5,616, decorated with small projective representations.
- No other bridges, and whole-number sizes. The hard part, 165 pages, shows there are no other bridges, using measurements at the boundary of symmetric spaces and Bruhat–Tits buildings, rather than Popa’s deformation and rigidity methods. Then a short argument finishes: if a copy stretched by t equals the original, there is a bridge with sizes t and 1/t. Both must be whole numbers, so t is one.
The numbers in the video
- Ours: conjugates of a basic matrix in SL₃(ℤ) within search radius 0 to 4: 1, 7, 40, 204, 925; −I in SL₂(ℤ) has only itself.
- Ours: the index of the level-N congruence subgroup in SL₃(ℤ) for N = 2 to 8: 168, 5,616, 43,008, 372,000, 943,488, 5,630,688, 11,010,048.
Read the source
- Arithmeticity of Twisted Finite Correspondences for Lattices over Local Fields (PDF), 165 pages.
- The Arithmetic Category and Stable Recovery of Lattice Factors (PDF), 25 pages.
- The collection’s catalog, where this is result 286.
The 2 papers in this result
OpenAI groups related papers, such as alternative proofs, supporting results and consequences, into one result; we label the first paper it lists the main paper and the others companions.
Transcript
Read the full script
0:00The question
Every group has a von Neumann algebra, an algebra of operators that records how the group acts on itself. When every element other than the identity has infinitely many conjugates, this algebra is a factor, with a trace that measures the size of its projections continuously. Our computer checks SL₃(ℤ): the conjugates of a basic matrix keep multiplying, 1, 7, 40, 204, 925 as the search radius grows, while minus the identity in SL₂(ℤ) has no conjugate but itself. Around 1980, Connes asked whether such a factor remembers the group, for rigid groups with Kazhdan’s property T, like SL₃(ℤ). Amenable groups of this kind give no hope: they all produce the same factor. In August 2026, counterexamples among property T groups showed that the answer is no in general, one from OpenAI and one by Zhou with AI help. Lattices like SL₃(ℤ) remained open.
1:04The claims
Result 286 is two OpenAI preprints, 190 pages, dated September 23, 2026. For such groups, commensurable with lattices in products of simple groups with property T, over the reals, the complex numbers or the p-adic numbers, the first paper classifies every finite-index bimodule into a twisted group factor of any group with infinite conjugacy classes: each one is built from finite-index subgroup isomorphisms and finite-dimensional projective representations. The second paper deduces that these factors remember everything. If a stretched copy of one twisted group factor equals another, the stretch is one, the groups are isomorphic and the twists agree. So SL₃(ℤ) and its relatives would be W*-superrigid, and their factors would have trivial fundamental group. Not claimed: lattices in groups without property T, like SL₂(ℤ); fields of positive characteristic; or property T groups in general.
2:09How the proof goes
Think of bimodules as bridges, each with a left and a right size. The obvious ones come from matching finite-index subgroups, such as the congruence subgroup of level 3, which our computer finds has index 5,616 in SL₃(ℤ), decorated with small projective representations. The hard part, 165 pages, shows there are no other bridges, using measurements at the boundary of symmetric spaces and Bruhat–Tits buildings, rather than Popa’s deformation and rigidity methods. Then a short argument finishes: if a copy stretched by t equals the original, there is a bridge with sizes t and 1/t. Both must be whole numbers, so t is one.
2:55Where it stands
The preprints are unrefereed and unchanged since release. In Lean, only some algebra of the obvious bridges is formalized, about 2,300 lines not linked to the item. OSVerify has not reviewed them. Cyril Houdayer, who had worked on this problem, wrote of being genuinely thrilled by the announced solution and wanting to understand the proof. We found no published check. If it holds, then for SL₃(ℤ) and its relatives, the operator algebra is a perfect fingerprint of the group.
Spot a mistake?
This episode simplifies on purpose, but it shouldn’t be wrong. If something is, open an issue on GitHub or leave a comment on YouTube. Corrections are noted here with a date.