Same group, two prices?
An AI preprint claims a group without fixed price, answering a question of Gaboriau
Covers result 259 in OpenAI’s catalog: its one paper. See its connections to other results and the work it cites.
3:42. English captions available. The video loads from YouTube when you press play.
Chapters
The claim, in plain words
Let a group act on a probability space, shuffling its points while preserving measure. To record which points share an orbit, connect them with measurable edges and pay for the average number of edges per point. The lowest price you can approach is the cost of the action.
A free action of the integers costs 1: link each point to the next. Every free action of an infinite amenable group costs 1, from the work of Ornstein and Weiss. Gaboriau showed in 2000 that a free group on n generators has cost n for every free action. In these examples the price does not depend on the action, a property called fixed price. Gaboriau asked whether every group has fixed price, and wrote in 2000 that no group without it was known. Later results pointed the same way: groups with property (T) have cost 1, and higher-rank lattices and products of infinite groups have fixed price 1.
Result 259 is one OpenAI preprint, 27 pages, dated October 5, 2026. It writes down an explicit group: generators a, b₁ through b₉₉, and t, where t commutes with every b and with the word a b₁ a b₂ … a b₉₉ a. The paper claims the Bernoulli action, which labels each group element with an independent random number between 0 and 1, has cost at least 1 plus a fixed positive gap. The same action with a clock of M positions added, for M above 100, has cost at most 1 + 99/M, which tends to 1. So this group does not have fixed price.
What it doesn’t claim
How large the true difference in cost is. The gap the proof guarantees is tiny: with the paper’s constants, our arithmetic gives a proven gap below 2 × 10⁻¹⁹, and the clock needs more than 6 × 10²⁰ positions to provably undercut it. The paper also says nothing about groups with property (T).
Where it stands
- Posted as a preprint
- OpenAI, 27 pages, dated October 5, 2026
- Peer reviewed
- Not yet
- Checked in Lean
- No
Last checked October 19, 2026. The preprint has not been revised. There is no Lean formalization, and OSVerify has not reviewed it. The preprint folder carries an unusual note: the manuscript was developed from an intermediate argument, and the source record does not establish that it was the final revision. We found no specialist comment on this result.
How the proof goes
- A clock that makes edges cheap. The long word moves the clock by 100. Paying for a-edges at only 99 clock positions is enough, because those force the rest, one position at a time.
- The lower bound: counting generators. A cheap way to connect orbits of the Bernoulli action would give, for large random finite models, generating sets of some explicit groups that are too small. Random permutations and a graph-shortening argument from 1996 show they cannot be.
The numbers in the video
- Ours: we built and checked the forced paths at 101 and 103 clock positions. At 150 positions, one path would need about 10¹⁷ paid a-edges, but cost only counts edges per point.
- Ours: the usual homology lower bound for cost gives zero for this group, so the proof has to be combinatorial.
Read the source
- A Group Without Fixed Price (PDF), 27 pages.
- The collection’s catalog, where this is result 259.
The paper
OpenAI groups related papers into one result; this one has a single paper.
Transcript
Read the full script
0:00The question
Let a group act on a probability space, shuffling its points while preserving measure. Each point’s orbit is the set of places the group can send it. To record which points share an orbit, connect them with edges, chosen measurably, and pay for the average number of edges per point. The lowest price you can approach is the cost of the action. For a free action of the integers, one edge per point suffices: link each point to the next one in the shuffle, so the cost is one, and Ornstein and Weiss’s work gives cost one for every free action of every infinite amenable group. Gaboriau showed in 2000 that a free group on n generators has cost n for every free action. In these examples, the price does not depend on the action, a property called fixed price. Gaboriau asked whether every group has fixed price; in 2000 he wrote that no group without it was known. 25 years of results pointed the same way: groups with property (T) were shown to have cost one, and lattices in higher rank and products of infinite groups fixed price one.
1:05The claim
Result 259 is one OpenAI preprint, 27 pages, dated October 5, 2026. It writes down an explicit group: generators a, b₁ through b₉₉, and t, where t commutes with every b and with one long word, a b₁ a b₂ … a b₉₉ a. That is 101 generators and 100 relations. It claims two essentially free actions with different costs. The Bernoulli action, which labels each group element with an independent random number between zero and one, has cost at least one plus a fixed positive gap. The Bernoulli action with a clock of M positions added, for M above 100, has cost at most one plus 99/M, which tends to one. So this group does not have fixed price. The proven gap is tiny. With the paper’s constants, our arithmetic gives a proven gap below 2 × 10⁻¹⁹, and the clock needs more than 6 × 10²⁰ positions to provably undercut it. A strict inequality is all the question asks.
2:12How the proof goes
The cheap actions come from the clock. The long word moves the clock by 100, so paying for a edges at only 99 clock positions forces the rest, one position at a time, like dominoes. Our computer built and checked these paths at 101 and 103 positions; at 150, one path would need about ten to the seventeen paid a edges, but cost only counts edges per point. The lower bound turns the question into counting generators. A cheap way to connect orbits of the Bernoulli action would give, for large random finite models, generating sets of some explicit groups that are too small. Random permutations and a graph-shortening argument from 1996 show they cannot be. By our reckoning, the usual homology bound is useless here; for this group it gives zero.
3:02Where it stands
The preprint is unrefereed and unchanged since release. It is not formalized in Lean, and OSVerify has not reviewed it. The folder carries an unusual note: the manuscript was developed from an intermediate argument, and the source record does not establish that it was the final revision. We found no specialist comment on this result. If it holds, cost is not a property of a group alone: the same group can be cheap to wire in one action and strictly more expensive in another, by a margin the proof can only guarantee to be about one part in 10¹⁹.
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.