Coin flips up to the knife edge?

0:00The question

Put a magnet, pointing up or down, on every vertex of an infinite tree where each vertex has d neighbors. In the free Ising state, each spin agrees with its neighbor with a bias set by the temperature: the correlation across one edge is the hyperbolic tangent of beta, and across n edges it is that number to the n-th power. Can this random magnet be built from coin flips, by one fixed rule that treats every vertex alike, with no root? That is a factor of IID. At high temperature, up to the hyperbolic tangent of beta equal to one over d minus one, finite random clusters do it. At low temperature it is impossible: Allan Sly proved this in 2009. Later, a theorem of Backhausz, Szegedy and Virág capped how strongly any factor of IID can correlate distant vertices, and above the threshold where the hyperbolic tangent of beta equals one over the square root of d minus one, the Ising correlations break that cap. In between, Russell Lyons wrote that the question was open, and in 2020, Nam, Sly and Zhang built a factor partway into the gap, for large d, and conjectured it should work right up to the threshold. The threshold is a knife edge. On the tree with three neighbors, it sits at 0.707. At distance r there are three times two to the r minus one vertices, each with squared correlation two to the minus r, so every distance contributes exactly 3/2, forever. Our computer checked, on the binary tree with the same branching, what this means for the root: just above the threshold, distant spins keep information about it; just below, it dies exponentially; exactly at the threshold it dies, but only like one over the distance.

1:51The claim

Result 236 is one OpenAI preprint, 50 pages, dated September 26, 2026. It claims that for every d at least three and every beta at least zero, the free ferromagnetic Ising state on the d-regular tree is a factor of IID exactly when the hyperbolic tangent of beta is at most one over the square root of d minus one, including equality. The new direction is the construction; the impossibility above the threshold was already known. The rule uses no root and is equivariant under every tree automorphism. If correct, this proves the Nam, Sly and Zhang conjecture, and settles the ferromagnetic case of Lyons’s question. Not claimed: the antiferromagnetic case, a finitary rule, or anything about other graphs.

2:43How the proof goes

Building on Nam, Sly and Zhang, and like its companion, result 231, the construction listens to the field through noise: watch every spin through its own Brownian channel, and solve the equation for the observations driven by fresh noise made from the vertex labels. Here the hard part is that the correlations are too strong: the response to new evidence is not uniformly bounded, so the equation is not inverted step by step. Instead, the paper runs two copies of the observations driven by the same noise and shows they must agree. Below the threshold a Gronwall estimate does it. At the threshold, where every distance shell contributes equally, it squeezes decay out of information in side branches and out of cancellations along paths, using the spectral radius of the tree, and closes with an Osgood-type inequality. A last step upgrades almost-sure symmetry to symmetry on every input. As an illustration, our computer ran the noisy-observation sampler on a 10-vertex tree at the threshold, 40,000 times from fresh noise. The correlations between spins at distances one to four came out 0.71, 0.50, 0.35 and 0.25, matching the target powers. On a finite tree this is easy; the theorem is about the infinite tree, where the whole difficulty lies.

4:06Where it stands

The preprint is unrefereed and unchanged since release. In Lean, one comparator states the full if-and-only-if from first principles, including the known converse; about 890 files with no unproved steps. It checks the almost-sure symmetry, not the every-input upgrade; we did not build it. Not among those OSVerify reviewed, and we found no competing claim and no public reaction. If it holds, the free ferromagnetic Ising magnet on a regular tree can be built from coin flips exactly up to the knife edge, and at the knife edge itself.