The exact factor-of-IID threshold for free Ising spins on trees

The explanation

Take a tree where every point, called a vertex, has exactly three neighbors, and the branching never stops. Put a spin, +1 or −1, on every vertex. Start at one vertex with a fair coin. Each neighbor then copies its parent's spin with probability (1 + θ)/2, and flips it otherwise. Keep going outward along every branch. Here θ = tanh β, and β is the inverse temperature. A large β means spins agree more often.

Try θ = 0.7. A child agrees with its parent 85% of the time. Two spins n steps apart have correlation θ to the power n. So at distance 10 the correlation is 0.7^10, about 0.028.

This random picture is the free Ising state on the tree. Its rules mention a starting vertex, but the final law does not depend on which one. The question is whether you can build it with no starting vertex at all. You give every vertex its own independent random number, a label. Then one fixed rule turns all the labels into spins. The rule must treat every vertex alike: rotate or reflect the tree, and the output moves with it. Such a rule is called a factor of IID, short for independent, identically distributed labels.

The paper studies trees where every vertex has d neighbors, with d at least 3. It claims a factor of IID exists exactly when tanh β ≤ 1/√(d−1)Source 1. For d = 3 that cut-off is θ = 1/√2, about 0.7071. So θ = 0.7 is just inside, and θ = 0.72 is just outside.

Some of this was known. Past the cut-off, no factor exists, by an argument of Sly recorded by LyonsSource 2. Well below it, a simpler construction from finite clusters worksSource 3. In 2022, Nam, Sly and Zhang built a factor for large d and small θ. Their method used random equations driven by independent Brownian motions, a model of continuous random noise. They left the full range as a conjecture. The paper claims the positive direction, including the cut-off itselfSource 4. So it claims to settle their conjecture for every d ≥ 3Source 5.

The cut-off has a second meaning. It is the point where distant spins stop carrying information about one spinSource 6. But that fact alone does not give a rule built from labelsSource 7.

The paper follows the observation method of Nam, Sly and Zhang. Imagine watching every spin through noise that slowly fades. Remove the part of what you see that you could already predict. What remains is fresh noise, called innovations, independent at each vertexSource 8. Labels can produce such noise directly. The hard step is to show that the innovations alone determine the spins.

The paper's test runs two copies of the observations with the same innovations. It shows the copies must agree. Below the cut-off, a size-weighted sum shrinks fast at each distance, and that gives agreement easily. At the cut-off it does not shrink at allSource 9. For d = 3 there are 3 × 2^(n−1) vertices at distance n. Multiply by θ^(2n) with θ = 1/√2. You get 1.5 at every distance n. The paper replaces the lost decay with three new estimatesSource 10. They end in a growth inequality that forces the gap between the copies to be zeroSource 11.

The result has limits. It does not say how many labels decide one spinSource 12. It gives no bounds that hold across all d and β at onceSource 13.

How sure is this? Lean, a computer proof checker, has a formal version of the main statement. It covers every d ≥ 3 and β ≥ 0, including equality at the cut-off. The scope note says finitary coding and coding-radius bounds are not included. No specialist review of the paper has been published.

What it doesn't show. This does not say how many labels are needed to decide one spin.