The sharp IID-factor threshold for free tree Ising spins

Source and verification

This article explains family 236 of the October 6, 2026 OpenAI collection. Research conclusions are attributed to its manuscripts. The review covers statements, introductions, and selected proof mechanisms; it is not an independent verification of the complete proofs.

Overview

The free Ising state on a tree has a simple root-based broadcast description. Family 236 claims a sharp criterion for generating the entire law without a root from independent vertex labels. The equality case requires more than nonreconstruction or a weak limit of samplers.

🏷️ Preliminaries: the free state and independent labels

On the infinite $d$-regular tree, $d\ge 3$, put $b=d-1$ and $\theta =\tanh \beta$. Choose a fair spin at one vertex and transmit independently along edges, making a child agree with its parent with probability $(1+\theta )/2$.

This gives the free zero-field Ising state. Its distribution is independent of the chosen root: on a finite connected subtree, it is the fair initial-spin probability times the product of the symmetric edge-transition probabilities.

A factor of IID is a measurable function of independent uniform vertex labels whose output commutes with tree automorphisms. The rule itself has no distinguished root or end. It need not be finitary: a spin may depend on infinitely many input labels.

The tree’s usual uniqueness condition and its reconstruction condition differ. The range $\theta \le 1/b$ is smaller than the nonreconstruction range $\theta \le 1/\sqrt{b}$. Uniqueness is therefore not the threshold asserted here.

🏷️ The claimed sharp factor criterion

Free Ising factor threshold

For each fixed $d\ge 3$ and $\beta \ge 0$, the manuscript claims that the free state is a factor of IID exactly when

$$\tanh \beta \le (d-1{)}^{-1/2}.$$

It proves the positive direction including equality and uses the known strict converse. The spin factor can be chosen equivariant on every label input. No coding-radius or computational bound is claimed. (OpenAI, 2026)

Nonreconstruction says distant spins do not reveal a fixed spin. It does not construct the whole process from independent randomness. A sequence of factor laws can also have a weak limit that is not itself a factor, so merely approaching the critical parameter from below does not resolve equality.

🏷️ The correlation obstruction above threshold

Along a path of length $r$, the Markov property gives

$$\mathbb{E}[{\sigma }_o{\sigma }_v]={\theta }^r.$$

The source uses a general correlation bound for centered variance-one invariant factors on the regular tree:

$$\begin{array}{c}|\mathrm{Corr}(X_o,X_v)|\\ \le \left(1+\frac{d-2}{d}r\right)b^{-r/2}.\end{array}$$

For Ising spins this would require

$$(\theta \sqrt{b}{)}^r\le 1+\frac{d-2}{d}r.$$

If $\theta \sqrt{b}>1$, the left side grows exponentially and the right side only linearly, giving the obstruction. At equality the calculation gives no contradiction, which pinpoints the endpoint left to the construction.

🏷️ Observations, innovations, and recovery

On an auxiliary probability space, sample the desired spins and independent Brownian motions, then observe

$$X_v(t)=t{\sigma }_v+B_v(t).$$

Let ${\mathcal{F}}_s$ contain all observations through time $s$, and put $m_v(s)=\mathbb{E}[{\sigma }_v\mid {\mathcal{F}}_s]$. Subtract this integrated posterior mean to form innovations:

$$W_v(t)=X_v(t)-{\int }_0^t{m}_v(s)ds.$$

The innovations are independent Brownian motions. This fact alone does not say that the observations, or the spins, can be recovered from them: the original construction used additional randomness in $\sigma$.

The manuscript takes two observation processes $X,X^{\prime }$, independently sampled conditional on the same full innovation array. It proves that these conditional copies remain compatible with the past filtration, so the shared innovations are still Brownian in the joined process. It then proves equality of the two copies.

For each coordinate,

$$\begin{array}{rl}& \mathbb{E}[(X_v(t)-X_v^{\prime }(t){)}^2\mid W]\\ & =2\mathrm{Var}(X_v(t)\mid W).\end{array}$$

Equality forces the conditional variance to vanish. Thus $X$ is a measurable function of $W$, and $X_v(n)/n\to {\sigma }_v$ recovers the spins. Independent uniform labels can encode these Brownian paths.

🏷️ The critical uniqueness estimate

Below the threshold, spatial decay makes the comparison between copies contract. At equality, a crude absolute-value bound loses the cancellations needed for integrability. The proof instead averages along paths, uses a spectral cancellation estimate, and controls the probability of small effective fields.

These estimates produce an Osgood-type comparison for the mean-square discrepancy: the comparison retains sufficient control near zero to force discrepancy zero from equal initial data. The observation-law conditioning is retained throughout; this is not a claim that every abstract infinite-dimensional drift equation has a uniform Lipschitz inverse.

The final measurable version makes the factor equivariant everywhere. The documented formalization selection uses the almost-sure equivariance formulation and does not include finitary coding.

🏷️ Key steps and proof outlines

At the threshold $b{\theta }^2=1$, where $b=d-1$ and $\theta =\tanh \beta$, an ordinary contraction disappears. The manuscript recovers uniqueness by combining small-time posterior information with a logarithmic differential inequality. Here are the estimates that close that endpoint, and how uniqueness becomes an equivariant sampler (OpenAI, 2026).

Tree messages factor the posterior response

Plant observations $X_v(t)=t{\sigma }_v+B_v(t)$. At time $t$, Bayes’ formula gives the finite-tree Ising law with external field $X_v(t)$: the Gaussian likelihood is $\exp ({\sum }_v{\sigma }_v{X}_v(t)-t|D|/2)$, and its last term is spin-independent. Differentiating a posterior mean with respect to an external field gives the corresponding covariance.

For the component on the $v$ side of a deleted edge $\{v,w\}$, let $h_{v\to w}$ be the inverse hyperbolic tangent of the posterior mean using only observations in that component. It is not conditioned on the opposite spin. Define

$$\psi (z)=\mathrm{atanh}(\theta \tanh z),a(z)=\frac{1-{\tanh }^{2}z}{1-{\theta }^2{\tanh }^{2}z}.$$

Summing over a child spin gives the cavity recursion

$$h_{v\to w}=X_v+\sum _{u\sim v,u\ne w}\psi (h_{u\to v}),{\psi }^{\prime }(z)=\theta a(z).$$

The response along a path consequently contains a factor $\theta$ and an attenuation factor $a$ for each edge. At criticality, a sphere of radius $r$ has $d{b}^{r-1}$ vertices, and $d{b}^{r-1}{\theta }^{2r}=d/b$. The bare squared response has no summable radial decay. The additional attenuation is indispensable.

The critical information estimate for a component field $Y$ is $\mathbb{E}{\tanh }^{2}Y\asymp \sqrt{t}$ at small time. Integrating intact side branches along a path gives attenuation $e^{-c\sqrt{t}r}$. The integration order matters: conditional on the path spins, a selected side branch is independent of the remaining shift, and later path factors do not use that branch. Removing the branches successively multiplies their losses. A bound for an isolated factor would not justify this product when factors share observations.

Conditional copies must retain a Brownian driving noise

Let $W=X-\int mdt$ be the innovations and $K(w,dx)$ the conditional observation law given the entire innovation array. Sample two observations independently from $K(w,\cdot )$, with the same $w$. Agreement of these copies will imply that $K$ is a point mass.

Conditioning on a whole future noise path might seem to spoil a stochastic comparison. The needed compatibility identity is

$$\mathbb{E}[F(X_{\le t})\mid W]=\mathbb{E}[F(X_{\le t})\mid W_{\le t}].$$

To prove it, test against products of functions of the noise past and future increments. In the original observation law, those future increments are independent of the observation past; causality makes the innovation past measurable there. In the conditional-copy law, conditional independence factors the two past tests, and the displayed identity makes both factors depend only on $W_{\le t}$. Thus future noise increments remain independent of the joined past. The source also approximates stochastic integrands by bounded causal step processes to transport the equations to this enlarged filtration; equality of pair laws alone is not used as a substitute for that check.

Higher moments improve the dangerous coefficient at zero

Let $V(t)=\mathbb{E}(h_o(t)-h_o^{\prime }(t){)}^2$ for the two full-site fields. For a cavity edge, write $D=\log a(Y)-\log a(Y^{\prime })$ and $S(t)=\mathbb{E}{D}^2$. The crude estimate $S\le CV$ would leave a noise contribution of order $V/t$, which cannot start a uniqueness comparison at zero.

The improvement uses $(\log a{)}^{\prime }(0)=0$. First the source proves, with one constant $K$ for all integers $p\ge 1$,

$$\mathbb{E}|Y{|}^p\le e^{K{p}^2}{t}^{p/4}.$$

The second-moment case follows from the critical information estimate and Gaussian tails of the own observation. For $p\ge 3$, condition on the root spin so child branches are independent. Spin symmetry leaves their absolute moments unchanged. Expanding the cavity recursion, the terms concentrated on one child contribute $b{\theta }^p\mathbb{E}|Y{|}^p$; their coefficient is strictly below one even at criticality. Mixed terms use lower moments, and their exponent sums obey $\sum r_i^2\le p^2-2(p-1)$. A sufficiently large fixed $K$ absorbs the multinomial factor and closes induction.

Choose $p$ proportional to $\log (1/t)$ in Markov’s inequality. Then

$$\mathbb{P}(|Y|>t^{1/8})\le e^{-c{\log }^2(1/t)}.$$

On the event that both fields are at most $t^{1/8}$, $|(\log a{)}^{\prime }|\le 2t^{1/8}$ gives a contribution $C{t}^{1/4}V$. On its complement use the smaller of the global bound $CV$ and the tail bound. With $K_T$ larger than the bounded discrepancy on a fixed time horizon and $I=1+\log (K_T/V)$, this yields

$$S(t)\le C_{T}V(t)\left(t^{1/4}+\frac{I(t)}{{\log }^2(1/t)}\right).$$

No independence of the two copies’ fields is assumed in this estimate.

A discrepancy-dependent cutoff and Osgood’s argument

Weighted path estimates have ordinary errors decaying like $e^{-c\sqrt{t}r}$ and exceptional errors decaying like $e^{-\alpha b^{\gamma r}}$. Choose the radius cutoff

$$R=\lceil C(t^{-1/2}+\log I(t))\rceil$$

at small positive time. Beyond it the exceptional term is at most $V(t{)}^2$ times a summable radial function. Before it use direct bounds. Summing $r{e}^{-c\sqrt{t}r}$ costs $O(t^{-1})$; the corresponding drift sum costs $O(t^{-3/4})$. Insert the improved estimate for $S$ precisely in the dangerous $S/t$ term. The resulting variance inequality is

$$\begin{array}{rl}{V}^{\prime }(t)& \le k_T(t)V(t)\left(1+\log \frac{K_T}{V(t)}\right),\\ k_T(t)& \le C_T\left(1+t^{-3/4}+\frac{1}{t{\log }^2(1/t)}\right)\end{array}$$

near zero, with $k_T$ bounded away from zero time. Both singular terms are integrable. If $V$ became positive on an interval starting at $a$, then $V(a)=0$ and the chain rule would give $(\log I{)}^{\prime }\ge -k_T$. Integrating would bound $\log I(u)$ as $u\downarrow a$, contradicting $V(u)\to 0$. Therefore $V$ vanishes. This is the critical endpoint argument; below the threshold the strict radial contraction already supplies the simpler comparison.

From zero conditional variance to a factor on every input

Intersect equality of the fields over countably many vertices and rational times, then use continuity. The shared innovation equation implies equality of the observation paths. Conditional independence gives

$$\mathbb{E}(X_v(q)-X_v^{\prime }(q){)}^2=2\mathbb{E}\mathrm{Var}(X_v(q)\mid W)=0.$$

Thus the conditional law is a point mass, since continuous paths are determined by rational values. Dyadic interpolation of conditional coordinate expectations constructs a Borel recovery map; integer-time slopes recover the spins by Gaussian tails and Borel–Cantelli.

Finally, almost-sure equivariance separately for each automorphism does not justify an uncountable intersection. Average each local conditional spin expectation over the finite automorphism group of its rooted ball. These local rules are equivariant on every input. Their limsup followed by a common sign threshold is still pointwise equivariant, and martingale convergence shows that it agrees almost surely with the recovered spin configuration. This supplies the exact equivariance required of the factor.

See Also

• Family 231 inverts innovations by a uniform response estimate, rather than the critical conditional-copy argument used here.
• Family 229 explains posterior information and thresholds for three- and four-state channels.

🏷️ Manuscripts and verification scope

• The sharp factor-of-IID threshold for the free Ising model on regular trees — pinned manuscript and companion source.

The family scope page describes selected formalization coverage. The selected statement covers the sharp threshold including equality with almost-sure equivariance for each fixed automorphism. It does not assert finitary coding or coding-radius estimates. No Lean build, axiom audit, or manuscript-to-formal-statement equivalence check was performed for these notes.

OpenAI collection provides prerequisites and the reading sequence; coverage register records completed and pending families.

🐻  OpenAI 2026. The sharp factor-of-IID threshold for the free Ising model on regular trees.