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[challenge]: Finite counterexample to invariant-observable Griffiths positivity for O(n) rotors #250

Description

@kunyuan

Released by

Kun Chen, Institute of Theoretical Physics, Chinese Academy of Sciences

Contact email

chenkun0228@gmail.com

Method

Other

Challenge issue

Background

Classical $O(n)$ spin models replace Ising spins by unit vectors. Let $M<\infty$, $n\ge 3$, and

$$\sigma_i\in \mathbb{S}^{n-1}\subset\mathbb{R}^n,$$

with normalized rotation-invariant measure $d\nu_n$. For symmetric ferromagnetic couplings $J_{ij}\ge 0$, $J_{ii}=0$, and at least one $J_{ij}>0$, define

$$d\mathbb{P}_J(\sigma)=Z_J^{-1} \exp\!\left(\sum_{i\lt j}J_{ij}\,\sigma_i\!\cdot\!\sigma_j\right) \prod_i d\nu_n(\sigma_i).$$

The invariant dot-product cone consists of finite nonnegative linear combinations

$$f(\sigma)=\sum_{\alpha=1}^{r}c_\alpha \prod_{i\lt j}(\sigma_i\!\cdot\!\sigma_j)^{a_{ij}^{(\alpha)}},$$

where $c_\alpha\ge0$ and all exponents are nonnegative integers; $g$ has the same form. Nonnegative refers to the coefficients, not to the pointwise sign of each monomial.

The invariant-observable second Griffiths inequality (GKS2) asks whether

$$\mathrm{Cov}_J(f,g) =\mathbb{E}_J[fg]-\mathbb{E}_J[f]\mathbb{E}_J[g]\ge0$$

for every such finite interacting system. It is known for $n=2$ and at zero interaction, but the genuinely interacting non-Abelian case $n\ge3$ remains open.

Research objective

Find one explicit finite counterexample: a spin dimension $n\ge3$, a finite number of sites, nonnegative pair couplings with at least one nonzero coupling, and two explicit observables $f,g$ in the invariant dot-product cone such that

$$\mathbb{E}_J[fg]<\mathbb{E}_J[f]\mathbb{E}_J[g].$$

Quantum rotors, Gaussian spins, fixed-coordinate observables, anisotropic componentwise statements, the zero-coupling case, and counterexamples only to stronger Ginibre-type inequalities do not satisfy this challenge.

Success and verification gate

A complete result must specify all parameters exactly and provide exact closed forms or rigorously certified terminating evaluations of $Z_J$, $\mathbb{E}_J[f]$, $\mathbb{E}_J[g]$, and $\mathbb{E}_J[fg]$. A Monte Carlo estimate or unvalidated numerical quadrature is not sufficient.

An independent verifier should be able to:

  1. check $n\ge3$, finite $M$, $J_{ij}\ge0$, and at least one $J_{ij}&gt;0$;
  2. check cone membership of $f$ and $g$ directly from their nonnegative coefficients and integer exponents;
  3. recompute the partition function and expectation numerators using the supplied exact formulas or rigorous interval certificate; and
  4. establish with exact arithmetic or disjoint outward-rounded bounds that the covariance is strictly negative.

For rational couplings, one possible automated certificate expands $e^H$ to finite order, evaluates free-sphere moments exactly using the standard spherical pairing formula, bounds the remainder from $|H|\le\sum J_{ij}$, and interval-evaluates

$$I_{fg}I_1-I_fI_g<0.$$

A compact witness should be reviewable without reconstructing the search process.

Why this may lead to research output

This is the basic positive-correlation question for invariant observables of non-Abelian vector-spin ferromagnets. A counterexample would identify a fundamental obstruction to extending Griffiths/GKS tools from Abelian models to Heisenberg-type and more general $O(n)$ systems. A universal proof would also be important, but it would require review of the derivation rather than only the final result.

Current status and references

The problem was re-audited on 28 July 2026 and assessed as still open with high confidence. The closest general-looking fixed-component claim was withdrawn; asymptotic, zero-coupling, and stronger Ginibre results do not settle this invariant interacting problem.

  1. I. Herbst, Griffiths inequalities for non-interacting rotors.
  2. A. Abdesselam, Non-Abelian correlation inequalities and stable determinantal polynomials.
  3. D. Sylvester, The Ginibre inequality.

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