Heisenberg ferromagnets: spontaneous order, Bloch's law, and spherical magnetization

Source and verification

This article explains family 271 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.

🏷️ Spin systems, pressure, and spontaneous order

Fix a spin $S\in \{\frac{1}{2},1,\frac{3}{2},\dots \}$. A site carries the finite-dimensional Hilbert space ${\mathbb{C}}^{2S+1}$, with basis $e_m$, $-S\le m\le S$. The matrix $S^z$ has eigenvalue $m$ on $e_m$, while $S^{+}{e}_m=\sqrt{S(S+1)-m(m+1)}{e}_{m+1}$, with $S^{+}{e}_S=0$. Set $S^{-}=(S^{+}{)}^{\ast }$, $S^x=(S^{+}+S^{-})/2$, and $S^y=(S^{+}-S^{-})/(2i)$. A subscript indicates the site where a matrix acts.

On a finite box $\Lambda \subset {\mathbb{Z}}^3$, let $J(z)\ge 0$ be symmetric, finitely supported, and zero at the origin. Its nonzero displacement vectors must generate ${\mathbb{Z}}^3$. Define

$$\begin{array}{rl}{H}_{\Lambda }^J& =-\frac{1}{2}\sum _{x,y\in \Lambda }J(y-x){\mathbf{S}}_x\cdot {\mathbf{S}}_y,\\ M_{\Lambda }& =\sum _{x\in \Lambda }{S}_x^z.\end{array}$$

The factor $1/2$ counts each unordered bond once. The nearest-neighbor model has $J(\pm e_j)=1$. The negative sign favors alignment. Temperature is $T={\beta }^{-1}$, in units where the lattice spacing, Boltzmann constant, and Planck constant are one.

The Gibbs expectation of an observable $A$ is the normalized trace against $e^{-\beta H}$. In a field $h$, replace $H$ by $H-hM$. The thermodynamic pressure and spontaneous magnetization are

$$\begin{array}{rl}{p}_{\beta }(h)& =\underset{\Lambda \uparrow {\mathbb{Z}}^3}{\lim }\frac{\log \mathrm{Tr}{e}^{-\beta (H_{\Lambda }^J-h{M}_{\Lambda })}}{\beta |\Lambda |},\\ m_{S,J}(\beta )& =p_{\beta }^{\prime }(0+).\end{array}$$

Finite-range interactions give the pressure limit along cubes; convexity and $\Vert M_{\Lambda }\Vert \le S|\Lambda |$ give the right derivative. A rotation reversing $S^z$ makes $p_{\beta }$ even. Thus every finite zero-field box has zero mean magnetization, although the limiting pressure can have a corner at zero and $m_{S,J}(\beta )>0$.

An infinite-volume state assigns compatible positive, normalized expectations to local observables and their norm limits. Equilibrium can be expressed by the KMS condition: for observables $A,B$, the function $\omega (A{\tau }_t(B))$ has a bounded continuous extension to the closed strip, analytic in $0<\mathrm{Im}z<\beta$, with upper boundary $\omega ({\tau }_t(B)A)$. Here ${\tau }_t$ is the infinite-volume zero-field time evolution. This formulation lets one specify an ordered state even when its rotational average has zero mean.

The needed background is finite-dimensional quantum mechanics, convex functions, Markov chains, heat kernels, and basic operator limits. The central difficulties concern probability and analysis; no development of spin representation theory is needed below.

🏷️ Four conclusions in one family

Results claimed by the manuscripts

For nearest-neighbor coupling in every dimension $d\ge 3$, every fixed spin has, at all sufficiently low positive temperatures, a translation-invariant zero-field KMS state with $\omega (S_0^z)\ge S/4$. (OpenAI, 2026)

In dimension three, the more general finite-range model above satisfies Bloch’s law, with the exact coefficient determined by its dispersion matrix. (OpenAI, 2026a) The nearest-neighbor companion identifies the first lattice correction. (OpenAI, 2026c) A further nearest-neighbor companion identifies the zero-field distribution at each sufficiently low fixed positive temperature as a uniform direction with a single deterministic magnitude. (OpenAI, 2026d)

The ordering statement gives existence and a positive lower bound. Bloch’s law quantifies approach to full polarization as $T\downarrow 0$. The spherical law describes a symmetric state at fixed $T>0$ and growing volume. All keep $S$ fixed. The lattice correction and spherical law are not asserted here for arbitrary finite-range $J$.

🏷️ The spin-wave coefficient

A magnon is a spin-lowering excitation above the fully polarized state. Write $\epsilon ={\epsilon }_{S,J}$ for its one-particle dispersion and $D=D_{S,J}$ for its quadratic coefficient:

$$\begin{array}{rl}\epsilon (k)& =S\sum _{z}J(z)(1-\cos (k\cdot z)),\\ D& =\frac{S}{2}\sum _{z}J(z)z{z}^{\mathsf{T}},\\ \epsilon (k)& =k^{\mathsf{T}}Dk+O(|k{|}^4).\end{array}$$

The matrix $D_{S,J}$ is positive definite. The stronger assumption that the support generates the lattice also ensures that the dispersion has no other zero on the momentum torus. An ideal gas of magnons has occupation density obtained by integrating $(e^{\beta \epsilon (k)}-1{)}^{-1}$ against $dk/(2\pi {)}^3$.

Expand the Bose factor as ${\sum }_{n\ge 1}{e}^{-n\beta \epsilon (k)}$. Each term is a return probability $p_{n\beta }(0)$ for the walk whose jump rate by $z$ is $SJ(z)$. Rescaling $k=u/\sqrt{t}$ gives

$$\begin{array}{rl}{p}_t(0)& \sim \kappa t^{-3/2},\\ \kappa & =\frac{1}{(4\pi {)}^{3/2}\sqrt{\det D_{S,J}}}.\end{array}$$

The Gaussian bound $p_t(0)\le C(1+t{)}^{-3/2}$ justifies summation after rescaling. Consequently the ideal density has leading coefficient $\kappa \zeta (3/2)$. The finite-range manuscript claims that the interacting magnetization has exactly this coefficient:

$$\underset{\beta \to \infty }{\lim }{\beta }^{3/2}(S-m_{S,J}(\beta ))=\kappa \zeta (3/2).$$

For nearest neighbors $D_{S,J}=SI$. For anisotropic nearest-neighbor couplings $J(\pm e_j)=a_j>0$, its determinant is $S^3{a}_1{a}_2{a}_3$. Thus anisotropy changes the coefficient through the product of the three diffusion scales. The Gaussian calculation alone is not a proof of the interacting theorem. Nor does a zero-field free-energy asymptotic justify differentiation at a possibly singular field value. (OpenAI, 2026a)

🏷️ Exchange cycles and the pin identity

Represent a spin-$S$ site by $\ell =2S$ spin-$1/2$ slots, restricted to their symmetric subspace. Poisson exchanges across slot edges have rate $J(y-x)/2$. Uniform permutations of the slots at each site join the endpoints of the time interval of length $\beta$. The resulting trajectories form cycles; each cycle has one up or down color. Summing colors gives weight two per cycle.

A pin is a slot-time point required to have up color. Cycles meeting the pin set $E$ then have weight one; cycles missing $E$ retain weight two. Boundary pins extend throughout the time circle. Denote by $d_E(i)$ the down probability at an unpinned point $i$.

Add $i$ to the pins and use this changed probability law. Let $r$ be the probability that its line returns to $i$ at a strictly positive elapsed time before hitting $E$. For a fixed cycle picture, this is precisely the event $A$ that the cycle through $i$ misses $E$. Removing the new pin doubles the picture weight on $A$ and leaves it unchanged elsewhere. Hence the normalizing constants have ratio $1+r$, and

$$\begin{array}{rl}{\mathbb{P}}_E(A)& =\frac{2r}{1+r},\\ d_E(i)& =\frac{r}{1+r}.\end{array}$$

The last factor $1/2$ comes from the fair color of a cycle missing the pins. This exact identity converts magnetization into a return problem. It would be incorrect to compute $r$ under the original $E$-law.

The companion pin estimate takes a nonempty finite sparse set $T$ at one time slice, disjoint from $E$. Its simultaneous avoidance probability is at most $(2r(E,T){)}^{|T|}$, where $r(E,T)$ is the mean fraction of lines returning to $T$ at positive elapsed time before hitting $E$, under the larger pin set $E\cup T$. A finite downward induction begins with all grid points pinned. The hypothesis for larger pin sets controls the obstacle geometry used to remove pins. Joint bounds are essential; a small mean obstacle density alone does not exclude connected barriers. (OpenAI, 2026b) (OpenAI, 2026a)

🏷️ Diffusion in the exposed environment

Expose the cycles missing the augmented pins. The remaining slots are called holes. Their one-particle evolution combines heat exchanges, transport when an exposed strand swaps with a hole, and killing at pins. A conditional first-moment identity from the manuscripts replaces the next-pin permutation by a Markov walk.

The exposed environment repeats every period. The walk uses fresh transition randomness on each traversal of that same environment. This does not make successive environments independent. The one-period kernel can be nonsymmetric even though the original interaction is symmetric.

Sparse-pin estimates support two analytic tools: local Nash inequalities spread mass, and finite-energy flows through surviving sites control its pairing with a test function. For Bloch’s law one first obtains a rough avoidance bound, improves the exposed-slot density to $O({\beta }^{-3/2})$, and then compares a fixed number $K$ of returns with the ideal walk. Splitting a return at its midpoint combines a small encounter probability with the heat-kernel factor $O({\beta }^{-3/2})$.

Arbitrarily late returns need an additional estimate. The manuscript bounds their contribution by $C{K}^{-1/2}{\beta }^{-3/2}+o_K({\beta }^{-3/2})$, with $C$ independent of $K$. Finite return sums then squeeze the coefficient between partial sums of $\kappa \sum n^{-3/2}$ plus a vanishing tail. One takes $\beta \to \infty$ at fixed $K$, and only afterward $K\to \infty$. The auxiliary three independent interaction directions supply lattice geometry; the full jump kernel supplies the coefficient. (OpenAI, 2026a)

🏷️ A capacity identity for the sharper comparison

The first lattice correction requires an error smaller than ${\beta }^{-5/2}$. Its manuscript compares all returns at once. The following finite-dimensional lemma explains why two residual estimates suffice. (OpenAI, 2026c)

Let $U$ be a nonnegative doubly substochastic matrix acting forward on columns, with zero rows and columns at killed coordinates. It is an ${\ell }^2$ contraction. Fix a surviving coordinate $i$, write $\eta =1_{\{i\}}$, and set $A=I-U$. For a real vector $z$, define its energy by $e_z=⟨z,Az⟩\ge 0$, using counting measure. Let $q$ be the probability of returning to $i$ after at least one step before killing, and $f_j$ the probability of hitting $i$ from $j$, with $f_i=1$ and $f=0$ at killed coordinates. First-step conditioning gives $A^{\ast }f=(1-q)\eta$.

Choose a real trial vector $h$ with $h_i=1$, zero on killed coordinates, and a scalar $c$. Define

$$\begin{array}{rl}{\mu }_{+}& =Uh-h+c\eta ,\\ {\mu }_{-}& =U^{\ast }h-h+c\eta ,\\ y& =f-h,e_y=⟨y,Ay⟩\ge 0.\end{array}$$

Because $y_i=0$, subtracting the adjoint residual equation and pairing with $y$ gives $e_y=⟨y,{\mu }_{-}⟩$. Pairing $A^{\ast }f$ with $h$ gives the second exact identity:

$$q-(1-c)=⟨h,{\mu }_{+}⟩+⟨y,{\mu }_{+}⟩.$$

Suppose $\Vert h{\Vert }_{\infty }\le M$, $X,ϵ\ge 0$, and both residuals satisfy $|⟨{\mu }_{\pm },z⟩|\le X\sqrt{e_z}+ϵ$ for vectors zero at killing coordinates with $\Vert z{\Vert }_{\infty }\le 1+M$. Then $y$ is admissible. The inequality $e_y\le X\sqrt{e_y}+ϵ$ implies $\sqrt{e_y}\le X+\sqrt{ϵ}$, so

$$\begin{array}{rl}& |q-(1-c)|\\ & \le |⟨h,{\mu }_{+}⟩|+2X^2+2ϵ.\end{array}$$

No eventual-killing assumption is needed: a trajectory may stay forever in a class disjoint from $i$. The adjoint residual controls the error in the hitting potential; the forward residual then controls the return probability. Symmetry of $U$ is never assumed.

The trial vector comes from the full free return potential $v(z)={\ell }^{-1}{\sum }_{n\ge 1}{p}_{n\beta }(z-x)$, normalized by $c=(1+v(x){)}^{-1}$ and spatially cut off. On a high-probability geometric event, the manuscript proves that the expected scalar residual and $X^2$ are $o({\beta }^{-5/2})$, while the additive residual error $ϵ$ is rapidly small. Its exceptional-event probability also is $o({\beta }^{-5/2})$. Since $v(x)=O({\beta }^{-3/2})$, both rational conversions, $1-c=v(x)/(1+v(x))$ and $d=r/(1+r)$, cost only $O({\beta }^{-3})$.

The improvement rests on a signed cancellation. Removing an edge to an obstacle creates one source; transporting the hole when the obstacle jumps creates another. Along an isolated strand $Z_s$, their signed sum for a comparison profile ${\phi }_s$ is

$${\phi }_v(Z_v)-{\phi }_u(Z_u)+{\int }_u^v{\gamma }_s(Z_s)ds,$$

where ${\gamma }_s=\mathcal{L}{\phi }_s-{\partial }_s{\phi }_s$ is the spatial-cutoff error. Continuous sources supply the time derivative, jumps supply spatial increments, and the time-seam averaging conserves total mass. Thus they telescope. The signed mass depends on displacement; its total variation counts every jump. The proof keeps those quantities separate when routing sources through the holes. The residual estimates, including their geometric inputs, are substantial analytic claims of the source, not consequences of the elementary capacity lemma alone.

🏷️ The first lattice correction

For nearest neighbors put $t=\beta S$. The one-dimensional return integral is

$$I(t)=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }{e}^{-2t(1-\cos k)}dk.$$

With $k=u/\sqrt{t}$, the exponent is $-u^2+u^4/(12t)+O(u^6/t^2)$. Gaussian integration uses $\int u^4{e}^{-u^2}du=3\sqrt{\pi }/4$ and gives $I(t)=(4\pi t{)}^{-1/2}(1+1/(16t)+O(t^{-2}))$. A small rescaled neighborhood supports the expansion; the bound $1-\cos k\ge c{k}^2$ controls its complement. The remainder is uniform for $t\ge 1$.

The three-dimensional return is $I(t{)}^3$. Summing at times $nt$ produces a summable remainder $O(t^{-7/2}\sum n^{-7/2})$. Combining this ideal calculation with the interacting comparison yields

$$\begin{array}{rl}S-m_S(\beta )=& \frac{\zeta (3/2)}{8{\pi }^{3/2}}{t}^{-3/2}\\ & +\frac{3\zeta (5/2)}{128{\pi }^{3/2}}{t}^{-5/2}\\ & +o(t^{-5/2}).\end{array}$$

This term comes from the quartic correction to the lattice dispersion. It is not a calculation of the later spin-wave interaction term. (OpenAI, 2026c)

🏷️ Pressure increments and the order of limits

A positive field can be represented exactly by extra pins. Given a colored picture with $N_{\downarrow }$ down slots, independently select each time-zero slot with probability $b=1-e^{-\beta h}$ and require all selected slots to be up. The probability of that event is $(1-b{)}^{N_{\downarrow }}=e^{-\beta h{N}_{\downarrow }}$, the field weight after extracting the all-up factor.

After normalization, the marginal selection law is generally not Bernoulli: it is weighted by the pinned partition function. Conditional on the picture and colors, however, up slots are selected independently and down slots cannot be selected. Therefore any prescribed slot has selection probability at most $b$. A union bound makes each fixed comparison neighborhood pin-free with probability tending to one as $h\downarrow 0$ at fixed $\beta$.

Boundary pinning changes the logarithmic partition function by at most a constant times the boundary size. Uniform bounds on pinned down probabilities therefore bound derivatives of the boundary-pinned finite-volume pressure. Integrate these bounds from $0$ to $h$, divide by $h$, and take volume to infinity. Only then send $h\downarrow 0$. This proves a bound on $p_{\beta }^{\prime }(0+)$ without interchanging a limit and a derivative. Low temperature is taken last. (OpenAI, 2026a) (OpenAI, 2026c)

The existence proof uses a related construction at fixed sufficiently large $\beta$. Its pin estimate supplies a free zero-field spectral tail of size at least $\frac{1}{2}{e}^{-C|\partial {\Lambda }_N|}$ for $M_N\ge S|{\Lambda }_N|/2$. A field $h_N=N^{-1/2}$ makes the probability of $M_N<S|{\Lambda }_N|/4$ at most

$$2\exp \left(C|\partial {\Lambda }_N|-\frac{\beta h_{N}S}{4}|{\Lambda }_N|\right).$$

The negative volume term grows like $N^{d-1/2}$, exceeding the surface term of order $N^{d-1}$. Translation averages and a subsequence give a state with mean spin at least $S/4$. To identify it as equilibrium for the original zero-field dynamics, the source separately proves convergence of local dynamics deep inside translated boxes and passes both boundary values of the KMS strip to this same state. Positive magnetization by itself would not discharge that equilibrium obligation. (OpenAI, 2026b)

🏷️ A fixed radius in the symmetric state

For the spherical law, use even periodic cubes with $V=L^3$, nearest-neighbor coupling, and fixed sufficiently large finite $\beta$. Write $M_L^a={\sum }_x{S}_x^a$ and $m=m_S(\beta )>0$. The conclusion is

$$\begin{array}{c}\underset{L\to \infty }{\lim }⟨e^{{\sum }_a{t}_a{M}_L^a/V}⟩=\frac{\sinh (m|t|)}{m|t|},\\ t\in {\mathbb{R}}^3,\end{array}$$

with value one at $t=0$. The expression uses each self-adjoint linear combination separately; it does not presume a joint measurement of the three noncommuting components. (OpenAI, 2026d)

Rotation symmetry alone permits a mixture of radii. The source needs two further inputs. First, a spatial comparison bounds the weighted mean of squared local box magnetizations by the squared global magnetization plus $C_{\beta }/\log R+o_L(1)$. It follows from exponential bounds for cellwise zero-mean spin tests and the exact weighted variance identity between parent and child boxes. The scale $R$ stays fixed while $L\to \infty$, then grows.

Second, every translation-ergodic equilibrium state must have magnetization length $m$. The manuscript proves this through zero-free matrix products, replica estimates, and harmonic limits of pressure tangents. The concluding contradiction is elementary once those inputs are available: a competing length $a<m$ supplies the affine branch $xa-a^2$. The limiting tangent equals $xm-m^2$ for large positive $x$; real analyticity extends that identity to all $x>0$. But the competing branch is larger when $0<x<m+a$, because their difference is $(m-a)(m+a-x)>0$.

These two inputs force the global squared magnetization to have lower limit at least $m^2$. To finish, let $j_L$ be the total-spin label in the symmetric Gibbs state. Conditional on $j_L=j$, one component is uniform on $-j,-j+1,\dots ,j$, and the sum of the three squared components equals $j(j+1)$. A fixed positive field and an exponential Markov bound show $\mathbb{P}(j_L/V>m+ϵ)\to 0$. Since $0\le j_L/V\le S$, every limiting radius law is supported on $[0,m]$ and has second moment at least $m^2$. It must be the point mass at $m$.

The conditional component transforms are now Riemann sums converging to $\frac{1}{2}{\int }_{-1}^1{e}^{m|t|u}du$. This is the stated spherical transform. In particular the average of $V^{-2}{\sum }_{x,y}⟨{\mathbf{S}}_x\cdot {\mathbf{S}}_y⟩$ tends to $m^2$, although every one-component mean in the symmetric finite box is zero.

🏷️ Key steps and proof outlines

The long-return estimate completes a square in hitting energy

For the finite-range Bloch law, let $U$ be the exposed one-period substochastic kernel and $Q=(U+U^{\ast })/2$. Let $\eta$ be unit mass at the target, and $f$ its hitting probability, with $f=1$ at that target. At a fixed integer $K\ge 1$, set

$$v=\sum _{n=1}^{K-1}{Q}^n\eta ,y=f-\eta -v,E_y=⟨y,(I-Q)y⟩\ge 0.$$

The finite telescoping identity

$$(I-Q)v=Q\eta -Q^K\eta$$

turns the hitting-energy inequality into

$$q_i\le \sum _{n=1}^{2K-1}⟨\eta ,Q^n\eta ⟩+2⟨Q^K\eta ,y⟩-E_y.$$

This is a finite expansion, so it does not assume an infinite Green series converges or every trajectory is eventually killed.

Split the final period at times in its middle third. On the regular geometric event, the propagated mass has already diffused for order $K\beta$, and its ${\ell }^2$ norm is at most $C(K\beta {)}^{-3/4}$. The source’s flow estimate pairs this spread mass with the propagated test $y$, using the test’s slice Dirichlet energy. The time integral of that energy is bounded by the original $E_y$. Consequently the two directional pairings are bounded by a constant times

$$K^{-1/4}{\beta }^{-3/4}\sqrt{E_y}.$$

Completing the square against $-E_y$ gives $C{K}^{-1/2}{\beta }^{-3/2}$. The source controls discarded geometries by $o_K({\beta }^{-3/2})$. The leading constant remains independent of $K$, which is precisely what permits taking the low-temperature limit first and then sending $K$ to infinity (OpenAI, 2026a).

The lattice correction uses a strictly ordered two-source expansion

The finer correction pairs a forward comparison error with a backward profile. The order of instantaneous exchanges matters. For chronological operators $K_j:V_{j-1}\to V_j$, take comparison vectors $a_j,b_j$ with $a_0=b_n=h$, and put

$$p_j=K_j{a}_{j-1}-a_j,q_j=K_j^{\ast }{b}_j-b_{j-1}.$$

Write $P_{r,l}=K_r\cdots K_{l+1}$, with empty product the identity. The forward error is ${\sum }_j{P}_{n,j}{p}_j$. Reverse telescoping gives

$$P_{n,j}^{\ast }h=b_j+\sum _{k>j}{P}_{k-1,j}^{\ast }{q}_k.$$

Their scalar pairing is therefore exactly

$$\sum _j⟨p_j,b_j⟩+\sum _{j<k}⟨P_{k-1,j}{p}_j,q_k⟩.$$

There is no same-operation term $j=k$: the forward defect is inserted after its operation, while the reverse defect lies before that operation. Continuous intervals contribute the corresponding variation-of-constants integrals. This identity separates direct signed-source cancellation from interactions between sources at distinct times, and prevents a time-seam contribution from being transported across the seam twice. The later scalar and energy estimates exploit this separation to obtain the error scale required by the lattice correction (OpenAI, 2026c).

Positive curvature propagates a small susceptibility bound

The spherical-law manuscript first works with a normalized logarithmic Laplace transform $F$ in a small field. Its zero-free trace argument makes $F$ holomorphic in a fixed strip, while positivity of the underlying spectral measure gives

$$F(x)\ge 0,\mathrm{Re}F(x+iy)\le F(x).$$

For a strip height $a>0$, let $g_a(x)=F(x)-\mathrm{Re}F(x+ia)\ge 0$. The source proves

$$F^{\prime \prime }=J_a\ast g_a,J_a=\frac{2}{a^2}(h_a\ast h_a),h_a(x)=\frac{\pi }{2a}{\mathrm{sech}}^2(\pi x/a).$$

Fourier transformation of the harmonic equation gives the multiplier ${\xi }^2/(\cosh (a\xi )-1)$; the displayed positive convolution kernel is its inverse transform. Since $|h_a^{\prime }|\le (2\pi /a)h_a$, the kernel satisfies the same relative derivative bound. Positivity of $g_a$ then yields

$$0\le F^{\prime \prime }(x)\le e^{2\pi |x|/a}{F}^{\prime \prime }(0).$$

Thus a power-small variance at zero remains small on an interval of length proportional to $\log R$. Integrating twice gives exponential-moment bounds, and exponential Markov gives the cellwise tail estimate.

To compare many children of a spatial box, the proof maximizes over signs of their deviations from the parent mean. There are at most $\exp (CV/r^3)$ sign tests at child scale $r=R^{9/10}$, while each failure costs $\exp (-cV{R}^{-5/2})$. Since $R^{-27/10}=o(R^{-5/2})$, the union bound survives. The exact weighted variance identity then telescopes these comparisons across scales. This is the mechanism behind the spatial error $C_{\beta }/\log R$ used to rule out a mixture of magnetization radii (OpenAI, 2026d).

The ordered limiting state satisfies both KMS boundary identities

The spontaneous-order paper must retain equilibrium while removing its selecting field. For a local observable $B$, translations a distance at least $\sqrt{N}$ from the box boundary have dynamics converging uniformly on compact real-time intervals to the infinite-volume dynamics. The field contribution is just a local rotation of angle $-h_{N}t$, because $[H_N,M_N]=0$; its norm error is $O(h_N|t|)$. The excluded boundary layer has vanishing volume fraction.

For the translated finite-volume Gibbs averages, trace cyclicity gives the lower strip boundary ${\omega }_N(A{\alpha }_t(B))$ and upper boundary ${\omega }_N({\alpha }_t(B)A)$. Both have the common bound $\Vert A\Vert \Vert B\Vert$. The source applies the maximum principle to $e^{-ϵz^2}{F}_N(z)$ on finite rectangles, sends their width to infinity, and then sends $ϵ$ to zero. This makes the same bound hold throughout the strip despite volume-dependent elementary interior bounds. Multiplication by $e^{-z^2}$ then turns local boundary convergence into a uniform Cauchy estimate.

Both boundary values consequently pass to the already selected magnetized state along the same subsequence. The analytic strip obtained this way proves its zero-field KMS condition; selecting a high-magnetization tail alone would not do so (OpenAI, 2026b).

🏷️ Connections and scope

The Haldane family studies antiferromagnetic spin-one chains and a spectral gap, whereas this family studies ferromagnetic order at positive temperature in dimensions at least three. The different signs, dimensions, and observables prevent transferring either result to the other model. The Bose-gas family also separates an occupation observable from scalar energy estimates and preserves the thermodynamic limit before an asymptotic regime.

The exposition checks the coefficient calculations, the singleton pin identity, the finite capacity lemma, field-mixture normalization, selected pressure-limit arguments, the vanishing-field tail selection, and the final radius-concentration argument against the sources. It does not independently establish every sparse-pin, routing, residual, complex zero-free, or KMS compactness estimate in these four manuscripts.

🏷️ Manuscripts and verification scope

• Bloch’s Law for Finite-Range Heisenberg Ferromagnets in Three Dimensions — pinned manuscript and companion source.
• The first lattice correction to Bloch’s law — pinned manuscript and companion source.
• The spherical magnetization law for the three-dimensional quantum Heisenberg ferromagnet — pinned manuscript and companion source.
• Spontaneous magnetization in the quantum Heisenberg ferromagnet — pinned manuscript and companion source.

The family scope page describes selected formalization coverage. Despite the scope page’s broad title, its listed manuscript and stated coverage concern spontaneous magnetization, the zero-field KMS condition, and convergence of finite-volume dynamics. It does not describe formalization of the sharp Bloch coefficient, the first lattice correction, or the spherical law. 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 2026a. Bloch’s Law for Finite-Range Heisenberg Ferromagnets in Three Dimensions.

🐻  OpenAI 2026b. Spontaneous magnetization in the quantum Heisenberg ferromagnet.

🐻  OpenAI 2026c. The first lattice correction to Bloch’s law.

🐻  OpenAI 2026d. The spherical magnetization law for the three-dimensional quantum Heisenberg ferromagnet.