Bloch’s Law for Finite-Range Heisenberg Ferromagnets in Three Dimensions

Proof. This is the exchange representation of Tóth [17], with the higher-spin symmetric-slot construction of Nachtergaele [13, Section 2.2]; see also [15, Lemma 2.1]. At one site let \psi_k be the normalized sum of the elementary spin-\tfrac12 tensors having k up spins, 0\le k\le\ell. The sum of the slot z-spins acts on \psi_k by k-\ell/2, and the sum of the raising operators acts by \sqrt{(\ell-k)(k+1)}\,\psi_{k+1}. These are the spin-S matrices under k=S+m. Thus the physical space is the range of

where \mathcal U denotes tensor permutation on the slot space.

The spin-\tfrac12 identity \boldsymbol s_i\cdot\boldsymbol s_j =\frac12\mathcal U_{(ij)}-\frac14I gives the lifted Hamiltonian

The full sum over all slot pairs above each bond commutes with every within-site permutation, and hence with \mathcal S_\Lambda. The Poisson product \varphi of the exchanges satisfies

For example, the first equality follows from the matrix evolution with generator \sum_ec_e(\mathcal U_e-I); no commutation of individual transpositions is needed. Averaging the seam completions inserts \mathcal S_\Lambda. Taking the trace after multiplication by e^{\beta h\widehat M} proves (2.1): in the elementary up/down basis, a tensor permutation has a nonzero diagonal entry exactly when the coloring is constant on each cycle. The same calculation with any function of \widehat M proves the spectral-law assertion.□

Proof. There are finitely many marks per period, their times are distinct, and seam completions preserve sites. Every visit to the boundary therefore contains a time interval of positive length and is detected by a cut from the dense union. There are finitely many cycles, so one finite cut set eventually detects every boundary-touching cycle. The asserted eventual equalities follow. Since 1\le2^{n_{D_n}}\le2^{|V_\Lambda|}, bounded convergence applies both to the weights and their positive normalizers. This gives convergence of the densities in L^1(\mathbb P_{\rm b}), and also the assertion about observables. This argument is at fixed volume.□

Proof. Inserting a transposition cuts and rejoins two strands. It joins two cycles or splits one cycle into two, leaving the other cycles unchanged, and preserves the union of the affected strands and their pin incidences. Joining cycles cannot increase n_D. On splitting, a cycle avoiding D produces two such cycles, while a cycle meeting D has at least one descendant that still meets D. Thus n_D increases by at most one, and the weight increases by at most two. This statement applies outside the null set of coincident marks or deterministic pin times, and also to continuous boundary pins.

The Poisson insertion identity, with the seam fixed, is

Indeed, in the Poisson series the k-mark term has k!/(k-j)! ordered choices for the distinguished marks. Cancelling that factor against 1/k! and summing the remaining marks proves the identity. Nonnegativity justifies it even for infinite integrals. The insertion bound and division by \mathcal Z_D prove (2.5).□

Proof. For T=\{i\}, write r=r(E,\{i\}) and A=\{i\in\mathcal B_E\}. A line returns to i before E exactly on A, so r(E,\{i\})=\mathbb P_P(A). Moreover 2^{n_E}=2^{n_P}(1+\ind{A}), giving \mathcal Z_E=\mathcal Z_P(1+r) and \mathbb P_E(A)=2r/(1+r). Equation (2.4) proves the singleton identity. This calculation applies directly to finite and allowed pin sets.

For the simultaneous bound, we use three precise finite-set facts from [15, Lemmas 3.2, 3.4, and 3.6]. A nonzero polynomial is stable if it has no zero when every variable has positive imaginary part. The facts are as follows.

1. (i)

   Variable identification preserves stability. Partial derivatives, real specialization, and coefficientwise limits of uniformly bounded degree preserve stability up to the zero polynomial. In particular, after identifying two variables as t, taking the coefficient of t preserves stability up to zero.

2. (ii)

   If a permutation \pi of a finite set V is a product of independent Bernoulli transpositions and deterministic permutations, then

   \mathbb E\prod_{j\in V}(x_j+y_{\pi(j)})

   is stable. This also holds for independent products of finite-time Poisson exchanges with arbitrary nonnegative rates and uniform permutations on specified subsets.

3. (iii)

   If a random subset A of a finite set has a stable generating polynomial \mathbb E\prod_{j\in A}y_j, then for every subset B,

   \Pr(B\subseteq A)\le\prod_{j\in B}\Pr(j\in A).

These facts belong to the stable-polynomial method for negative dependence developed in [3]. They concern arbitrary finite sets, with no spatial hypothesis. To check the continuous-time scope in (ii), subdivide a finite interval into small slices and replace each edge clock in each slice by its independent indicator of at least one mark. Apply these Bernoulli transpositions in a fixed order. The products agree with the Poisson products whenever no slice has two marks in total; the probability of disagreement tends to zero. A uniform permutation is likewise a distributional limit of complete-graph Poisson exchanges: the unit-time transition matrix on its finite permutation group has all entries positive and preserves the uniform law, so it contracts total variation distance to that law. Thus the rates c_e=J_e/2 meet exactly the hypotheses of (ii).

For this bound, first suppose that E is finite. Cut the time circle at zero and at every pin time in P. Let V be the disjoint union of the slot layers at these cuts, and let \pi follow a line from one cut to the next cyclic cut. Include the seam completion in the interval ending at zero. The interval maps are independent and use disjoint sets of input x variables and output y variables. Their stable symbols therefore multiply to give the stable polynomial

Delete an unpinned point q\in V\setminus P by setting x_q=y_q=t and taking the linear coefficient in t. For a fixed permutation with predecessor i and successor j at a nonfixed q, the two affected factors are (x_i+t)(t+y_j), whose linear coefficient is x_i+y_j. This splices i\to q\to j to i\to j. A fixed point instead contributes the factor 2t and hence the factor two. Deleting all nonpins gives

where \sigma is the permutation taking each pin to its next pin on the same cycle. Each cycle avoiding P contributes exactly one factor two, when its final vertex is deleted. The closure facts give stability; the positive coefficients exclude the zero polynomial.

Differentiate (2.8) in every x_j with j\in P\setminus T, set every remaining x variable to zero, and divide by \mathcal Z_P. The result is the stable generating polynomial

It is nonzero because G_T(\boldsymbol1)=1, and it represents the subset \sigma(T) because its indices are distinct. Counting the outputs lying in T gives

Fact (iii) and the arithmetic–geometric mean inequality now imply

The event T\subseteq\sigma(T) is equivalent to T\subseteq\mathcal B_E. Indeed |\sigma(T)|=|T|, so the first event says \sigma(T)=T; iterating \sigma then meets only pins of T on every cycle through T, and meets all the pins on each such cycle. The converse follows by the same cycle description.

Deleting m_T pins frees at most m_T cycles. Thus, pointwise,

For A=\{T\subseteq\mathcal B_E\}, this yields

For E=F\cup D_0, replace D_0 by the finite cuts of Lemma 2.2, obtaining E_n and P_n=E_n\cup T. The simultaneous-free event and the weights converge almost surely. The return events converge as well: follow each line from i\in T until its first positive visit to T. This horizon is finite because the line eventually returns to i on its finite permutation cycle. Every boundary visit before that horizon contains a positive-duration interval and is eventually detected. Thus \ind{\mathcal R_i(E_n,T)}\to\ind{\mathcal R_i(E,T)}. Lemma 2.2 now passes the finite-pin inequality to the limit.□

Proof. We give the rate-dependent construction underlying [15, Proposition 4.1]. Fix A before sampling the base picture. On every edge e take two independent rate-c_e clocks. Accept a mark from the first family when at least one endpoint contains an A-label just before the mark; accept one from the second family when neither endpoint does.

This produces the base exchange process. Indeed, if c_{\rm tot}=\sum_ec_e>0, the superposed clock has rate 2c_{\rm tot}. At each of its times, conditionally on the past, the accepted-edge or rejection outcome has probability c_e/(2c_{\rm tot}) for edge e and probability 1/2 for rejection. These probabilities do not depend on the past or the superposed waiting times. Iterated conditioning and Poisson splitting give independent accepted edge clocks of rates c_e. If c_{\rm tot}=0, the statement is immediate.

The A-label histories use only first-family clocks. Given their positions just before the seam, revealing their images under each uniform site permutation leaves a uniform bijection of the remaining slots, independently from site to site. The histories, including these revealed images, therefore do not use the second clock family or the residual completions. Conditional on those histories, the second clocks restricted to the now fixed hole intervals remain independent Poisson clocks, while the cross jumps give exactly the forced transports. This proves the base conditional kernel \mathbb Q_{\mathbf a} on F_A.

On F_A, the labels in A form whole cycles missing P. Requiring G_P makes them precisely all such cycles. Consequently each complete picture contributes to exactly one term in

where a summand is zero off F_A. Apply the base conditional kernel to this finite sum and divide by \mathcal Z_P to obtain (2.10). Integrating out the remainder gives (2.11), which gives zero outer mass to g_P=0. Dividing by g_P on its positive set proves the asserted conditional law. All history spaces here are finite-jump path spaces on finite state spaces, with finite endpoint data; they are standard Borel spaces and admit the conditional kernels used above.□

Proof. For a transposition matrix T_{ij},

Every minor of I+u(T_{ij}-I) is affine in u, since a term using two columns of the rank-one perturbation has determinant zero. Thus

which proves the identity for one Bernoulli transposition. Multiplicativity of exterior powers and independence preserve it under products. Deterministic permutations satisfy it directly, and continuity of the minors on the finite matrix space passes it to distributional limits.

For the hole picture, subdivide each interval between prescribed changes and use the Bernoulli approximation already described. Forced transports are deterministic bijections. The residual seam completions are uniform bijections, each expressible as a fixed bijection composed with a uniform permutation; the latter has the complete-graph exchange approximation described above. Identify every cut layer with a fixed set of |H_t| elements. The interval maps act independently on separate copies of this set; their direct sum followed by the deterministic cyclic shift of the layers is the layered permutation. A transposition in one block is a transposition on the full layered set. This verifies the required product representation, before imposing G_P.□

Proof. The principal-minor expansion and (2.12) give, for every fixed matrix D,

Choose D=\operatorname{diag}(W,I_B) with an arbitrary complex I\times I matrix W. If a cycle lies in B, its indicator vector is fixed by D\Pi, so the determinant vanishes. On G_I, \Pi_{BB} is nilpotent and its resolvent follows the successive nonpins on a permutation cycle. Block elimination therefore gives

Consequently

Setting W=0 proves (2.13) without any invertibility assumption. When its value is positive, I-K_{BB} is invertible; taking the Schur complement in (2.16) and dividing by that determinant yields

Replace W by tW and compare linear coefficients. The identity \det(I-tWA)=1-t\Tr(WA)+O(t^2) shows that \Tr(W\mathbb E[\Sigma\mid G_I])=\Tr(WH) for every W. Matrix units recover every entry, proving (2.14).

Finally, K is nonnegative and doubly stochastic. Its restriction K_{BB} is substochastic. If some state of B could never reach I, its reachable set would contain a closed class and force the eigenvalue one for K_{BB}, contradicting invertibility. Finiteness then supplies an integer a and \varepsilon>0 such that every state exits B within a steps with probability at least \varepsilon. Hence \|K_{BB}^{an}\|_{1\to1}\le(1-\varepsilon)^n and

The first term in H is a direct hit of I, and K_{IB}K_{BB}^nK_{BI} records a first hit after n+1 intervening states of B. This proves the asserted Markov interpretation.□

Proof. First use finitely many pins and take cuts at all pin times and zero. The unconditioned layered hole permutation under \mathbb Q_{\mathbf a} satisfies Lemma 2.6. Conditional on every cycle hitting the pins, Proposition 2.7 identifies its next-pin first moment with independent transitions of its mean matrix. Those transitions are exactly the fresh walk observed at the cuts. Together with Proposition 2.5, this proves the finite-pin assertion.

For an allowed E=F\cup D_0, fix an exposure for which the disintegration holds and g_P>0. Approximate the boundary by increasing dense finite cuts and write E_n=F\cup D_{0,n}, P_n=E_n\cup T. The cuts include zero and all times in P_n. Under the fixed unconditioned law \mathbb Q_{\mathbf a}, let G_n mean that every remaining cycle meets P_n, and put g_n=\mathbb Q_{\mathbf a}(G_n).

Every boundary visit of the remaining picture has positive residence time. Its Poisson marks almost surely avoid the finitely many prescribed changes; seam completions preserve sites; and exposed histories avoid the boundary, so forced transports cannot cause an instantaneous boundary excursion. Consequently

For all sufficiently large n, Proposition 2.7 is therefore applicable with the pin layer I=P_n. If A_n is the event that the next P_n-pin after i is in T, it gives

No assertion about the possibly singular earlier n is needed.

In a fixed remaining picture the first positive return from i to T is finite. Detecting all boundary visits before that return shows that \ind{A_n} decreases to \ind{\mathcal R_i(E,T)}. Bounded convergence on the left of (2.19), including its positive limiting denominator, gives the return probability under \mathbb Q_{\mathbf a}(\,\cdot\mid G_P).

On the right use one uncensored fresh trajectory for all n. The events decrease; every one requires \tau_T^+<\infty. Any boundary visit before this finite time is detected by a dense cut, again because it has positive duration. Their intersection is therefore exactly \{\tau_T^+<\tau_E\} up to a null set. Continuity from above proves convergence to (2.17). The disintegration identifies the two limits, and averaging under the exposure marginal (2.11) proves (2.18).□

Proof. We give the transfer argument of [15, Proposition 5.6], followed by the lattice averaging proof of Nash’s inequality; the energy-to-diffusion method originates in [14]. First work in one coset and write its bar coordinates simply as sites. At each unblocked site w, choose a slot where v is maximal and call this value V(w). Each blocked site of B_{R+3}(z) has a surviving nearest neighbor: otherwise its six neighbors would contradict local regularity. Choose one, denoted \phi(w), and define

Then f is supported on B_{R+1}(z), and

For a lattice edge with a nonzero endpoint value of f, both values come from maximizing slots at surviving sites at distance at most three from each other, within B_{R+3}(z). The local-connectivity input with b=44 joins these sites by a path of fixed length and displacement avoiding blocked sites. Lift it to their maximizing slots. Telescoping along the lifted path and using Cauchy–Schwarz bounds the squared difference across the original edge by a fixed multiple of the sum of squared slot differences on that path. Any slot edge can occur in only a bounded number of these paths, since their original edges lie in a fixed neighborhood of it. Hence

For completeness, average f over its translates by \{0,\ldots,b-1\}^3, and denote the result by A_bf. Telescoping coordinate translations and using the triangle inequality in \ell^2 gives \|f-A_bf\|_2^2\le Cb^2\mathcal E_{\Z^3}(f). The averaging kernel has 2-norm b^{-3/2}, so \|A_bf\|_2^2\le b^{-3}\|f\|_1^2. Therefore

For f\ne0, take b=\lceil(4\|f\|_1^2/\|f\|_2^2)^{1/3}\rceil. The second term is at most half the left side; since \|f\|_1\ge\|f\|_2, rounding changes only the constant. Absorbing this term proves \|f\|_2^{10/3}\le C\mathcal E_{\Z^3}(f)\|f\|_1^{4/3}. The zero case is immediate. Equations (3.10)–(3.11) give (3.9) for one coset with \mathcal D_t^\gamma. Finally, writing v_\gamma for the restriction to coset \gamma, use

and (3.6) to obtain the assertion on all cosets.□

Proof. In each coset sum the unit flows of Lemma 3.4 with the coefficients supplied by the restrictions of the \mu_i, and call the result F^\gamma. Its divergence is the sum of these restrictions. The supports of the masses are finite. If |z_i-z_j|_\infty>4R, the two source sites in their respective supports have distance at least |z_i-z_j|_\infty/2; otherwise use the uniform constant overlap bound. Equation (3.15) and the mass bound at each center give

The first count is at most CL R^\alpha. Divide the second sum into annuli 2^n4R<|z_i-z_j|_\infty\le2^{n+1}4R; its bound is CL\sum_{n\ge0}(2^nR)^{\alpha-q}\le C, because q>\alpha and R\ge1. Consequently \sum_\gamma\sum_e|F^\gamma(e)|^2\le CmR^\alpha. Using (3.12), Cauchy–Schwarz on the disjoint coset edge sets, and (3.6), we get

This includes the empty set I, for which the flow is zero.□

Proof. A blocked component meeting B_{10R}(z) has at most five sites. Indeed, an exploration of six sites starting at a point in that ball would place them all in a radius-40 ball, contradicting regularity and M_*\ge48. Thus such a component has diameter at most 32. Its radius-four thickening has bounding-box diameter at most 40, so \operatorname{Box}_C^+ has diameter at most 42.

If a component avoids B_{10R}(z), put b_C=\min_{w\in C}|w-z|_\infty>10R. Remote regularity, applied at a minimizer, gives

after increasing the fixed threshold R_0. The minimizing site belongs to the box, so the entire box lies outside B_{9R}(z).

Now let e be a target slot edge with h\le R. Both its endpoint sites lie in B_{2R+1}(z). If a routed original edge e' uses e, route containment in Lemma 3.4 says that this occurrence is either on e' itself or in an incident enlarged box. That box cannot be remote, and therefore has diameter at most 42. Since it contains an original endpoint of e', the edge e' is within a fixed distance of e. Thus only a bounded number of original edges contribute to e, and their source distances differ from h by at most a fixed constant. Each route is simple and uses e at most once. Summing (3.14) over these edges gives (3.17); the bounded connector changes only the constant. This argument also covers routes between adjacent maximal hulls, since their middle portion is the original edge with reversed orientation.□

Proof. Work first on one coset, with the restriction of \mu, and set F=\sum_a\mu(a)F_a. Then \Div F=\mu. For each source at y_a, split its flow at distance s=\sqrt\tau into near and far parts. The parts need not have prescribed divergences; their sum is the original flow.

Index each edge by one endpoint site, with bounded multiplicity, and put \bar\mu(w)=\sum_{a:y_a=w}\mu(a). Since at most \ell slots lie over a site, \|\bar\mu\|_2\le\sqrt\ell\|\mu\|_2. As s\le R, Lemma 3.6 bounds the near part pointwise by a convolution with a nonnegative lattice kernel

Here changing the minimum distance to an edge into the distance from its indexed endpoint costs at most one. Young’s inequality yields

For the far part of each individual flow, count at most C(1+h)^2 slot edges at distance h, use the exponent 2 until R, and then use the exponent k. This gives

The last comparison uses exactly R\ge\tau\ge1 and 49/50>1/2. The triangle inequality in edge \ell^2, with \sum_a\mu(a)\le1, bounds \|F_{\mathrm{far}}\|_2\le C\tau^{-1/4}. Together with (3.19), this proves \sum_e|F(e)|^2\le C\tau^{-1/2}.

Apply this construction on every coset. Each restriction satisfies the same mass and 2-norm bounds, so the sum of all coset flow energies is at most C|\mathfrak C|\tau^{-1/2}. Summation by parts (3.12), Cauchy–Schwarz, and (3.6) now prove (3.18).□

Proof. Let q be the grid time preceding t, taking q=t when t\in\mathcal G. The circular interval (q,t] has length at most p^{12}. A site blocked at t was either blocked at q or has an incident inter-site mark in this interval. The seam does not change the number of holes at any site. The total base intensity of all slot edges incident to a site is bounded by a constant depending only on S,J. Thus Lemma 2.3 bounds the probability of any mark incident to six prescribed sites by Cp^{12}. Since a six-site set is sparse, L\ge6 and (4.2) give p^6+Cp^{12}\le Cp^6.

For a sparse set of n\ge1 sites, either at least \lceil n/2\rceil were blocked at q, or at least that many have incident marks. The first event costs at most 2^np^{\lceil n/2\rceil}, by (4.2) for the subsets. In the second event there are at least j=\lceil n/4\rceil distinct marks in the union of the edge stars: one mark is incident to at most two of the selected sites. If N counts the marks in that union, factorial domination gives

Here n/j\le4 and j!\ge(j/e)^j. Both event bounds are at most (C\sqrt p)^n after enlarging C. This argument does not require independence between blocking and intervening marks, nor any bipartite property of the interaction graph.□

Proof. Fix an integer b>100 so large that b^\alpha>4, and set h_j=16b^j for j\ge0. An admissible depth-j witness rooted at w assigns a coordinate site x_v to each vertex of the complete rooted ordered binary tree with every leaf at depth j, with x_\varnothing=w. At a vertex having k\ge1 levels below it, require that the first child has the same site as its parent, that the second child’s site is within distance h_k+8 of the parent, and that the leaf sets under the two children have mutual distance at least h_k/4. Only the leaves are required to be blocked. In depth zero the sole leaf is the root.

A blocked path from w to distance at least h_j supplies such a witness. For j\ge1, truncate the path at its first site at distance at least h_j. The first child crossing starts at w; the second starts at the first path site at distance at least h_j/2 from w. That second start is at distance at most h_j/2+8, and the remaining path reaches distance at least h_j/2-8 from it. Each crossing can therefore be truncated on first reaching distance h_{j-1} from its own start. These truncated crossings lie in the two start-centered balls of radius h_{j-1}+8, whose mutual distance is at least

Recursing inside these crossings constructs the children and keeps all their descendant leaves in the indicated balls.

Every admissible witness has 2^j distinct leaves and its leaf set is sparse. Indeed, a ball of radius h<h_k/8 cannot meet both branches of any split of scale at least h_k; hence it contains at most 2^{k-1} leaves. For h_{k-1}/8\le h<h_k/8, with 1\le k\le j, this is compatible with the sparse bound because

Below h_0/8 there is at most one leaf. At and above h_j/8, the same calculation allows all 2^j leaves. The depth-zero case is immediate. These observations cover every center and every real radius, including the scale endpoints.

Let A_j bound the number of admissible assignments rooted at a specified site. Ignoring restrictions between the two subtrees gives

Thus A_j\le C_b^{2^j}. This numerical overcount does not declare the unrestricted assignments admissible: only the separated, and therefore sparse, witnesses enter the probability union bound. For each such witness, (4.4) bounds the probability of its blocked leaves by (C\sqrt p)^{2^j}. Choose \beta_0 so large that C_bC\sqrt p\le e^{-2} for all \beta\ge\beta_0. Then

For h_j\le h<h_{j+1}, we have 2^j\ge(h/(16b))^\lambda, where \lambda=\log_b2>0. Increasing C for 1\le h<16 proves (4.5).□

Proof. For each coset there are O(R^3) centers in B_{11R}(z). A ball of the fixed radius M_* has a fixed number of six-site subsets. Equation (4.3) and a union bound give CR^3p^6 for failure of the first regularity condition. Failure of the second is bounded by Lemma 4.2 and

For the last inequality, count at most C(1+n)^2 sites at distance n and absorb both this polynomial and the tail sum into a smaller exponential rate. Finally sum over the finitely many cosets.□

Proof. Sample the actual picture under \mathbb P_P, thereby giving the exposure its correct marginal. Independently supply fresh Poisson clocks for each traversal piece. Accept these clocks where both endpoints are holes. Forced transports use actual picture marks; fresh uniform completions give the seam transitions. This generates the fresh walk conditional on the exposure, in either direction. Its expected mass loss from localization is at most its probability of exiting \mathfrak B_R(z).

An exit requires at least R/B_* inter-site steps. Before the first exit, one of the pieces must therefore make at least

steps, starting at a site in \mathfrak B_R(z). Within a single piece the actual marks in this list are distinct: physical time runs injectively through at most one period, also for a reverse piece or a piece crossing the seam. The fresh marks are likewise distinct, and seam completions cause no inter-site step.

For a fixed starting site, ignore slot compatibility to overcount all lists of j steps. There are a bounded number of possible next sites, slot edges, and actual-or-fresh designations at each step. Lemma 2.3, combined with independence of the fresh Poisson family, bounds the mixed factorial measure by the product of the actual intensity multiplied by two and the fresh intensity. On unwrapping the chosen piece, the chronological times range over a simplex of volume at most \beta^j/j!. After summing edge choices and designations, the expected number of lists is consequently at most

Choose a so large that j\ge a\beta/(3B_*) for large \beta makes this quantity at most e^{-c\beta}, by j!\ge(j/e)^j. Sum over the at most 2K+2 pieces and the O(R^3) possible starting sites; the polynomial factor is absorbed by reducing c and using a constant C_K.

All factorial estimates were applied unconditionally under \mathbb P_P, within a single piece at a time. Actual marks may be reused on later traversals; no independence of those uses was asserted. For the full-lattice walk the same argument uses only fresh marks, and gives the final assertion.□

Proof. This is Nash’s dissipation argument, applied only during regular times; see [14, pp. 935–936]. Let q=\|v\|_2^2 for an evolving localized mass. At a regular smooth time, Proposition 3.3 and (3.5) imply

At other smooth times and at instantaneous operations the norm does not increase. If it becomes zero, the conclusion is immediate. The input and output masses are at most one, by contraction.

The subinterval has only two possible partial end pieces; all intermediate pieces have length \beta. Its total nonregular time is therefore at most (b'/\beta+2)\beta/100. For b'\ge\beta/16, its regular time is at least 67b'/100. Integrating the displayed differential inequality proves (4.11). The same count holds after reversing the subinterval, and the reversed instantaneous factors are contractions. For (4.12), split at the temporal midpoint: an entry is the inner product of a forward mass from its starting slot and an adjoint mass from its ending slot. Each propagates for b'/2\ge\beta/16, so Cauchy–Schwarz and (4.11) give the result.□

Proof. As in [15, Lemma 4.5], construct f from finite hitting horizons. Set f_0=\eta, and define f_{n+1}=1 on T and f_{n+1}=U^*f_n off T. Positivity and substochasticity give 0\le f_n\le f_{n+1}\le1. Their coordinatewise limit has the stated hitting interpretation and satisfies U^*f=f off T, with zero boundary coordinates. No eventual absorption assumption is needed.

Since T is present only at the cut, a strictly positive return to T occurs at a positive integer turn. Removing the extra pins of E\setminus D_0 from the stopping rule yields q_i(\mathbf a)\le(U^*f)_i for i\in T. Harmonicity off T and f=1 on T give

The quadratic forms of U, U^*, and Q agree on real vectors. Finally, \langle\eta,g\rangle=0; expanding f=\eta+g gives (4.13).□

Proof. As u decreases, B_d(u)^*y applies successively more adjoint factors from the endpoint. Equation (3.5) identifies its continuous squared norm loss with twice the displayed Dirichlet energy; instantaneous factors cannot increase the norm. Integrating over the subinterval I gives the first inequality. The second follows from the exact identity

since \langle y,U_d^*y\rangle=\langle y,Qy\rangle. Supremum norm contraction proves the final assertion.□

Proof. First fix a grid \mathcal G as above and prove \mathsf I(E) by downward induction on the finite number of grid pins outside D_0. The assertion is vacuous when all grid points are pinned. Suppose it is proved for every strict superset of E among these pin sets, and fix an admissible sparse test T for E. Then P=E\cup T satisfies \mathsf I(P), so all the preceding probability estimates apply under \mathbb P_P. Use K=1, R=R_1, and the vectors of Lemma 4.6.

The first term in (4.13) obeys

Indeed, propagate a unit mass from each i\in T in either direction, localizing at z_i=\bar x_i. Lemma 4.4 bounds the expected discarded mass by Ce^{-c\beta}. On an exposure good for z_i,R, Proposition 4.5 bounds each one-period entry by C\beta^{-3/2}, and sparseness allows at most L(1+R)^\alpha target points of T in that region. On a bad exposure the total output mass is at most one, and its probability is at most 100\delta_R. Sum over i and average the directions.

It remains to absorb the linear term by its accompanying energy. Write E_g=\langle g,A_Qg\rangle, use the splits above with y=g, and let C_{d,R,z_i}(u) denote the starting portion localized at z_i. Define a mass on the split layer by

The unchanged pairing \langle U_d\eta,g\rangle =\langle C_d(u)\eta,Y_d(u)\rangle is constant in u. Since 0\le g\le1, the suffix has supremum norm at most one. Localization, deterministic-time regularity, and Fubini therefore give

No good-exposure indicator is needed here: the next estimate uses only regularity at the split slice.

Each retained summand of M_d(u) has mass at most one and is supported above \mathfrak B_R(z_i); the list of centers, including repetitions, has the growth bound of T. Proposition 3.5, followed by Cauchy–Schwarz in u and Lemma 4.7, gives pointwise

Because |I|=\beta/3 and E_g\ge0, completing the square in \sqrt{E_g} yields

Now 2\langle Q\eta,g\rangle =\langle U_+\eta,g\rangle+\langle U_-\eta,g\rangle. Use the fresh-return identity (2.18), (4.13), and (4.16)–(4.17) to conclude

Here R=\lceil2a\beta\rceil, so the largest polynomial term is O(\beta^{-99/100}); the error is uniform in every induction datum. Increase \beta_1, independently of the grid and its size, so that 2r(E,T)\le p. Proposition 2.4 now gives \mathbb P_E(T\subseteq\mathcal B_E)\le(2r(E,T))^{m_T} \le p^{m_T}, completing the induction.

For arbitrary allowed E and a test at any deterministic time, choose a finite grid containing zero, that test time, and all the finitely many extra-pin times, then refine it to mesh at most p^{12}. The uniform grid result proves (4.15). The same choice of grid for an arbitrary allowed P removes the hypothesis \mathsf I(P) from the regularity and good-exposure bounds.□

Proof. Lemma 4.6 gives

Every Q^n\eta is a nonnegative mass of total at most one. Consequently \|v\|_\infty\le K-1 and \|y\|_\infty\le K; the slightly weaker displayed bound also covers all later uses. The contraction and symmetry of Q give E_y\ge0. The finite telescoping identity A_Qv=Q\eta-Q^K\eta shows that substitution of g=v+y leaves the linear term 2\langle Q^K\eta,y\rangle. To compute the constant term, put a_n=\langle\eta,Q^n\eta\rangle. By symmetry,

This proves (5.1). No infinite Green series or assumption of eventual boundary absorption is used.□

Proof. Set w=Q^{K-1}\eta and I=[\beta/3,2\beta/3]. Then

For d\in\{+,-\} and almost every u\in I, split the last traversal as in Lemma 4.7:

Both vectors in the resulting pairing

are on the same hole layer at physical time t_d(u)=\theta+u or \theta-u\pmod\beta. The suffix remains killed only at the boundary. With zero extension to the slice graph, Lemma 4.7 gives

In particular the energy on the right is the original one-period quadratic form, with no localization error.

Localize the mass in the first argument throughout its propagation from i. Namely, put

where the subscript R means killing on leaving \mathfrak B_R(z), in addition to boundary killing. Expanding the powers of Q and Q_R gives convex combinations of matching direction words. Lemma 4.4, applied to the whole history of each word, therefore yields

Let G denote the event that the exposure is good for z,R, and retain only regular split slices in the time average

The original pairing is independent of u. Since \mu_d(u) has mass at most one, (5.3), (5.4), (4.6), and (4.9) imply

These are unconditional expectations under the exposure law. In particular no independence between the propagated mass, the suffix, and the regularity events is asserted or needed.

On a retained slice, every component of \mu_d(u) has elapsed propagation length

Proposition 4.5 and convexity of the norm give \|\mu_d(u)\|_2\le C\tau^{-3/4}, with a constant independent of K. The mass is at most one and is supported over \mathfrak B_R(z). Since 1\le\tau\le R, all hypotheses of Proposition 3.7 hold. Denote its constant for this fixed norm bound by C_{\rm sp}; it too is independent of K. Hence

Cauchy–Schwarz in u, followed by (5.3), gives

This also holds on G^c, where F_d=0. Completing the square therefore gives, pointwise in the exposure,

Take expectations and add the two errors in (5.5). This proves (5.2) and identifies a leading constant that is independent of K.□

Proof. Take K=1 in Lemma 5.1 and Proposition 5.2. By the fresh-walk identity,

The two one-turn localization losses are at most Ce^{-c\beta} in expectation. On a good exposure, the localized diagonal is at most C\beta^{-3/2} by Proposition 4.5; on its complement it is at most one. The complement has probability o(\beta^{-3/2}) by (4.9). This bounds \mathbb E_PQ_{ii} and proves (5.6). For j\notin D, apply this estimate with E=D, using the time of j as the cut, and use Proposition 2.4:

For j\in D, the left side is zero.□

Proof. At a zero of \phi, each k\cdot v with J(v)>0 belongs to 2\pi\mathbb Z. Since these v generate \mathbb Z^3, this forces k\in2\pi\mathbb Z^3. Thus zero is the only zero in the integration cube. Moreover,

The matrix is positive definite because the support spans \mathbb R^3. Compactness away from zero now gives \phi(k)\ge c|k|^2 throughout [-\pi,\pi]^3. Taking absolute values in (6.1) proves the heat kernel upper bound, including t\le1 by enlarging the constant. For the diagonal, substitute k=u/\sqrt t and use dominated convergence with the bound e^{-c|u|^2}. The limiting integral is

This proves (6.2).

Averaging a uniform site permutation makes the slot index uniform conditional on the site. The inverse of a uniform permutation has the same property. Every complete traversal contains one seam: if the cut is zero, it is at the end of a forward traversal and at the start of a reverse traversal. Free exchange transitions preserve this conditional uniformity, since the rate to each target slot is the same and the exit rate does not depend on the starting slot. Thus after the first seam, the slot index is uniform conditional on the site. The site process runs for total elapsed time n\beta, proving the exact equality in (6.3). In particular, seam factors at a forward and reverse junction are two fresh averages; they do not cancel as fixed permutations would. The local limit is uniform over the finite set 1\le n\le2K.

Killing restricts a path event, proving (6.4). Under the interior hypothesis it cannot act before the full ideal path leaves \mathfrak B_R(z). The ideal localization estimate in Lemma 4.4, with the same radius and at most 2K traversals, bounds this exit probability by C_K e^{-c\beta}. This proves (6.5).□

Proof. Midpoint reduction. Split the elapsed word at time n\beta/2, writing

Here \mathsf C is the starting portion. If the split is at an instantaneous operation, assign that operation to one portion exactly once. At a junction retain the order of the individual seam factors. For \eta=\ind{\{i\}}, put

After extending hole vectors by zero, all four are nonnegative masses of total at most one on the same full slot layer. Consequently

The second mass is propagated backward from the terminal copy of i; it is not a separately sampled exposure. This is the role of the matching layer in Figure 1.

Let G be the good-exposure event for z,R. Each half has elapsed length n\beta/2\ge\beta/2, so Proposition 4.5, including its adjoint statement, gives \|\mu_1\|_\infty+\|\mu_2\|_\infty\le C\beta^{-3/2} on G. The ideal halves obey this bound without a good-exposure condition: each slot probability is bounded by its full site probability in (6.2). A half need not contain a seam for this upper bound. The elementary pairing inequality gives

Indeed on G subtract the two inner products by changing their first and second arguments successively; on G^c both diagonal entries lie in [0,1]. No independence of the two halves is used. Since \mathbb P_P(G^c)=o_K(\beta^{-3/2}), it remains to prove

Coupling conditional on the exposure. Both masses are generated from i along a deterministic schedule of full or partial traversals. For j=2, reverse and transpose the suffix factors. We give the coupling for either schedule. Sample the entire actual picture with law \mathbb P_P and expose \mathcal B_P. Independently of that entire picture, sample a full unkilled ideal path on the infinite slot lattice, with fresh clocks and seam permutations on each traversal. Its bounded jump rate \lambda_J ensures that it is well defined. Conditional on the exposure, use the same fresh clock on each slot edge whenever its endpoints are holes in the cube. At a prescribed exposed jump, perform the corresponding hole transport. At a seam, share the uniform site permutation if the site has no exposed slots; otherwise use an appropriate fresh uniform completion on its remaining slots. These rules give exactly the hole kernel conditional on the exposure. They also apply to reverse pieces: heat generators are symmetric, transports are inverted, and inverse uniform bijections are uniform. Repeated traversals share the exposure but use independent fresh transitions.

Kill both paths outside \Lambda^\circ or \mathfrak B_R(z). Until they separate these killings agree. There are no interaction bonds from \Lambda^\circ to the exterior of \Lambda, by the choice of boundary thickness. A discrepancy before killing therefore requires an exposed slot at a site whose bar-coordinate is within distance 2B_* of the full ideal path, at some traversed time. Include both sides of jumps and instantaneous operations in this event. This covers a missing clock edge, a forced transport affecting the common path, and a seam completion different from a full site permutation.

Averaged encounter bound. We bound the probability of this encounter unconditionally. Partition the schedule into O_K(\beta) deterministic intervals of elapsed length at most one, each within a single directed traversal piece; assign endpoint operations once. Take h_\beta=\lceil\log\beta\rceil. In one such interval let w be the bar-coordinate of the full ideal path at its initial time. Since each jump changes bar-coordinate by at most B_*, the probability of displacement greater than B_*h_\beta during the interval is at most

For a deterministic w, Corollary 5.3, applied with pin set P and summed over slots, gives

At a seam the set of sites occupied by exposed slots is the same on both sides, so this estimate covers either endpoint convention.

If this initial event is absent but an exposed strand subsequently visits \mathfrak B_{B_*(h_\beta+2)}(w) in the interval, trace that strand back toward the initial slice. The distance between the two regions forces at least h_\beta inter-site steps. Count the first h_\beta steps from a visit. Their actual marks are distinct: this interval lies in one traversal, and for large \beta its length is less than \beta, so unwrapped physical time is monotone and injective modulo \beta. There are at most C(h_\beta+1)^3 starting sites in the smaller region and a bounded number of next-site and slot-edge choices per step. By the factorial measure domination of Lemma 2.3, the probability of such a list is at most

Discarding slot compatibility only enlarges the count. A seam can change a slot index but no site, so it merely contributes a bounded slot choice; unwrapping preserves the simplex volume. Reversing a traversal reverses the order of the times and gives the same bound. Thus (6.12) uses factorial domination under the full picture law, without conditioning that domination on the exposure.

The random coordinate w depends only on the independent full ideal path. The bounds (6.11) and (6.12), uniform for deterministic w and deterministic initial times, therefore apply to this random w. A union bound over the intervals bounds the encounter probability by

The first term is O_K(\beta^{-1/2}(\log\beta)^3); the factorial term decays faster than every fixed inverse power of \beta. Actual marks may reappear in later traversals, but the count used distinctness only within one interval. The union bound needs no independence across these intervals.

Add a death state for killing. Conditional on each exposure, the \ell^1 distance between the two live output masses is at most twice the failure probability of this coupling. Average over the picture and use (6.13). This proves (6.9) for both schedules and, by (6.8), proves (6.6).□

Proof. Suppose first that i\notin E, and take the exposure law \mathbb P_P, P=E\cup\{i\}. Lemma 5.1 and Proposition 5.2 give

Expand Q^n as the equally weighted average of the 2^n direction words. Replacing each word by its localized version costs at most C_Ke^{-c\beta} in expected diagonal mass, by Lemma 4.4. Proposition 6.2 and Lemma 6.1 then imply

There are only finitely many words at fixed K, so all comparison errors enter o_K(1). They do not alter the constant multiplying K^{-1/2}.

For the lower bound assume (6.15). At fixed exposure, run the forward fresh walk with boundary and ball killing, and continue it after visits to i. Let A_n be the event that it is alive and at i at elapsed time n\beta. Before any such visit it has stayed in the ball and hence avoided E. Thus q_i\ge\Pr(\bigcup_{n=1}^K A_n\mid\text{exposure}). The fresh transitions on successive turns give

The Bonferroni lower bound therefore reads

On good exposures each product is O(\beta^{-3}) by Proposition 4.5; on the complement each is at most one, and that complement has probability o_K(\beta^{-3/2}). The expected pair sum is consequently o_K(\beta^{-3/2}). For the first sum apply Proposition 6.2 to the all-forward word and use the interior comparison (6.5). Averaging (6.17) gives

Finally, Proposition 2.4 and Corollary 5.3 give, uniformly,

This changes only the normalized o_K(1) errors. Choose a common threshold \beta_K and a nonnegative error function tending to zero that dominates all these uniform error magnitudes. This proves both estimates for r and d. If i\in E, then d_E(i)=0, so the upper bound still holds; the lower-bound hypotheses already exclude this case.□

Proof. Factor e^{\beta hSV} from the unrestricted color sum. Its picture weight, relative to the base law, becomes

Given the picture, the colors of different cycles are independent and the probability that a cycle is up is at least 1/2. If b_\partial counts the cycles meeting D_0, the corresponding normalized picture expectation therefore satisfies

A cycle visiting the boundary either contains a time-zero boundary slot or arrives there at a mark incident to the boundary. Thus

where N_\partial is the number of such marks. Insertion of a mark changes at most two cycle factors in W_h, each between one and two, so it increases W_h by at most a factor of four. The Poisson insertion argument of Lemma 2.3 gives \mathbb E_hN_\partial\le C\beta|\partial\Lambda|. Taking logarithms in (7.2) yields

For fixed \beta, the right side tends to zero.□

Proof. For a colored picture let D_{\rm down} count its down slots at time zero. Then M=SV-D_{\rm down}. Start with independent selection of all time-zero slots with probability b(h). The probability that only up slots are selected is (1-b(h))^{D_{\rm down}}=e^{-\beta hD_{\rm down}}. Consequently the joint weight obtained by multiplying the zero-field colored-picture weight by the independent selection weight and by \ind{F\text{ contains only up slots}} has exactly the desired field-tilted colored-picture marginal, after normalization. Conditioning instead on F leaves unit weight for each coloring satisfying the pins D_0\cup F. Summing these colorings gives the picture factor 2^{n_{D_0\cup F}}, as required.

Conditional on the picture and its colors in this joint law, down slots are never selected, while up slots are selected independently with probability b(h). This proves (7.3). Finally, differentiation of the finite-volume field tilt gives (p_N^+)'(h)=V^{-1}\mathbb E M; conditioning on F gives (7.4).□

Proof of Theorem 1.1. Integrate (7.6) and (7.7) from 0 to h>0, and divide by h. This bounds

First send N\to\infty, using Proposition 7.1 at both endpoints and \rho_N\to1. Then send h\downarrow0. For these fixed \beta,K, continuity of b at zero gives

By the definition of the right derivative,

Thus the thermodynamic limit and the zero-field derivative have been handled through pressure increments. Now let \beta\to\infty with K fixed and then K\to\infty. Both sides of (7.8) converge to \kappa\zeta(3/2), proving the stated law.□