Spontaneous magnetization in the quantum Heisenberg ferromagnet

Proof. Let \psi_k be the normalized sum of the \binom\ell k elementary tensor vectors having exactly k up spins, for 0\le k\le\ell. These form an orthonormal basis of the symmetric subspace, and

For 0\le k<\ell, each elementary vector with k+1 up spins occurs k+1 times in \sum_j s_j^+\psi_k. Consequently

Also \widehat S^+\psi_\ell=0. Under m=k-S, the squared coefficient is S(S+1)-m(m+1). The lowering, x, and y matrices follow by adjunction and the definitions. Thus e_m\mapsto\psi_{S+m} gives the claimed unitary identification.

The average over the finite permutation group is self-adjoint and idempotent, and its range is exactly the invariant subspace, proving the projection formula. Conjugation by a within-site permutation merely reorders the summands of each \widehat S_x^a. It therefore commutes with \widehat H_\Lambda and \widehat M_\Lambda. Equivalently, in the slot-edge expression, conjugation permutes the complete collection of \ell^2 slot pairs above every incident lattice edge. This last observation is essential: an individual inter-site transposition need not commute with \mathcal S_x.

Finally, on two spin-\tfrac12 factors, direct calculation in the up/down basis gives

Summing this identity over the slot edges proves (2.1) and the asserted restrictions.□

Proof. Let \varphi_t be the permutation generated by the inter-site marks up to time t, without the endpoint permutations. Independent increments, or the Poisson expansion in time-ordered marks, give

Indeed, over an interval of length \dd t, the expectation is I+\frac12\sum_{\{i,j\}}(\mathcal U_{(ij)}-I)\dd t+o(\dd t); the resulting constant-coefficient matrix differential equation gives the displayed exponential. Averaging the independent endpoint permutations now gives

Lemma 2.1 identifies the trace on the range of \mathcal S_\Lambda with the physical trace. Moreover, \widehat M_\Lambda commutes with every slot permutation, since it is the sum of the same one-slot observable over all slots. Therefore (2.4) yields

To calculate the last trace, index the elementary up/down tensor basis by \eta\in\{-1,1\}^{V_\Lambda}. The diagonal entry of \mathcal U_{\pi_\omega} at \eta is one exactly when \eta is constant on every cycle of \pi_\omega, and is zero otherwise. For such a basis vector the eigenvalue of \widehat M_\Lambda is \frac12\sum_C\varepsilon_Ca_C. Hence

Taking \Phi=1 proves (2.2); dividing (2.5) by it proves (2.3). In particular, taking \Phi to be the indicator of any spectral subset proves the claim about the spectral distribution, not only its moments.□

Proof. In one period the number of Poisson marks is finite almost surely. Their times are distinct and avoid the countable collection of deterministic cut times and pin times, almost surely. A line that visits a boundary site occupies that site on a time interval of positive length: inter-site marks do not coincide, and endpoint permutations preserve the site. This remains true for a visit crossing the time seam. A dense collection of cuts detects every such visit, with every slot included at each boundary cut.

There are only finitely many cycles. For a fixed picture, choose one detecting cut for each cycle that meets D_0 and take the largest of their finite indices. After this index every cycle has the same incidence with D_n as with D. This proves both eventual equalities. Also w_{D_n}\downarrow w_D and w_{D_n}\le 2^{|V_\Lambda|}. Dominated convergence gives Z_{D_n}\to Z_D and

which proves total-variation convergence. Applying dominated convergence to w_{D_n}\Phi_n proves (2.8). The dominating constant here depends on the fixed volume; no uniform-in-volume bound on the total loop weight is asserted.□

Proof. Inserting one inter-site transposition into a picture cuts two directed strands and reconnects them with their outgoing portions exchanged. If they lie on different cycles, those cycles join; if they lie on one cycle, that cycle splits into two. The union of their slot-time strands, and hence all pin incidences on that union, is retained. On joining, the number of cycles missing D cannot increase. On splitting, it increases by at most one: a cycle missing D gives two such cycles, while a cycle meeting D has at least one descendant still meeting D. Thus, away from coincident mark times,

For finite pins coincidences with pin times have product-intensity measure zero; including D_0 does not change the surgery argument. The endpoint permutations are held fixed throughout this insertion.

The identity used to integrate (2.10) is the iterated Mecke formula [26]; its factorial form is stated in [21, Theorem 4.4]. We derive it here under the base Poisson law, with the pin weight retained inside the integrand. Write \lambda=\mu(X_\Lambda). For a fixed endpoint permutation, expand the Poisson measure as

In the weighted sum over ordered distinct n-tuples, the k-mark term has k!/(k-n)! choices of the distinguished marks. Relabel them u_1,\ldots,u_n and sum the remaining k-n marks. Averaging also over the independent endpoint permutations gives the exact identity

All terms are nonnegative, so this manipulation is justified also when an integral is infinite. By (2.10), the expectation inside the integral is at most 2^n Z_D. Division by Z_D proves (2.9).□

Proof. Let F_n be stable polynomials of bounded degree converging coefficientwise to a nonzero polynomial F. If F vanished at a point of a product of upper half-planes, choose a complex line through that point on which the restriction of F is not identically zero. A sufficiently small disk in this line lies in the product of upper half-planes. The restrictions of F_n converge uniformly on the disk. On a smaller circle avoiding the isolated zeros of the restriction of F, the argument principle (or Rouché’s theorem) forces F_n to have a zero inside for large n. This is impossible. The existence of the line follows, for example, by choosing a direction where the first nonzero homogeneous term of the Taylor expansion of F at the purported zero does not vanish.

Specialization at a\in\R is the limit of specialization at a+\mathrm i/n, so the first assertion applies. Identification of variables is immediate from the definition.

For differentiation in t, let k be the degree of F in t. The case k=0 gives the zero polynomial. If k\ge1, the leading coefficient is the nonzero coefficientwise limit

It is therefore stable. Consequently, after fixing all other variables in \mathbb H, the resulting polynomial in t still has degree k. Write its roots, with multiplicity, as \zeta_1,\ldots,\zeta_k; each has imaginary part at most zero. For t\in\mathbb H,

has strictly negative imaginary part. It cannot vanish, proving stability of the derivative. Finally, taking the linear coefficient after identification is differentiation in t followed by t=0.□

Proof. Fix every other variable in \mathbb H and write the two-variable section as

where a,b,c,e can be complex. For every z\in\mathbb H,

Indeed, if c+ez\ne0, the root -(a+bz)/(c+ez) in the w variable cannot belong to \mathbb H; if c+ez=0, the displayed quantity is zero. The swapped section q(w,z) is also stable, so

Put b_u=(1-u)b+uc and c_u=(1-u)c+ub. Direct expansion gives

If 0<u<1 and b\ne c, the right-hand side is strictly positive. In particular c_u+ez\ne0, and the root in w of the mixed section lies strictly below the real axis. If u\in\{0,1\} or b=c, the section is the original section or its swap. These cases also exclude zeros in \mathbb H^2. Since the other variables were arbitrary, the mixed polynomial is stable; multiaffinity is preserved directly.□

Proof. For the identity permutation the polynomial is \prod_{i\in V}(x_i+y_i), a product of nonvanishing upper-half-plane sums. Composing the output with a transposition either leaves the polynomial unchanged or interchanges the corresponding two y variables. Averaging an independent Bernoulli choice is precisely the operation in Lemma 3.3. Deterministic permutations only relabel variables. Repetition proves the first assertion. Convergence of permutation laws gives coefficientwise convergence of their symbols, whose degrees are bounded and whose value at all variables equal to 1 is 2^{|V|}. Lemma 3.2 therefore proves stability of every such limit, with zero excluded.

For a finite-time Poisson exchange process, partition its time interval into slices of length at most \delta. On each edge and slice record whether the Poisson clock has at least one mark; these indicators are independent Bernoulli variables. Apply the indicated transpositions in a fixed order within a slice. This coupled approximation agrees with the actual process whenever every slice has at most one mark in total. If the total rate is \lambda and the interval length is t, the probability of disagreement is at most C\lambda^2(t+\delta)\delta, which tends to zero. This constructs the required limit of Bernoulli-product laws.

To approximate a uniform permutation on a subset A\subseteq V, run Poisson transpositions at positive rates on every pair in A. For |A|\le1 there is nothing to prove. Otherwise the unit-time transition matrix on the finite group of permutations of A has all entries positive: every target permutation is a finite word in these transpositions, and that word can occur in the prescribed order within unit time with no other marks. The uniform distribution is invariant, since each step is multiplication by a random group element. If the group has M elements and the minimum transition entry is \eta>0, the transition matrix has a common uniform part of mass M\eta. Splitting off that part shows that distance from the uniform law contracts by at most 1-M\eta at each unit step (and is already zero if M\eta=1). Hence the law converges to uniform as time tends to infinity. At each finite time the small-slice construction gives a Bernoulli-product approximation. Taking a sequence with both errors tending to zero proves the asserted representation of the uniform law. For a finite independent product, approximate its factors independently; their product laws then converge to the desired product law.□

Proof. Consider first one deterministic permutation. If q is not a fixed point, write i=\pi^{-1}(q) and j=\pi(q); the two factors involving y_q and x_q become

Their linear coefficient is x_i+y_j. Thus deleting q splices the directed edges i\to q\to j into i\to j. This includes a two-cycle, for which i=j. If q is a fixed point, its factor is x_q+y_q=2t, giving a factor 2. Successive deletion leaves the next-pin permutation on every cycle meeting P. A cycle missing P eventually becomes a fixed point and contributes exactly one factor 2. Linearity gives (3.8); Lemma 3.2 gives stability up to zero. Its nonnegative coefficients and positive value at all surviving variables equal to 1 exclude zero.

In each product on the right of (3.8), differentiation in x_i, i\in P\setminus T, selects the x_i term from exactly those factors. Setting every remaining x variable to zero, including those indexed by T, selects y_{\sigma(i)} from each remaining factor. Division by Z_P gives (3.9). Derivatives and real specialization preserve stability up to zero; G_T(\boldsymbol1)=1 excludes zero. Since \sigma is a permutation, the product contains m=|T| distinct y variables, so this is indeed a multiaffine subset generating polynomial with nonnegative coefficients.□

Proof. Write X_j=\1_{\{j\in A\}}. Conditioning on B\subseteq A gives, for the indicators outside B, the generating polynomial

It is nonzero and stable by Lemma 3.2. Under this conditional law, keep only two variables z,w and specialize all others to 1. The resulting stable polynomial has the form a+bz+cw+ezw, where a,b,c,e\ge0 and a+b+c+e=1. Applying (3.5) with real coefficients gives

Thus bc\ge ae. The covariance of the corresponding presence indicators equals

Whenever the second indicator has positive probability of being one, conditioning it to be one therefore cannot increase the probability that the first is one.

Order J=\{j_1,\ldots,j_k\}. If any prefix-presence event has probability zero, (3.10) is immediate. Otherwise, for each h, remove the conditions X_{j_{h-1}}=1,\ldots,X_{j_1}=1 one at a time. The preceding conditional covariance inequality gives

Multiplying the conditional probabilities proves (3.10). Applying the same argument to the conditional generating polynomial proves the final assertion.□

Proof of Proposition 3.1. Suppose first that E is finite, and use the layered construction above with P=E\cup T. By Lemmas 3.5 and 3.6, followed by the arithmetic–geometric mean inequality and (3.4),

If one of the factors is zero this conclusion is interpreted directly; no conditioning on a zero-probability event is needed.

By (3.3), the event on the left is H=\{T\subseteq\mathcal B_E\}. It remains to compare its probability under the two pin laws.

For clarity, write w_D=2^{n_D}, where n_D is the number of cycles missing D, and Z_D=\mathbb E_{\mathrm b}w_D. Deleting the m pins in T changes the number of unpinned cycles by the number of cycles that meet T but miss E. That number lies between 0 and m. Hence, picture by picture,

We obtain

This proves the finite-pin case with the required normalization.

Finally let E=F\cup D_0 with F finite. Choose increasing finite sets of boundary cuts with dense union, and let D_{0,n} contain all boundary slots at those cuts. Put E_n=F\cup D_{0,n} and P_n=E_n\cup T. Then E_n and T are disjoint and the finite-pin inequality applies. As in Lemma 2.3, almost every picture has finitely many marks, and every boundary visit contains an interval of positive duration. Thus every boundary-touching cycle is eventually detected, and

The weights are bounded by 2^{|V_\Lambda|} in this fixed volume, and all their normalizers are positive, so the corresponding normalized probabilities converge by dominated convergence.

The same reasoning applies to the return test, with a useful finite horizon. Starting from i\in T, follow the uncensored line until its first positive return to T. This time is finite, since the line eventually returns to i on its finite permutation cycle. If the line meets F before that return, every finite-cut test fails. If it first meets the boundary before that return, one of its positive-duration boundary visits before the return contains a cut from the dense union, so all sufficiently fine tests fail. If it meets neither, every test succeeds. Since T is disjoint from D_0, a boundary visit cannot first occur only at the return point. Consequently, for every i\in T,

Bounded convergence with the weights w_{P_n} now gives r(E_n,T)\to r(E,T). Taking limits in \mathbb P_{E_n}(T\subseteq\mathcal B_{E_n})\le(2r(E_n,T))^m proves (3.2) for continuous boundary pins. Only scalar probabilities and bounded weights are passed to this limit; no polynomial with infinitely many variables is required.□

Proof. We first fix A under the base law and verify the conditional construction before imposing any cycle condition. For each slot edge e, take two independent rate-1/2 Poisson clocks N_e^{(1)},N_e^{(2)}, all independent across edges. Accept a mark of N_e^{(1)} exactly when at least one endpoint contains an A-label immediately before it. Accept a mark of N_e^{(2)} exactly when neither endpoint contains an A-label.

This construction has the base exchange law. To see this directly, if there are b slot edges, the superposed clock has rate b, and its next mark has an independent uniform type among the 2b clocks. For each edge exactly one of its two types is accepted, whatever the past configuration. Thus, conditionally on the past, the next superposed mark is accepted on any specified edge with probability 1/(2b), and is rejected with probability 1/2. Iterated conditioning shows that these accepted-edge or rejection outcomes have the fixed product distribution and are independent of the superposed waiting times. Poisson splitting therefore yields independent rate-1/2 accepted processes on the edges. If there are no edges the assertion is immediate.

The motions of the A-labels before the endpoint permutations use only the first clock family: second-family marks always exchange two other labels. At each site, condition a uniform endpoint permutation on its images of the A-labels that are there just before the endpoint. Each bijection between the remaining source and target slots has the same number, namely one, of extensions with those prescribed images. The conditional completion is therefore uniform. Completions at different sites remain independent, since the endpoint permutations were independent and each revealed restriction concerns its own site.

It follows that the full specified histories, including endpoint images, are determined by the first clock family and these revealed restrictions, without using the second family or the remaining completions. More explicitly, the pre-completion locations of the A-labels depend only on the first clock family. The endpoint permutations are independent of both clock families, so conditioning on these locations and the selected endpoint images leaves the second clock family independent. Conditional on each revealed source and target restriction, the remaining bijection is uniform over its allowed completions. Conditional on the histories, the second-family clocks are still independent Poisson clocks. Restricting them to the now fixed intervals with two hole endpoints gives the independent exchanges in \mathbb Q_{\mathbf a}. The cross jumps and the remaining endpoint completions give exactly its other prescribed operations. Although conditioning on histories may leave unobserved first-family marks, those marks are rejected whenever they have two hole endpoints and cannot affect the remaining picture. This proves the claimed base conditional kernel on F_A.

On F_A, the labels of A form a union of whole time cycles missing P. Their number is c_A(\mathbf a), which is determined by the endpoint mapping of A. Requiring in addition that every remaining cycle hit P makes A exactly the initial labels of all the missing cycles. Therefore for each base picture exactly one initial set satisfies both requirements. If n_P is the number of cycles missing P, the pointwise identity is

where each summand is defined as zero off F_A. Apply the base conditional kernel on each branch of this finite sum. Dividing by Z_P=\mathbb E_{\mathrm b}2^{n_P} gives (4.3). Integrating the remaining picture gives (4.4). Exposures with g_P(\mathbf a)=0 consequently have zero outer measure; on its positive set, division by g_P(\mathbf a) gives the asserted conditional law.

For finite pins the same argument applies without change. For pins including D_0, avoidance of the boundary is a measurable property of the finite-jump histories. Thus the argument applies directly to that case as well.□

Proof. If T_{ij} transposes two coordinates, then

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

This is (4.6) for one Bernoulli transposition. For independent factors A,B satisfying the identity,

Deterministic permutations satisfy the identity as well, so induction proves it for finite products. All the matrix entries and minors are bounded continuous functions on the finite permutation space, which proves passage to limits.

We verify the limit representation for the hole picture. Between successive exposed jumps and endpoint times, the hole set is fixed and the available exchanges have constant rates. Their permutation law has the Bernoulli-product approximation proved in Lemma 3.4. The forced transports are deterministic bijections between equally sized hole sets.

The same lemma supplies this approximation for every uniform completion. When its source and target slot sets differ, precede or follow a uniform permutation of one fixed set by a deterministic bijection.

For completeness, the layered permutation has the same product structure. The total number of holes is constant, so identify every cut layer with one fixed set of that size. The independent interval maps act on separate layer blocks; their direct sum, followed by the fixed cyclic shift of the layers, is the layered permutation. Each random operation within a block is a transposition on the full layered set, and each fixed transport is a permutation there. Independence of exchanges on disjoint intervals and of the completions proves the required representation.□

Proof. For a fixed matrix D, the principal-minor expansion and multiplicativity of exterior powers give

Take D=\operatorname{diag}(W,I_Q). If a permutation cycle lies entirely in Q, its indicator vector v, extended by zero on J, satisfies D\Pi v=v, even for arbitrary W. Thus the determinant on that picture is zero. On G_J, \Pi_{QQ} is nilpotent: a sufficiently long path confined to Q would contain a cycle there. Its inverse resolvent is a finite sum, and block elimination gives

Here \det(I-\Pi_{QQ})=1, and each term of the resolvent follows the permutation through a specified number of intermediate nonpins. Consequently the preceding determinant identity reads

At W=0, this proves (4.7) without any invertibility assumption. If its value is positive, I-K_{QQ} is invertible. The Schur complement on the right of (4.10), followed by division by that positive determinant, gives (4.9).

To extract the first moment, use \det(I-tWA)=1-t\operatorname{Tr}(WA)+O(t^2). The linear terms in (4.9) yield

Taking the matrix units for W gives each entry of (4.8).

Finally, K is nonnegative and doubly stochastic, as the mean of permutation matrices. Its restriction K_{QQ} is substochastic. If a state in Q could not reach J by any positive-probability path, its reachable set would be a nonempty closed subset of Q. The restriction to that set would have column sums one and eigenvalue one, forcing I-K_{QQ} to be singular. Therefore every state in Q has a positive path to J. Since there are finitely many states, some integer n and number \varepsilon>0 make the probability of leaving Q within n steps at least \varepsilon, uniformly over starting states in Q. It follows that

The first term of H records a direct next-step hit of J, and the term K_{JQ}K_{QQ}^jK_{QJ} records a first hit after j+1 intervening steps in Q. This proves its Markov interpretation and completes the proof.□

Proof. Fix an exposure for which Proposition 4.1 holds and g_P(\mathbf a)>0. Choose increasing finite deterministic time sets \mathcal T_n with dense union, and let D_{0,n} contain all boundary slots at the times in \mathcal T_n. Put

The cuts used below include all times occurring in P_n, as well as time 0 if necessary. An extra unpinned cut merely subdivides an interval and does not change which cycles hit P_n.

Under \mathbb Q_{\mathbf a}, let G_n be the event that all hole cycles meet P_n, and set g_n=\mathbb Q_{\mathbf a}(G_n). The events G_n increase. Almost surely the one-period hole picture has only finitely many random exchanges, in addition to the finitely many prescribed changes. Its spatial visits have positive residence time: independent Poisson jumps do not coincide with any prescribed change, and within-site completions do not change the site. The exposed histories avoid D_0, so their cross jumps cannot create an instantaneous boundary excursion either. Every boundary visit by a hole cycle is therefore detected by some cut in the dense union. There are finitely many cycles, whence

In particular, g_n>0 for all sufficiently large n. No invertibility is required for the earlier indices.

For such an n, Lemma 4.2 applies to the unconditioned layered hole permutation. With J=P_n, Proposition 4.3 identifies each next-pin probability under \mathbb Q_{\mathbf a}(\,\cdot\mid G_n) with the corresponding first-hit probability for its matrix K. The latter is precisely the walk just defined, observed at the finitely many cuts: interval choices are independent, including on later passes through the same interval. Thus, if A_n is the event in a fixed hole picture that the next P_n-pin after i belongs to T,

On the picture side, a cycle through i\in T returns to T after finitely many periods, regardless of which other pins it meets. Detection of boundary visits before that return shows that \1_{A_n} decreases to the event A that the next full P-pin is in T. Together with (4.12) and bounded convergence, this gives

Dividing by the positive limiting denominator yields the return probability under \mathbb Q_{\mathbf a}(\,\cdot\mid G_P).

For the walk side, use a single uncensored fresh-turn trajectory for all the tests. Its first positive T-hitting time \tau_T^+ is then fixed, independently of n. The events on the right of (4.13) decrease. A path in their intersection has \tau_T^+<\infty, since this is already part of each event. If that path had visited the boundary before \tau_T^+, the positive residence time of the visit would make one of the dense cuts detect it before \tau_T^+. Thus their intersection is exactly \{\tau_T^+<\tau_E\}, up to a null set. Continuity from above gives convergence of the probabilities. Finally Proposition 4.1 identifies the limiting conditional picture law with the conditional law under \mathbb P_P. This proves (4.11).□

Proof. Set f_0=\1_T and recursively define

where the multiplication by the indicator is pointwise. Nonnegativity and the substochasticity of U^* show inductively that 0\le f_n\le f_{n+1}\le1. The Markov property identifies f_n, off T, with the probability of reaching T within the next n periods before killing. Thus its pointwise limit f has the asserted probability interpretation. There are finitely many coordinates, so passage to the limit in (4.17) gives (4.15). A boundary starting point is killed immediately, and consequently every f_n and f is zero there.

If h is any nonnegative solution of (4.15), then h\ge f_0 and induction in (4.17) gives h\ge f_n for every n; hence h\ge f. This proves minimality. In particular f is zero on any closed class that cannot reach T. No assumption of certain absorption by the boundary is needed.

Since T occurs at a single time modulo the period, a strictly positive return from that cut to T occurs after an integer number of periods. Conditioning on the first period therefore shows that the relaxed return probability from i\in T is (U^*f)_i. By Lemma 4.4, the conditional probability q_i(\mathbf a) contributing to (4.1) is the return probability with all of E retained. Hence

The vectors are real, so transposition in the scalar quadratic form and harmonicity off T give

Averaging under the actual exposure marginal gives (4.16). Measurability causes no additional limiting issue: finite-time transition probabilities are measurable functions of the finite-jump exposure, and f is the increasing limit of the finite-horizon functions in (4.17).□

Proof. Let q be the preceding grid time and I=(q,t], interpreted cyclically, so that |I|\le p^{4d}. On-site permutations do not change the number of holes at a site. Consequently a site blocked at t was either blocked at q or has an incident inter-site mark in I. This remains true when I crosses the time seam. By Lemma 2.4, the expected number of marks in the edge star of a prescribed site during I is at most 2d\ell^2|I|. If all 2d prescribed sites are blocked at t, either they were all blocked at q, or one of these finitely many edge stars contains a mark. Their set is sparse since L\ge2d. The union bound and (5.5) give p^{2d}+C p^{4d}\le Cp^{2d}.

Now let n=|A| and assume A is sparse. If every site of A is blocked at t, either at least \lceil n/2\rceil of its sites were blocked at q, or at least \lceil n/2\rceil have an incident mark in I. For the first alternative, summing (5.5) over subsets costs at most 2^n p^{\lceil n/2\rceil}. For the second, one of the two lattice parities supplies at least k=\lceil n/4\rceil sites with incident marks. Edge stars of distinct sites of the same parity are disjoint. If N_z counts marks in the star of z during I, then for any such prescribed k-set A', the product \prod_{z\in A'}N_z counts choices of k distinct marks. The factorial bound therefore yields

Summing over at most 2^{n+1} possible choices bounds the second alternative by 2^{n+1}(C p^{4d})^{\lceil n/4\rceil}. As 4d\lceil n/4\rceil\ge dn\ge n/2, both alternatives are bounded together by (C_0\sqrt p)^n, after enlarging C_0. These are separate event bounds combined by a union bound; no independence between old blocking and intervening marks is required.□

Proof. Fix an integer b>100 so large that b^\alpha>4, and put h_j=16b^j. We first describe exactly which site lists will be used as witnesses. Their separated binary-tree structure is also used in multiscale cascading-event arguments; see, for example, [34, Section 3]. Here the leaf probabilities will come directly from (5.7). A depth-j witness rooted at z is an assignment of sites x_w to the vertices w of a rooted ordered binary tree of depth j, with x_\varnothing=z. If a vertex w has k\ge1 levels below it, require

where \mathcal L_{wa} is the set of leaf sites descended from child wa. Call such an assignment admissible. Its leaves are distinct by the separation condition. Depth zero has just the prescribed root as its leaf.

A blocked crossing from z to distance at least h_j produces an admissible witness all of whose leaves are blocked. For j=0 use the root alone. For j\ge1, truncate the crossing at its first exit to distance h_j; it is contained in B_{h_j+8}(z). One child crossing starts at z. The other starts at the first site whose distance from z is at least h_j/2, hence at distance at most h_j/2+8. The remaining part of the parent crossing reaches distance at least h_j/2-8 from this second start. Each child can thus be truncated at its first exit to distance h_{j-1}, and is contained in its start-centered ball of radius h_{j-1}+8. The distance between these two balls is at least

Recurse on the two truncated child crossings. Every descendant leaf remains in its child’s truncated crossing, so all the separation conditions in (5.9) hold.

Every admissible witness has a sparse leaf set. Indeed, for 1\le k\le j, a ball of diameter strictly less than h_k/4 meets at most one child at every split of scale at least h_k. It therefore contains at most 2^{k-1} leaves. If h_{k-1}/8\le h<h_k/8, this bound is at most L(1+h)^\alpha, since

For h<h_0/8 there is at most one leaf, and for h\ge h_j/8 the same calculation allows all 2^j leaves. The depth-zero case is immediate. This verifies (5.2) for all centers and real radii, including the scale endpoints.

Let A_j be an upper bound on the number of admissible assignments rooted at any prescribed site. At the top split the first child root is fixed and the second has at most (C_dh_j)^d choices. Ignoring restrictions between the two child assignments gives the numerical overcount

In particular

Only admissible assignments enter the probability union bound. The unrestricted product used to bound their number need not itself consist of sparse witnesses.

For each admissible assignment, Lemma 5.1 bounds the probability that all its leaves are blocked by (C_0\sqrt p)^{2^j}. Choose p_0\le1 once and for all so that C_bC_0\sqrt{p_0}\le e^{-2}. For every p\le p_0,

For h\ge16, choose j\ge0 with h_j\le h<h_{j+1}. Then 2^j\ge(h/(16b))^\gamma, where \gamma=\log_b2>0. This proves (5.8); increasing C covers 1\le h<16. No constant in this last step changes when p is decreased further.□

Proof. There are at most C R^d centers in B_{11R}(x), and a ball of radius M_* has a fixed number of 2d-element subsets. The union bound and (5.6) therefore bound failure of the first condition by C R^d p^{2d}. By Lemma 5.2, failure of the second has probability at most

For the last inequality, count at most C(1+n)^{d-1} sites at distance n and absorb this polynomial and the tail summation into half of the exponential. Since \nu\gamma>0 is fixed, the resulting constants are uniform. Add the two bounds.□

Proof. First suppose F is a set of fewer than 2d sites in the whole lattice. The complement of a box containing F is connected when d\ge2, so \Z^d\setminus F has only one possible infinite component. It has no finite component: from a vertex of such a component the first exit along each of the 2d positive and negative coordinate rays must lie in F, giving 2d distinct deleted sites. Hence \Z^d\setminus F is connected.

The confinement radius can be uniform when the two endpoints have bounded separation. Otherwise, after translating the first endpoint to 0, there would be sets F_n with |F_n|<2d and surviving endpoints v_n\in B_K(0), admitting no path inside B_n(0). Pass to a subsequence with v_n=v and with the indicators of F_n converging pointwise on \Z^d. The limiting set F_\infty has fewer than 2d sites and avoids both endpoints. By the preceding paragraph there is a finite path from 0 to v avoiding F_\infty. For all sufficiently large n, this path lies in B_n(0) and avoids F_n, a contradiction. Choose a uniform radius \rho\ge K+1.

For the asserted local version apply this conclusion to F'=F\cap B_\rho(u), a set of fewer than 2d sites in the whole lattice. The resulting path lies in B_\rho(u), where F' and F agree. Erasing loops gives the claimed length bound.□

Proof. We transfer v to a lattice function while controlling energy, then prove the lattice inequality by averaging. At every surviving site z, choose an available slot maximizing v, and let V(z) be this maximum. Every blocked site in B_{R+3}(x) has a surviving nearest neighbor: otherwise its 2d neighbors alone would violate the first regularity condition in the ball centered there. Choose one such neighbor \phi(z) and define

The function w is supported in B_{R+1}(x). Since a site’s value is copied only by its nearest neighbors,

For a lattice edge \{z,z'\} with a nonzero endpoint value, both endpoints belong to B_{R+2}(x). Each of w(z),w(z') is the value at a chosen maximizing slot over a surviving site, say u,u', at distance at most one from the corresponding endpoint. Thus |u-u'|_\infty\le3 and u,u'\in B_{R+3}(x). Lemma 5.5 joins them through surviving sites within a fixed radius of u, with a fixed length bound. Lift this path using the chosen maximizing slot at every site. Every lifted edge is available because all pairs of available slots across a lattice edge are joined. If u=u', the difference is zero. Otherwise telescoping and Cauchy–Schwarz show that (w(z)-w(z'))^2 is at most a fixed constant times the sum of (v_i-v_j)^2 along the lifted path. For any fixed slot edge, only a bounded number of the original lattice edges can have a path using it: all of their endpoints lie within a fixed radius of that edge. Hence, summing,

For completeness, write \mathcal E_{\Z^d}(w) for the sum on the left of (5.16). For an integer k\ge1, let A_kw be the average of translates of w by the points of \{0,\ldots,k-1\}^d. Telescoping a translate along coordinate directions, followed by Cauchy–Schwarz, gives

The averaging kernel has counting 2-norm k^{-d/2}; the convolution inequality therefore gives \|A_kw\|_2^2\le k^{-d}\|w\|_1^2. Consequently

If w\ne0, choose k=\lceil(4\|w\|_1^2/\|w\|_2^2)^{1/d}\rceil. The second term is at most \|w\|_2^2/2, and \|w\|_1^2/\|w\|_2^2\ge1 bounds the effect of rounding. Absorption yields

For w=0 the inequality is immediate. Finally use (5.15) and (5.16) to obtain (5.14).□

Proof. Distinct components have mutual distance at least nine, so their radius-four thickenings are disjoint. The radius-four lattice cubes about successive points of a component intersect; their union is connected. Every finite complementary component has a neighbor in A_C, which proves that filling preserves connectivity. A vertex outside the coordinate bounding box has a coordinate ray avoiding that box, so it belongs to U_C. This proves finiteness and the claimed containment.

For laminarity, take two disjoint connected thickenings A_C,A_{C'}. The second lies in a single component of the complement of the first. If it lies in a finite component, then U_C is an infinite connected subset of the complement of A_{C'}, whence U_C\subset U_{C'} and H_{C'}\subset H_C. The reversed containment is treated identically. In the remaining case each thickening lies in the infinite complementary component of the other. Then H_C avoids A_{C'} and is connected; since it contains A_C\subset U_{C'}, it lies wholly in U_{C'}. Hence H_C and H_{C'} are disjoint.

We next prove connectivity of the exterior boundary without a planar separation argument. Let \mathcal E_C be the edges between H_C and U_C, and join two of these edges when they lie on an elementary lattice plaquette. Every plaquette meets \mathcal E_C an even number of times, and all cut edges of one plaquette belong to the same component of this edge relation. Thus each component separately meets every plaquette evenly. It then meets every finite closed lattice walk evenly: modulo two, such a walk is a sum of elementary plaquettes, as follows by commuting successive steps in different coordinate directions and canceling inverse steps. If the cut had two components, take one edge from each, join their inner endpoints within H_C and their outer endpoints within U_C. The resulting closed walk uses exactly one cut edge from each chosen component, a contradiction. The cut is therefore connected by plaquettes. The outer endpoints of cut edges on a common plaquette have sup distance at most one. This proves the asserted connectivity of S_C.

An inner cut endpoint u must belong to A_C: otherwise a complementary pocket containing u would be adjacent to the infinite component through v, which is impossible. Since v\notin A_C and |u-v|_\infty=1, the distances to C are exactly four and five. For another component C', separation gives \dist(v,C')\ge4 and \dist(u,C')\ge5. To interpolate a sup step between two points of S_C, change the differing coordinates one at a time. Every intermediate vertex is within sup distance one of its starting point, hence has distance at least four from C and at least three from every other component. All these vertices survive. They also remain in the coordinate bounding box of S_C, which is contained in the unit enlargement stated above.□

Proof. First we quantify the hull sizes. A blocked component meeting B_{10R}(x) has fewer than 2d vertices. Otherwise, starting at a vertex there, a connected exploration would find 2d vertices within 2d-1 component steps, all in a ball of radius 8(2d-1)\le M_*, contradicting regularity. Such a component has diameter at most 8(2d-2), and its hull has diameter at most

For a component avoiding B_{10R}(x), let b_C=\min_{z\in C}|z-x|_\infty>10R. Applying remote regularity at a minimizer gives

This also bounds the diameter of its coordinate bounding box. Since \nu<1, increasing a fixed lower threshold for R makes the right side, plus any of the fixed enlargements used above, at most b_C/10. All sites of these enlarged boxes then have distance comparable to b_C from every y\in B_R(x). In particular, no such remote box contains y.

Apply the routes following (6.2), using fixed choices of representatives and paths. The preceding size bounds show that the route of a lattice edge e lies within distance

of that edge. For a local hull this is the fixed bound D_{\mathrm{loc}}+2. For each remote hull incident to e, an endpoint of e belongs to its bounding box, so its b_C is comparable to 1+\dist(e,y); now use (6.5). The same reasoning applies when the edge joins two hulls. These comparisons are uniform over all injection sites y\in B_R(x).

Choose one canonical available slot c(z) at each unblocked site z. Lift every routed site path to these canonical slots. All its edges exist because every pair of available slots across a lattice edge is present. The map of the original lattice vertices into slots is \psi(z)=c(\rho(z)). Its preimages are finite: outside the finitely many finite hulls it is injective, and only those hulls can supply additional preimages.

We also need a bounded path from the specified slot a to \psi(y). If y is in a hull, that hull is local and the separation between y and \rho(y) is at most D_{\mathrm{loc}}+1. Both sites are unblocked, so Lemma 5.5 joins them through survivors in a fixed neighborhood of y. Its assumptions hold because y\in B_R(x), the neighborhood has fixed radius, and R dominates that radius. Lift this path, starting at a. If \rho(y)=y but a\ne c(y), use a two-edge path through a surviving neighbor of y. Such a neighbor exists, since blocking all 2d neighbors would violate local regularity. Thus in all cases there is a simple slot path Q_a from a to \psi(y) of bounded length and displacement. The choice K_{\rm conn}\ge8(2d-2)+12 in the local-connectivity construction covers these requirements before M_*, R, and finally p are fixed.

Start with an explicit unit flow F_y on the full lattice. In the positive orthant based at y, give each site y+n, n\in\Z_{\ge0}^d, |n|_1=N, mass

and send the fraction (n_j+1)/(N+d) along its edge in direction e_j. Put zero on all other edges. At a site y+n of level N\ge1, the incoming mass is

which equals its outgoing mass. The outgoing mass at y is one. Consequently

For each lattice edge, send its signed F_y-value along its lifted simple path. To justify this infinite sum and estimate it, fix a target slot edge f and write h=\dist(f,y). If the route of an original edge e uses f, put t=\dist(e,y). The displacement bound gives

Because \nu<1, this implies 1+t\asymp1+h, with uniform constants: above a fixed threshold the error is at most (1+t)/2, and below that threshold all quantities are bounded. It follows also that e lies within distance C(1+h)^\nu of f. There are at most

such lattice edges. Each routed path is simple, so any one of them uses f at most once. Summing (6.7) over this finite set gives a well-defined flow J_a^0 with

The count (6.8) already counts all contributions to f; no path-length factor is present.

Local finiteness of the path sum and finiteness of the vertex preimages justify the divergence calculation at any slot b:

Here one fixed orientation is used for each original lattice edge; the signed route is always from \psi(u) to \psi(v). Adding the unit path flow from a to \psi(y) therefore gives

The connector has fixed bounded support near y, so enlarging C preserves the envelope (6.4). Finally, 2k>d for d\ge3, and the number of slot edges at radius h is at most C(1+h)^{d-1}. Squaring the envelope and summing proves finite energy. Equivalently, this is a unit flow to infinity: its net outward flux across any finite vertex set containing a is one.□

Proof. Write H=1+|y-y'|_\infty. For bounded H the assertion follows by summing the squared envelopes and Cauchy–Schwarz. Otherwise, in the ball of radius (H-1)/3 about either center, the other envelope is at most CH^{-k}, whereas the sum of the nearer envelope is at most CH^{d-k}. These two regions cost CH^{d-2k}. In the remaining region the two distances are comparable. Dyadic annuli of radii comparable to 2^jH, j\ge0, cost at most

since 2k>d. Changing a site distance to the distance from an incident edge changes it by at most one, and the number of slot edges per site is bounded by 2d\ell^2. Thus the same estimates apply to the stated edge sum.□

Proof. The empty case uses J=0. Otherwise, for each slot a in the support of \mu_i, take a unit flow J_{i,a} from Proposition 6.2, using anchor x_i, and set

This is a finite linear combination, and its divergence is as asserted. Put q=2k-d=d-2-2d\nu>\alpha. By Lemma 6.3, the absolute overlap of J_{i,a} and J_{j,b} is at most C(1+|y_a-y_b|_\infty)^{-q}, where y_a,y_b are the sites of their injection slots. If |x_i-x_j|_\infty>4R, then |y_a-y_b|_\infty\ge |x_i-x_j|_\infty/2, uniformly over both supports. For the remaining anchor pairs use the uniform constant overlap bound. The mass bound of one per anchor therefore gives

The first term is at most CL R^\alpha. Splitting the second into annuli 2^j4R<|x_j-x_i|_\infty\le2^{j+1}4R bounds it by

because R\ge1 and q>\alpha. This proves (6.10), including repeated anchor sites since the growth assumption counts slots.

Finally, finite support of g permits summation by divergence without a term at infinity:

Cauchy–Schwarz and \sum_e|\nabla_e g|^2=2\mathcal D(g) give (6.11). In particular it applies to functions that vanish on the killed boundary and outside the physical box, with exactly the energy normalization in (5.12).□

Proof. The nonoverlapping initial and ending portions factor U=BMA, with A=C_{2s}, B^*=C_{2s}^{\rm rev}, and M the intervening propagation. At \beta=4s, assign every shared-cut jump once according to the preceding convention and take the intervening identity on that cut. All factors are contractions, so

This proves (7.6). Longer initial portions are obtained by left multiplication by further contraction factors. For the ending adjoints the same fact follows after reversing the factor order, using (7.5). It proves both monotonicity assertions.□

Proof. First sample a complete actual exchange picture with law \mathbb P_P and expose its missing cycles. Conditional on that exposure, realize the unconditioned one-particle propagation of Proposition 4.1 using independent fresh rate-1/2 Poisson clocks on all slot edges, accepting a fresh mark only when its two endpoints are holes. The forced moves use marks of the actual picture. Completion randomness changes slots only within a site. This construction is for the unconditioned fresh walk; its exposure still has the marginal induced by the full pin law.

The difference in (7.9), summed over output slots, is bounded by the sum of the probabilities that these individual walks exit their respective balls before elapsed time u. Exiting requires at least R nearest-neighbor spatial steps. The first R such steps give a chronological list of marks, each from either the actual picture or the fresh family. For a fixed list of sites, slot edges, and choices of source, the actual marks used are distinct: the time interval has length at most 2s<\beta, so its image on the time circle does not use a mark twice. This remains true if the interval crosses time zero, and if it is traversed backwards. Fresh marks in the list are distinct as well.

Suppose a source choice uses k actual marks and R-k fresh marks. By Lemma 2.4, the joint factorial measure of the actual marks is bounded by 2^k times their base product intensity. Independence of the fresh family then bounds the mixed measure by

on the prescribed edge coordinates. All source choices are integrated over the same ordered simplex 0<t_1<\cdots<t_R<u, of volume u^R/R!. At each spatial step there are at most 2d\ell^2 choices for the next site and the slot edge, even if one ignores constraints from the current slot. Summing source choices and edge choices therefore bounds the expected number of such lists, for one initial point, by

The existence of an exit path is bounded by this count; the counted lists need not themselves be realizable walk histories. In particular, the restrictions imposed by the exposure can be ignored when taking this upper bound. No factorial bound conditional on a fixed exposure has been used.

Since R!\ge(R/e)^R, choosing C_*>4eK bounds the last expression by 2^{-R}\le e^{-cs}. Sum over i\in T to obtain (7.10). Transposition merely reverses the chronological lists and the forced transports. The same mixed factorial estimate applies under the original law \mathbb P_P, which proves the reverse assertion too.□

Proof. Put q_i(u)=\|\eta_{i,u}\|_2^2. All localized factors are contractions, so 0\le q_i\le1 and q_i is nonincreasing, including at jumps. Between jumps the additional killing gives the Dirichlet generator with zero values outside B_R(x_i) as well as on the physical boundary. Consequently, at every regular smooth time, (7.13) and \|\eta_{i,u}\|_1\le1 imply

At other smooth times q_i'\le0. As long as q_i>0, the function q_i^{-2/d} thus increases at a fixed positive rate during regular time and cannot decrease at other times or jumps. If q_i reaches zero, it stays zero. It follows in either case that

The expected amount of nonregular time in [0,s] is at most s\delta by (7.12) and Fubini. Markov’s inequality bounds the probability of its exceeding s/2 by 2\delta. This proves (7.15). For reverse propagation, t(v) simply runs in the opposite direction; the same fixed-time estimate (7.12) gives the same integral bound, without changing the probability law.

At any slot above z, only anchors with x_i\in B_R(z) can contribute to \zeta_u. Their number is at most L(1+R)^\alpha\le C R^\alpha by (7.1). Hence pointwise Cauchy–Schwarz and summation give

Since 0\le\zeta_u\le\eta_u\le1,

Use Lemma 7.2 and (7.15) to obtain (7.16).□

Proof. For s\le u\le2s, let

By (7.9), 0\le\mu_u\le\eta_u. The nonnegative discarded mass is bounded by the localization error plus the masses attached to nonregular anchors. Since 0\le g_u\le1 and each local mass is at most one, Lemma 7.2 and (7.12) give

This uses only pointwise bounds, so dependence of g_u on the same environment causes no change.

At a fixed regular slice, each summand of \mu_u meets exactly the support and mass hypotheses of (7.14). Choose a reference orientation for the available edges, with \operatorname{div}J equal to outgoing minus incoming flow. Since g_u is finitely supported, summation by parts has finitely many nonzero terms and gives

Cauchy–Schwarz, (7.14), and \sum_e|\nabla_e g_u|^2=2\mathcal D_{t(u)}(g_u) imply the deterministic inequality

It suffices that a flow exists at each time: the resulting inequality contains only measurable propagated vectors and energies, so no measurable or continuous choice of flows is required.

To make the remainder in weighted Young’s inequality integrable at the final time, use an averaging density that vanishes linearly there; its tail weight then vanishes quadratically. Average (7.8) over u\in[s,2s] with probability density

Since \mathcal L is nondecreasing, \mathcal L(2s)\ge\int_s^{2s}w(u)\mathcal L(u)\,\mathrm du. Tonelli’s theorem shows that the positive energy in this average contains

the additional contribution from [0,s] is nonnegative. At each s\le u<2s, weighted Young’s inequality gives

The same-time energy has therefore absorbed the retained pairing. The remaining integral is finite, because

The endpoint u=2s has measure zero; the quotient has the constant value 4/s^2 on [s,2s).

Taking expectation, using (7.20), and then Lemma 7.3, we conclude that

Here R\ge1 allows the unweighted \delta error to be included in R^\alpha\delta. Every step applied to either ordered propagation, which proves both assertions.□

Proof. Fix all auxiliary constants in the stated order and choose p_* so that every 0<p\le p_* meets the final requirements. The set of interior slot points on a fixed finite grid is finite. Induct on the number of these points missing from E. When none is missing there is no disjoint nonempty test, so \mathsf I_p(E) holds vacuously.

Assume the assertion for pin sets with fewer missing grid points than E, and let T be a test for E. Since T is nonempty and disjoint from E, the set P=E\cup T has strictly fewer missing points. The induction hypothesis therefore supplies \mathsf I_p(P), exactly the hypothesis used in Proposition 5.4 and Proposition 7.4. The two leakage estimates and Lemma 7.1 give

By Lemma 4.5, or (7.2), it follows that r\le\varepsilon(p). Proposition 3.1 now yields

This proves the induction step for every allowed T, and finiteness of the grid completes the induction. A singleton interior slot test at time zero satisfies the growth bound, so taking E=D_0 gives (7.27). A boundary slot belongs to D_0, and its missing-cycle probability is zero. This covers all slots.□

Proof. Let n_0 be the number of cycles meeting D_0, and let J_N count inter-site marks incident to a boundary site during one period. Every cycle meeting the boundary either contains a time-zero boundary slot or uses an inter-site mark incident to the boundary. Indeed, in the absence of a time-zero boundary slot on that cycle, a boundary visit must begin by a spatial jump; within-site endpoint permutations do not change the site. Each mark has two strands and can witness at most two cycles. Consequently, picture by picture,

There are at most 2d\ell^2|\partial\Lambda_N| slot edges incident to the boundary. Their base rate is 1/2. The first factorial bound of Lemma 2.4, applied with no pins, therefore gives

Given a picture, G_N has probability 2^{-n_0}. Jensen’s inequality now proves (8.1), for example with C_{\beta,d,S}=(\log2)(\ell+4d\beta\ell^2).

Multiplying the zero-field picture weight 2^{\#\mathrm{cycles}} by 2^{-n_0} leaves precisely the weight for cycles missing D_0. This proves the assertion about the conditional picture; conditioning fixes only the colors of the other cycles. For a time-zero slot i, its conditional mean spin is thus

For an unpinned slot the probability on the right is at most p by the singleton case of Theorem 7.5; for a boundary slot it is zero. Summing over the \ell V_N slots gives

Here M_N denotes the random variable in the coloring representation. If q=\Pr(M_N/V_N\geq S/2\mid G_N), its upper bound M_N/V_N\leq S implies

Multiply by \Pr(G_N) and use the exact spectral interpretation of the unconditioned coloring law to obtain (8.2).□

Proof. Each interaction term commutes with the sum of the two endpoint z-spins. This follows directly from

Thus [H_N,M_N]=0. For any h>0, the spectral law in the Gibbs state of H_N-hM_N is exactly

Bound the numerator on m<SV_N/4, and retain only m\geq SV_N/2 in the denominator. Lemma 8.1 then gives

For these cubes,

At h=h_N, the exponent in (8.5) tends to -\infty: its negative term has order N^{d-1/2}, whereas its positive term has order N^{d-1}. Hence, writing q_N=\mu_{N,h_N}(({-\infty},SV_N/4)), we have q_N\to0. The spectral bounds -S\leq M_N/V_N\leq S imply

In particular this estimate uses a single explicit volume-field sequence for the fixed \beta.

Choose a countable norm-dense set in the union of local matrix algebras, closed under rational complex linear combinations and adjoints. The bound |\omega_N(A)|\leq\|A\| permits diagonal subsequence selection on that set. Norm approximation then gives pointwise convergence on \mathcal A. The resulting functional \omega is linear, normalized and positive, since these properties hold for each \omega_N and pass to pointwise limits. It is therefore a state.

For every fixed y\in\Z^d and A\in\mathcal A,

This proves translation invariance of \omega. Finally,

Taking the limit in (8.6) proves (8.4).□

Proof. First fix t. If (8.7) failed at that time, there would be \varepsilon>0 and cubes \Gamma_k whose distances from L to their complements tend to infinity, with \|\tau_t^{\Gamma_k}(B)-\tau_t(B)\|\geq\varepsilon. Such cubes contain centered boxes with radii tending to infinity. We can therefore select a subsequence that is increasing by inclusion and exhausts \Z^d: after any finite selected cube, a sufficiently late cube contains it and the next prescribed centered box. This contradicts the norm convergence along increasing boxes in the definition of \tau_t.

To make this convergence uniform in time, write \Phi(\{u,v\})=-\mathbf S_u\cdot\mathbf S_v. Only edges meeting L contribute to [H_\Gamma,B], so for every \Gamma

Consequently \|\tau_t^\Gamma(B)-\tau_u^\Gamma(B)\|\leq c_B|t-u|. Passing to the infinite-volume limit gives the same bound for \tau_t(B). A finite mesh of [-T,T], together with the fixed-time conclusion, proves (8.7).

For x\in I_N, set \Gamma=\Lambda_N-x. Then \dist_\infty(L,\Gamma^c)\to\infty uniformly over these x, since L is fixed. For all sufficiently large N, L\subset\Gamma. Define the local rotation

The commutation [H_N,M_N]=0 and translation of the finite Hamiltonian give the exact formula

The norm effect of this rotation is bounded by

Use the isometry of \tau_t^\Gamma, followed by (8.7), to prove (8.8). Finally, the excluded sites lie in a layer of thickness at most \lfloor\sqrt N\rfloor, whose fraction of the cube is O(N^{-1/2}).□

Proof. Fix n, and let M_n<\infty bound |F_n| on the strip. For \varepsilon>0, apply the maximum principle to e^{-\varepsilon z^2}F_n(z) on [-R,R]+i[0,\beta]. Its modulus is at most Ce^{\varepsilon\beta^2} on the horizontal sides and at most M_ne^{-\varepsilon R^2+\varepsilon\beta^2} on the vertical sides. Send R\to\infty while holding a point z fixed, then send \varepsilon\downarrow0. This proves |F_n(z)|\leq C. Only the already assumed finite bound M_n was needed to remove the vertical sides.

Now set G_n(z)=e^{-z^2}F_n(z). The common bound just proved implies

Local uniform convergence of the two boundary sequences and the Gaussian tail therefore make (G_n) uniformly Cauchy on both horizontal edges. Write

The maximum principle on the same rectangles, now applied to G_n-G_m, has horizontal bound \delta_{n,m} and vertical bound 2Ce^{-R^2+\beta^2}. Letting R\to\infty shows

Hence G_n converges uniformly on the closed strip to a continuous function G, analytic in the interior. Multiplication by e^{z^2} gives local uniform convergence of F_n on the closed strip to F=e^{z^2}G. Its claimed boundary values and the bound |F|\leq C follow by taking pointwise limits.□

Completion of the proof of Theorem 1.1. Let \omega and N_j be as in Proposition 8.2. For local observables A,B, define the entire functions

Trace cyclicity gives both identities

These formulas also hold when the translated supports cross the boundary of \Lambda_N. To see this without a support restriction, take a finite set containing \Lambda_N and both translated supports. On its matrix algebra the density of \rho_N is e^{-\beta K_N}/\Tr(e^{-\beta K_N}) tensored with a normalized identity. Equivalently, it is the Gibbs density for the operator K_N\otimes I on that larger algebra. The dynamics is generated by that same operator, so finite-dimensional trace cyclicity proves (8.12) there as well.

For real t, the isometry of \alpha_t^N bounds both right-hand sides by \|A\|\|B\|. Before using any common strip bound, we also have the finite bound

Indeed, the real-time unitary factors have norm one, and \|e^{\pm yK_N}\|\leq e^{\beta\|K_N\|}. The first part of Lemma 8.4 therefore applies and gives the uniform bound

We next identify the two boundary limits. Given 0<T<\infty, Lemma 8.3 supplies numbers \varepsilon_N(T)\to0 bounding the norm error in (8.8) on I_N. On its complement the error is at most 2\|B\|, since both real-time dynamics are isometric. It follows that, uniformly for |t|\leq T,

For example, the comparison term in the first line is exactly V_N^{-1}\sum_x\rho_N(T_x(A)T_x(\tau_t(B))), which equals \omega_N(A\tau_t(B)); the second line retains the opposite operator order. Thus both error bounds tend to zero.

The maps t\mapsto A\tau_t(B) and t\mapsto\tau_t(B)A are norm-continuous. Their images of a compact time interval are norm-compact. Pointwise convergence \omega_{N_j}(C)\to\omega(C) is uniform over each such image: use a finite norm net and the common functional norm one. Therefore, along the already chosen subsequence,

locally uniformly for real t. Lemma 8.4 now gives a bounded continuous function F_{A,B} on the closed strip, analytic in its interior, with precisely these two boundary values and

Since the strip limit uses the full state subsequence, the bilinearity of (8.10) passes to F_{A,B}.

It remains to obtain the asserted condition on the entire algebra. Let A_n,B_n be local observables converging in norm to arbitrary A,B\in\mathcal A. Bilinearity and (8.16) imply

Hence these functions converge uniformly on the closed strip to a bounded continuous F_{A,B}, analytic in the interior. The same inequality proves independence of the approximating sequences and the bound \sup|F_{A,B}|\leq\|A\|\|B\|. Since states have norm one and \tau_t is isometric, norm approximation passes both boundary identities to the limit, uniformly in real t:

This is the full \beta-KMS condition (1.2) for the zero-field dynamics \tau defined in Section 1.1. The state is translation invariant and satisfies \omega(S_0^z)\geq S/4>0 by Proposition 8.2. Since \beta was any fixed finite number at least \beta_0(d,S), Theorem 1.1 follows.□