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 \[\mathbb E_{\mathrm b} \left|\frac{w_{D_n}}{Z_{D_n}}-\frac{w_D}{Z_D}\right| \longrightarrow0,\] which proves total-variation convergence. Applying dominated convergence to \(w_{D_n}\Phi_n\) proves (10). 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, \[ w_D\!\left(\omega+\sum_{j=1}^n\delta_{u_j}\right) \le 2^n w_D(\omega). \tag{12}\] 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 (12) is the iterated Mecke formula (Mecke 1967); its factorial form is stated in (Last and Penrose 2017, 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 \[e^{-\lambda}\sum_{k=0}^{\infty}\frac1{k!} \prod_{j=1}^k\mu(\,\mathrm du_j).\] 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 \[\begin{align*} &\mathbb E_{\mathrm b}\left[ w_D(\omega) \sum_{(u_1,\ldots,u_n)\in(\omega_{\mathrm m})^n_{\ne}} F(u_1,\ldots,u_n)\right]\\ &\hspace{8mm}= \int_{X_\Lambda^n}F(u_1,\ldots,u_n) \mathbb E_{\mathrm b}\!\left[ w_D\!\left(\omega+\sum_{j=1}^n\delta_{u_j}\right)\right] \prod_{j=1}^n\mu(\,\mathrm du_j). \end{align*}\] All terms are nonnegative, so this manipulation is justified also when an integral is infinite. By (12), the expectation inside the integral is at most \(2^n Z_D\). Division by \(Z_D\) proves (11). ◻

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 \(a\in\mathbb R\) 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 \[\lim_{R\to\infty}(\mathrm iR)^{-k} F(z_1,\ldots,z_{n-1},\mathrm iR).\] 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 \(t\in\mathbb H\), \[\frac{\partial_t F}{F} =\sum_{j=1}^k\frac1{t-\zeta_j}\] 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 \[q(z,w)=a+bz+cw+ezw,\] where \(a,b,c,e\) can be complex. For every \(z\in\mathbb H\), \[ N(z):=\operatorname{Im}\bigl((a+bz)\overline{(c+ez)}\bigr)\ge0. \tag{17}\] 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 \[N^\tau(z):=\operatorname{Im}\bigl((a+cz)\overline{(b+ez)}\bigr)\ge0.\] Put \(b_u=(1-u)b+uc\) and \(c_u=(1-u)c+ub\). Direct expansion gives \[\begin{align*} \operatorname{Im}\bigl((a+b_uz)\overline{(c_u+ez)}\bigr) &=(1-u)N(z)+uN^\tau(z) +u(1-u)|b-c|^2\operatorname{Im}z. \tag{18}\end{align*}\] 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 8. 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 7 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 \[(x_i+t)(t+y_j).\] 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 (20); Lemma 7 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 (20), 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 (21). 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 \(X_j=\mathbf 1_{\{j\in A\}}\). Conditioning on \(B\subseteq A\) gives, for the indicators outside \(B\), the generating polynomial \[\frac{\left.\left(\prod_{j\in B}\partial_{y_j}\right) G(\boldsymbol y)\right|_{y_j=1\ (j\in B)}} {\Pr(B\subseteq A)}.\] It is nonzero and stable by Lemma 7. 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 (17) with real coefficients gives \[0\le\operatorname{Im}\bigl((a+bz)\overline{(c+ez)}\bigr) =(bc-ae)\operatorname{Im}z.\] Thus \(bc\ge ae\). The covariance of the corresponding presence indicators equals \[e-(b+e)(c+e)=ae-bc\le0.\] 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, (22) 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 \[\Pr\bigl(X_{j_h}=1\mid X_{j_1}=\cdots=X_{j_{h-1}}=1\bigr) \le \Pr(X_{j_h}=1).\] Multiplying the conditional probabilities proves (22). Applying the same argument to the conditional generating polynomial proves the final assertion. ◻

Proof of Proposition 6. Suppose first that \(E\) is finite, and use the layered construction above with \(P=E\cup T\). By Lemmas 10 and 11, followed by the arithmetic–geometric mean inequality and (16), \[ \mathbb P_P\bigl(T\subseteq\sigma(T)\bigr) \le\prod_{j\in T}\mathbb P_P(j\in\sigma(T))\le r^m. \tag{23}\] If one of the factors is zero this conclusion is interpreted directly; no conditioning on a zero-probability event is needed.

By (15), 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, \[w_P\le w_E\le 2^m w_P, \qquad Z_E\ge Z_P>0.\] We obtain \[\mathbb P_E(H) =\frac{\mathbb E_{\mathrm b}(\mathbf 1_Hw_E)}{Z_E} \le 2^m\frac{Z_P}{Z_E}\mathbb P_P(H) \le 2^m r^m.\] 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 4, 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 \[w_{E_n}\longrightarrow w_E,\qquad w_{P_n}\longrightarrow w_P,\qquad \mathbf 1_{\{T\subseteq\mathcal B_{E_n}\}} \longrightarrow\mathbf 1_{\{T\subseteq\mathcal B_E\}}.\] 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 \(i\in T\), \[\mathbf 1_{\mathcal R_i(E_n,T)}\longrightarrow \mathbf 1_{\mathcal R_i(E,T)} \quad\text{almost surely}.\] 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 (14) 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 \[ 2^{n_P} =\sum_{A\subseteq V_\Lambda} \mathbf 1_{F_A}\,2^{c_A}\,\mathbf 1_{G_P}, \tag{28}\] 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 (26). Integrating the remaining picture gives (27). 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 \[T_{ij}-I=-(e_i-e_j)(e_i-e_j)^*.\] 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 \[\bigwedge\nolimits^k((1-u)I+uT_{ij}) =(1-u)\bigwedge\nolimits^kI+u\bigwedge\nolimits^kT_{ij}.\] This is (29) for one Bernoulli transposition. For independent factors \(A,B\) satisfying the identity, \[\mathbb E\bigwedge\nolimits^k(AB) =\mathbb E\bigwedge\nolimits^kA\, \mathbb E\bigwedge\nolimits^kB =\bigwedge\nolimits^k(\mathbb EA\,\mathbb EB) =\bigwedge\nolimits^k\mathbb E(AB).\] 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 9. 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 \[\begin{align*} \mathbb E\det(I-D\Pi) &=\sum_{k=0}^{|V|}(-1)^k \operatorname{Tr}\left(\bigwedge\nolimits^kD\, \mathbb E\bigwedge\nolimits^k\Pi\right)\\ &=\det(I-DK). \end{align*}\] 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 \[\begin{align*} \det(I-D\Pi) &=\det(I-\Pi_{QQ})\, \det\left(I-W\left[ \Pi_{JJ}+\Pi_{JQ}(I-\Pi_{QQ})^{-1}\Pi_{QJ}\right]\right)\\ &=\det(I-W\Sigma). \end{align*}\] 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 \[ \mathbb E[\mathbf 1_{G_J}\det(I-W\Sigma)] =\det\begin{pmatrix} I-WK_{JJ}&-WK_{JQ}\\ -K_{QJ}&I-K_{QQ} \end{pmatrix}. \tag{33}\] At \(W=0\), this proves (30) without any invertibility assumption. If its value is positive, \(I-K_{QQ}\) is invertible. The Schur complement on the right of (33), followed by division by that positive determinant, gives (32).

To extract the first moment, use \(\det(I-tWA)=1-t\operatorname{Tr}(WA)+O(t^2)\). The linear terms in (32) yield \[\operatorname{Tr}\bigl(W\mathbb E[\Sigma\mid G_J]\bigr) =\operatorname{Tr}(WH) \qquad\text{for every }W.\] Taking the matrix units for \(W\) gives each entry of (31).

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 \[\|K_{QQ}^{kn}\|_{1\to1}\le(1-\varepsilon)^k, \qquad (I-K_{QQ})^{-1}=\sum_{j=0}^{\infty}K_{QQ}^j.\] 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 12 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 \[E_n=E_{\mathrm f}\cup D_{0,n},\qquad P_n=E_n\cup T.\] 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 \[ \mathbf 1_{G_n}\uparrow\mathbf 1_{G_P}\quad \mathbb Q_{\mathbf a}\text{-almost surely},\qquad g_n\uparrow g_P(\mathbf a)>0. \tag{35}\] In particular, \(g_n>0\) for all sufficiently large \(n\). No invertibility is required for the earlier indices.

For such an \(n\), Lemma 13 applies to the unconditioned layered hole permutation. With \(J=P_n\), Proposition 14 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 \(T\), \[ \frac{\mathbb Q_{\mathbf a}(G_n\cap A_n)}{g_n} =\mathbb P_{\mathbf a}^{\mathrm{walk}} (\tau_T^+<\tau_{E_n}). \tag{36}\]

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 \(\mathbf 1_{A_n}\) decreases to the event \(A\) that the next full \(P\)-pin is in \(T\). Together with (35) and bounded convergence, this gives \[\mathbb Q_{\mathbf a}(G_n\cap A_n) \longrightarrow\mathbb Q_{\mathbf a}(G_P\cap A).\] 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 (36) 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 12 identifies the limiting conditional picture law with the conditional law under \(\mathbb P_P\). This proves (34). ◻

Proof. Set \(f_0=\mathbf 1_T\) and recursively define \[ f_{n+1}=\mathbf 1_T+\mathbf 1_{H_\theta\setminus T}\,U^*f_n, \tag{40}\] 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 (40) gives (38). A boundary starting point is killed immediately, and consequently every \(f_n\) and \(f\) is zero there.

If \(h\) is any nonnegative solution of (38), then \(h\ge f_0\) and induction in (40) 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 15, the conditional probability \(q_i(\mathbf a)\) contributing to (24) is the return probability with all of \(E\) retained. Hence \[q_i(\mathbf a)\le (U^*f)_i\qquad(i\in T).\] The vectors are real, so transposition in the scalar quadratic form and harmonicity off \(T\) give \[\begin{align*} \langle f,(I-U)f\rangle &=\langle (I-U^*)f,f\rangle\\ &=\sum_{i\in T}\bigl(1-(U^*f)_i\bigr) \le\sum_{i\in T}(1-q_i(\mathbf a)). \end{align*}\] Averaging under the actual exposure marginal gives (39). 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 (40). ◻

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 5, 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 (45) 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 (45) 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 \[\mathbb P_P(N_z\ge1\text{ for all }z\in A') \le \mathbb E_P\prod_{z\in A'}N_z \le (2d\ell^2|I|)^k.\] 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, (Sznitman 2012, sec. 3). Here the leaf probabilities will come directly from (47). 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 \[ x_{w0}=x_w,\qquad x_{w1}\in B_{h_k+8}(x_w),\qquad \operatorname{dist}_\infty(\mathcal L_{w0},\mathcal L_{w1}) \ge h_k/4, \tag{49}\] 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 \[h_j/2-2(h_{j-1}+8)\ge h_j/4.\] Recurse on the two truncated child crossings. Every descendant leaf remains in its child’s truncated crossing, so all the separation conditions in (49) 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 \[L(1+h)^\alpha\ge L(2b^{k-1})^\alpha \ge 2^{k-1}.\] 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 (42) 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 \[A_0=1,\qquad A_j\le(C_dh_j)^d A_{j-1}^2.\] In particular \[2^{-j}\log A_j \le d\sum_{k=1}^j2^{-k}\log(C_dh_k) \le \log C_b<\infty, \qquad A_j\le C_b^{2^j}.\] 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 17 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 \(p\le p_0\), \[ \mathbb P_P\bigl(z\in\mathcal O_t,\ \operatorname{rad}_z\mathcal C_t(z)\ge h_j\bigr) \le \exp(-2\cdot2^j). \tag{50}\] 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 (48); 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 (46) therefore bound failure of the first condition by \(C R^d p^{2d}\). By Lemma 18, failure of the second has probability at most \[C\sum_{z:|z-x|_\infty>10R} \exp\bigl(-c(1+|z-x|_\infty)^{\nu\gamma}\bigr) \le C\exp(-c'R^{\nu\gamma}).\] 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 \(\mathbb Z^d\setminus F\) 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 \(\mathbb Z^d\setminus F\) 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 \(\mathbb Z^d\). 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 \[w(z)= \begin{cases} V(z),&z\notin\mathcal O_t,\\ V(\phi(z)),&z\in\mathcal O_t\cap B_{R+3}(x),\\ 0,&z\in\mathcal O_t\setminus B_{R+3}(x). \end{cases}\] The function \(w\) is supported in \(B_{R+1}(x)\). Since a site’s value is copied only by its nearest neighbors, \[ \|v\|_2^2\le\ell\|w\|_2^2, \qquad \|w\|_1\le(2d+1)\|v\|_1. \tag{55}\]

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 21 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, \[ \sum_{\{z,z'\}:|z-z'|_1=1}(w(z)-w(z'))^2 \le C\mathcal D_t(v). \tag{56}\]

For completeness, write \(\mathcal E_{\mathbb Z^d}(w)\) for the sum on the left of (56). 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 \[\|w-A_kw\|_2^2\le C_d k^2\mathcal E_{\mathbb Z^d}(w).\] 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 \[\|w\|_2^2\le C_d k^2\mathcal E_{\mathbb Z^d}(w) +2k^{-d}\|w\|_1^2.\] 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 \[\|w\|_2^{2+4/d} \le C_d\mathcal E_{\mathbb Z^d}(w)\|w\|_1^{4/d}.\] For \(w=0\) the inequality is immediate. Finally use (55) and (56) to obtain (54). ◻

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 \(\mathop{\mathrm{dist}}(v,C')\ge4\) and \(\mathop{\mathrm{dist}}(u,C')\ge5\). 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 \[D_{\mathrm{loc}}=8(2d-2)+8.\] 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 \[ \operatorname{diam}H_C\le 2(1+b_C)^\nu+8. \tag{61}\] 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 (58), using fixed choices of representatives and paths. The preceding size bounds show that the route of a lattice edge \(e\) lies within distance \[ C(1+\mathop{\mathrm{dist}}(e,y))^\nu \tag{62}\] 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 \(1+\mathop{\mathrm{dist}}(e,y)\); now use (61). 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 21 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 \(y+n\), \(n\in\mathbb Z_{\ge0}^d\), \(|n|_1=N\), mass \[a_N=\binom{N+d-1}{d-1}^{-1},\] 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 \[a_{N-1}\frac{\sum_j n_j}{N+d-1} =a_{N-1}\frac{N}{N+d-1}=a_N,\] which equals its outgoing mass. The outgoing mass at \(y\) is one. Consequently \[ \operatorname{div}F_y=\mathbf 1_{\{y\}},\qquad |F_y(e)|\le C(1+\mathop{\mathrm{dist}}(e,y))^{1-d}. \tag{63}\]

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 \(h=\mathop{\mathrm{dist}}(f,y)\). If the route of an original edge \(e\) uses \(f\), put \(t=\mathop{\mathrm{dist}}(e,y)\). The displacement bound gives \[|h-t|\le C(1+t)^\nu+C.\] 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 \[ C(1+h)^{d\nu} \tag{64}\] such lattice edges. Each routed path is simple, so any one of them uses \(f\) at most once. Summing (63) over this finite set gives a well-defined flow \(J_a^0\) with \[|J_a^0(f)|\le C(1+h)^{1-d+d\nu}.\] The count (64) 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 \(b\): \[\begin{align*} \operatorname{div}J_a^0(b) &=\sum_{e=(u,v)}F_y(e) \bigl(\mathbf 1_{\{\psi(u)=b\}}-\mathbf 1_{\{\psi(v)=b\}}\bigr)\\ &=\sum_{z:\psi(z)=b}\operatorname{div}F_y(z) =\mathbf 1_{\{\psi(y)=b\}}. \end{align*}\] 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 \[J_a=J_a^0+\mathbf 1_{Q_a},\qquad \operatorname{div}J_a=\mathbf 1_{\{a\}}.\] The connector has fixed bounded support near \(y\), so enlarging \(C\) preserves the envelope (60). 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 \[C\sum_{j\ge0}(2^jH)^{d-2k}\le CH^{d-2k},\] 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 24, using anchor \(x_i\), and set \[J=\sum_{i\in I}\sum_a\mu_i(a)J_{i,a}.\] This is a finite linear combination, and its divergence is as asserted. Put \(q=2k-d=d-2-2d\nu>\alpha\). By Lemma 25, 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 \[\sum_eJ(e)^2\le C\sum_{i\in I} \left[\#\{j\in I:|x_j-x_i|_\infty\le4R\} +\sum_{\substack{j\in I\\|x_j-x_i|_\infty>4R}} (1+|x_j-x_i|_\infty)^{-q}\right].\] 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 \[CL\sum_{j\ge0}(2^jR)^{\alpha-q}\le C,\] because \(R\ge1\) and \(q>\alpha\). This proves (66), including repeated anchor sites since the growth assumption counts slots.

Finally, finite support of \(g\) permits summation by divergence without a term at infinity: \[\left\langle\sum_i\mu_i,g\right\rangle =\sum_e J(e)(g(e^-)-g(e^+)).\] Cauchy–Schwarz and \(\sum_e|\nabla_e g|^2=2\mathcal D(g)\) give (67). In particular it applies to functions that vanish on the killed boundary and outside the physical box, with exactly the energy normalization in (52). ◻

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 \[\langle v,Uv\rangle =\langle B^*v,MAv\rangle \le\|B^*v\|_2\|Av\|_2 \le\tfrac12\bigl(\|B^*v\|_2^2+\|Av\|_2^2\bigr).\] This proves (73). 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 (72). 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 12 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 (76), 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 5, 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 \[1^k(1/2)^{R-k}\,\mathrm dt_1\cdots\mathrm dt_R\] 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 \[ (2d\ell^2)^R(1+1/2)^R\frac{u^R}{R!} \le\frac{(2Ks)^R}{R!},\qquad K=3d\ell^2. \tag{78}\] 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 (77). 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, (80) and \(\|\eta_{i,u}\|_1\le1\) imply \[ q_i'(u)=-2\mathcal D_{t(u)}(\eta_{i,u}) \le-c q_i(u)^{1+2/d}. \tag{84}\] 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 \[\int_0^s\mathbf 1_{\{t(v)\text{ regular for }i\}}\,\mathrm dv\ge s/2 \quad\Longrightarrow\quad q_i(u)\le C s^{-d/2}\quad(s\le u\le2s).\] The expected amount of nonregular time in \([0,s]\) is at most \(s\delta\) by (79) and Fubini. Markov’s inequality bounds the probability of its exceeding \(s/2\) by \(2\delta\). This proves (82). For reverse propagation, \(t(v)\) simply runs in the opposite direction; the same fixed-time estimate (79) 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 (68). Hence pointwise Cauchy–Schwarz and summation give \[\|\zeta_u\|_2^2\le C R^\alpha\sum_{i\in T}q_i(u).\] Since \(0\le\zeta_u\le\eta_u\le1\), \[\|\eta_u\|_2^2-\|\zeta_u\|_2^2 =\sum_k(\eta_u(k)-\zeta_u(k))(\eta_u(k)+\zeta_u(k)) \le2\|\eta_u-\zeta_u\|_1.\] Use Lemma 28 and (82) to obtain (83). ◻

Proof. For \(s\le u\le2s\), let \[\mu_u=\sum_{i:\,t(u)\text{ regular for }i}\eta_{i,u}.\] By (76), \(0\le\mu_u\le\eta_u\). 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 28 and (79) give \[ \mathbb E_P\langle\eta_u-\mu_u,g_u\rangle \le C m(e^{-cs}+\delta). \tag{87}\] 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 (81). 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 \[\langle\mu_u,g_u\rangle =\sum_{e=(j,k)}J(e)(g_u(j)-g_u(k)).\] Cauchy–Schwarz, (81), and \(\sum_e|\nabla_e g_u|^2=2\mathcal D_{t(u)}(g_u)\) imply the deterministic inequality \[ |\langle\mu_u,g_u\rangle| \le\bigl(A\mathcal D_{t(u)}(g_u)\bigr)^{1/2}, \qquad A=C m R^\alpha. \tag{88}\] 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 (75) over \(u\in[s,2s]\) with probability density \[w(u)=\frac{2(2s-u)}{s^2},\qquad W(u)=\int_u^{2s}w(v)\,\mathrm dv=\frac{(2s-u)^2}{s^2}.\] 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 \[2\int_s^{2s}W(u)\mathcal D_{t(u)}(g_u)\,\mathrm du;\] the additional contribution from \([0,s]\) is nonnegative. At each \(s\le u<2s\), weighted Young’s inequality gives \[ 2w(u)\sqrt{A\mathcal D_{t(u)}(g_u)} \le2W(u)\mathcal D_{t(u)}(g_u)+\frac{A w(u)^2}{2W(u)}. \tag{89}\] The same-time energy has therefore absorbed the retained pairing. The remaining integral is finite, because \[ \int_s^{2s}\frac{w(u)^2}{W(u)}\,\mathrm du=\frac4s. \tag{90}\] The endpoint \(u=2s\) has measure zero; the quotient has the constant value \(4/s^2\) on \([s,2s)\).

Taking expectation, using (87), and then Lemma 29, we conclude that \[\begin{align*} \mathbb E_P\mathcal L(2s) &\ge m -C m\bigl[e^{-cs}+R^\alpha(s^{-d/2}+\delta)\bigr] -C m(e^{-cs}+\delta)-C mR^\alpha/s\\ &\ge m\left[1-C\left(e^{-cs} +R^\alpha(s^{-d/2}+\delta+s^{-1})\right)\right]. \end{align*}\] 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 20 and Proposition 30. The two leakage estimates and Lemma 27 give \[\mathbb E_P\langle f,(I-U)f\rangle \ge m(1-\varepsilon(p)).\] By Lemma 16, or (69), it follows that \(r\le\varepsilon(p)\). Proposition 6 now yields \[\mathbb P_E(T\subseteq\mathcal B_E) \le(2r)^m\le(2\varepsilon(p))^m\le p^m.\] 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 (94). 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, \[n_0\leq \ell|\partial\Lambda_N|+2J_N.\] 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 5, applied with no pins, therefore gives \[\mathbb E_{\varnothing}J_N \leq 2d\beta\ell^2|\partial\Lambda_N|, \qquad \mathbb E_{\varnothing}n_0 \leq(\ell+4d\beta\ell^2)|\partial\Lambda_N|.\] Given a picture, \(G_N\) has probability \(2^{-n_0}\). Jensen’s inequality now proves (95), 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 \[\frac12\bigl(1-\mathbb P_{D_0}(i\in\mathcal B_{D_0})\bigr).\] For an unpinned slot the probability on the right is at most \(p\) by the singleton case of Theorem 31; for a boundary slot it is zero. Summing over the \(\ell V_N\) slots gives \[\mathbb E\left[\frac{M_N}{V_N}\,\middle|\,G_N\right] \geq S(1-p).\] 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 \[S(1-p)\leq Sq+\frac S2(1-q),\qquad q\geq1-2p>\frac12.\] Multiply by \(\Pr(G_N)\) and use the exact spectral interpretation of the unconditioned coloring law to obtain (96). ◻

Proof. Each interaction term commutes with the sum of the two endpoint \(z\)-spins. This follows directly from \[\mathbf S_x\cdot\mathbf S_y =S_x^zS_y^z+\tfrac12(S_x^+S_y^-+S_x^-S_y^+).\] Thus \([H_N,M_N]=0\). For any \(h>0\), the spectral law in the Gibbs state of \(H_N-hM_N\) is exactly \[\mu_{N,h}(\,\mathrm dm) =\frac{e^{\beta hm}\mu_N(\,\mathrm dm)} {\int e^{\beta hu}\mu_N(\,\mathrm du)}.\] Bound the numerator on \(m<SV_N/4\), and retain only \(m\geq SV_N/2\) in the denominator. Lemma 32 then gives \[ \mu_{N,h}(({-\infty},SV_N/4)) \leq 2\exp\left\{ C_{\beta,d,S}|\partial\Lambda_N|-\frac{\beta hS}{4}V_N \right\}. \tag{99}\] For these cubes, \[\frac{|\partial\Lambda_N|}{V_N} =1-\left(\frac{2N-1}{2N+1}\right)^d \leq\frac{2d}{2N+1}.\] At \(h=h_N\), the exponent in (99) 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 \[ \rho_N^{\mathrm{in}}(M_N/V_N) \geq\frac S4(1-q_N)-Sq_N =\frac S4-\frac{5S}{4}q_N. \tag{100}\] 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 \(y\in\mathbb Z^d\) and \(A\in\mathcal A\), \[|\omega_N(T_y(A))-\omega_N(A)| \leq\frac{|(\Lambda_N+y)\mathbin\triangle\Lambda_N|}{V_N}\|A\| \longrightarrow0.\] This proves translation invariance of \(\omega\). Finally, \[\omega_N(S_0^z) =\frac1{V_N}\sum_{x\in\Lambda_N}\rho_N(S_x^z) =\rho_N^{\mathrm{in}}(M_N/V_N).\] Taking the limit in (100) proves (98). ◻

Proof. First fix \(t\). If (101) 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 \(\mathbb Z^d\): 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 \(\Gamma\) \[\|[H_\Gamma,B]\|\leq 2\|B\|\sum_{\substack{e\text{ nearest-neighbor edge}\\e\cap L\ne\varnothing}} \|\Phi(e)\|=:c_B<\infty.\] 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 (101).

For \(x\in I_N\), set \(\Gamma=\Lambda_N-x\). Then \(\mathop{\mathrm{dist}}_\infty(L,\Gamma^c)\to\infty\) uniformly over these \(x\), since \(L\) is fixed. For all sufficiently large \(N\), \(L\subset\Gamma\). Define the local rotation \[\gamma_\theta^L(B) =e^{i\theta\sum_{u\in L}S_u^z}B e^{-i\theta\sum_{u\in L}S_u^z}.\] The commutation \([H_N,M_N]=0\) and translation of the finite Hamiltonian give the exact formula \[ T_{-x}\bigl(\alpha_t^N(T_x(B))\bigr) =\tau_t^\Gamma\bigl(\gamma_{-h_Nt}^L(B)\bigr). \tag{103}\] The norm effect of this rotation is bounded by \[\|\gamma_{-h_Nt}^L(B)-B\| \leq h_N|t|\left\|\left[\sum_{u\in L}S_u^z,B\right]\right\|.\] Use the isometry of \(\tau_t^\Gamma\), followed by (101), to prove (102). 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 \[|G_n(t+iy)|\leq Ce^{-t^2+y^2} \quad(0\leq y\leq\beta).\] Local uniform convergence of the two boundary sequences and the Gaussian tail therefore make \((G_n)\) uniformly Cauchy on both horizontal edges. Write \[\delta_{n,m}= \max_{y\in\{0,\beta\}}\sup_{t\in\mathbb R} |G_n(t+iy)-G_m(t+iy)|; \qquad \delta_{n,m}\longrightarrow0.\] 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 \[\sup_{z\in\mathcal S_\beta}|G_n(z)-G_m(z)|\leq\delta_{n,m}.\] 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. Let \(\omega\) and \(N_j\) be as in Proposition 33. For local observables \(A,B\), define the entire functions \[ F_N^{A,B}(z)=\frac1{V_N}\sum_{x\in\Lambda_N} \rho_N\bigl(T_x(A)\alpha_z^N(T_x(B))\bigr). \tag{104}\] Trace cyclicity gives both identities \[\begin{align*} F_N^{A,B}(t) &=\frac1{V_N}\sum_{x\in\Lambda_N} \rho_N\bigl(T_x(A)\alpha_t^N(T_x(B))\bigr), \tag{105}\\ F_N^{A,B}(t+i\beta) &=\frac1{V_N}\sum_{x\in\Lambda_N} \rho_N\bigl(\alpha_t^N(T_x(B))T_x(A)\bigr). \tag{106}\end{align*}\] 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 \(e^{-\beta K_N}/\mathop{\mathrm{Tr}}(e^{-\beta K_N})\) 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 (106) 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 \[|F_N^{A,B}(t+iy)| \leq \|A\|\|B\|e^{2\beta\|K_N\|} \quad(t\in\mathbb R,\ 0\leq y\leq\beta).\] Indeed, the real-time unitary factors have norm one, and \(\|e^{\pm yK_N}\|\leq e^{\beta\|K_N\|}\). The first part of Lemma 35 therefore applies and gives the uniform bound \[ |F_N^{A,B}(z)|\leq\|A\|\|B\| \quad(z\in\mathcal S_\beta,\ N\geq1). \tag{107}\]

We next identify the two boundary limits. Given \(0<T<\infty\), Lemma 34 supplies numbers \(\varepsilon_N(T)\to0\) bounding the norm error in (102) 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 \(|t|\leq T\), \[\begin{align*} \left|F_N^{A,B}(t)-\omega_N(A\tau_t(B))\right| &\leq\|A\|\varepsilon_N(T) +2\|A\|\|B\|\frac{|\Lambda_N\setminus I_N|}{V_N}, \tag{108}\\ \left|F_N^{A,B}(t+i\beta)-\omega_N(\tau_t(B)A)\right| &\leq\|A\|\varepsilon_N(T) +2\|A\|\|B\|\frac{|\Lambda_N\setminus I_N|}{V_N}. \tag{109}\end{align*}\] 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, \[\begin{align*} F_{N_j}^{A,B}(t)&\longrightarrow\omega(A\tau_t(B)),\\ F_{N_j}^{A,B}(t+i\beta)&\longrightarrow\omega(\tau_t(B)A) \end{align*}\] locally uniformly for real \(t\). Lemma 35 now gives a bounded continuous function \(F_{A,B}\) on the closed strip, analytic in its interior, with precisely these two boundary values and \[ \sup_{z\in\mathcal S_\beta}|F_{A,B}(z)|\leq\|A\|\|B\|. \tag{110}\] Since the strip limit uses the full state subsequence, the bilinearity of (104) 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 (110) imply \[\sup_{z\in\mathcal S_\beta} |F_{A_n,B_n}(z)-F_{A_m,B_m}(z)| \leq \|A_n-A_m\|\|B_n\| +\|A_m\|\|B_n-B_m\|.\] 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 \(t\): \[F_{A,B}(t)=\omega(A\tau_t(B)),\qquad F_{A,B}(t+i\beta)=\omega(\tau_t(B)A).\] This is the full \(\beta\)-KMS condition (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 33. Since \(\beta\) was any fixed finite number at least \(\beta_0(d,S)\), Theorem 1 follows. ◻