How to read this file. This is a study guide to one research paper: the proof that uniform odd integer Lipschitz surfaces on the triangular lattice with two-arc boundary data converge to a fixed multiple of the Dirichlet Gaussian free field. It is written for a patient high-school reader. Each block starts with "Exercise:" followed by a short question, then a self-contained answer. A "Proof:" sketch inside an answer can be skipped on a first read and returned to later. Named results carry a short nickname in parentheses, e.g. "Lemma (Grey Code)"; find the full statement by searching for the same nickname. Most notation is defined at first use; you meet first Z (integers), R (real numbers), T (the triangular lattice), D (a smooth planar domain), G_D (the Green function of the Dirichlet Laplacian) and Phi_D (the Dirichlet Gaussian free field on D).
Sources. The paper being explained is:
OpenAI 2026, The Gaussian free field limit of integer Lipschitz heights with two-arc boundary data. Preprint; arXiv-style source folder "The-Gaussian-free-field-limit-of-integer-Lipschitz-heights-with-two-arc-boundary-data".
Two companion papers of the same group are cited later:
OpenAI 2026, The Gaussian free field limit of the balanced six-vertex model with variance multiplier 1/arcsin(c/2).
OpenAI 2026, Gaussian free-field limits of weighted integer Lipschitz heights.
Exercise: What is the paper about, in one paragraph?
The paper shows that the random odd-integer Lipschitz surface on the triangular lattice, with a hard rule on every edge and a fixed boundary pattern of +1 on one boundary arc and -1 on the other, has, at long scales, the same fluctuations as the Dirichlet Gaussian free field, up to a fixed positive constant sigma. The proof is quantitative and covers arbitrary smooth simply connected planar domains. The result settles the FIELD-part of Schramm's Problem 2.2. The interface question (where is the level set h=0?) is not touched.
Exercise: What is a Lipschitz height and why is the model "hard"?
A Lipschitz surface H on the triangular lattice assigns an integer value H(v) to each vertex v, with |H(v) - H(w)| <= 1 on every edge (v,w). "Hard" means that this inequality is a strict constraint, not a soft penalty: configurations that violate it have probability zero. Most Gaussian-based models are soft; the hard constraint is exactly what makes this problem delicate, because the model is not a small perturbation of a Gaussian.
The rescaling between the odd-integer model h and the unit-integer model H is H = (h-1)/2. So |h(x)-h(y)| in {0,2} is the same as |H(x)-H(y)| <= 1.
Exercise: What is a "scaling limit"?
Let delta > 0 be the mesh of a triangular lattice (edge length delta). Attach an independently constructed height function H_delta(.) to that mesh so that the surface covers a fixed physical domain D at finer and finer resolution. Extend H_delta affinely on every small triangle, and smooth against a fixed compactly supported test function f by setting H_delta(f) = the integral of H_delta(x) f(x) dx. The family of random variables {H_delta(f)} as delta decreases to 0 has a scaling limit when, for each fixed f, H_delta(f) converges in distribution to a limit. More strongly, finite vectors of test functions converge jointly. The limit is itself a random distribution, i.e. a random object that only takes values after testing against a smooth compactly supported function.
Exercise: What is the Dirichlet Gaussian free field on a planar domain D?
Let G_D(x,y) be the Green function of the Dirichlet Laplacian on D, i.e. the integral kernel of the inverse (-Delta_D)^{-1} with zero boundary values. The Dirichlet Gaussian free field Phi_D is a centred Gaussian random distribution with covariance
E[Phi_D(f) Phi_D(g)] = double integral over D x D of f(x) G_D(x,y) g(y) dx dy,
for smooth compactly supported f, g. In two dimensions the Green function has a logarithmic singularity on the diagonal: G_D(x,y) = (1/2 pi) log(1/|x-y|) + a function that is harmonic near y. This logarithmic blow-up is why point values of Phi_D are infinite; only integrals against smooth tests are finite. Think of it as the "random surface whose gradient looks like a two-dimensional noise", with correlations 1/|x-y|.
Reference for the Dirichlet GFF: Sheffield 2007, Gaussian free fields for mathematicians: https://arxiv.org/abs/math/0312099.
Exercise: What is a "two-arc boundary condition"?
Fix a bounded smooth simply connected planar domain D and two distinct marked boundary points a and b. The two points split the boundary circle into two open arcs A+ and A-. Prescribe h = +1 along A+ and h = -1 along A-. The marks a,b themselves receive no value. This is precisely Schramm's boundary pattern; the +1 arc and the -1 arc sit opposite each other, and the only freedom is how the surface bridges them in the interior.
Exercise: What is the "uniform odd height law" precisely?
Approximate D by polygonal domains D_delta made of small triangular cells of the triangular lattice of mesh delta. Boundary vertices split into A_{+,delta} and A_{-,delta}, with no vertex on a marked edge. Weigh every odd integer height function h on the vertices of D_delta satisfying
h = +1 on A_{+,delta},
h = -1 on A_{-,delta},
|h(x)-h(y)| in {0,2} on every edge,
with equal probability. This is a uniform counting law, not a product of independent random variables. The uniform odd height law is the basic object of the paper.
Exercise: What is the main theorem (nickname "Two-Arc GFF Limit")?
Fix a bounded smooth simply connected planar domain D with two distinct marked boundary points. Fix also simply connected triangular-lattice polygonal approximations D_delta whose marked boundary parameterizations converge uniformly to those of D, with edge-midpoint marks. Sample the odd-height law h_delta uniformly. Then there exists a constant sigma > 0, depending only on the unit triangular-lattice model, such that
h_delta => g_bar + sigma Phi_D,
weakly, in the sense of distributions on D. Here g_bar is the harmonic extension to D of the boundary data +1 on A+ and -1 on A-. The convergence is in distributional topology D'(D). All finite joint test-function integrals converge jointly to the corresponding Gaussian vectors, and all mixed moments converge. The centered version is
(h_delta - E[h_delta]) / sigma => Phi_D.
The constant sigma is specified by an absolutely convergent finite-volume series; in the paper's spectral convention sigma = 4 pi sqrt(c_*), with c_* defined by the normalization series in Proposition (Normalization Series).
Exercise: What is g_bar and why is it the mean?
The harmonic function g_bar: D -> R is the unique bounded harmonic function with boundary values +1 on the open arc A+ and -1 on the open arc A-. In two dimensions this is exactly the "electrostatic potential of two charged boundary arcs", so g_bar(x) is close to +1 near A+ and close to -1 near A-, with a smooth transition in between. The theorem says the mean of h_delta tends to g_bar, i.e. E[h_delta(f)] -> integral g_bar f, for every smooth compactly supported test f.
Exercise: Why does this settle part of Schramm's Problem 2.2?
Schramm's 2006 problem collection asks about scaling limits of the uniform integer Lipschitz height function and of its zero-level interface. Question 2.2 has two parts: a field part (does the height function itself converge to a universal random surface?) and an interface part (does the level set {h = 0} converge to a curve, and which one?). Theorem (Two-Arc GFF Limit) proves the field part for the integer model on smooth simply connected domains. The interface part is a separate open problem, and the real-valued Lipschitz model of Problem 2.3 is also separate.
Schramm 2007, Conformally invariant scaling limits: an overview and a collection of problems: https://arxiv.org/abs/math/0602151v2, DOI 10.4171/022-1/20.
Exercise: What does "sigma is specified by an absolutely convergent finite-volume formula" mean?
The constant sigma is universal, i.e. independent of D, of a, b, and of the mesh. It is not some unknown "fudge factor". Proposition (Normalization Series) writes c_* = sigma^2 / (16 pi^2) as an explicit infinite series of finite-volume expectations of height differences squared, times explicit Fourier coefficients of a fixed cutoff. Each term is a finite sum of prescribed Fourier coefficients and ratios of finite integer counts. The series converges absolutely. So sigma can in principle be computed to any accuracy by finite calculations, even though the paper does not evaluate it numerically.
Exercise: What makes the proof hard?
Three things. First, the hard local constraint does not look Gaussian: there is no linear structure to exploit. Second, the boundary conditions are prescribed on arcs, so you cannot even compare the surface directly with a planar or half-planar reference. Third, the height fluctuations are logarithmic, so the model is critical and no ordinary "small noise" expansion applies.
The proof uses a spin representation, loop-and-cut techniques, reflection positivity, a discrete midpoint inequality, rare-pin comparisons, and boundary estimates. Each of these is the object of one chapter of the guide.
Exercise: What does the proof strategy look like, at a high level?
The goal is to identify every subsequential limit of the height field. The key intermediate identity is
E[H(Delta phi) . H(f_1) . ... . H(f_q)] = 0,
for smooth compactly supported phi, f_1, ..., f_q with pairwise disjoint supports (including q = 0). This identity says each "absolute-height moment distribution" is harmonic in each variable away from collisions. Boundary traces and collision residues then uniquely identify those moments as Gaussian.
Get there in three steps. (1) Reflection positivity on a stationary plane version of the model gives a spectral representation of the covariance and forces the plane covariance to be c_*/|p|^2 with c_* > 0. (2) Rare-pin comparisons and the "arch estimate" at a constant boundary arc give boundary traces, which pin down the additive constants and supply the harmonicity identity. (3) Harmonic moments with logarithmic collision singularities and local mixing to the plane law identify each singular coefficient; the moment recursion that results is exactly the Gaussian Wick formula.
The three steps occupy Sections 2-10 of the paper and correspond to the three main parts of this guide.
Exercise: What is the triangular lattice T and its dual honeycomb lattice T*?
The triangular lattice T is the set of points of the plane of the form u*(1,0) + v*(1/2, sqrt(3)/2) for integers u,v. Each vertex has six nearest neighbours at lattice distance one, and the plane is tiled by small equilateral triangles of side one. Its dual lattice T* is the honeycomb lattice: one vertex for each small triangle, edges between triangles that share a side. A path on T* is therefore a "wall" on T (a wall that runs between two adjacent height vertices and never crosses one). This duality is used all through the paper: contours (loops) live on the honeycomb dual, and the height values live on the triangular primal.
Exercise: How does a height function decompose into spin fields B and W (Lemma Grey Code)?
To every vertex v with integer height H(v), assign two spins B(v), W(v) in {-1,+1} by reading H(v) mod 4 through the Gray-code table
H(v) = 0 (mod 4): B = +1, W = +1;
H(v) = 1 (mod 4): B = +1, W = -1;
H(v) = 2 (mod 4): B = -1, W = -1;
H(v) = 3 (mod 4): B = -1, W = +1.
The pair (B,W) is called a coherent pair when B and W never BOTH change across a single edge. On every edge (x,y) of a coherent pair the integer step
dH(x,y) = ( W(x) B(y) - B(x) W(y) ) / 2
is well defined (it is 0 or +/-1). It is the "Gray-code increment" formula. Its name is (Lemma Grey Code). A local check shows that, on every elementary triangle, the step formula sums to zero, so the increments define a single-valued integer function up to an additive constant, provided the domain is simply connected.
Exercise: How does the spin picture recover the original height?
Fix a basepoint v_* and a height value H_0. (Lemma Grey Code) says: there is a bijection between admissible integer height functions H on a finite simply connected triangular domain with H(v_*) = H_0, and coherent spin pairs (B,W) whose Gray state at v_* matches H_0 (mod 4). Moreover, uniform counting of admissible heights, with prescribed boundary values, is the same measure as uniform counting of admissible spin pairs, with the boundary spins fixed consistently. This is why the spin language is not a change of model: it is exactly the same probability law.
Exercise: What is the loop representation (Lemma Loop Representation)?
Work on a finite simply connected triangular domain with a constant boundary height (all boundary vertices at one integer value). That is, work on the "flat" case. Say that an edge of T is "stepped" when H jumps by +1 or -1 across it. (Lemma Loop Representation) says three things. First, the stepped edges form disjoint closed honeycomb loops (no branching). Each loop separates a higher-level region from a lower-level region. Second, given the loop geometry, the inward jump of every loop is an independent fair sign. Third, the total weight of an unoriented loop configuration is 2 raised to the number of loops; summing the fair signs gives the height configuration back. So the flat height model is really a loop O(2) model with edge weight one. The same holds inside any bounded face of a connected, constant-height lattice skeleton.
Exercise: What does the loop representation buy you?
A local hard constraint becomes a purely combinatorial description in terms of disjoint loops with fair orientations. Loops can be estimated with existing tools: FKG, positive association for the height, loop lower and upper estimates from the Glazman-Manolescu paper. Every powerful estimate about the flat height model ultimately goes through the loop representation.
Glazman-Manolescu 2021, Uniform Lipschitz functions on the triangular lattice have logarithmic variations: DOI 10.1007/s00220-020-03920-z, arXiv:1810.05592v3.
Exercise: What is positive association, and does the constrained spin law satisfy it (Lemma Spin FKG)?
Positive association for a finite random configuration means: for any two increasing events A and B (increasing = event stays true whenever you raise coordinates), Prob[A and B] >= Prob[A]*Prob[B]. Equivalently the FKG lattice condition: for any two configurations X, X', Prob[X] Prob[X'] <= Prob[X and X'] Prob[X or X'] where "and"/"or" are coordinatewise min/max. (Lemma Spin FKG) says that the B marginal of the constrained spin law (fix W constant on some edges, fix some B spins) satisfies the FKG lattice condition. Fixing more B spins only sharpens it. Conditional on B, the W field is i.i.d. uniform on the components cut by B-changing edges and constant-W edges.
Exercise: What is the proof idea of Lemma (Spin FKG)?
Encode W up to sign by a binary variable on every edge (0 = W does not change, 1 = W does change). This edge-labelling comes from vertex spins if and only if it satisfies one parity equation per triangle, over F2. Rank-nullity then says: the number of W configurations compatible with a given B equals 2^(1 + n(F_+) + n(F_-)) for two set functions n on sets of "same-B" edges. Rank is submodular, hence n is supermodular, and each of the two factors satisfies the FKG lattice condition mechanically. A deterministic indicator (fixing B spins) preserves an FKG condition even if some of the compared weights are zero. Once B is fixed, W is just independent fair signs on components, giving the second half.
Exercise: What is a "double path" and why is it important?
Look at the dual honeycomb lattice T*. For a dual edge crossing a primal pair (x,y), say it is "double-B of sign t" if B(x) = B(y) = t. A double path (or circuit) is a chain of such dual edges, all with the same sign t. (These are sometimes called "double-B paths".) At a dual vertex of a double path, three primal vertices lie around the honeycomb hexagon: all three have B = t. Such a chain is therefore pinned to a single "gray strip" of the integer lift. Crucially, crossing a double-B edge gives no constraint on W (since B is unchanged by hypothesis), so coherence is not violated. Double-B edges are the natural contours that do not "feel" the W field: this is why they make clean cut surfaces.
Exercise: What is the exact cut law (Lemma Double Cut)?
Let C be a simple double-B circuit of sign t (all along the circuit, adjacent primal vertices have B = t). Assume that no equality edge E_0 crosses C. Then, conditional on the exterior pair configuration and on B = t at the inner boundary layer of C, the interior law is uniform counting of coherent pairs with that boundary and with the E_0 constraints inside C. The same holds if C was selected by a stopping exploration from the exterior that only revealed exterior B values and the B = t cut sites. No W spins need to be revealed during the exploration.
Exercise: Why is the exact cut law true?
Every interaction edge crossing C has B equal on both sides, so the W-coherence condition is silent there. That means the total counting weight factorizes: exterior factor times interior factor, conditioned on boundary. In the stopping version, run a depth-first primal exploration from the exterior that stops when it would cross a good dual edge. If a double circuit does exist, the exploration cannot pass it. Bounded pockets are filled on the already-explored side and do not change the interior counting weight. The point is that the "cut" is exactly the surface where the exterior and interior are statistically independent given the boundary trace.
Exercise: What does a "height lift along a double path" look like?
Along any consecutive chain of vertices of a double-B path, all such vertices have B = t. Since B = t covers only two consecutive integer residues, the lift must stay in that pair of consecutive integers. (Lemma Double Cut) records this as one of its remarks: the inner face chain of a double circuit takes values in one pair of consecutive integers. This is what lets the interior law look like an ordinary "bounded-boundary" Lipschitz height problem: after subtracting the smaller integer of the pair, the boundary trace is again a two-value problem at target heights.
Exercise: What is "height association" and why is it free?
The set of admissible integer Lipschitz functions on a finite triangular domain is closed under taking coordinatewise min and coordinatewise max: if H and H' are both admissible, so is min(H,H') and so is max(H,H'). Boundary values that are coordinatewise ordered are preserved by min and max. (This is the elementary FKG observation of Fortuin-Kasteleyn-Ginibre 1971.) As a consequence, coupled boundary data can be coupled on the full interior order: if boundary data g <= g', then the corresponding interior laws can be coupled so that H_g <= H_g' everywhere. The paper builds this via coordinate-by-coordinate conditional quantile updates, driven by a stream of independent uniform random variables and a deterministic cyclic schedule. Each update is monotone in the height and in its uniform variable, so the resulting global map is monotone in the boundary data. Taking limits gives positive association and ordered couplings, including the spin-pin and W-equality versions.
Fortuin-Kasteleyn-Ginibre 1971, Correlation inequalities on some partially ordered sets: DOI 10.1007/BF01651330.
Exercise: What is the "circuit under the opposite cut condition" (Lemma Worst Cut)?
Inside a regular double-cut region of radius K*n (K a fixed large constant), fix B = -t on the inner boundary layer, and leave W free. (Lemma Worst Cut) says: with probability at least c > 0 (uniform in n, for n large), there is a double-B circuit of sign t between radii n and 2n inside the region. So even under the "wrong" boundary sign, a sign-t double circuit forms a constant fraction of the time. The strategy in the proof is: plant three separated enclosing loops so that an outer loop flips W but not B, and an inner loop flips W but not B; conditional on loop geometry, the loop signs are independent fair coins, so this configuration has fixed probability. Then cut at the resulting double-B of sign t.
Exercise: What is a "buffer" and what is Lemma (Buffer Circuits)?
A buffer is a small open region where a deterministic cut of known shape is guaranteed to lie, away from all pins and equality edges. Given a fixed positive-width polygonal collar or tube, with a buffer inside it, and conditioned on arbitrary B values away from the buffer, the probability of a double-B circuit around the hole (or a connected double-B path in the tube) is at least c > 0, where c depends only on the geometry and not on mesh, number of pins elsewhere, or their values. Additionally, a logarithmic stack of disjoint similar buffers has failure probability at most C (r/L)^alpha for an explicit exponent alpha > 0 (equation eq:stack-failure of the paper). This is the "long necklace" fact used every time the paper needs to squeeze a circuit past nothing in particular.
Exercise: What is the local-to-plane comparison (Proposition Plane Mixing)?
Free coherent pairs on expanding regular hexagons converge on cylinder events to a limit law mu_infinity. This mu_infinity is invariant under lattice translations, lattice symmetries, global sign changes of both spins, and colour swap. Two finite-plane comparisons follow. (i) If a bounded-boundary problem has a ball of radius L with no constraints inside, and you look at radius r << L, its law is within total variation C*(r/L)^alpha of mu_infinity. (ii) Plane height increment laws from mu_infinity have the same spatial symmetries, and every finite feasible B-pin plus W = +1 along finitely many connected paths has positive plane probability. This is the "faraway observations look like the plane" fact; it is the bridge between the finite-domain model and the reference m_infinity used throughout.
Exercise: What is the relative comparison for one obstacle (Lemma Relative Pin)?
Let mu and nu be finite plane pair laws, with same-fill neighbourhood U on one obstacle, same retained pins R inside U, and no constraints in the buffer collar. Allowed to differ outside U. Then for a further sign-t event A on that obstacle, c nu(A) <= mu(A) <= c^-1 nu(A), with c depending only on the collar and its buffers. Key point: c does not depend on the probability of A, the number of retained pins, or the values of the pins. Proof: sandwich A between the corresponding "worst-case boundary" probability q (upper) and c_1 q (lower), both computed from a deterministic double-cut and FKG monotonicity; q is common to mu and nu.
Exercise: What is (Proposition Pin Product)?
For a family of separated "obstacles" (each a bounded compact neighbourhood U_i with a buffer collar, disjoint from the others), define pin events P_i with probabilities p_i under mu_infinity. Then for some 0 < c <= C < infinity,
c * prod p_i <= mu_infinity(cap_i P_i) <= C * prod p_i.
The constants depend only on the collars, buffer radii, the number of obstacles, and the bound on W-subpath counts. They do NOT depend on the number or values of microscopic B pins, nor on the p_i themselves. This says that pinning events on separated obstacles are almost independent. The proof: first handle W-constancy (2^m label counting between constancy and all-plus). Then condition on W-constancy and impose the B pins obstacle by obstacle, using Lemma (Relative Pin); at each step, retained pins on the obstacle stay fixed. Finally restore the W-labels at fixed cost 2^{-sum m_i}.
Exercise: What is a "lattice smearing"?
A radius-a lattice smearing of a height H is a real linear statistic Y = sum over vertices x of w_x H(x), where the weights w_x satisfy two binding rules: sum of |w_x| <= A (L1 control) and sup |w_x| <= A a^{-2} (sup control), with A a fixed constant. Weights are allowed to be signed. If also sum w_x = 0, the smearing is called zero-total: it does not see the additive constant of H. Given a smooth compactly supported test density of size O(a^{-2}), affine interpolation of the height against it produces smearing weights of this form with the same total mass. This is why smearings are the right object: pointwise analysis of H is meaningless, but smearing bounds against a fixed A are uniform in mesh.
Exercise: Why are smearing moments the right technical object?
Continuous random surfaces of the model (like the GFF) do not have pointwise values; only their integrals against smooth tests make sense. If you can show that every radius-a smearing Y has bounded p-th moment for all p, uniformly in the mesh, then you have (i) tightness of the test statistics, (ii) convergence of their moments along limits, and (iii) ability to pass from "distributional limits" to "pointwise almost sure bounds". This is why the paper spends a big chunk of Section 4 proving smearing moment bounds.
Exercise: What is the Klartag-Lehec midpoint inequality?
In one dimension: let f, g >= 0 be finitely supported functions on the integers. Suppose there exist nonnegative functions u, v with f(x) g(y) <= u(floor((x+y)/2)) v(ceil((x+y)/2)) for all x, y in the support. Then (sum f)(sum g) <= (sum u)(sum v). This is a discrete relative of Prekopa-Leindler and holds for nonnegative functions including ones with zeros. In dimension d, apply coordinatewise rounding and repeat the scalar inequality once per coordinate. The formulation used in the paper is the nonnegative entropy-inequality version of Gozlan-Roberto-Samson-Tetali:
Klartag-Lehec 2019, Poisson processes and a log-concave Bernstein theorem: DOI 10.4064/sm180212-30-7, arXiv:1802.04176v2.
Gozlan-Roberto-Samson-Tetali 2019, Transport proofs of some discrete variants of the Prekopa-Leindler inequality: DOI 10.2422/2036-2145.201906_002, arXiv:1905.04038v1.
Exercise: What is the "rounded midpoint" of two integer configurations?
If x and y are integer arrays indexed by the same finite set of lattice sites, then floor-x-y and ceil-x-y are obtained by taking each coordinate separately and applying floor((x_i+y_i)/2) or ceil((x_i+y_i)/2). Both are again integer arrays.
Exercise: Why is the set of admissible heights stable under rounded midpoints (for the paper's use of Lemma (Midpoint Tails))?
Three things must survive rounding. (i) The hard Lipschitz condition |H(x)-H(y)| <= 1: if x and y satisfy it, then floor and ceil do too. (ii) Fixed integer boundary values or a fixed integer anchor: floors and ceilings preserve a fixed integral value. (iii) The rounding error of a linear statistic: |Y(rounded midpoint) - (Y(x)+Y(y))/2| is at most (1/2) sum |w_x| for a smearing. These three facts are what make Lemma (Midpoint Tails) applicable, because they let the rounded midpoint blow up a bounded central probability into a full Gaussian-like tail.
Exercise: What is Lemma (Midpoint Tails)?
Let S be a finite set of integer arrays with uniform counting measure. Let Y be a real linear statistic and M > 0. Partition S into at most m_* classes. Assume (i) if x, y are in the same class, both of their coordinatewise rounded midpoints belong to S (they may fall in a different class); (ii) Y(rounded midpoints) differs from (Y(x)+Y(y))/2 by at most M for x, y in the same class; (iii) for every class in use, P(|Y| <= M and x in that class) >= c > 0. Then for constants C, c' depending only on c and m_*,
P(|Y| > t M) <= C e^{-c' t} (t >= 1),
and ||Y||_{L^p} <= C_p M for all p < infinity. This is the "amplification" lemma.
Exercise: What is the proof idea of Lemma (Midpoint Tails)?
Dyadic amplification. Take intervals I_k = [2^k M - 4 M, 2^{k+1} M + 4 M]. The lower tail of Y and the two output arrays from the coordinatewise rounding both fall inside I_{k-1}, with a controlled margin. Apply the midpoint inequality to the indicator of a chosen central class (which carries at least p_k / m_* of the mass at scale I_k) and to the indicator of the central mass {|Y| <= M}. The result is a recursion p_{k-1}^2 >= c p_k / m_* on the tail probabilities. Iterate to get p_k <= C exp(-c' 2^k). Symmetric argument for the negative tail; integrate the resulting tail to get moments. The point is that a single fixed-mass central estimate plus a midpoint inequality produces exponential tails in one shot, without needing to check anything beyond admissibility.
Exercise: What is Lemma (Loop Shells)?
In a finite flat-boundary height problem, take an interior site x. Look at the level loops surrounding x whose graph inradius lies in a shell [d, 2d]. (Lemma Loop Shells) says: the count of such loops has bounded moments at every order, uniformly in the domain, and uniformly in d >= 1. The same holds if the shell is defined by Euclidean closest approach to x instead of graph inradius, and also in a bounded face of a connected constant-height skeleton. Idea: expose the loops from the outside in. Between successive exposed loops is a free interior; the probability of another loop at scale >= d is at most 1 - c_0 (imported Glazman-Manolescu). After each exposure the argument reapplies, so the loop count in [d, 2d] is dominated by 1 plus twice a Geometric(c_0) random variable, for a bounded number of sub-shells. Bounded distances are handled by a deterministic vertex bound.
Exercise: What is Proposition (Smearing Moments)?
Consider a uniform height problem with boundary values bounded in absolute value by K. Take a radius-a smearing centred at distance n from the boundary with 1 <= a <= c_0 * n. Then for all finite p,
||Y||_{L^p} <= C_p (1 + sqrt(log(n/a))).
Constants depend only on p, A, K, not on boundary regularity. Same for a bounded-boundary problem on a face chain of a dual cut or on a skeleton face. Consequence: normalized averages on balls of radius fixed fraction of their depth have bounded moments uniformly as the depth tends to zero, which is exactly what is needed at boundary points and marks later.
Exercise: What is the proof idea of Proposition (Smearing Moments)?
First reduce to zero boundary. Loop representation gives E[H(x)H(y)] = E[number of loops surrounding both x and y]. A loop that surrounds both and approaches x within distance 2d must have a "wall arm" from scale d to scale s = max(1, dist(x,y)). Such an arm is blocked by a double circuit of the same first spin just outside. Lemma (Buffer Circuits) then bounds the arm probability by C (d/s)^alpha. Summing the small shells and Cauchy-Schwarz gives E[H(x)H(y)] <= C (1 + log(n/s)). Summing against weights gives E[Y^2] <= C (1 + log(n/a)); Chebyshev gives a fixed central probability at scale M = C (1 + sqrt(log(n/a))); Lemma (Midpoint Tails) gives all higher moments. Bounded boundary data is handled by comparison with constant boundary -K and +K plus the triangle inequality in L^p.
Exercise: What is Proposition (Plane Moments)?
For every zero-total smearing in the plane law mu_infinity, for every finite p,
||Y||_{L^p(mu_infinity)} <= C_p,
uniformly in radius, location, and lattice rotation. Same bound holds in sufficiently larger finite free volumes. Idea: search from the outside for a double-B circuit in an annulus between K*a and 2 K*a; Lemma (Buffer Circuits) gives probability >= c > 0. On success the inner trace sits in one pair of consecutive integers, so the interior is a bounded-boundary problem solved by Proposition (Smearing Moments). Central probability at fixed M and Lemma (Midpoint Tails) then give all moments. Plane zero-total statistic is invariant to interior shift (sum w_x = 0), so the shift from "one pair to zero pair" is invisible.
Exercise: What is a "component switch" (Lemma Component Switch)?
Consider two nested double-B circuits with opposite signs, with a collar between them free of E_0 edges and pinned W sites. In this collar there is a component J of B-changing edges whose associated potential function psi_J takes exactly two values, differing by 2, one value on the inner region and one on the outer region. Because the collar is free, J's W label w is an independent fair coin. Flipping w changes every height inside the inner cut by +4 or -4 in absolute value, while leaving every height outside the outer cut unchanged. For a zero-total statistic Y with total weight lambda inside the inner cut, the two values of Y differ by 4|lambda|, so conditional on everything else its variance under the fair coin is at least 4 lambda^2. This "fair switch between opposite cuts" is how the paper manufactures independent fair bits inside cut regions.
Exercise: What is a "interval-compatible pin" pattern?
On obstacle i, a pin pattern is interval-compatible if along a connected chain the exact height is
H(v) in I_{i,v} + 4 t_i for v on the chain,
with prescribed integer intervals I_{i,v}, at least one of which is a singleton, and t_i in Z a common integer offset. The spin description of the event is: some B pins plus W = +1 along a bounded number of connected subpaths. The singleton makes t_i unique once a global lift is chosen. The patterns used in the paper have I_{i,v} of one or two consecutive integers. Reflections and translations preserve this definition.
Exercise: What is Proposition (Obstacle Moments)?
Fix finitely many separated interval-compatible obstacles with filled isolating neighbourhoods U_i. Put a test disk in a pin-free simply connected neighbourhood U_0 with room for a buffered cut inside U_0, and require that closures of U_i and U_0 be disjoint. Let Y be a zero-total smearing in the test disk. Then conditional on all pins,
||Y||_{L^p} <= C_p for every finite p,
uniformly in mesh and in the microscopic pin probabilities. The constants depend only on the macroscopic arrangement, the weight bound, the number of obstacles, and the W-subpath count bound. No uniformity as obstacle separations vanish is asserted here (see the next exercise).
Exercise: What is the proof idea of Proposition (Obstacle Moments)?
Anchor at obstacle 1 at t_1 = 0. Split arrays by the parities of t_2, ..., t_m mod 4. In one parity class, coordinatewise rounded midpoints preserve all of the interval pins (interval endpoints are integral), the hard Lipschitz bound, and the anchor. Rounding error of Y is bounded by A/2. To get a uniform central probability, find via Lemma (Buffer Circuits) a double-B cut around the test disk and two opposite-sign cuts per non-anchor obstacle, all with probability >= c_1 > 0 conditional on pins. The test search's interior is a bounded-boundary problem with moment bound C, so Chebyshev gives a fixed-mass central event. Then, conditional on B and other W labels, flipping each obstacle collar's fair coin via Lemma (Component Switch) preserves Y and everything else but changes t_i by an odd integer; it acts on the parity classes exactly transitively, so each class gets at least 2^{-(m-1)} c_2 of the central mass. That is the class-central-mass hypothesis of Lemma (Midpoint Tails), with at most m_* = 2^{m-1} classes, so we get all moments.
Exercise: What is Corollary (Equal-Offset Moments)?
Same obstacles as in Proposition (Obstacle Moments), but now impose the extra event E = {t_i = t_1 for every i}. The obstacles may vary or approach one another, provided they stay away from a fixed test disk and pin-free buffer. Suppose the exterior search finds a surrounding double-B cut in this buffer with probability >= c uniformly in the varying arrangements, and E can be checked in the exterior of the selected cut. Then the zero-total test in that disk has moments bounded uniformly in those arrangements. A sufficient condition: cut probability >= c_0 conditional on pins alone, P(E | pins) -> 1 uniformly on the varying family, and exterior measurability of E.
Exercise: Why is Corollary (Equal-Offset Moments) important even though Proposition (Obstacle Moments) already gives conditional moments?
Because (Obstacle Moments) required the obstacle SEPARATIONS not to vanish. Along the boundary comparison, gaps will shrink to zero, and the obstacles are pinned boundary arcs that may almost touch. Equal offsets buy uniformity in that regime: after anchoring, all t_i = 0, the pinned constraints are fixed integer intervals, so midpoint rounding needs no parity classes at all, and the argument collapses to a single-class application of Lemma (Midpoint Tails).
Exercise: What is "reflection positivity" and what does it give you here (Proposition Reflection Positivity)?
Fix a row ell of triangular-lattice sites parallel to a nearest-neighbour direction. Let Theta_ell denote spatial reflection across ell. Let A_ell be the set of cylinder functions supported in one closed half-plane with increments evaluated along paths in that half-plane. The bilinear form
_ell = E[ conj(Theta_ell F) G ]
on A_ell is positive semidefinite. In its quotient completion: (i) normal translation by sqrt(3) farther into the half-plane is a positive self-adjoint contraction S; (ii) tangential translation by one lattice unit is a commuting unitary operator. On mesh delta the normal step is sqrt(3) * delta.
Exercise: Why is the plane model reflection-positive across a row of sites?
Key geometric fact: no nearest-neighbour edge jumps ACROSS a row of sites. So the row is the separating surface. Conditional on the spin values at the row (and on any auxiliary phase values at one anchor on the row), the two half-planes are independent and reflected-identical. Therefore
E[conj(Theta_ell F) F] = E[ | E[F | row data] |^2 ] >= 0
for every F supported in one half-plane. Approximation and Cauchy-Schwarz give that this holds on the full L^2 closure. This is the site-reflection version of the classical mechanism of Froehlich-Israel-Lieb-Simon.
Froehlich-Israel-Lieb-Simon 1978, Phase Transitions and Reflection Positivity I: DOI 10.1007/BF01940327.
Usui 2012, A note on reflection positivity and the Umezawa-Kamefuchi-Kaellen-Lehmann representation: arXiv:1201.3415v3.
Exercise: Why "site reflection" rather than "edge reflection"?
A site-reflection line is itself a row of vertices: reflecting across it maps vertices to vertices, and no edge crosses the line. An edge-reflection line would either be a line through edge midpoints (and would make spins interact across the line) or not respect the triangular lattice. Site reflection is the one that produces a positive self-adjoint contraction acting on increments; edge reflection, if attempted, breaks positivity by mixing vertices that share an edge. The exact halfway-line choice matters for the proof of the positive normal transfer. The composition of two site reflections at y=0 and y=sqrt(3)/2 is a pure normal translation by sqrt(3), with no tangential correction: this is why the normal step is sqrt(3) in mesh units.
Exercise: What is a "auxiliary phase field" and why allow one (equation eq:auxiliary-phases)?
For each real theta, add an independent Haar-uniform unit-circle phase U_theta times exp(i theta (H(u) - H(o))) at each lattice site u, with o an arbitrary fixed anchor. Because the Haar distribution on the unit circle is invariant under multiplication by any unit complex number, replacing the anchor o by another site has the effect of multiplying all phases by one fixed unit complex number, which leaves their joint law unchanged. This makes the phase field statistically invariant under spatial translations and reflection, and independent of the height increments. Reflection positivity, stationarity, and the positive transfer still hold for the joint law (Proposition Reflection Positivity). The phases are used later to Fourier-transform offsets t_i in Z that arise from separating boundary pin arcs, and to control their equality via a positivity argument that would fail without the phase augmentation.
Exercise: What is the "increment spectral measure" (Lemma Increment Spectral)?
Let V_u be the unitary lattice-translation operators on the plane law. Let E(dp) be their joint spectral resolution on the dual lattice T^perp = { p in R^2 : p . u in 2 pi Z for all u in T }. The increment cocycle xi_u = H(u) - H(0) satisfies xi_{u+v} = xi_v + V_v xi_u and (V_v - I) xi_u = (V_u - I) xi_v. There exists a nonnegative measure C, locally finite on the punctured cotorus T^perp minus {0}, such that for every finitely supported real array a with sum a_u = 0,
E[ ( sum a_u H(u) )^2 ] = integral_{T^perp minus {0}} | sum a_u e^{i p . u} |^2 C(dp).
The measure C is invariant under lattice reflections and rotations. There is no additional random linear tilt.
Exercise: Why is there no linear tilt in the spectral representation?
Because the spectral measure's potential 0-mass would be an additive linear functional: if the invariant subspace of translations has a nonzero projection P_0 xi_u, then P_0 xi_u = A . u for a random vector A in L^2. Smearing a zero-integral f at scale L against the height gives a statistic whose invariant part is L * A . m with m = first moment of f. Plane zero-total moment bounds make this norm bounded uniformly in L. Two choices of f with independent first moments force A = 0. So there is no tilt.
Exercise: What is Lemma (Annulus Mass)?
The spectral measure C has uniformly bounded annular mass: C{ r <= |p| <= 2 r } <= K, for all 0 < r < 1/4. Consequently for every eta > 0, integral over 0<|p|<1/4 of |p|^eta C(dp) < infinity. The proof: use the uniform L^2 bound of plane zero-total smearings, plus a pair of real zero-integral smooth functions f_1, f_2 whose Fourier transforms have no common zero on the unit annulus, at every dyadic scale. The rest is local finiteness of C.
Exercise: What is the "Cauchy mixture" representation (Proposition Cauchy Mixture)?
Restrict the lattice translations to the rectangular sublattice generated by (1,0) and (0,sqrt(3)). Project C to the reciprocal rectangle R = [-pi, pi) x [-P/2, P/2) with P = 2 pi/sqrt(3). Call the result C'. Then there is a nonnegative measure nu(dk, ds) on the horizontal circle times [0, infinity] such that
C'(dk, dl) = integral_{[0,infinity]} Q_s(dl) nu(dk, ds), k != 0,
where Q_s is the periodized Cauchy probability density on the vertical circle of length P. Moreover, integral_{0<|k| ( (a + i z) / (a - i z) )^2 has boundary real part chi(a, l) on the upper half-plane; Poisson formula at z = is gives the unperiodized identity; periodization gives an O(a) error and the tail is O(a^2).
Exercise: Why the FOURTH angular harmonic specifically?
Because the lattice has sixfold rotational symmetry. Rotate the plane by k * pi/3 for k = 0, 1, ..., 5, and sum chi(k, l) at the six rotated points. The result is zero: the fourth harmonic is the smallest even harmonic annihilated by the sixfold rotation sum. C is invariant under these rotations, so the integral of chi against C over a full radial annulus vanishes identically. That makes the comparison of a Cauchy-kernel integral against the "dispersion defect" D(k,s) usable: any nonzero total from the D-comparison must be exactly balanced by boundary or periodization effects.
Exercise: What is Proposition (Summable Dispersion Defect)?
The dispersion defect D(k,s) is summable against nu:
integral_{0<|k| 0. Additionally there is a finite constant K_1 with
nu{ r <= |k| <= 2 r } <= K_1
for all sufficiently small r > 0.
The argument uses the vanish-on-annulus identity, the (D + O(|k|)) representation of chi's Q_s average, the bounded endcap idea, and dyadic annular estimates. In words: the deviation between the normal "wavelength" s of each Cauchy kernel and the "transverse rate" |k| is summable over all scales. This is the paper's analytic core: after rescaling, the spectral mass can only sit on the relation s = |k|, i.e. on the dispersion curve of a massless field.
Exercise: What is Lemma (Endcap Estimate)?
There is a finite constant K_3 such that
C{ r < |p| < 2 r, |k| < eps } <= K_3 * (eps / r)
uniformly for 0 < eps < r/10 and 0 < r < 1/4. In particular, C has no mass on the line k = 0 in the punctured small disk. Idea: rotate the region by pi/3, then use the Cauchy-density bound on a short vertical interval at each fixed horizontal coordinate; the horizontal band mass bound (Prop Dispersion Defect) closes it.
Exercise: What is Lemma (Spectral Subsequences)?
Rescale the spectral measure near the origin: replace p -> p / delta for small delta and call the result C_delta. Every vague subsequential limit of C_delta on a fixed compact away from 0 has the form
M(dp) = c * dp / |p|^2
for some finite c >= 0 (allowed to depend on the subsequence). Idea: rescale the Cauchy mixing variables; Prop (Dispersion Defect) forces the mixing to concentrate on the graph s = |k|; periodized Cauchy converges after rescaling to the ordinary Cauchy kernel; aliias contributions vanish because their preimages avoid the fixed physical small disk. So the rescaled M has |p|^2 M invariant under vertical translations on k != 0; taking all three lattice rotations gives two nonparallel directional derivatives zero; hence |p|^2 M is local Lebesgue, hence constant. So M = c dp / |p|^2 on the whole punctured plane.
Exercise: What is Proposition (Angular Sum Rule) and how does it fix c_*?
Choose any even C^infty cutoff q: [0,infinity) -> [0,1] with q = 1 on [0, 1/2] and q = 0 outside [0, 1). Define dyadic horizontal bands w_j(k) = q(2^j k) - q(2^{j+1} k) and set
I_j = integral q(4 |p|) * chi(k, l) * w_j(k) * C(dp).
Then the series sum_j |I_j| converges. There is a single constant c_* >= 0 such that EVERY vague limit in Lemma (Spectral Subsequences) has coefficient c_*: c_* is the same no matter which subsequence you choose. And the sum rule is
sum_j I_j = -(pi/2) * c_*.
The value is independent of the choice of q.
How: telescoping reduces the partial sums to one integral whose integrand contains a difference q(|p|/eps) - q(|k|/eps) times chi. Vague convergence uses the endcap bound for uniform tails. Then polar-coordinate integration gives c_* * integral_0^{2pi} cos(4 u) * log|cos u| du = c_* * (-pi/2), using the known fourth Fourier coefficient of log|1 + e^{2iu}|. Since the LHS is fixed, c_* is fixed: no subsequence can produce a different c_*.
Exercise: What is Theorem (Universal Plane Covariance)?
The rescaled spectral measures converge vaguely on the punctured plane to c_* * dp / |p|^2, with 0 < c_* < infinity. Writing v = (2 pi)^2 c_*, we get that for real f, g in C_c^infty(D) of integral zero,
lim_delta E[ H_delta(f) H_delta(g) ]
= c_* * integral f_hat(zeta) conj(g_hat(zeta)) / |zeta|^2 dzeta
= ( v / (2 pi) ) * double integral f(x) log(1/|x-y|) g(y) dx dy.
The same conclusion holds for tests converging in C_c^infty with support in a common compact set. In words: at every scaling limit the plane covariance of H is the Dirichlet Green kernel, up to the constant v.
Exercise: Why is 0 < c_* < infinity, i.e. why can c_* not vanish?
Finiteness is already in the annulus-mass bound. Positivity is the more delicate point: some subsequence might in principle push c_* down to zero. The paper proves 0 < c_* by an explicit lower bound. Take two disjoint disks with nonnegative smooth unit-mass tests and set f = test_1 - test_2 (so the integral of f is 0). By Lemma (Buffer Circuits), with probability at least c > 0, two nested opposite-sign B double cuts surround one disk and avoid the other. Lemma (Component Switch) applies: the fair coin in the collar acts on H_delta(f) by flipping its sign, so the conditional variance is 4. Averaging gives a uniform positive lower bound on E[H_delta(f)^2]. Passing to the covariance limit forces c_* > 0.
Exercise: What is Corollary (Reflection Null Vector)?
Let phi in C_c^infty be supported strictly on one side of a line parallel to a lattice-reflection direction. Choose site rows ell_delta converging to that line, still separated by a positive margin from supp phi. Put X_delta = H_delta(Delta phi) (the Laplace insertion of H_delta against phi). Then the reflection norm of X_delta at ell_delta tends to zero:
|| X_delta ||_{ell_delta} -> 0.
The same holds under lattice translations of the placement and under arbitrary auxiliary phase augmentation. The reflection norm here is the norm-squared E[ (Theta_{ell_delta} X_delta) X_delta ]. It is NOT the ordinary variance of X_delta; the ordinary variance can be strictly positive.
Exercise: How does Corollary (Reflection Null) follow from Theorem (Plane Covariance)?
By Theorem (Plane Covariance), the reflection-norm-squared at row ell_delta converges to v * integral of nabla phi . nabla(phi o Theta) over the plane. Because phi is supported strictly on one side of the reflection line, phi and its reflection have disjoint supports, so the integral splits (via Green's identity) into nabla phi paired with the REFLECTED image; that pairing is exactly zero because the two functions live on disjoint supports. Phases do not change the expectation of a statistic that does not use them. Reflection positivity guarantees the squared reflection norms are nonnegative, so we get actual convergence to zero.
Exercise: What is the explicit cutoff q(r) in Proposition (Normalization Series)?
Even smooth q: [0,infinity) -> [0,1] with q(r) = 1 for r <= 1/2, q(r) = 0 for r >= 1, and smooth on (1/2, 1) via the exponential bump
q(r) = exp(-1/(1 - r)) / ( exp(-1/(1 - r)) + exp(-1/(r - 1/2)) )
for 1/2 < r < 1. The cutoff is even in the sense q(|k|), and it satisfies a "scale-localization" role: q(2^j k) - q(2^{j+1} k) is a dyadic band selector for the horizontal coordinate k.
Exercise: What is the multiplier F_j(p)?
For p = (k, l) put
F_j(p) = q(4 |p|) * cos(4 arg p) * [ q(2^j k) - q(2^{j+1} k) ],
extended by zero outside |p| < 1/4, with F_j(0) = 0. This is a smooth function on the reciprocal torus T^perp. It is smooth at the apparent angular singularity because cos(4 arg p) is continuous around the origin when p is expressed in Euclidean polar coordinates, and because the bracket vanishes when |k| <= 2^{-j-2}. It is even in p (because cos(4 arg) is even in p in the sense cos(4 arg(-p)) = cos(4 arg p) and the dyadic band selector and q(4|p|) are even in |p| and k = -k respectively).
Exercise: What are the Fourier coefficients d_ju of F_j?
d_ju = ( A_Tri / (2 pi)^2 ) * integral F_j(p) e^{-i p . u} dp,
where A_Tri = sqrt(3)/2 is the area of a fundamental unit cell of T. The coefficients are real and satisfy d_{j,-u} = d_{ju} because F_j is real and even. This integral is over a fundamental representative of the torus T^perp, whose Euclidean area is (2 pi)^2 / A_Tri.
Exercise: What are the radii R_js and hexagons H_R (eq:normalization-radii)?
For j, s >= 0, R_js = 2^{(j+s+4)^2}, which is a super-exponential dependence on j+s. H_R is the triangular hexagon {u in T : dist_T(0,u) <= R}. The pair (R_js, H_R_js) is the "finite volume at scale (j, s)" that serves as a controlled approximation to an infinite coherent-pair average.
Exercise: What is < . >_R?
This is uniform averaging over pairs (B,W) on H_R that are coherent (both spins never change on the same edge), with no boundary conditions. It is the natural "free" finite-volume average at radius R.
Exercise: What is b_js (eq:finite-band-coefficient)?
b_js = sum_{u in T : |u| <= sqrt(R_js)} d_ju * < ( sum_{xy in gamma_{0u}} (W(x) B(y) - B(x) W(y))/2 )^2 >_{R_js},
where gamma_{0u} is any lattice path in H_R_js from 0 to u. The Gray-code curl identity makes this independent of the path. Because the mean of the squared height difference is path-independent, this b_js is well defined.
Exercise: What is Proposition (Normalization Series)?
The series of the indicated terms is absolutely convergent:
sum_{j>=0} |b_j0| + sum_{j, s >= 0} |b_{j,s+1} - b_js| < infinity.
Its value satisfies
c_* = (1/pi) * sum_{j>=0} [ b_j0 + sum_{s>=0} ( b_{j,s+1} - b_js ) ] > 0,
and
sigma = 4 pi sqrt(c_*).
So c_* is a strictly positive number, defined via Chapter 6's construction, without reference to D, a, b, or of a "scaling limit".
Exercise: How is Proposition (Normalization Series) proved?
Write V(u) = E[(H(u) - H(0))^2] for the plane height difference, and set b_j = sum_u d_ju V(u). The Fourier smoothness of F_j and the quadratic growth V(u) <= K |u|^2 make the series defining b_j absolutely convergent for each fixed j. The increment spectral identity (Lemma Increment Spectral) gives V(u) = 2 * integral (1 - cos(p . u)) C(dp). Because sum_u d_ju = F_j(0) = 0 and sum_u d_ju cos(p . u) = F_j(p), the interchange of sum and integral is legitimate even though C has infinite mass near 0, and yields b_j = -2 I_j. From Proposition (Angular Sum Rule), sum_j |b_j| < infinity and sum_j b_j = pi c_*. Now b_js approximates b_j in the finite hexagon: the finite-variance error decays like |u|^2 * (|u|/R)^alpha, and Fourier decay of d_ju comes from differentiating F_j. Choosing R_js = 2^{(j+s+4)^2} makes the finite-volume error summable, telescoping gives b_j0 + sum_s (b_{j,s+1} - b_js) = b_j, and the alternating series bound converts into absolute convergence. Positivity of c_* was proved in Theorem (Plane Covariance), and the formula sigma = 4 pi sqrt(c_*) uses v = (2 pi)^2 c_* and the fact that the odd height is twice the integer height (plus 1, which is removed by centering).
Exercise: What is Remark (Cell Area) about?
There is no hidden "discrete cell factor" in the normalization. If lambda_u is the unit-mesh nodal hat, then integral lambda_u = A_Tri, so the physical interpolation weight of h_delta against f at mesh-delta resolution is
a_{delta,u}(f) = delta^2 * integral lambda_0(y) f(delta u + delta y) dy.
The nodal hat mass A_Tri delta^2 cancels the site density (A_Tri delta^2)^{-1} when taking the transform at p = delta * zeta. Its limit is f_hat(-zeta), with no prefactor. The reciprocal-cell factor (A_Tri / (2 pi)^2) in d_ju is already folded into the Fourier coefficients used in the equation b_j = -2 I_j. The only remaining comparison is between c_* * integral |f_hat(zeta)|^2 |zeta|^{-2} dzeta and the conventional inverse-Laplacian covariance (2 pi)^{-2} * integral |f_hat(zeta)|^2 |zeta|^{-2} dzeta, giving v = (2 pi)^2 c_*.
Exercise: What is a "local increment function" (used in Section 7)?
It is a measurable function of the height increments on lattice edges inside a disk K, with a fixed, arbitrarily small enlargement of K allowed for affine interpolation. If f in C_c^infty(K) has integral zero, then H_delta(f) is a local increment function: its interpolation coefficients sum to zero, and paths joining its vertices can be chosen within the enlarged disk.
Exercise: What is the test setup for Section 7 (eq:pins-insertion)?
Write X_delta = H_delta(Delta phi) for a compactly supported smooth real phi, and F_delta = prod_{j=1}^q F_{j,delta}, where each F_{j,delta} is a local increment function in a disk K_j. Suppose supp phi is contained in the interior of K_0, and that K_1, ..., K_q are disjoint from K_0 with positive clearance (they may overlap each other). Assume a uniform bound sup |F_delta| < infinity. The case q = 0 means F_delta = 1. This is the standard "one Laplace insertion times bounded local observations" pattern.
Exercise: What are boundary "windows" and why introduce them?
Fix a finite set L of horizontal levels. Require that every level meeting dD is a regular value of the y-coordinate on dD, and that no level contains a marked point. Its intersections with dD are finite. Around each intersection choose a small open boundary arc, called a window. Windows are disjoint, avoid marks, and are small enough that the portions of boundary between successive windows contain nonempty exact-value arcs. Windows are needed because the two-arc boundary condition is discontinuous at the marks: at a place where the boundary crosses a level, imposing B=+ everywhere on the window gives a "bracket" that is strictly ordered with the true g_delta.
Exercise: What are the upper and lower brackets of the height law (Lemma Boundary Brackets)?
Off windows, impose the exact Gray-code pins of g_delta. On windows, replace the exact pins: the upper bracket imposes only B = +1 on every window, and the lower bracket only B = -1. Anchor at an exact-value vertex v_* by H(v_*) = g_delta(v_*). Let mu^{+,W}_delta and mu^{-,W}_delta be the resulting laws inside D_delta. Lemma (Boundary Brackets) says:
(i) Under the upper bracket, H in {0, 1} on every window, and H = g_delta off windows;
(ii) Under the lower bracket, H in {-2, -1} on every window, and H = g_delta off windows;
(iii) Conditional on the trace, the interior is a uniform height Gibbs law with that trace;
(iv) There is a coupling of the true law mu^g_delta, mu^{+,W}_delta, and mu^{-,W}_delta with H^-_delta <= H^g_delta <= H^+_delta at every vertex, hence after affine interpolation.
Consequently both bracket means lie between -2 and 1, and smearing estimates apply to both brackets uniformly in number and size of windows.
Exercise: What is the proof idea of Lemma (Boundary Brackets)?
Look at the Gray-code table: B = + corresponds to residues 0 and 1; B = - corresponds to residues -1 and -2. So an upper bracket at a window (B = +) restricts the height at each of its vertices to the pair {0, 1}. An adjacent exact value of 0 is compatible with the {0, 1} pair, and an adjacent exact value of -1 forces the first window vertex to be 0 (so the pair remains {0, 1}). Similarly the lower bracket yields {-2, -1} on the window. Because the boundary is a connected chain whose exact-value arcs are separated by windows, and each window's pair propagates across that chain by an adjacent-pair argument, we get (i) and (ii). Gibbsianness of the interior is Lemma (Grey Code). Ordering: upper trace dominates g_delta and g_delta dominates lower trace, so by height association (constructive FKG, Fortuin-Kasteleyn-Ginibre) we get a coupled triple with the stated order.
Exercise: Why introduce brackets if the true boundary law is different?
The bracket laws are computationally tractable. The bracket identity (Proposition Bracket Identity) shows that the SAME local increment identity holds under both brackets. Then the boundary arch estimate (Theorem Boundary Arches) verifies that, in the limit, both brackets converge to the true law in the specific local sense required. This is the standard two-sided squeeze: brackets bracket the truth, and an order-preserving upgrade collapses the bracket gap in the limit.
Exercise: What is the auxiliary phase field Z_theta (eq:haar-phase) used for in Section 7?
Z_theta(v) = U_theta * exp(i theta (H(v) - H(o))),
with U_theta uniform on the unit circle, independent across sites. Because the Haar measure on the unit circle is invariant under multiplication by any unit complex number, replacing the anchor o by another vertex multiplies all Z_theta by one fixed unit complex number, leaving their joint law unchanged. The phases are used to Fourier-transform the boundary pin offsets t_i in Z (they encode the equality "t_i = t_1" as a balanced integral over the phase frequency). Proposition (Reflection Positivity) confirms that the joint height-plus-phases law still satisfies reflection positivity, stationarity, and has a positive normal transfer.
Exercise: What is the bilinear form used in the reflection argument (eq:pins-reflection-product)?
Fix a site-reflection row vartheta. On functions supported in its closed positive half-plane, put
_vartheta = E[ conj(vartheta A) * B ],
where vartheta A denotes the reflection of A. Because Reflection Positivity guarantees _vartheta >= 0 for all A in the natural domain, this is a positive semidefinite bilinear form. Quotient by the null space to get a Hilbert space. Normal translation through tau_delta = sqrt(3) * delta is then a positive self-adjoint contraction S_delta, which has a well-defined spectral representation, and S_delta^z is a positive self-adjoint operator for Re z >= 0.
Exercise: What is Lemma (Normalized Reflection Bound)?
Suppose the obstacle events P_1, ..., P_m and the support of A are strictly on one side of a reflection row. The obstacles have disjoint filled isolating neighbourhoods, and every unbounded insertion in A has a pin-free filled test neighbourhood disjoint from them. Let p_i = P(P_i) and put
V = A * prod_{i=1}^m ( indicator(P_i) / p_i ).
Assume A is either bounded or a bounded factor times one smooth zero-total height insertion in such a disk (phases allowed). Then ||V||_vartheta <= C, uniformly as delta -> 0. Uniformity holds on any compact family of arrangements preserving the stated positive separations.
Exercise: What is the proof idea of Lemma (Normalized Reflection Bound)?
Direct expansion:
||V||_vartheta^2 = E[ indicator(P and vartheta P) * conj(vartheta A) * A ] / J_delta^2,
where J_delta = prod p_i. The reflected obstacle family is separated from the original by the row margin. So the joint event P and vartheta P is a family of 2m separated obstacle events with individual probabilities p_1, ..., p_m, p_1, ..., p_m. Proposition (Pin Product) gives E[1_{P and vartheta P}] <= C J_delta^2. For bounded A this proves the claim. For one zero-total insertion Y, Condition (Obstacle Moments) gives E[ |Y vartheta Y| | P and vartheta P ] <= C, and a Cauchy-Schwarz factor cancels the squared denominator J_delta^2.
Exercise: What does the paper's figure "pin-moving" illustrate?
Three moves in succession, each across a strict projection gap: (1) move the above-band supports T up by +e and the below-band supports B down by -e; (2) move the combined right family T union R by +u and combined left family B union L by -u; (3) move combined below-family Q_- by +u'. The three projection directions are e = (0,1), u = (sqrt(3)/2, 1/2), u' = (-sqrt(3)/2, 1/2). Because e.u = 1/2 and u.u' = -1/2, each successive move separates its group from K_0 in a new direction. After all moves the moving groups are strictly above K_0 in every projection and the setup is ready for a reflection argument. The construction is schematic: distances are chosen successively, each in an open interval of allowed values.
Exercise: What is Lemma (Analytic Translation)?
Partition a finite arrangement into a low group and a high group strictly separated by projection on some lattice-reflection normal n. Factors have the allowed form (at most one unbounded zero-total insertion). Move the high group rigidly by t n, keeping the low group fixed, and normalize the plane expectation by the product of the individual obstacle probabilities:
M_delta(t) = normalized plane expectation at translation t n.
If M_delta(t_delta) -> 0 for every t in a nonempty open interval I subset (0, infinity), and lattice-rounded t_delta -> t, then M_delta(0) -> 0. Similarly if the low group is moved in direction -n. Also holds for convergent variations of the placements, preserving the clearances.
Exercise: What is the proof idea of Lemma (Analytic Translation)?
Back off the high group by eta in (-n) direction (chosen so that the projection gap stays positive). Then choose a site-reflection row in the gap, and get a positive-side Hilbert vector A_delta from the low group, and a normalized high-group factor B_delta. Uniform reflection-norm bounds (Lemma Normalized Reflection) give ||A_delta||_vartheta, ||B_delta||_vartheta <= C. Holomorphic interpolation:
f_delta(z) = ,
which is analytic in Re z > -eta_delta. At lattice parameters t = m * tau_delta >= 0, the exponent is a positive integer, so f_delta(t) = M_delta(t). Uniform boundedness in Re z > -eta/2 makes Cauchy's estimate uniform; passing to a local uniform limit along a subsequence gives a holomorphic f that vanishes on the open interval I, hence (identity theorem) on the whole common half-plane, hence f_delta(0) -> 0. Successive application removes one translation direction at a time.
Exercise: What are the three normal directions in (eq:three-normals) and why use all three?
e = (0, 1), u = (sqrt(3)/2, 1/2), u' = (-sqrt(3)/2, 1/2).
These are the outward normals to the three distinct lattice-reflection directions (the three directions of rows of sites). No single reflection direction has enough clearance to move an arbitrary set of boundary obstacles past K_0: two may be separated by a projection that is close to zero in one frame, or by a piece of boundary crossing one frame unfavourably. Using all three lets the proof pick, at each successive step, an unambiguous projection gap: T above and B below first in e, then T union R above and B union L below in u, then Q_- can be moved in u'. Because e.u = 1/2 and u.u' = -1/2, the moves compose cleanly.
Exercise: What is Proposition (Gapped Identity)?
Fix windows, gaps, and tests satisfying the paper's clearances. Let P = intersection of P_i and J_delta = prod P(P_i). For any product Psi_delta of unit-modulus phase factors at the obstacle representatives v_i:
E[ X_delta * F_delta * Psi_delta * 1_P ] / J_delta -> 0.
Same holds with P replaced by P and E, where E is the event that all offsets t_i are equal:
E[ X_delta * F_delta * 1_{P and E} ] / J_delta -> 0.
The result holds for every convergent sequence of lattice-rounded placements with the stated strict separations.
Exercise: What is the proof idea of Proposition (Gapped Identity)?
Case A: phases only. Move all pins and spectators strictly above K_0 using the three-direction schedule. At the final placement, put a reflection row between K_0 and all other supports. Reflection Cauchy-Schwarz plus Lemma (Normalized Reflection) bound the normalized expectation by C * ||X_delta||_vartheta. By Corollary (Reflection Null) the right side tends to 0. Use Lemma (Analytic Translation) successively to remove each translation move, going backwards to the original placement.
Case B: equal offsets. On P,
D_i = (H(v_i) - g_i - H(v_1) + g_1) / 4 = t_i - t_1 in Z.
Write 1_E as a Fourier integral over s in [0, 2 pi]^{m-1}:
e^{i sum s_i D_i} = prod_{i>=2} ( e^{-i s_i (g_i - g_1)/4} * Z_{s_i/4}^{(i)}(v_i) * conj( Z_{s_i/4}^{(i)}(v_1) ) ),
using an independent phase field for each i. The Haar factors cancel with their conjugates, giving an identity of functions of the original increments. Fourier orthogonality then expresses 1_E as (2 pi)^{-(m-1)} * integral over s of e^{i sum s_i D_i} ds. Apply Case A for each fixed s, integrate, use dominated convergence (uniform in s and delta) to conclude.
Exercise: What is Lemma (Gap Filling)?
Fix a small gap size a around each selected intersection. Let mu_{delta,a} be the plane law conditioned on the gapped pins. Let mu^=_{delta,a} = mu_{delta,a}(. | E) when E has positive probability. Let mu_{delta,0} be the filled bracket law. Then for some beta > 0 and a function epsilon(a) -> 0 as a -> 0:
(i) limsup_delta mu_{delta,a}(E^c) <= epsilon(a);
(ii) limsup_delta TV(mu^=_{delta,a}, mu_{delta,0}) restricted to joint increment information in the test disks <= epsilon(a);
(iii) uniform p-th moments of X_delta under both mu^=_{delta,a} and mu_{delta,0}, in the sense of a bound on limsup_delta independent of a.
Exercise: What is the proof idea of Lemma (Gap Filling)?
Compute E^c. Around each gap there is a buffer annulus from scale C*a to scale r, containing only favorable B pins of the window sign. Lemma (Buffer Circuits) gives a double circuit of that sign around the gap, except with probability C (a/r)^beta + o_delta(1). Crossings of the two adjoining pinned arms both have the window sign at their incident primal sites. Consecutive double-circuit edges form a connected chain of same-sign vertices, so the same lifted pair {residue, residue + 1} is forced on both arms, hence the offsets t_i agree. The gap adjacency graph (arcs around the boundary circle) is connected, so a union bound over the fixed number of gaps gives (i). For (ii), couple the gapped law and the filled law by successively filling one gap: the B marginals before and after are ordered by Lemma (Spin FKG), the stopping coupling (Lemma Double Cut) is run outwards from the gap, and the more adverse law's double circuit becomes a common circuit before the search leaves the window neighbourhood with probability C (a/r)^beta + o_delta(1). A label-energy comparison at fixed 2^{-M} handles the W-plus labels. For (iii), a fixed annular buffer keeps the pin-free setup required by Corollary (Equal-Offset Moments).
Exercise: What is Proposition (Bracket Identity)?
Fix a window system mathcal W for a finite grid mathcal L. Take either the upper bracket mu^{+,mathcal W} or the lower bracket mu^{-,mathcal W}. Let K_0, ..., K_q in the interior of D be disks with K_0 at positive distance from every K_j, j >= 1. Let X_delta and F_delta be as in the Section 7 setup. Suppose two levels ell_- < ell_+ in mathcal L satisfy:
(i) K_0 is strictly between the two levels;
(ii) every level avoids every spectator disk K_j with positive margin;
(iii) after opening sufficiently small gaps, every pin arc and spectator disk in the middle band belongs to one of the two strictly separated u-projection families L, R of (eq:middle-projections).
Then
lim_delta E_{mu^{+,mathcal W}_delta} [ H_delta(Delta phi) * F_delta ] = 0,
and similarly for mu^{-,mathcal W}_delta. In particular the identity holds for products of bounded continuous functions of finitely many smooth zero-total tests in each K_j, and passes to any subsequential distributional limit of the bracket laws. The geometric hypotheses can be arranged locally for any fixed bound q on the number of spectator disks.
Exercise: How does Proposition (Bracket Identity) follow from the gapped identity and gap filling?
At fixed gaps, Proposition (Pin Product) gives a positive lower bound on P(P) / J_delta. Lemma (Gap Filling) gives P(E | P) >= 1/2 at small mesh. Dividing the gapped equal-offset identity by P(P) / J_delta yields a limit 0 for E_{mu^=_{delta,a}}[X_delta F_delta]. Truncate X_delta at scale T. The total variation bound of Lemma (Gap Filling) then compares the equal-offset law and the filled law on any bounded function of joint test increments, with error 2 * T * epsilon(a). The truncation errors are of order T^{1-p} times a p-moment bound. Let a -> 0 first and T -> infinity second, giving the bracket identity.
Exercise: What is a "region" and a "usable boundary" in Section 8?
A skeleton is a finite connected union of lattice edges. The free sites of a region lie in bounded faces of its skeleton. Every neighbor of a free site is either free or a fixed skeleton site; the fixed skeleton sites adjacent to free sites are the usable boundary. Conditional on values at the usable boundary, the interior is uniform among the admissible integer Lipschitz extensions. Initially the skeleton S_0 is the polygon boundary (dD_delta). Later it may be S_0 union finitely many extra simple lattice circuits meeting it, cutting off tiny boundary patches. Usable boundary sites are always real; only the "outside the region" part is declared artificially open. At threshold a the open sites are
{free or usable v : H(v) >= a} union {v strictly outside and not usable}.
The choice of which points are artificial is a technical trick: it lets us join an actual high path to the exterior through "nothing", without perturbing the actual height on any real site.
Exercise: What is a Q_r box and an "arch separator"?
At a boundary point p, work in tangent-inward-normal coordinates, so the boundary is locally horizontal and the interior is above. The box Q_r is p + [-r, r]^2. An open (at threshold a) site circuit surrounding Q_r is an arch separator: the part of the circuit in the domain joins the "clean boundary" on the two sides of Q_r, and the remaining part of the circuit can run through artificially open sites to close up. The point of an arch is that it separates the free sites near p from the remote boundary data: on the "outer" side of the arch, everything is fixed; on the "inner" side, the height is a Gibbs interior.
Exercise: What is the synthetic two-interval construction?
Given k in Z and a connected skeleton S, choose one of the two Gray colours, call it t, and one of its signs, s. Prescribe sign s on the selected "high" connected portions of S and sign -s on the rest ("low" portions). Require the complementary colour v = c_k at a bounded number b of transition edges, where the transition is the local step k-2 <-> k-1. Anchor a lift at one transition. Then, inside the free region, any admissible synthetic height K satisfies
K in {k-1, k} on the high portions,
K in {k-3, k-2} on the low portions.
This is a genuine height configuration once you sum the Gray-code increments; it is called "synthetic" because it artificially prescribes a two-spin pattern that need not match any pinning pattern of the actual model. Conditional on t, the W-labels on components of t-changing edges are independent fair coins; requiring b labels costs at most the factor 2^{-b}. Height comparison (Fortuin-Kasteleyn-Ginibre monotone coupling) gives K(v) <= H(v) on free sites, provided the actual trace has eta >= k on the high portions and eta >= k-2 everywhere else.
Exercise: What is Theorem (Boundary Arches)?
There are M > 1, c > 0, C > 0 and a geometric tolerance with the following property. Fix a positive integer N and scales
16 r < ell_1 < ... < ell_N < R/10, ell_{i+1} >= M ell_i.
Suppose that, in Q_R, the limiting boundary is a single graph y = f(x) with f(0) = f'(0) = 0, ||f'||_infty smaller than the geometric tolerance, interior at positive normal side, and the polygon boundary is a sufficiently fine uniform parametric approximation of this arc with the same orientation, with approximation error divided by ell_1 tending to zero. Allow a fixed finite number of boundary-attached patch replacements, all sufficiently small. Suppose the actual usable boundary data obey
eta >= k-2 everywhere,
eta >= k on the usable boundary in Q_R minus Q_r,
with k in Z the target integer threshold. Then on sufficiently fine mesh,
P( an arch at threshold k surrounds Q_r in Q_R ) >= 1 - C e^{-c N}.
The constants do not depend on the boundary heights, the number of boundary vertices, or the number of tiny replacements. There is a reversed statement with eta <= k+2, clean data eta <= k, and separator at height at most k.
Exercise: What is Lemma (Many Preliminary Pairs)?
Under the hypotheses of Theorem (Boundary Arches), with probability at least 1 - C_0 e^{-c_0 N}, at least c_0 N of the selected scales have both an open site circuit at threshold k-1 in Q_{ell_i} minus Q_{ell_i/2} and an open site circuit at threshold k-1 in Q_{7 ell_i} minus Q_{5 ell_i}, each surrounding its inner box. IDEA. Use the anchored two-interval comparison, with the high interval [k-1, k] on the two clean portions of the base arc running from just inside the innermost comparison strip to beyond the outermost strip, and the low interval elsewhere. The transition buffers are disjoint from all the strips and their surrounding buffers. Height comparison gives the synthetic K <= H. Then, before imposing complementary labels, every buffer pin is favourable for a t=s double circuit, so Lemma (Buffer Circuits) gives probability >= a > 0 of a double circuit in a slightly narrower strip. Choose M large enough that the buffers of successive scales are disjoint. Conditional on each preceding outcome, the probability of a fresh success is still >= a > 0. For independent success indicators I_i with these conditional probabilities, E exp(-lambda sum I_i) <= (1 - a + a e^{-lambda})^N, so sum I_i >= a N/2 except with probability C e^{-c N}. Adding at most four complementary labels costs a factor 2^4. Each double circuit that meets the high part of the skeleton yields an actual surrounding site circuit at the anchored pair {k-1, k} of the synthetic height. The event "at least a given number of pairs exist" is increasing in the height, so height comparison transfers its probability to the actual height law.
Exercise: What is Lemma (Extremal Circuit Stopping)?
In a deterministic annular strip, good surrounding simple site circuits have inner and outer envelopes: for a given successful strip, there is a unique innermost good simple circuit and a unique outermost good simple circuit. Each can be selected by revealing the strip from one side: the inner by revealing from the inner side, the outer from the outer side. In both cases, the selection and the revealed values are measurable with respect to the sites on the circuit and on the tested side. No height on the untested side is conditioned by the selection. Consequence used later: in each successful preliminary block, we select an innermost C_i^- from the inner strip and an outermost C_i^+ from the outer strip, and conditional on these selections and the exterior configuration, the intervening free regions have independent Gibbs interiors with their revealed boundary values.
Exercise: What is the proof idea of Lemma (Extremal Stopping)?
If two surrounding simple circuits share at least two vertices, their union is two-vertex-connected: after deleting any single vertex, both remaining pieces stay connected and still share a vertex. The inner boundary of that union and the outer boundary are both simple cycles. So the inner envelope and the outer envelope are well-defined simple circuits surrounded by good vertices and staying inside the strip. If the two circuits share only one vertex, they cannot cross just once (Jordan curve), and since both surround the inner box, their interiors are nested. Iterate this envelope operation on the finite strip: a unique innermost and unique outermost good circuit exist. The selection is measurable from one side because "circuit C is the innermost good circuit" is determined by whether a better good circuit exists in the interior of C; a better one would have to have an inner envelope with C that is itself strictly better and lies entirely on the inner side. The outer version is identical with directions reversed.
Exercise: What is Lemma (Boundary Attachment)?
Consider a retained middle problem at scale ell. Fix a boundary location near (-2 ell, 0) or near (2 ell, 0). Choose fixed constants 0 < d/W_0 << 1 and 0 < W_0/ell << 1. If the local boundary and any tiny replacements are within a sufficiently small fraction of d of the horizontal boundary line, then, with probability at least a_0 > 0 (uniform in the exposed boundary values of the middle problem), there is a path of sites at height at least k that enters from the clean boundary band and reaches depth d inside the box of half-width W_0. Its boundary end is connected via good fixed or artificial sites to the exterior below, within a slight enlargement of the box.
Exercise: What is the proof idea of Lemma (Boundary Attachment)?
Work near the origin. Choose W_0 as a small fraction of ell, then d much smaller than W_0. The two preliminary circuits are outside the local box. A site x at (0, 3d) is free in that face.
SYNTHETIC MEAN. Construct the two-interval synthetic height K with high interval [k-1, k] on a base subarc of width roughly W_0/4 near the origin and low interval [k-3, k-2] elsewhere, with at most two transition labels. Height comparison gives K <= H. Let A be the event that a primal path of heights >= k joins the high patch to depth d inside the attachment box. A is increasing, so we only need P(A) > 0 in the synthetic law.
SYMMETRY. Annular buffers between radii d and a small fraction of W_0 contain only favourable t=s pins. The stack estimate gives a double t=s circuit around x that meets the high patch except with probability O(e(d/W_0)). On that circuit, complementary labels in the inner face are all free, so flipping them flips K -> 2k - 1 - K inside. Conditional mean at x is exactly k - 1/2.
FAILURE BOUND. Uniform bound on |E[K(x)] - k| <= C_1, using the component-potential decomposition.
LOW BARRIERS. Around horizontal offsets +/- 2 W_0/3, put annuli whose pins favour t=-s. With failure probability O(e(d/W_0)) each, those annuli contain double t=-s circuits meeting the low skeleton and reaching above depth 3d. Let L be the event that both barriers exist. They have K <= k-2 in the synthetic problem.
DETECT THE SIDE. Condition on the synthetic skeleton trace and follow every dual level path at threshold k-1/2 that enters through a skeleton edge. The open paths cut the bounded face into cells, each labelled high or low by the threshold sign. Let U be the event that x lies in a high cell. Comparison with a constant ceiling gives E[K(x) | trace] <= k - 1 + P(U | trace), so P(U) >= 1/2 - O(e(d/W_0)).
BELL. If L occurs but A does not, a high lip from either endpoint on the high patch cannot cross a low barrier (heights k vs k-2), and cannot reach depth d without giving A. So every open path stays between the barriers and above depth 2d, which puts x in the low cell, contradicting U. Hence P(A) >= 1/2 - O(e(d/W_0)); choose d/W_0 so this is >= 1/4. A is increasing, so height comparison transfers positivity to the actual law.
Exercise: What is Lemma (Arch Upgrade)?
For each successful preliminary block, conditional on the extremal-stopping sigma-field, the probability of an arch separator at threshold k in its untouched middle is at least an absolute constant a_1 > 0. The "middle" is the untouched region Q_{5 ell_i} minus Q_{ell_i} between the two selected preliminary circuits. IDEA. Two attachments at horizontal positions about +/-2 ell_i each have uniform positive probability by Lemma (Boundary Attachment); those events are increasing in the actual height. Bulk path estimates from below come from a constant-boundary-k-2 comparison, in which the height contours are closed loops and the loop signs are independent fair signs; applying Lemma (Buffer Circuits) at four concentric buffers gives four nested enclosing unoriented loops with positive probability. Conditional on the loop geometry, the height just outside the first chosen loop is k-2 plus a sum of other fair loop signs; with probability at least 2^{-5} all four chosen signs are positive and the outer height is k-2 or above. So the inner face path of the fourth loop is at least k. Height positive association then multiplies the attachment events with the bulk events to get a_1 > 0. The actual arch is built by tubes: a horizontal crossbar near depth d/2 joins the left and right attachments; tubes at horizontal +/- 2 ell connect up and down; a bottom tube and the local boundary close the circuit. The whole construction stays strictly between Q_{ell} and Q_{5 ell}. Every artificial site used here was already outside the original region or in a removed boundary patch.
Exercise: What is Corollary (Boundary Robustness)?
The arch estimate remains valid conditional on any exterior sigma-field, whenever the remaining free heights have a uniform Gibbs law and their actual usable boundary data obey the two-sided bounds. It also remains valid after a fixed finite number of boundary-attached patches have been removed, provided the same data bounds hold on the new usable boundary. For a prescribed error, first fix finitely many arch scales and their comparison boxes; then make the replacement diameters, boundary approximation errors, and mesh sufficiently small. The success constants of the resulting blocks have no factor depending on the number of microscopic pins or the fixed number of replacements.
Exercise: What does the paper's figure "boundary-arches" show?
The picture is drawn in tangent coordinates at the boundary point p. The four labelled squares (Q_{ell_i/2}, Q_{ell_i}, Q_{5 ell_i}, Q_{7 ell_i}) determine the two searched strips between successive square boundaries. The attachment boxes (near horizontal +/- 2 ell_i) and the path widths are enlarged for legibility. Two shaded strips contain the preliminary separators; their facing sides leave the retained middle Gibbs region unsearched. The blue route uses the two attachments, horizontal crossbars near depth d/2, and bulk paths at positive clearance from the boundary in the actual height law. The right-side panel is a schematic of the synthetic low-barrier argument used to prove one attachment: a central high patch, two low lateral barriers, and a bulk path that must be in the middle cell or contradict U. That synthetic picture is only used inside the proof; the low barriers are not imposed in the actual law. The dashed lower route uses the artificial exterior. All displayed paths are corridors, not a prescribed realization of the random separator.
Exercise: Why is the arch estimate the missing piece in the boundary squeeze?
The bracket laws sandwitched the true law between H^- and H^+, but the sandwich was extremely loose: it allowed every boundary window to take the extreme values +1 or -2. The arch estimate says that, locally near a clean boundary arc of value k, the true law (and the bracket laws) actually looks like a clean height-k boundary a fixed fraction of the time. This is enough to show: (i) the "bracket means" converge to the harmonic function m; (ii) the brackets themselves converge to the true law; (iii) local absolute-height identities hold at the true law, not just at the brackets.
Exercise: What is Proposition (Tightness and Convergence of Moments)?
The true fields and all bracket fields are tight in H^{-t}_loc(D) for every t > 1. For K contained in D, p < infinity, and tests supported in K,
sup_delta ||H_delta(f)||_{L^p} <= C_{K,p} * ||f||_infty.
The constants are uniform over bracket laws. Along any weak subsequence, all finite joint test moments converge. Same for any fixed finite collection of fields in a coupling.
Exercise: What is the tightness proof structure?
Cover K by finitely many balls whose radii are a small fixed fraction of their distance to dD. On each ball, the smearing estimate (Proposition Smearing Moments) gives the L^p bound of the local statistic; a partition of unity sums them. For the Sobolev statement, place a cutoff chi in a fixed square (regarded as a flat torus), use the normalized Fourier basis e_n, and note that ||chi * e_n||_infty is bounded independently of n. Then
sup_delta E ||chi H_delta||_{H^{-s}}^2 <= sum_n (1+|n|^2)^{-s} sup_delta E |H_delta(chi e_n)|^2 < infinity,
which is (eq:sobolev-bound). For s > 1, the embedding H^{-s} -> H^{-t} (t > s) is compact (its Fourier multiplier tends to zero; truncation gives finite-rank approximations). So tightness in H^{-t} follows. A nested countable family of cutoffs on an exhaustion gives tightness in H^{-t}_loc(D), hence in distributions. Moment convergence uses H"older's inequality at order p(1+eta) to give uniform integrability of products, and the same moment inequalities pass to limits by truncation.
Exercise: What is Proposition (Boundary Trace and Mixed Boundary Moments)?
Both limiting bracket means are the harmonic extension m of g. Wait: here m is the harmonic extension of g, i.e. the "integer height" version. In terms of H = (h-1)/2, g = 0 on A+, g = -1 on A-. So m is the (unique bounded) harmonic function with m = 0 on A+ and m = -1 on A-.
Consequences:
(i) H^- = H = H^+ almost surely in the ordered coupling.
(ii) E[H] = m.
(iii) MIXED boundary trace: for p on an open boundary arc on which g = k (k is 0 or -1), x -> p from within D, and Y_x = H(rho_{x, kappa d(x)}), if F is a product of a fixed finite number of height averages supported away from a neighbourhood of p, then E[(Y_x - k) * F] -> 0. The same holds when those averages approach finitely many other distinct nonmarked boundary points, with their radii a fixed small fraction of their depths (points and count fixed before depths shrink).
Exercise: What is the proof idea of Proposition (Boundary Trace)?
Two searches, done first at finite mesh with a fixed grid and windows.
CENSOR: around each window, choose a small clean neighbourhood in the constant arc. Its values are all at least k_i - 2, and those in the censor-neighbourhood off the window are at least k_i. Around each windowed exceptional piece, use Theorem (Boundary Arches) at level k_i to produce an arch separator; discard the windowed patch; retain the part away from the window, which is untouched (so Gibbs). The new usable separator sites have height >= k_i, so we have upgraded the window neighbourhood to a clean boundary piece that is at least k_i.
FINAL ARCH: in the retained clean region, apply corollary (Boundary Robustness) to get an outermost arch at level k in a smaller neighbourhood of p, chosen from the remote side. The untouched region now contains the testing ball and has usable boundary at height >= k. Height comparison gives conditional mean at least k for Y_x. The symmetric construction (using censor and final arches in the decreasing direction) gives the matching upper bound.
The H"older bound on the failure event controls the loss from failure. Then bounded harmonic uniqueness with the two marks (harmonic measure of shrinking neighbourhoods tends to zero) gives m^- = m^+ = m, and the order H^- <= H^+ with equal expectations gives H^- = H^+ a.s.
For the mixed assertion, no censoring is needed: on the lower good event E_-, choose the outermost lower arch from the exterior; on E_+, choose the outermost upper arch. The estimates pass to expectations with H"older weighting; then tightness and truncation give the limit form.
Exercise: What is Lemma (Deterministic Boundary Reference Averages)?
Fix a compact subarc of an open constant boundary arc, with value k. Then there exist finite averages of comparable-depth unit smearings near that subarc which converge to k in every finite L^p of the true limiting field. Concretely: choose N distinct points p_1, ..., p_N in the subarc, let x_i(t) approach p_i normally at depth t, and put
A_{N,t} = (1/N) * sum_i H(rho_{x_i(t), kappa d(x_i(t))}).
For all small t the balls are disjoint. Then
limsup_{t -> 0} E |A_{N,t} - k|^2 <= C/N,
sup_{N,t} ||A_{N,t} - k||_{L^q} <= C_q.
Consequently lim_{N -> infinity} limsup_{t -> 0} ||A_{N,t} - k||_{L^p} = 0 for every finite p. This is what pins the additive constant of H: near any clean boundary arc of value k, a large finite average of nearby boundary values must converge to k.
Exercise: What is the proof idea of Lemma (Boundary References)?
Each Y_i(t) = H(rho_{x_i(t), kappa d(x_i(t))}) has bounded second moment (Prop Smearing Moments). For i != l, apply the mixed boundary trace with exterior factor Y_l(t) - k to get E[(Y_i(t) - k)(Y_l(t) - k)] -> 0. Expanding the square of the average and passing to the depth limit at fixed N gives the first estimate. Minkowski in L^q gives the second. A diagonal choice of finite depths as N increases gives the claimed reference sequence. We have NOT asserted that one comparable-depth average, or a fixed finite collection, becomes deterministic; only that a large averaging does.
Exercise: What is Proposition (Moment Harmonicity)?
Define distributional moments T_j on D^j by
= E[prod_i H(f_i)], T_0 = 1.
Then T_j is a smooth function on the configuration space C_j = {(x_1, ..., x_j) in D^j : x_i != x_l for i != l}, and it is harmonic in each variable separately: Delta_{x_i} T_j = 0 on C_j.
Exercise: What is the proof idea of Proposition (Moment Harmonicity)?
Start from the true local increment identity E[H(Delta phi) F] = 0 (which came out of the bracket identity combined with the boundary trace). Approximate each spectator H(f_i) as
H(f_i) = c_i k + H(f_i - c_i rho_i) + (c_i / N) sum_{n,l} H(rho_{i,n,l} - rho_{i,n,l+1}) + c_i (A_{N,t} - k),
where rho_{i,n,l} is a chain of unit-mass smooth averages from the observation point rho_i to the boundary reference point x_n(t). The nonconstant terms have zero total integral and are supported in small disks along the chain; the final reference term tends to zero in every L^p. Expand the product; each term has at most j-1 disks and satisfies the bracket identity by choosing the level separations far from those disks. So each finite-approximation term contributes zero expectation. Then pass to the reference limit, and apply distributional harmonicity of the sum. Elliptic regularity upgrades distributional harmonicity to smoothness on C_j.
Exercise: What is Lemma (No Distributions Supported on Collision Diagonals)?
The smooth function T_j on C_j is locally integrable on D^j and represents its full moment distribution: there is no additional contribution supported on the collision diagonals. For j >= 2, K in D, and distinct x_1, ..., x_j in K with d_* = min_{i dD, including at the two marks. At a nonmarked point p of an arc of value k,
lim_{x -> p} T_j(x, y_2, ..., y_j) = k * T_{j-1}(y_2, ..., y_j).
Exercise: What is the proof idea of Lemma (No-Contact)?
Radially mollify every variable at the same radius r, and denote the result by T_j^{(r)}. For centers in K, and r small, Holder plus smearing bounds give
|T_j^{(r)}(x_1, ..., x_j)| <= C_{K,j} * (1 + |log r|)^{j/2}.
When d_* > 2r, the balls of the mollification are disjoint, and repeated mean-value property says T_j^{(r)} = T_j EXACTLY on that region. Taking r to be a fixed small fraction of d_* gives the off-diagonal bound. The function (1 + log_+(1/d_*))^{j/2} is integrable on K^j in two dimensions: it is bounded by (1 + sum of logs of |x_i - x_l|)^{j/2}, and powers of logs are integrable. The collision neighbourhood {d_* <= 2r} has volume O(r^2); the mollified integral there tends to 0; and the off-diagonal integral there tends to 0 by integrability. So T_j^{(r)} -> T_j in L^1 on compact sets, and since it also converges distributionally to T_j, the two are the same distribution. Boundary bound uses mixed boundary trace with the partition argument on disjoint balls and mean-value. Boundary value at a clean arc applies the mixed boundary trace with F = prod Y_i and mean-value once more.
Exercise: What is Lemma (Log Singularity)?
Fix a collision-free list (y, z) of interior points. In a neighbourhood of x = y that contains none of the points of z,
T_j(x, y, z) = L(y, z) * log(1/|x - y|) + R(x; y, z),
where R extends harmonically across x = y in the first variable. L is smooth on the parameter space (y, z). On compact subsets there is a uniform radius on which R and its first-variable derivatives on a smaller disk are uniformly bounded.
Exercise: What is the proof idea of Lemma (Log Singularity)?
By Moment Harmonicity and No-Contact, x -> T_j(x, y, z) is a harmonic function on a punctured disk around y, with at most polylog growth at the puncture. The polar Fourier expansion of a harmonic function on an annulus around y has: a constant angular part a + b log r, and, in each angular mode n >= 1, radial factors r^n and r^{-n}. A coefficient of r^{-n} would grow faster than any power of log(1/r) as r -> 0, which contradicts the polylog growth bound. So all negative coefficients vanish; the remaining positive modes extend harmonically to the centre. This yields the log + smooth decomposition.
For uniform control, use the parameter formula
L(y, z) = -(1/(2 pi)) * integral_{|w - y| = a} partial_{n_w} T_j(w, y, z) ds_w,
obtained by outward differentiation in the first variable. The integral of the derivative of the removable remainder is zero, and the integral of log(1/|w - y|) is -2 pi. Parametrize w = y + a e^{i theta}; the integrand is smooth in all parameters on a collision-free compact set. Subtract the log term and use the Poisson formula to get R, with uniform derivative bounds. This gives all the smoothness and uniformity assertions.
The comparison-with-plane technique used in the proof is close to Kenyon's identification of dimer-height moments via collision poles:
Kenyon 2001, Dominos and the Gaussian Free Field: DOI 10.1214/aop/1015345599, arXiv:math-ph/0002027v1.
And the harmonicity + collision asymptotic method is a predecessor in the full-plane six-vertex proof:
Duminil-Copin-Kozlowski-Lammers-Manolescu 2026, Gaussian free field convergence of the six-vertex model with -1 <= Delta <= -1/2: arXiv:2603.06268v1 (see Section 7.2).
Exercise: What is "stopped comparison with the plane"?
It is the coupling device that lets us compare a local collision statistic with the plane law. Fix an interior center y_0 and fixed spectator balls away from y_0. For two smooth zero-integral tests alpha, beta in disjoint bounded disks centred at y_0, put
alpha_r(x) = r^{-2} * alpha((x - y_0) / r), beta_r(x) = r^{-2} * beta((x - y_0) / r).
Their supports lie in B(y_0, C r). For any product F_delta of height averages on the spectator balls, and with H_pl denoting a plane copy,
E[H_delta(alpha_r) H_delta(beta_r) F_delta] - E_plane[H_pl(alpha_r) H_pl(beta_r)] * E[F_delta]
-> 0 as r -> 0 (mesh limit first at every fixed r).
Exercise: How is stopped comparison proved?
Choose a fixed disk B(y_0, R_0) inside D disjoint from the spectators. In the spin representation, the boundary pin is a W-constancy along one connected boundary path; imposing it and then its plus label leaves the B marginal unchanged. The double-cut stack estimate (as in Proposition Plane Mixing) applies in the pin-free interior disk. Search from the outside inward for a double-B circuit between radii r^{1/2} and R_0; failure probability at most a positive power of r. On success the circuit surrounds a disk of radius comparable to r^{1/2} and the stopping exploration has not examined its interior. Reveal the absolute heights at spectator sites via paths that avoid the searched disk (they exist after shrinking R_0). Then the edge-increment formula summed from the fixed boundary anchor gives those heights using only exterior spins. Conditional on the exterior, the interior is a free-W double-cut law. Couple it to a plane copy with the standard worst-boundary coupling, again with failure probability at most a positive power of r. On the matching event, the two zero-total tests agree; the sum-zero removes the additive constant. All estimates pass to unbounded spectator products by Holder: L^q bounds of the two local smearings times the failure indicator give an overall O(r^{gamma/4} * (1 + |log r|)) error.
Exercise: What is Proposition (Collision Coefficient)?
For j >= 2 and distinct (y, z),
L(y, z) = (v / (2 pi)) * T_{j-2}(z).
Convention T_0 = 1. No centering is made in this identity.
Exercise: What is the proof idea of Proposition (Collision Coefficient)?
Pick two tests alpha, beta so that J = integral integral alpha(u) beta(w) log(1/|u - w|) du dw is nonzero. (Take the first-coordinate derivatives of two small bumps on a horizontal line, with small support: the mixed derivative of log(1/|.|) is nonzero and keeps its sign.) Fix disjoint radial unit mollifiers rho_z at the points of z; F = product of resulting height averages, or F = 1 when j = 2. Separate harmonicity gives E[F] = T_{j-2}(z). By the plane covariance theorem, the plane limit on the RHS of the stopped comparison equals (v J)/(2 pi). Combined with E[F] = T_{j-2}(z), we get
lim_{r -> 0} E[H(alpha_r) H(beta_r) F] = (v J / (2 pi)) T_{j-2}(z).
From the parameter form of Lemma (Log Singularity), the same LHS can be evaluated as a double integral with the explicit form
L(y_0 + r w, z) * (log(1/r) + log(1/|u - w|)) + R(y_0 + r u; y_0 + r w, z).
The log(1/r) coefficient integrates to zero against alpha (since alpha has integral zero). By continuity of L, the remaining log integral converges to L(y_0, z) * J. The remainder integrates to O(r) (use the uniform first-derivative bound and subtract a reference point). So the LHS limit is L(y_0, z) * J. Equating gives L(y_0, z) * J = (v J / (2 pi)) * T_{j-2}(z). Divide by J.
Exercise: What is Theorem (Gaussian Moment Recursion)?
For distinct points (x, y_1, ..., y_{j-1}),
T_j(x, y) = m(x) * T_{j-1}(y) + v * sum_{i=1}^{j-1} G_D(x, y_i) * T_{j-2}(y minus y_i),
where G_D(x, y) = (1/(2 pi)) log(1/|x - y|) + (harmonic in y). For j = 1 the assertion is T_1 = m. As locally integrable distributions, these T_j are the moments of
m + sqrt(v) * Phi_D,
where Phi_D is the zero-boundary GFF on D with covariance G_D. Consequently the limiting field H has that law.
Exercise: What is the proof idea of Theorem (Gaussian Moments)?
Fix distinct spectator points y. Define
U(x) = T_j(x, y) - [m(x) T_{j-1}(y) + v sum_i G_D(x, y_i) T_{j-2}(y minus y_i)].
Away from the y_i's, U is harmonic in x by Moment Harmonicity. At each y_i the log coefficient cancels by Proposition (Collision Coefficient). By Lemma (Log Singularity) the harmonic extension across y_i is allowed. So U is harmonic on all of D. Boundary control: T_j bounded (No-Contact), the mean m bounded, and the Green terms tend to 0 at the boundary. On each open arc of value k (0 or -1), the boundary trace lemma gives U -> 0. Bounded harmonic uniqueness with the two exceptional marks (compare U to the potential of a shrinking neighbourhood around each mark) forces U = 0. So the recursion holds everywhere.
Starting from T_0 = 1 the recursion gives the partial pairing formula
T_j(x_1, ..., x_j) = sum_{pairings P} prod_{pairs {a,b} in P} v G_D(x_a, x_b) * prod_{c not in union P} m(x_c).
This is precisely the uncentered multivariate Gaussian moment formula for m + sqrt(v) Phi_D. Each displayed product is locally integrable on D^j. Agreement off the collision diagonals plus the no-diagonal-content lemma gives agreement of the full moment distributions.
Exercise: What is the determinacy step?
For a real test f, let Z = H(f) - integral m f and put s^2 = v * integral integral f(x) G_D(x,y) f(y) dx dy. The moment formula gives E Z^{2n} = (2n - 1)!! s^{2n} and all odd moments zero. If s = 0 then Z = 0 a.s. Otherwise Cauchy-Schwarz (or Gaussian-style moment bounds) gives E |Z|^n <= s^n (2n)^{n/2}, so sum_n |t|^n E|Z|^n / n! converges for every real t. The exponential and characteristic-function series integrate term by term, giving E[e^{i t Z}] = e^{-s^2 t^2 / 2}. That is the characteristic function of a centered Gaussian with variance s^2. Extend to linear combinations and invoke Cramer-Wold. A countable dense collection of tests determines the Borel law in the local Sobolev space (Prop Tightness). So the random distribution itself has the asserted Gaussian law.
Exercise: What is Corollary (Field Limit)?
For every t > 1,
H_delta => m + sqrt(v) Phi_D in H^{-t}_loc(D),
H_delta - E[H_delta] => sqrt(v) Phi_D in H^{-t}_loc(D).
All finite joint test moments converge. In particular
(h_delta - E[h_delta]) / sigma => Phi_D in H^{-t}_loc(D),
with sigma = 2 sqrt(v) = 4 pi sqrt(c_*) > 0. sigma has the absolutely convergent finite-volume expression of Proposition (Normalization Series) and is independent of D, a, b. The uncentered odd-height limit has mean 2 m + 1, the harmonic extension of its prescribed values +1 on A+ and -1 on A-.
Exercise: What is the proof idea of Corollary (Field Limit)?
Every subsequence admits a further limit by Tightness, and Theorem (Gaussian Moments) identifies every such limit with the same law. Tightness + uniqueness gives full convergence. Uniform integrability (from the L^p bounds) gives convergence of joint moments. That yields E[H_delta(f)] -> integral m f for each smooth compactly supported f. There is also deterministic local Sobolev convergence of the means (Fourier expansion plus dominated convergence on summable Sobolev weights), so we can subtract the actual discrete expectation in H^{-t}. Affine interpolation preserves h_delta = 2 H_delta + 1 exactly; centering removes the constant +1 term, and the multiplication by 2 gives the factor sigma = 2 sqrt(v).
Exercise: What is the connection between this paper and its two companion papers?
The paper explicitly separates itself from two companion results. (i) OpenAIBalancedSixVertex2026 handles the balanced six-vertex model at a=b=1 with c in [sqrt(3), 2] on the square lattice; the plane state is a GFF limit with variance multiplier 1/arcsin(c/2). Its plane-state analysis does not identify the two-arc triangular-lattice law or its normalization. (ii) OpenAIWeighted2026 handles weighted integer Lipschitz heights with zero boundary; it proves the GFF limit together with moment estimates for each fixed edge weight in [1/sqrt(2), 1]. Neither field theorem is a premise for the other. The present paper's two-arc boundary theorem and normalization series are retained separately.
OpenAI 2026, The Gaussian free field limit of the balanced six-vertex model with variance multiplier 1/arcsin(c/2).
OpenAI 2026, Gaussian free-field limits of weighted integer Lipschitz heights.
Exercise: What is the significance of the constant c_* being positive?
If c_* = 0, the plane covariance of H would vanish in the scaling limit, and we would just see the deterministic harmonic function m plus zero. The positive lower bound on c_* (Section 5) rules out this degenerate alternative: the model really has nonvanishing logarithmic fluctuations, and so really is a nontrivial random surface. The positivity is proved by an explicit nondegenerate event (two nested opposite-sign double-B cuts, plus the corresponding component switch) whose probability is bounded below in the mesh.
Exercise: What is left open by this paper?
Four questions:
(a) The zero-level interface: does {h = 0} converge to a curve, and if so to which one? This is the second half of Schramm's Problem 2.2.
(b) Non-simply-connected domains: the paper only handles smooth simply connected domains. Annuli and general Riemann surfaces are not covered.
(c) Real-valued Lipschitz heights (Schramm's Problem 2.3): the current paper is about the hard-integer model. The continuous-height model is a different scaling regime.
(d) Numerical value of sigma: the paper gives an absolutely convergent finite-volume formula for c_* (and hence sigma), but does not evaluate it. One could imagine Monte Carlo computation of the first several terms b_js to get a decimal approximation.
Exercise: What is the cleanest one-paragraph summary of the whole paper?
Random odd integer Lipschitz heights on the triangular lattice, with +1 on one arc and -1 on the other arc of a smooth simply connected planar domain, scaled to zero mesh, converge weakly to (harmonic extension of the two-arc boundary data) + (universal positive constant) * (Dirichlet Gaussian free field). The universal constant has an explicit absolutely convergent finite-volume formula. There is no logarithmic rescaling, no non-universal prefactor, and no divergence; only the pointwise fluctuations diverge as expected of a log-correlated field. The result settles the field part of Schramm's Problem 2.2.
Exercise: What are all the named results, in order, for a quick lookup?
Main:
Theorem (Two-Arc GFF Limit)
Spin and loops:
Lemma (Grey Code)
Lemma (Loop Representation)
Lemma (Spin FKG)
Lemma (Double Cut)
Lemma (Worst Cut)
Lemma (Buffer Circuits)
Proposition (Plane Mixing)
Lemma (Relative Pin)
Proposition (Pin Product)
Moment estimates:
Lemma (Midpoint Tails)
Lemma (Loop Shells)
Proposition (Smearing Moments)
Proposition (Plane Moments)
Lemma (Component Switch)
Proposition (Obstacle Moments)
Corollary (Equal-Offset Moments)
Reflection positivity and spectrum:
Proposition (Reflection Positivity)
Lemma (Increment Spectral)
Lemma (Annulus Mass)
Proposition (Cauchy Mixture)
Lemma (Angular Poisson)
Proposition (Dispersion Defect)
Lemma (Endcap)
Lemma (Spectral Subsequences)
Proposition (Angular Sum Rule)
Theorem (Plane Covariance)
Corollary (Reflection Null)
Normalization:
Proposition (Normalization Series)
Remark (Cell Area)
Pins and bracket identity:
Lemma (Boundary Brackets)
Lemma (Normalized Reflection)
Lemma (Analytic Translation)
Proposition (Gapped Identity)
Lemma (Gap Filling)
Proposition (Bracket Identity)
Boundary arch:
Theorem (Boundary Arches)
Lemma (Many Preliminary Pairs)
Lemma (Extremal Circuit Stopping)
Lemma (Boundary Attachment)
Lemma (Arch Upgrade)
Corollary (Boundary Robustness)
Limits and moment structure:
Proposition (Tightness)
Proposition (Boundary Trace)
Lemma (Boundary References)
Proposition (Moment Harmonicity)
Lemma (No-Contact)
Gaussian identification:
Lemma (Log Singularity)
Proposition (Collision Coefficient)
Theorem (Gaussian Moments)
Corollary (Field Limit)