Planar random clusters, sharpness and the exact square-lattice critical point

Form the quotient vertex set $V_{\xi }$ by identifying each $\xi$-class. For an edge with distinct quotient endpoints $u,v$, let its incidence vector be $e_u+e_v\in {\mathbb{F}}_2^{V_{\xi }}$; a quotient loop has zero vector. In a component $C$ of the opened quotient graph $(V_{\xi },A)$, every edge vector has even coordinate sum, so their span lies in the even-sum subspace, of dimension $\mathrm{\mid }C\mathrm{\mid }-1$. Conversely choose a root $r\in C$. For every $v\in C\setminus \{r\}$, connectedness [F2] gives an open path from $r$ to $v$; summing its edge vectors over ${\mathbb{F}}_2$ cancels each internal vertex twice and gives $e_r+e_v$. These $\mathrm{\mid }C\mathrm{\mid }-1$ vectors are linearly independent by their distinct $v$-coordinates. Thus the incidence rank on $C$ is $\mathrm{\mid }C\mathrm{\mid }-1$, and summing over components gives $r(A)=\mathrm{\mid }{V}_{\xi }\mathrm{\mid }-k_{\xi }(A)$, including isolated vertices and quotient loops. If $U_A$ and $U_C$ are the spans of incidence vectors from edge sets $A$ and $C$, then $U_{A\cup C}=U_A+U_C$ and $U_{A\cap C}\subseteq U_A\cap U_C$. The subspace dimension formula therefore gives $r(A)+r(C)\ge r(A\cup C)+r(A\cap C)$, hence $k_{\xi }(A\cup C)+k_{\xi }(A\cap C)\ge k_{\xi }(A)+k_{\xi }(C)$.

Fix the exterior edge states outside a finite $F\subseteq E$. For an interior edge set $A\subseteq F$, the total number of components of the combined graph is the number of exterior-only components disjoint from the boundary of $F$, plus the number of components of the graph on $F$ after identifying its boundary vertices that are connected by the fixed exterior open edges or by $\xi$. The first summand and all exterior edge factors are constant in $A$; they cancel when the conditional weights are normalized. Hence the conditional law on $F$ is exactly the finite FK law with this induced boundary partition.

Since $\mathrm{\mid }A\cup C\mathrm{\mid }+\mathrm{\mid }A\cap C\mathrm{\mid }=\mathrm{\mid }A\mathrm{\mid }+\mathrm{\mid }C\mathrm{\mid }$, the factors $p^{\mathrm{\mid }A\mathrm{\mid }}(1-p{)}^{\mathrm{\mid }E\setminus A\mathrm{\mid }}$ have equality in the lattice condition. The component inequality in step 1.1 and $q\ge 1$ give the displayed log-supermodular inequality; every mass is positive by $0<p<1$ and $q\ge 1$.

Given all edges except $e$, let $s$ be the conditional number of components when $e$ is closed. If its endpoints are already connected, opening it leaves $s$ unchanged, so open-to-closed odds are $p\mathrm{/}(1-p)$ and the open probability is $p$. If the endpoints are not connected, opening it changes the component factor from $q^s$ to $q^{s-1}$, so the odds are $p\mathrm{/}[q(1-p)]$ and the open probability is $p\mathrm{/}(p+q(1-p))$. Both expressions increase with $p$; opening other edges or making the boundary partition coarser can only turn the second case into the first.

We prove positive association for a strictly positive log-supermodular measure on $\{0,1{\}}^n$ by induction on $n$. It is trivial at $n=0$. For the last coordinate $s\in \{0,1\}$, each conditional measure ${\mu }_s$ on the first $n-1$ coordinates remains log-supermodular, so induction makes each ${\mu }_s$ positively associated. The ratio $R(x)=\mu (X_n=1,x)\mathrm{/}\mu (X_n=0,x)$ is increasing in $x$: for $x\le y$, apply log-supermodularity to $(y,0)$ and $(x,1)$ to obtain $R(y)\ge R(x)$. Thus for every increasing $h$, ${\mathbb{E}}_{{\mu }_1}h={\mathbb{E}}_{{\mu }_0}(hR)\mathrm{/}{\mathbb{E}}_{{\mu }_0}R\ge {\mathbb{E}}_{{\mu }_0}h$ by positive association under ${\mu }_0$. For increasing $f,g$, their conditional means $F_s=\mathbb{E}[f\mid X_n=s]$ and $G_s=\mathbb{E}[g\mid X_n=s]$ therefore satisfy $F_1\ge F_0$ and $G_1\ge G_0$. The conditional association and the nonnegative covariance of two increasing functions of the single Bernoulli coordinate give $\mathbb{E}(fg)\ge {\sum }_s\mu (X_n=s)F_s{G}_s\ge \mathbb{E}f\mathbb{E}g$.

For two parameter/boundary choices ordered by $p_1\le p_2$ and by refinement ${\xi }_1\le {\xi }_2$, run random-scan heat-bath chains from the all-closed configuration, choosing the same edge and uniform random number at each update. Each update resamples one edge from its conditional FK law, so it preserves the corresponding FK law. Step 2.2 makes every coupled update preserve coordinatewise order. To prove marginal convergence, fix $j\in \{1,2\}$ and couple two copies of the chain with this same fixed kernel $(p_j,q,{\xi }_j)$, one from any configuration and one from its stationary FK law. If there are $r>0$ labels, a block of $r$ updates chooses them in a prescribed order with probability $r^{-r}$. If all its uniforms exceed $p_2<1$, every label is closed at the block end in both copies, regardless of their initial configurations. This event has probability $\delta =r^{-r}(1-p_2{)}^r>0$. Once these same-kernel copies meet they remain equal under shared updates. Independence of the update blocks bounds their disagreement probability after $k$ blocks by $(1-\delta {)}^k$, proving convergence of each marginal to its FK law. No absorbing equality is claimed between the different-kernel chains. For every increasing real $f$ on the finite configuration space, their order at every time gives $\mathbb{E}f(X_t^{(1)})\le \mathbb{E}f(X_t^{(2)})$; taking the two marginal limits proves stochastic domination. When $r=0$ there is one configuration and the conclusion is immediate.

For the spanning subgraph $(V,A)$, write $F_A$ for its faces. Starting with $A=E$, the dual open set is empty, so its components are the individual faces of $G$. Delete primal edges one at a time. If a deleted edge is a bridge of the current subgraph, its component count increases by one while its edge count decreases by one; [F3] therefore leaves the face count unchanged. By [F4] its two local sides lie in one current face, hence in one current dual component by induction, so adding its dual edge does not change the dual component count (the edge may be a loop in the original dual or may join vertices already connected by previously opened dual edges). If it is not a bridge, its component count is unchanged while its edge count decreases by one; [F3] therefore lowers the face count by one. By [F4], the edge has two distinct incident current faces, hence two distinct current dual components by induction; opening its dual edge merges exactly those components. Induction over the finitely many deleted edges proves that dual components correspond to faces of the current primal subgraph; in particular $k_{G^{\ast }}(A^{\ast })=\mathrm{\mid }{F}_A\mathrm{\mid }$.

Apply [F3] to $(V,A)$, whose number of components is $k_G(A)$. It gives $\mathrm{\mid }V\mathrm{\mid }-\mathrm{\mid }A\mathrm{\mid }+\mathrm{\mid }{F}_A\mathrm{\mid }=1+k_G(A)$. Substituting step 1.1 and rearranging yields $k_G(A)=k_{G^{\ast }}(A^{\ast })+\mathrm{\mid }V\mathrm{\mid }-\mathrm{\mid }A\mathrm{\mid }-1$.

Let $a=\mathrm{\mid }A\mathrm{\mid }$ and $m=\mathrm{\mid }E\mathrm{\mid }$. By step 2.1, the ratio of primal and dual unnormalized weights is $\frac{p^a(1-p{)}^{m-a}{q}^{k_G(A)}}{(p^{\ast }{)}^{m-a}(1-p^{\ast }{)}^a{q}^{k_{G^{\ast }}(A^{\ast })}}=q^{\mathrm{\mid }V\mathrm{\mid }-1}{\left(\frac{p}{q(1-p^{\ast })}\right)}^a{\left(\frac{1-p}{p^{\ast }}\right)}^{m-a}$. The defining relation for $p^{\ast }$ gives $p\mathrm{/}[q(1-p^{\ast })]=(1-p)\mathrm{/}{p}^{\ast }$, so this ratio is $q^{\mathrm{\mid }V\mathrm{\mid }-1}((1-p)\mathrm{/}{p}^{\ast }{)}^m$, independent of $A$.

Since $A\mapsto A^{\ast }$ is a bijection from primal edge subsets to dual edge subsets, normalizing the proportional weights in step 3.1 proves the asserted equality of probability laws. The factor $q^{\mathrm{\mid }V\mathrm{\mid }-1}$ is the contribution of the unique exterior face under this free dual convention; the identity is not a criticality argument.

Fix $n<m$ and condition the law on $E_m$ on every edge outside $E_n$. By [F1], the conditional law on $E_n$ is an FK law whose boundary partition records connections made by those fixed exterior edges. This partition is at least as wired as the free partition, so averaging the conditional laws gives ${\varphi }_{m,p,q}^0{\mathrm{\mid }}_{E_n}{\ge }_{st}{\varphi }_{n,p,q}^0$. For the wired law on $E_m$, each induced partition on $\mathrm{\partial }{\mathrm{\Lambda }}_n$ is at most as wired as the one-class wired partition, so the same conditional comparison gives ${\varphi }_{m,p,q}^1{\mathrm{\mid }}_{E_n}{\le }_{st}{\varphi }_{n,p,q}^1$.

For each finite edge set $F\subset E_{\mathrm{\infty }}$ and each $S\subseteq F$, the event that every edge in $S$ is open is increasing. Its free probabilities therefore have a bounded monotone limit by step 1.1, and its wired probabilities also have a bounded monotone limit. Every atom on $F$ is recovered from these all-open probabilities by finite inclusion-exclusion: if $O$ is the prescribed open set and $C=F\setminus O$ the prescribed closed set, then $P(\omega {\mathrm{\mid }}_F=O)={\sum }_{T\subseteq C}(-1{)}^{\mathrm{\mid }T\mathrm{\mid }}P({\omega }_e=1\text{ for all }e\in O\cup T)$. Hence every finite-dimensional atom probability converges for both sequences.

The limits in step 2.1 are consistent under restriction, because the finite-volume atom probabilities are consistent whenever the smaller edge set lies inside $E_n$. Under [A1], use a fixed explicit enumeration of the countable edge set to apply the finite-alphabet extension in [F2]. Each limiting consistent family extends uniquely to a probability measure on the cylinder-generated Borel sigma-algebra; the same cylinder limits make it the unique subsequential limit of the corresponding sequence. Therefore the free and wired cylinder limits exist and are unique.

Delete loops from each finite path until it is simple, preserving its endpoints. In the square grid, two unit edges can meet only at a shared endpoint, so if the paths do not share a vertex their polygonal realizations are disjoint compact sets. Their distance is then positive.

Suppose the paths are disjoint and denote the simple left-to-right path by $\gamma$. Perturb its finitely many interior vertices and boundary contacts by less than one third of the distance in step 1.1, pushing every nonendpoint boundary contact slightly into the rectangle. Since $\gamma$ is a finite simple polygonal arc, choose disjoint small neighborhoods of its nonadjacent segments; local polygonal detours inside those neighborhoods preserve simplicity, keep its endpoints on the left and right sides, and preserve disjointness from the other path. The resulting simple arc $\tilde{\gamma }$ meets the rectangle boundary only at its two endpoints.

Join the endpoints of $\tilde{\gamma }$ by a simple polygonal arc lying outside the rectangle and running around its top side. Their union is a simple polygonal closed curve by the boundary-contact property in step 2.1. By [F1], its bounded component contains the part of the rectangle immediately below the top side and above $\tilde{\gamma }$, while the bottom side lies in the unbounded component. A bottom-to-top path staying in the rectangle must therefore cross this Jordan curve; it cannot meet the outside closing arc, so it must meet $\tilde{\gamma }$. This contradicts the preserved disjointness in step 2.1. Thus the original paths intersect; grid edges intersect only at shared vertices, proving the claim.

The elementary integral is well-defined: adjoining an endpoint splits an interval into two whose lengths sum to its original length, so finite common refinement leaves each integral unchanged. Indicators of finite-box sets, their fixed-coordinate sections, and their finite products and differences are rectangular step functions. Common refinement proves linearity, positivity and the integral triangle inequality. It also proves equality of iterated and full integrals, because both sum the same finitely many cell values times products of lengths; outer grid endpoints have zero length. A translation preserves side lengths, so a piecewise translation bijectively exchanging finitely many boxes preserves these integrals. These are all the continuous-cube integration rules used below.

For $1\le p\le 2$ and real $u,v$, we prove $(u^2+(p-1)v^2{)}^{1\mathrm{/}2}\le ((\mathrm{\mid }u+v{\mathrm{\mid }}^p+\mathrm{\mid }u-v{\mathrm{\mid }}^p)\mathrm{/}2{)}^{1\mathrm{/}p}$. At $p=1$ this follows from the triangle inequality, and at $p=2$ it is equality. For $1<p<2$ signs can be removed. If $\mathrm{\mid }v\mathrm{\mid }>\mathrm{\mid }u\mathrm{\mid }$, interchange their absolute values: the right side is unchanged and the squared left side increases by $(2-p)(v^2-u^2)$. It therefore suffices to take $u>0$ and $0\le t=v\mathrm{/}u\le 1$, apart from the trivial zero case. For $t<1$ put $H(t)=((1+t{)}^p+(1-t{)}^p)\mathrm{/}2$. The convexity of $s^{p-2}$, obtained by two differentiations, gives $H^{\mathrm{\prime }\mathrm{\prime }}(t)\ge p(p-1)$. Since $H(0)=1$ and $H^{\mathrm{\prime }}(0)=0$, integration twice gives $H(t)\ge 1+p(p-1)t^2\mathrm{/}2$. Concavity of $s^{p\mathrm{/}2}$ gives $(1+(p-1)t^2{)}^{p\mathrm{/}2}\le 1+p(p-1)t^2\mathrm{/}2$. Multiply by $u^p$ and take the $p$th root; continuity includes $t=1$. This proves the claimed two-point inequality.

On a uniform bit $x\in \{-1,1\}$ define $T_{\rho }(u+vx)=u+\rho vx$, for $0\le \rho \le 1$. Step 1.2 with $p=1+{\rho }^2$ proves $\mathrm{\parallel }{T}_{\rho }h{\mathrm{\parallel }}_2\le \mathrm{\parallel }h{\mathrm{\parallel }}_p$. To extend this to $d$ bits, first note that finite-sum Minkowski gives $\mathrm{\parallel }h{\mathrm{\parallel }}_{L_y^2(L_x^p)}\le \mathrm{\parallel }h{\mathrm{\parallel }}_{L_x^p(L_y^2)}$ for finite probability spaces and $1\le p\le 2$: its $p$th-power form is the triangle inequality for ${\mathbb{E}}_x\mathrm{\mid }h(x,\cdot ){\mathrm{\mid }}^p$ in $L_y^{2\mathrm{/}p}$. For completeness, finite weighted Minkowski at exponent $r>1$ follows by multiplying $\mathrm{\mid }z+w\mathrm{\mid }\le \mathrm{\mid }z\mathrm{\mid }+\mathrm{\mid }w\mathrm{\mid }$ by $\mathrm{\mid }z+w{\mathrm{\mid }}^{r-1}$, summing, applying Hölder and dividing by $\mathrm{\parallel }z+w{\mathrm{\parallel }}_r^{r-1}$ if nonzero. Weighted Hölder follows by normalizing the two norms and summing Young's inequality $ab\le a^r\mathrm{/}r+b^{r^{\mathrm{\prime }}}\mathrm{/}{r}^{\mathrm{\prime }}$, which follows by minimizing $a^r\mathrm{/}r-ab$ in $a\ge 0$; zero norms and $r=1$ are immediate. If $T_X,T_Y$ contract $L^p$ into $L^2$ on two blocks, apply the $X$ contraction pointwise in $y$, the mixed-norm inequality, and then the $Y$ contraction pointwise in $x$ to obtain $\mathrm{\parallel }{T}_X{T}_{Y}h{\mathrm{\parallel }}_2\le \mathrm{\parallel }{T}_{Y}h{\mathrm{\parallel }}_{L_y^2(L_x^p)}\le \mathrm{\parallel }{T}_{Y}h{\mathrm{\parallel }}_{L_x^p(L_y^2)}\le \mathrm{\parallel }h{\mathrm{\parallel }}_p$. Induction therefore gives $\mathrm{\parallel }{T}_{\rho }h{\mathrm{\parallel }}_2\le \mathrm{\parallel }h{\mathrm{\parallel }}_{1+{\rho }^2}$ on every finite Boolean cube, where $T_{\rho }$ applies the one-bit operator in each coordinate.

Put $N=2^m$ for $m\ge 1$ and partition $[0,1{)}^n$ into the dyadic cells $Q_a={\prod }_i[a_i\mathrm{/}N,(a_i+1)\mathrm{/}N)$, with $0\le a_i<N$. On $Q_a$ let ${\mathrm{\ell }}_m=f(a\mathrm{/}N)$, $u_m=f((a+\mathbf{1})\mathrm{/}N)$ and $g_m=N^n{\int }_{Q_a}f$. Faces at coordinate one may be assigned arbitrary values for integration. Monotonicity gives ${\mathrm{\ell }}_m\le f,g_m\le u_m$. Telescope each difference $u_m-{\mathrm{\ell }}_m$ one coordinate at a time. For coordinate $i$, summation over $a_i$ at fixed remaining upper/lower corner coordinates telescopes to a difference of values at coordinate one and zero, at most one. There are $N^{n-1}$ choices of the other indices, hence $\int (u_m-{\mathrm{\ell }}_m)\le n\mathrm{/}N$ and ${\delta }_m:=\int \mathrm{\mid }{g}_m-f\mathrm{\mid }\le n\mathrm{/}N$. Cell averaging gives $\int g_m=\int f=a$, while $\mathrm{\mid }{g}_m^2-f^2\mathrm{\mid }\le 2\mathrm{\mid }{g}_m-f\mathrm{\mid }$. Therefore $\mathrm{\mid }\mathrm{Var}(g_m)-a(1-a)\mathrm{\mid }\le 2{\delta }_m\to 0$. This uses an explicit numerical error bound, not a theorem interchanging a limit and an integral.

On $\{-1,1{\}}^d$ let $x^{(j)}$ flip bit $j$ and set $D_{j}h=(h(x)-h(x^{(j)}))\mathrm{/}2$. Suppose numbers $b_j\in [0,1]$ satisfy $\mathrm{\parallel }{D}_{j}h{\mathrm{\parallel }}_p\le b_j^{1\mathrm{/}p}\mathrm{/}2$ for every $1\le p\le 2$. We will prove $\mathrm{Var}(h)\le {\sum }_j{b}_j\mathrm{/}(1+\log (1\mathrm{/}{b}_j))$, interpreting the summand at zero as zero. For $S\subseteq \{1,\dots ,d\}$ define ${\chi }_S(x)={\prod }_{j\in S}{x}_j$. Independence and zero mean of each bit show these $2^d$ functions are orthonormal. Since the function space has dimension $2^d$, they form a basis; write $h={\sum }_S\widehat{h}(S){\chi }_S$. Then $D_{j}h={\sum }_{S\ni j}\widehat{h}(S){\chi }_S$ and $T_{\rho }{\chi }_S={\rho }^{\mathrm{\mid }S\mathrm{\mid }}{\chi }_S$. Parseval and finite-sum integration yield the exact identity $\mathrm{Var}(h)=2{\sum }_j{\int }_0^1\mathrm{\parallel }{T}_{\rho }{D}_{j}h{\mathrm{\parallel }}_2^2\frac{d\rho }{\rho },$ because the coefficient of each $\widehat{h}(S{)}^2$ for nonempty $S$ is $2\mathrm{\mid }S\mathrm{\mid }{\int }_0^1{\rho }^{2\mathrm{\mid }S\mathrm{\mid }-1}d\rho =1$. These are ordinary one-variable Riemann integrals of finite polynomials after cancellation of the factor $\rho$; no singularity at zero occurs.

Identify each cell index $a_i$ with its $m$ binary digits, most significant first; the uniform distribution of the $nm$ bits is the uniform distribution of the equal-volume cells. Regard $g_m$ as a real function on this finite Boolean cube. For $1\le k\le m$, let $R_{ik}$ flip the $k$th binary digit of coordinate $i$ on $[0,1)$: in every prefix interval of length $2^{1-k}$ translate its lower half up by $2^{-k}$ and its upper half down by $2^{-k}$. Define it arbitrarily at coordinate one. This is a finite piecewise translation preserving elementary integrals, and it pairs the level-$m$ cells with that digit flipped. Let $b_{ik}=\int \mathrm{\mid }f-f\circ R_{ik}\mathrm{\mid }$. The bit derivative of $g_m$ on a cell is the cell average of $(f-f\circ R_{ik})\mathrm{/}2$. For such an average $z=\mathrm{\mid }Q{\mathrm{\mid }}^{-1}{\int }_Q(f-f\circ R_{ik})$, the finite triangle inequality gives $\mathrm{\mid }z\mathrm{\mid }\le \mathrm{\mid }Q{\mathrm{\mid }}^{-1}{\int }_Q\mathrm{\mid }f-f\circ R_{ik}\mathrm{\mid }\le 1$, so $\mathrm{\mid }z{\mathrm{\mid }}^p\le \mathrm{\mid }z\mathrm{\mid }$ for $p\ge 1$. Multiplying by $2^{-p}\mathrm{\mid }Q\mathrm{\mid }$ and summing cells yields $\mathrm{\parallel }{D}_{ik}{g}_m{\mathrm{\parallel }}_p^p\le 2^{-p}{b}_{ik}$, the hypothesis of step 3.1.

Write $D_{j}h=x_j{g}_j(x_{-j})$. The product noise operator gives $T_{\rho }{D}_{j}h=\rho x_j{T}_{\rho }{g}_j$, so step 2.1 and the norm hypothesis in step 3.1 imply $\mathrm{\parallel }{T}_{\rho }{D}_{j}h{\mathrm{\parallel }}_2^2\le {\rho }^2{b}_j^{2\mathrm{/}(1+{\rho }^2)}\mathrm{/}4$. For $b=b_j>0$ put $L=\log (1\mathrm{/}b)$. Its contribution to the integral in step 3.1 is at most $\frac{1}{2}{\int }_0^1\rho b^{2\mathrm{/}(1+{\rho }^2)}d\rho \le \frac{b}{4}{\int }_0^1{e}^{-Ls\mathrm{/}2}ds\le \frac{b}{1+L}\mathrm{.}$ The middle inequality substitutes $s=1-{\rho }^2$ and uses $2\mathrm{/}(1+{\rho }^2)\ge 1+(1-{\rho }^2)\mathrm{/}2$. The last integral is at most one, proving the last inequality when $0\le L\le 1$, and at most $2\mathrm{/}L$, proving it when $L\ge 1$. At $b_j=0$ the norm hypothesis gives $D_{j}h=0$. Summing establishes the variance estimate claimed in step 3.1.

Fix the coordinates other than $i$. The fiber indicator is an increasing zero-one function of one real variable, hence a threshold indicator except possibly at its threshold. If the endpoint values agree it is constant everywhere. Otherwise flipping the $k$th digit can change its value only in the single prefix interval of length $2^{1-k}$ containing the threshold. If the threshold is a prefix endpoint, disagreement at that endpoint has length zero; other intervals are constant. Thus the disagreement length is at most $2^{1-k}(f(x_{-i},1)-f(x_{-i},0))$. All these sections and disagreement sets are finite-box sets. Finite-sum iterated integration from step 1.1 gives $b_{ik}\le 2^{1-k}{F}_i$, and consequently ${\sum }_{k=1}^m{b}_{ik}\le 2F_i$ and $b_{ik}\le M$.

If $M=0$, steps 4.1–5.1 make every bit derivative of every $g_m$ zero, so every $g_m$ is constant, and step 2.2 gives $a(1-a)=0$. If $M>0$, apply the variance bound of step 4.2 to $g_m$, using steps 4.1–5.1. Each nonzero denominator is at least $1+\log (1\mathrm{/}M)$, giving $\mathrm{Var}(g_m)\le 2{\sum }_i{F}_i\mathrm{/}(1+\log (1\mathrm{/}M))$. Taking the limit with the explicit error in step 2.2 proves the first stated inequality.

If $0<M<n^{-1\mathrm{/}2}$, step 6.1 gives $a(1-a)\le 4nM\mathrm{/}\log n$, which is the required lower bound on $M$. If $M\ge n^{-1\mathrm{/}2}$, use $a(1-a)\le 1\mathrm{/}4$ and $\log n\le \sqrt{n}$ to get $a(1-a)\log n\mathrm{/}(4n)\le 1\mathrm{/}(16\sqrt{n})\le M$. The elementary bound $\log t\le \sqrt{t}$ for $t\ge 1$ follows by differentiating $(\log t)\mathrm{/}\sqrt{t}$: its maximum is $2\mathrm{/}e<1$. The case $M=0$ was settled in step 6.1. This proves the stated bound for every integer $n\ge 2$.

In the boundary of each face, every vertex occurrence occurs twice after applying ${\mathrm{\partial }}_1$, so ${\mathrm{\partial }}_1{\mathrm{\partial }}_2=0$. This includes identifications in a thin quotient: occurrences are added in ${\mathbb{F}}_2$, so repeated edges and endpoints cancel with their multiplicity rather than being removed beforehand. Both primal and dual graphs are connected, since horizontal and vertical steps modulo $m$ reach every coordinate pair. For $m=1$ each graph has a single vertex and is connected even though its edge labels are loops.

The labelled graph map sending $f(i,j)$ to $(i,j)$, $h(i,j{)}^{\ast }$ to $v(i,j-1)$, and $v(i,j{)}^{\ast }$ to $h(i-1,j)$ is an isomorphism from the dual graph to the primal graph. It is bijective on both vertices and edge labels and preserves both endpoint occurrences; it also retains all parallel labels and loops at $m=2$ and $m=1$. The induced bijection on edge subsets preserves their cardinalities and component counts. Consequently the ordinary primal and dual FK partition functions $Z$ and $Z^{\ast }$ are equal.

For any finite multigraph with $s$ vertices and $c$ components, its endpoint-incidence map over ${\mathbb{F}}_2$ has rank $s-c$. Indeed, within one component every edge vector has even coordinate sum, a subspace of dimension at most one less than the number of vertices. Fix a root in that component. For every other vertex, the sum of edge vectors along a path from the root to that vertex is the vector consisting of the root plus that vertex. These vectors are independent by their nonroot coordinates and attain that dimension. Summing over the disjoint components proves the rank formula; isolated vertices contribute zero rank and loops have zero incidence vector. In particular $\mathrm{rank}{\mathrm{\partial }}_1=\mathrm{\mid }V\mathrm{\mid }-1$, and restriction to $C_1(A)$ gives $\dim Z(A)=\mathrm{\mid }A\mathrm{\mid }-\mathrm{\mid }V\mathrm{\mid }+k(A)$.

Write a two-chain as $c={\sum }_f{c}_{f}f$. Its coefficient on a primal edge with incident face occurrences $f,g$ is $c_f+c_g$, including the zero sum when those faces agree. Thus ${\mathrm{\partial }}_{2}c=0$ exactly when the coefficients agree across every edge of the connected dual graph. Such coefficients are constant on all faces, and conversely every constant two-chain has zero boundary. Hence $\dim \ker {\mathrm{\partial }}_2=1$ and $\mathrm{rank}{\mathrm{\partial }}_2=\mathrm{\mid }F\mathrm{\mid }-1$. Together with $\mathrm{\mid }E\mathrm{\mid }=2m^2$ and $\mathrm{\mid }V\mathrm{\mid }=\mathrm{\mid }F\mathrm{\mid }=m^2$, step 2.1 gives $\dim H=(\mathrm{\mid }E\mathrm{\mid }-\mathrm{\mid }V\mathrm{\mid }+1)-(\mathrm{\mid }F\mathrm{\mid }-1)=2$. The dimension formulas used here follow by taking a basis of the kernel of a finite linear map and extending it to a basis of its domain: the images of the added vectors form a basis of its image. The corresponding quotient dimension is obtained by extending a basis of its subspace.

A two-chain $c$ satisfies ${\mathrm{\partial }}_{2}c\in C_1(A)$ exactly when its coefficients agree across every dual edge crossing a primal label outside $A$. These are precisely the labels of $A^{\ast }$. Its coefficients may therefore be chosen independently, and constantly, on the $k^{\ast }(A^{\ast })$ connected components of the dual spanning graph. The vector space $U_A:=\{c\in C_2:{\mathrm{\partial }}_{2}c\in C_1(A)\}$ has dimension $k^{\ast }(A^{\ast })$. The restriction of ${\mathrm{\partial }}_2$ to $U_A$ has image $\mathrm{im}{\mathrm{\partial }}_2\cap C_1(A)$ and kernel $\ker {\mathrm{\partial }}_2$, the one-dimensional space of constant chains. Therefore $\dim (\mathrm{im}{\mathrm{\partial }}_2\cap C_1(A))=k^{\ast }(A^{\ast })-1$.