Classical Potts random clusters and planar loop transfer

For fixed $\sigma$, summing $W(\sigma ,A)$ over all $A\subseteq E$ chooses, independently for each edge, either its absent factor $1$ or its present factor $v_e{\mathbf{1}}_{\{\sigma (u_e)=\sigma (w_e)\}}$. Distributivity of the finite product therefore gives ${\sum }_{A}W(\sigma ,A)=w_P(\sigma )$. Summing this equality over the finite set $[Q{]}^V$ gives ${\sum }_{\sigma ,A}W(\sigma ,A)=Z_P$.

Fix $A\subseteq E$. The product $W(\sigma ,A)$ is zero unless the spins agree across every edge in $A$. Equality propagates along each finite path, so the admissible configurations are exactly those constant on each connected component of $G_A$. Assigning one of $Q$ colors independently to each of its $k_G(A)$ components gives exactly $Q^{k_G(A)}$ such configurations, each with weight ${\prod }_{e\in A}{v}_e$. Hence ${\sum }_{\sigma }W(\sigma ,A)=w_{FK}(A)$, including when this weight is zero. Summing over $A$ gives ${\sum }_{\sigma ,A}W(\sigma ,A)=Z_{FK}$.

The common total is positive by [F1], so dividing $W$ by it defines a probability law. The finite marginal sums in steps 1.1 and 1.2 show that its spin and edge marginals are respectively $w_P\mathrm{/}{Z}_P$ and $w_{FK}\mathrm{/}{Z}_{FK}$, and that $Z_P=Z_{FK}$.

For a fixed $A$ with positive FK probability, every spin assignment constant on its components has the same joint weight ${\prod }_{e\in A}{v}_e$. By step 1.2 there are $Q^{k_G(A)}$ such assignments, so the conditional law is uniform on them; this is equivalent to assigning independent uniform colors to the components.

If $x,y$ are connected in $G_A$, step 2.2 makes their colors equal with conditional probability $1$. If they are in distinct components, their independent uniform colors agree with probability $1\mathrm{/}Q$. Averaging over the FK marginal from step 2.1 yields $\mathbb{P}(\sigma (x)=\sigma (y))={\mathbb{P}}_{FK}(x\leftrightarrow y)+Q^{-1}{\mathbb{P}}_{FK}(x\leftrightarrow ̸y)=Q^{-1}+(1-Q^{-1}){\mathbb{P}}_{FK}(x\leftrightarrow y)$.

Let $E_t=\{e_1,\dots ,e_t\}$ and let $n_0$ count vertices outside $B$ that are isolated in $G$. For each edge subset $C\subseteq E_t$, form the graph on the initially present boundary vertices and the endpoints introduced by time $t$, impose $\xi$, and retain the present edges in $C$. Let ${\pi }_t(C)$ be the resulting equivalence partition restricted to the active frontier, and let $c_t(C)$ count its equivalence classes containing no active vertex. The coefficient vector after time $t$ satisfies $W_t(\pi )=Q^{n_0}{\sum }_{\begin{array}{c}C\subseteq E_t\\ {\pi }_t(C)=\pi \end{array}}{Q}^{c_t(C)}{\prod }_{e\in C}{v}_e\mathrm{.}$ At $t=0$ the only subset is empty, every boundary vertex is active, their partition is $\xi$, and $c_0=0$. The initialized coefficient is $Q^{n_0}$, so the invariant holds.

Assume the invariant after $t-1$ edges. Introducing a new endpoint as a singleton agrees with adding that previously unseen vertex to the partial graph; a boundary endpoint is already present in the initial relation $\xi$. For each partial subset $C\subseteq E_{t-1}$, the two subsets at the next edge are exactly $C$ and $C\cup \{e_t\}$. The absent choice preserves the partition and weight; the present choice multiplies by $v_{e_t}$ and merges precisely the endpoint blocks. Thus $I+v_{e_t}{J}_{uv}$ gives the two required contributions, even when $u,v$ were already equivalent through $\xi$ or previous edges.

When vertices at their last incident edge are forgotten, no future edge is incident to them. If their block still has an active vertex, that equivalence class can connect to future edges through that vertex and is not yet closed. If removing a vertex empties its block, every boundary vertex in its $\xi$-class was already initially active and is now absent as well; no future edge can meet any member of the class. The class is therefore final, and the algorithm multiplies by $Q$ exactly once at this first emptying. This updates $c_t(C)$ and preserves the invariant. After the last edge, terminal cleanup applies the same criterion to the remaining blocks, which have no future incident edges.

For each full present-edge subset $C$, let $r_m(C)$ be the number of blocks still active just before terminal cleanup. Every equivalence class generated by $\xi$ and $C$ is counted exactly once: isolated vertices outside $B$ contribute $n_0$, classes closed earlier contribute $c_m(C)$, and the $r_m(C)$ remaining blocks each contribute one factor $Q$ in cleanup. Therefore $n_0+c_m(C)+r_m(C)=k_{\xi }(C)$, and summing the invariant from step 1.1 with these cleanup factors over $C\subseteq E$ gives the returned scalar ${\sum }_C{Q}^{k_{\xi }(C)}{v}_C=Z_{FK}^{\xi }$.

Each of the finitely many $m$ edges is processed once; every active frontier is finite and has finitely many set partitions, and terminal cleanup removes finitely many vertices. Hence the procedure terminates. For discrete $\xi$, [F3] identifies $Z_{FK}^{\xi }$ with $Z_{FK}$.

Fix $A\subseteq E$. The drawing of $G_A$ is a finite union of points and closed line segments, hence is closed in ${\mathbb{R}}^2$. Its complement is open, so each of its face components is open and polygonally connected by [F7]. The original faces are vertices of $G^{\ast }$. Two such faces lie in the same face of $G_A$ exactly when they are connected by dual edges corresponding to $E\setminus A$. A dual path crosses only absent primal edges and therefore stays in the complement of $G_A$. Conversely, a path between two original face points inside one face of $G_A$ can be chosen polygonal; perturb its finitely many segments to avoid primal vertices and to cross edge interiors transversely. Because it avoids $G_A$, every crossed edge lies in $E\setminus A$, so the crossed dual edges form a path in $(F(G),A^{\ast })$. Thus the faces of $G_A$ correspond bijectively to the components counted by $k^{\ast }(A)$.

Apply [F4] to the spanning plane subgraph $G_A$. Its vertex set is $V$, its edge set is $A$, its component count is $k(A)$, and by step 1.1 its face count is $k^{\ast }(A)$. Therefore $\mathrm{\mid }V\mathrm{\mid }-\mathrm{\mid }A\mathrm{\mid }+k^{\ast }(A)=1+k(A)$, or $k(A)-k^{\ast }(A)=\mathrm{\mid }V\mathrm{\mid }-\mathrm{\mid }A\mathrm{\mid }-1$.

View the fixed embedding on the sphere $S^2$ by adding the point at infinity. By [F4] the plane subgraph $G_A$ has finitely many faces; choose one point in each. For every vertex $x\in V$, including each vertex isolated in $G_A$, choose a closed polygonal disk $D_x$ so that the disks are pairwise disjoint and avoid all the chosen face points. Choose pairwise disjoint attachment intervals on each $\mathrm{\partial }{D}_x$, one for each edge incident to $x$ and in the cyclic order of the embedding. For each $e\in A$, choose a closed polygonal rectangular band around the part of its polygonal arc outside its endpoint disks; make the bands pairwise disjoint, disjoint from all nonendpoint disks, and attach them to the corresponding intervals. These choices are possible with sufficiently small radii and widths because the graph is finite and edge arcs meet only at shared endpoints. Let $N_A$ be the union of these disks and bands. Each band joins exactly its two endpoint disks, and distinct bands otherwise remain disjoint, so the connected components of $N_A$ correspond exactly to those of $G_A$, including isolated-vertex disks; hence there are $k(A)$. Its boundary is a finite disjoint union of simple polygonal circles. In particular, the one-vertex graph with no edges gives one disk and one boundary loop. At a primal-dual crossing, the two sides of an occupied-edge band join the medial half-edges as $(h_4,h_1)$ and $(h_2,h_3)$; when the edge is absent, there is no band there and the two face-side sectors join as $(h_1,h_2)$ and $(h_3,h_4)$. The cyclic ordering of the attachment intervals fixes these local continuations, so $\mathrm{\partial }{N}_A$ gives exactly the smoothings in the Statement and has $\mathrm{\ell }(A)$ loops. Inside each face $U$ of $G_A$, the disk-and-band pieces form collars along the facial boundary side occurrences: each band side is a polygonal strip, and intervening vertex-disk sectors join consecutive strips in cyclic order. Repeated vertices or edges in a facial boundary walk are treated as separate side occurrences, so a bridge's two sides remain distinct; an isolated vertex gives a punctured-disk collar in its unique incident face. Choose all radii and widths small enough that distinct collars meet only at their prescribed continuations within the same face. Trimming them leaves a nonempty core $U_A=U\setminus N_A$, since the selected point of $U$ lies outside $N_A$. The core is connected: for finite points use the polygonal path in the plane face $U$ supplied by [F7], perturb the path to meet the finitely many polygonal collar sides transversely, and replace each excursion into a collar by a parallel path just outside $N_A$ along that collar's connected $U$-facing boundary, with the vertex-disk sector prescribing the continuation at each turn. For the unbounded face, $N_A$ is bounded; outside a sufficiently large disk its complement contains the whole exterior, and a ray there joins finite points to the point at infinity in the one-point compactification. Thus the same connected core includes infinity. Each face therefore leaves exactly one nonempty component of $S^2\setminus N_A$. Since $N_A$ contains $G_A$, these trimmed faces are all the components of $S^2\setminus N_A$; by step 1.1 their number is $k^{\ast }(A)$. The regions of $S^2\setminus \mathrm{\partial }{N}_A$ are the interiors of the $k(A)$ components of $N_A$ together with those $k^{\ast }(A)$ complementary components, so there are $k(A)+k^{\ast }(A)$ regions.

Subdivide each of the disjoint polygonal boundary loops from step 2.2 at at least three distinct points and regard the result as a finite simple plane graph $H_A$, with one cycle component per loop. Each cycle has the same number of vertices and edges, so $\mathrm{\mid }V(H_A)\mathrm{\mid }=\mathrm{\mid }E(H_A)\mathrm{\mid }$, and $H_A$ has $\mathrm{\ell }(A)$ connected components. By [F4], its complement therefore has $\mathrm{\mid }F(H_A)\mathrm{\mid }=1+\mathrm{\ell }(A)$ regions. These are the regions counted in step 2.2, hence $k(A)+k^{\ast }(A)=1+\mathrm{\ell }(A)$ and $\mathrm{\ell }(A)=k(A)+k^{\ast }(A)-1$.

Adding the identities in steps 2.1 and 3.1 yields $2k(A)=\mathrm{\mid }V\mathrm{\mid }-\mathrm{\mid }A\mathrm{\mid }+\mathrm{\ell }(A)$. Since $Q\ge 2$, its positive square root $\delta =\sqrt{Q}$ is nonzero, and $Q^{k(A)}{\prod }_{e\in A}{v}_e=Q^{\mathrm{\mid }V\mathrm{\mid }\mathrm{/}2}(\sqrt{Q}{)}^{\mathrm{\ell }(A)}{\prod }_{e\in A}\frac{v_e}{\sqrt{Q}}$. This equality also holds when an activity is zero, because both corresponding terms vanish.

Each subset $A$ selects exactly one of the two noncrossing smoothings at every medial vertex, so summing the termwise identity of step 4.1 over $A\subseteq E$ gives the displayed loop formula. In any fixed rectangular decomposition, the local smoothings are Temperley–Lieb diagrams; [F5] fixes the left-above-right and $\delta$ conventions, and [F6] makes the stacked product independent of parentheses. Thus the local loop weight is exactly $\delta =\sqrt{Q}$, without invoking a trace or any representation-theoretic property.