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Planar FK duality with the free exterior-face convention

Statement

Let $G = (V, E)$ be a finite connected plane graph with a fixed embedding and $∣ V ∣ \geq 1$ . Use the free partition on all primal vertices, $B = V$ . Its plane dual multigraph $G^{*}$ has one vertex for each face, including the unbounded face, and one dual edge $e^{*}$ crossing each primal edge $e$ ; use the identity partition on the dual face-vertices. The unbounded face is one dual vertex, so this convention retains the exterior-face connection in the dual. For a primal open set $A \subseteq E$ , let $A^{*} : = {e^{*} : e \in E ∖ A}$ , so a dual edge is open exactly when its crossing primal edge is closed. For $q \geq 1$ and $p \in (0, 1)$ define $p^{*} \in (0, 1)$ by

$\frac{p^{*}}{1 - p^{*}} = q \frac{1 - p}{p} .$

Then complementing the primal edges sends the finite FK law $ϕ_{G, p, q}^{f r e e}$ to the dual FK law $ϕ_{G^{*}, p^{*}, q}^{f r e e}$ : the random dual edge set $A^{*}$ has exactly that distribution. The unnormalized weight ratio is the configuration-independent factor $q^{∣ V ∣ - 1} ((1 - p) / p^{*})^{∣ E ∣}$ ; the exponent $∣ V ∣ - 1$ includes the exterior-face contribution under this identity-partition convention.

Statement references

Finite FK random-cluster weights for real q and boundary partitions

Plane embeddings of finite simple graphs, their faces, facial boundary walks and lengths (counting a bridge twice), and planar graphs

The plane dual multigraph, with a vertex for each face and one crossing edge for each primal edge

For a plane graph with $c$ components, $∣ V ∣ - ∣ E ∣ + ∣ F ∣ = 1 + c$ , including the null graph

Face frontiers are unions of whole edges; a cycle edge borders two faces and a bridge borders one

Facts & Assumptions

Given: A finite connected plane graph with its fixed embedding, its dual multigraph including the unbounded face, $A \subseteq E$ , $q \geq 1$ , and $p \in (0, 1)$ .

[F1]

The finite FK weight is $p^{∣ A ∣} (1 - p)^{∣ E ∖ A ∣} q^{k (A)}$ ; components count isolated vertices (Finite FK random-cluster weights for real q and boundary partitions).

[F2]

The dual has one vertex per face and one edge crossing each primal edge; loops and parallel edges are permitted in the multigraph (Plane embeddings of finite simple graphs, their faces, facial boundary walks and lengths (counting a bridge twice), and planar graphs, The plane dual multigraph, with a vertex for each face and one crossing edge for each primal edge).

[F3]

For a plane graph with $c$ components, including isolated vertices, $∣ V ∣ - ∣ E ∣ + ∣ F ∣ = 1 + c$ (For a plane graph with $c$ components, $∣ V ∣ - ∣ E ∣ + ∣ F ∣ = 1 + c$ , including the null graph).

[F4]

A cycle edge has two distinct incident faces and a bridge has one face on both local sides (Face frontiers are unions of whole edges; a cycle edge borders two faces and a bridge borders one).

Proof

1.1F2F3F4algebrainduction

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 $k_{G^{*}} (A^{*}) = ∣ F_{A} ∣$ .

2.1F3step 1.1algebra

Apply [F3] to $(V, A)$ , whose number of components is $k_{G} (A)$ . It gives $∣ V ∣ - ∣ A ∣ + ∣ F_{A} ∣ = 1 + k_{G} (A)$ . Substituting step 1.1 and rearranging yields $k_{G} (A) = k_{G^{*}} (A^{*}) + ∣ V ∣ - ∣ A ∣ - 1$ .

3.1F1step 2.1algebra

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

4.1F1F2step 3.1algebra∎

Since $A \mapsto A^{*}$ 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 $q^{∣ V ∣ - 1}$ is the contribution of the unique exterior face under this free dual convention; the identity is not a criticality argument.

Planar FK duality with the free exterior-face convention

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources