Cellular component counts and self-dual FK likelihoods on a finite square torus

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 $\dim (\mathrm{im}{\mathrm{\partial }}_2\cap C_1(A))=k^{\ast }(A^{\ast })-1$.

Because ${\mathrm{\partial }}_1{\mathrm{\partial }}_2=0$, the kernel of $Z(A)\to H$ is exactly $\mathrm{im}{\mathrm{\partial }}_2\cap C_1(A)$. Its image has dimension $r(A)$ by definition, so steps 2.1 and 4.1 give $k^{\ast }(A^{\ast })-1=\mathrm{\mid }A\mathrm{\mid }-\mathrm{\mid }V\mathrm{\mid }+k(A)-r(A)$. The image is a subspace of the two-dimensional space $H$, so $r(A)\in \{0,1,2\}$. This proves the component identity without assuming that any open component is cellular or that any open boundary is a simple curve.

The dual cellulation is again a labelled square cellulation with $\mathrm{\mid }{V}^{\ast }\mathrm{\mid }=\mathrm{\mid }F\mathrm{\mid }=m^2$ and $\mathrm{\mid }{E}^{\ast }\mathrm{\mid }=2m^2$. Its endpoint and face-boundary maps have the same incidence descriptions used in the preceding matrix argument, including all thin-quotient occurrences. Applying the proved component identity to its open set $A^{\ast }$, whose complementary double-dual set is $A$, gives $k(A)-1=\mathrm{\mid }{A}^{\ast }\mathrm{\mid }-\mathrm{\mid }F\mathrm{\mid }+k^{\ast }(A^{\ast })-r^{\ast }(A^{\ast })$. Substitute $\mathrm{\mid }{A}^{\ast }\mathrm{\mid }=2\mathrm{\mid }V\mathrm{\mid }-\mathrm{\mid }A\mathrm{\mid }$, $\mathrm{\mid }F\mathrm{\mid }=\mathrm{\mid }V\mathrm{\mid }$, and the identity from step 5.1; the resulting equality is $k(A)-1=k(A)+1-r(A)-r^{\ast }(A^{\ast })$. Thus $r^{\ast }(A^{\ast })=2-r(A)$.

At the stated parameter, $p\mathrm{/}(1-p)=\sqrt{q}$. By [F1], $\mathrm{\mid }{A}^{\ast }\mathrm{\mid }=\mathrm{\mid }E\mathrm{\mid }-\mathrm{\mid }A\mathrm{\mid }$, and $\mathrm{\mid }E\mathrm{\mid }=2\mathrm{\mid }V\mathrm{\mid }$, the primal-to-dual weight ratio is $w(A)\mathrm{/}{w}^{\ast }(A^{\ast })=(p\mathrm{/}(1-p){)}^{2\mathrm{\mid }A\mathrm{\mid }-\mathrm{\mid }E\mathrm{\mid }}{q}^{k(A)-k^{\ast }(A^{\ast })}=q^{\mathrm{\mid }A\mathrm{\mid }-\mathrm{\mid }V\mathrm{\mid }+k(A)-k^{\ast }(A^{\ast })}$. The component identity in step 5.1 makes this exponent $r(A)-1$. This proves the asserted exact unnormalized ratio.

For a fixed dual configuration $B$, let $A$ be its unique complement-edge preimage. The definition of pushforward gives $\nu (B)=w(A)\mathrm{/}Z$, while ${\varphi }^{\ast }(B)=w^{\ast }(B)\mathrm{/}{Z}^{\ast }$. By steps 6.2 and 1.2 their ratio is $q^{r(A)-1}$. Step 6.1 gives $r(A)-1=1-r^{\ast }(B)$, proving the exact normalized likelihood. Since $0\le r^{\ast }(B)\le 2$ and $q\ge 1$, that likelihood belongs to $[q^{-1},q]\subseteq [q^{-2},q^2]$. Positivity in [F1] ensures every denominator is nonzero.

Multiply the weaker pointwise bounds of step 7.1 by ${\varphi }^{\ast }(B)$ and sum over $B\in \mathcal{E}$ to obtain the stated event comparison, including empty events. Any labelled graph isomorphism of the dual torus with the primal torus preserves the ordinary FK weights by the same cardinality and component-count argument as in step 1.2. Pushing both laws and the event through that bijection therefore preserves the event bounds. This completes all claims.