The Z-invariant Ising model via dimers

Abstract

The Z-invariant Ising model (Baxter in Philos Trans R Soc Lond A Math Phys Eng Sci 289(1359):315–346, 1978) is defined on an isoradial graph and has coupling constants depending on an elliptic parameter k. When $k=0$ the model is critical, and as k varies the whole range of temperatures is covered. In this paper we study the corresponding dimer model on the Fisher graph, thus extending our papers (Boutillier and de Tilière in Probab Theory Relat Fields 147:379–413, 2010; Commun Math Phys 301(2):473–516, 2011) to the fullZ-invariant case. One of our main results is an explicit, local formula for the inverse of the Kasteleyn operator. Its most remarkable feature is that it is an elliptic generalization of Boutillier and de Tilière (2011): it involves a local function and the massive discrete exponential function introduced in Boutillier et al. (Invent Math 208(1):109–189, 2017). This shows in particular that Z-invariance, and not criticality, is at the heart of obtaining local expressions. We then compute asymptotics and deduce an explicit, local expression for a natural Gibbs measure. We prove a local formula for the Ising model free energy. We also prove that this free energy is equal, up to constants, to that of the Z-invariant spanning forests of Boutillier et al.  (2017), and deduce that the two models have the same second order phase transition in k. Next, we prove a self-duality relation for this model, extending a result of Baxter to all isoradial graphs. In the last part we prove explicit, local expressions for the dimer model on a bipartite graph corresponding to the XOR version of this Z-invariant Ising model.

Appendices

Appendix A: Useful identities involving elliptic functions

In this section we list required identities satisfied by elliptic functions.

1.1 Identities for Jacobi’s elliptic functions

Change of argument Jacobi’s elliptic functions satisfy various addition formulas by quarter-periods and half-periods, among which (cf. [1, Table 16.8]):

Table 2 Addition formulas by quarter-periods and half-periods, taken from [1, 16.8]

1.2 Jacobi’s epsilon, zeta and related functions

The explicit expressions of the inverse operators of Theorems 11 and 37 involve the function $H$ defined as follows. In the remarks following Theorem 11, we also use the function $V$ mentioned below.

For $0<k^2<1$, let

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle H(u\vert k)=\displaystyle \frac{K'}{\pi }\Bigl \{\mathrm {E}\Bigl (\frac{u}{2}\Big \vert k\Bigr )+\frac{E'-K'}{K'}\frac{u}{2}\Bigr \},\\ \displaystyle V(u\vert k)=\displaystyle \frac{iK}{\pi }\Bigl \{\mathrm {E}\Bigl (\frac{u}{2}\Big \vert k\Bigr )-\frac{E}{K}\frac{u}{2}\Bigl \}, \end{array}\right. \end{aligned}$$

(66)

where $\mathrm {E}$ is Jacobi’s epsilon function: $\mathrm {E}(u)=\int _{0}^u {{\mathrm{ dn }}}^2(v\vert k)\mathrm {d}v$, see [1, 16.26.3]. For $k^2<0$, let

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle H(u\vert k)=H(k'u\vert k^*),\\ \displaystyle V(u\vert k)=V(k'u\vert k^*). \end{array}\right. \end{aligned}$$

(67)

Properties of these functions are presented below.

Lemma 42

The functions $H$ and $V$ admit jumps in the horizontal and vertical directions, respectively:

$$\begin{aligned} \left\{ \begin{array}{l} H(u+4K\vert k)-H(u\vert k)=1,\\ H(u+4i\mathfrak {R}K'\vert k)-H(u\vert k)=0, \end{array}\right. \qquad \left\{ \begin{array}{l} V(u+4K\vert k)-V(u\vert k)=0,\\ V(u+4i\mathfrak {R}K'\vert k)-V(u\vert k)=1. \end{array}\right. \end{aligned}$$

(68)

In the fundamental rectangle $[0,4K]+[0,4i\mathfrak {R}K']$, the function $H$ (resp. $V$) has a simple pole, at $2i\mathfrak {R}K'$, with residue $\frac{2\mathfrak {R}K'}{\pi }$ (resp. $\frac{2iK}{\pi }$). Moreover,

$$\begin{aligned} \lim _{k\rightarrow 0}H(u\vert k)=\frac{u}{2\pi },\qquad \lim _{k\rightarrow 0}V(u\vert k)=0. \end{aligned}$$

The following addition formulas hold:

$$\begin{aligned} H(v-u\vert k)= & {} H(v\vert k)-H(u\vert k)\nonumber \\&+\frac{k^2 \mathfrak {R}K'}{\pi } \left\{ \begin{array}{ll} \displaystyle {{\mathrm{sn }}}\Bigl (\frac{u}{2}\Big \vert k\Bigr ){{\mathrm{sn }}}\Bigl (\frac{v}{2}\Big \vert k\Bigr ){{\mathrm{sn }}}\Bigl (\frac{v-u}{2}\Big \vert k\Bigr ) &{}\quad \text {if }\; 0<k^2<1,\\ \displaystyle (-{k'}^2) {{\mathrm{sd}}}\Bigl (\frac{u}{2}\Big \vert k\Bigr ){{\mathrm{sd}}}\Bigl (\frac{v}{2}\Big \vert k\Bigr ){{\mathrm{sd}}}\Bigl (\frac{v-u}{2}\Big \vert k\Bigr ) &{}\quad \text {if }\; k^2<0, \end{array}\right. \end{aligned}$$

(69)

$$\begin{aligned} V(v-u\vert k)= & {} V(v\vert k)-V(u\vert k)\nonumber \\&+\frac{ik^2 K}{\pi } \left\{ \begin{array}{ll} \displaystyle {{\mathrm{sn }}}\Bigl (\frac{u}{2}\Big \vert k\Bigr ){{\mathrm{sn }}}\Bigl (\frac{v}{2}\Big \vert k\Bigr ){{\mathrm{sn }}}\Bigl (\frac{v-u}{2}\Big \vert k\Bigr ) &{}\quad \text {if }\; 0<k^2<1,\\ \displaystyle (-{k'}^2) {{\mathrm{sd}}}\Bigl (\frac{u}{2}\Big \vert k\Bigr ){{\mathrm{sd}}}\Bigl (\frac{v}{2}\Big \vert k\Bigr ){{\mathrm{sd}}}\Bigl (\frac{v-u}{2}\Big \vert k\Bigr ) &{}\quad \text {if }\; k^2<0. \end{array}\right. \nonumber \\ \end{aligned}$$

(70)

Proof

We first prove the lemma in the case $0<k^2<1$. All properties concerning $H$ are proved in [10, Lemmas 44 and 45]. The statements for $V$ follow similarly.

A slightly different proof would consist in using a reformulation of H and V in terms of Jacobi’s zeta functionZ:

$$\begin{aligned} H(u\vert k)=\frac{K'}{\pi }Z\Bigl (\frac{u}{2}\Big \vert k\Bigr )+\frac{u}{4K},\quad V(u\vert k)=\frac{iK}{\pi }Z\Bigl (\frac{u}{2}\Big \vert k\Bigr ). \end{aligned}$$

(71)

(Eq. (71) is a consequence of (66) and [1, 17.4.28 and 17.3.13].) Then we could use the numerous properties satisfied by Z (see in particular the addition formula [1, 17.4.35]) to derive Lemma 42.

In the case $k^2<0$, we use the definition (67) of $H$ and $V$, allowing to transfer all properties from the case $k^2>0$. In doing so we have to: consider $\mathfrak {R}K'$ in (68) because in the case $k^2<0$ the quarter-period $K'$ is non-real, see (46), and use the transformation of the ${{\mathrm{sn }}}$ function into the ${{\mathrm{sd}}}$ one under the dual transformation, see [1, 16.10]. $\square$

The Laplacian operator (6) uses the function $\mathrm {A}$, which satisfies the following properties.

Lemma 43

(Lemma 44 in [10]) The function $\mathrm {A}(\cdot \vert k)$ is odd and satisfies the following identities:

$$\begin{aligned} \bullet&\quad \mathrm {A}(v-u\vert k)=\mathrm {A}(v\vert k)-\mathrm {A}(u\vert k)-k'{{\mathrm{sc}}}(u\vert k){{\mathrm{sc}}}(v\vert k){{\mathrm{sc}}}(v-u\vert k), \end{aligned}$$

(72)

$$\begin{aligned} \bullet&\quad \mathrm {A}(u+2K\vert k)=\mathrm {A}(u\vert k). \end{aligned}$$

(73)

Appendix B: Some explicit integral computations

We gather here computations of some contour integrals appearing in the expression of the Kasteleyn operator, in the Fisher case of Sect. 3.

1.1 An important contour integral

The following result has been used when proving Theorem 11.

Lemma 44

One has

$$\begin{aligned} \frac{1}{2i\pi } \int _\Gamma \mathsf {f}(u+2K)\mathsf {f}(u)\mathrm {d}u=-\frac{1}{k'}, \end{aligned}$$

where $\Gamma$ is a vertical contour on ${\mathbb T}(k)$ winding once vertically on the torus and crossing the horizontal axis in the interval $[\alpha ,\alpha +2K]=\{x:\alpha \leqslant x \leqslant \alpha +2K\}$ and $\mathsf {f}(u)= {{\mathrm{nc}}}(\frac{u-\alpha }{2})$.

On the rectangle, the contour $\Gamma$ is supposed to cross the horizontal axis inside of the interval $[\alpha ,\alpha +2K]$. If the vertical contour crosses the horizontal axis in the other interval $[\alpha +2K,\alpha (+4K)]$, the integral is equal to $+\frac{1}{k'}$, as it corresponds to changing $\alpha$ into $\alpha +2K$ and ${{\mathrm{nc}}}(\frac{u-(\alpha +4K)}{2})=-{{\mathrm{nc}}}(\frac{u-\alpha }{2})$. Note further that Lemma 44 is independent of the choice of the angle $\alpha$ mod 4K, and the integrand $\mathsf {f}(u+2K)\mathsf {f}(u)$ is.

Proof

First, it follows from the change of variable $u\rightarrow u+2K$ and the above-mentioned property of ${{\mathrm{nc}}}$ that

$$\begin{aligned} \frac{1}{2i\pi } \int _\Gamma \mathsf {f}(u+2K)\mathsf {f}(u)\mathrm {d}u=-\frac{1}{2i\pi } \int _{\Gamma -2K} \mathsf {f}(u+2K)\mathsf {f}(u)\mathrm {d}u=\frac{1}{2i\pi } \int _{\widetilde{\Gamma }} \mathsf {f}(u+2K)\mathsf {f}(u)\mathrm {d}u, \end{aligned}$$

where $\widetilde{\Gamma }$ is the contour $\Gamma -2K$ crossed in the opposite direction of $\Gamma$. Further, using the $4iK'$-periodicity of the integrand, we deduce that

$$\begin{aligned} \frac{1}{2i\pi } \int _\Gamma \mathsf {f}(u+2K)\mathsf {f}(u)\mathrm {d}u=\frac{1}{2}\frac{1}{2i\pi } \int _{\mathscr { C}} \mathsf {f}(u+2K)\mathsf {f}(u)\mathrm {d}u, \end{aligned}$$

(74)

where $\mathscr { C}$ is the closed contour $(\Gamma -2iK')\bigcup (\widetilde{\Gamma }-2iK')\bigcup S_1\bigcup S_2$, $S_1$ and $S_2$ being the horizontal segments joining $(\Gamma -2iK')$ and $(\widetilde{\Gamma }-2iK')$, see Fig. 19.

Fig. 19

The contour $\mathscr { C}$ in (74) used in the proof of Lemma 44

The main point is that the contour integral in the right-hand side of (74) can be computed with the residue theorem: the only pole (of order 1) is at $\alpha$ and has residue $\frac{-2}{k'}$, see Table 2. Lemma 44 follows. $\square$

1.2 Inverse Kasteleyn operator at an edge

Here we compute the probability that a given edge $e={\mathbf x}{\mathbf y}$ of the isoradial graph $\mathsf {G}$ with rhombus half-angle $\theta _e$ appears in the high temperature contour expansion of the Ising model. In terms of dimers, it corresponds to the probability that the corresponding edge $\mathsf {e}=\mathsf {v}_j({\mathbf x})\mathsf {v}_\ell ({\mathbf y})=\mathsf {v}_j\mathsf {v}_\ell$ belongs to a dimer configuration of ${\mathsf {G}}^{\mathrm F}$. By Theorem 19 this probability is given by $\mathbb {P}_{\mathrm {dimer}}(\mathsf {e})=\mathsf {K}_{\mathsf {v}_{j},\mathsf {v}_{\ell }}\mathsf {K}^{-1}_{\mathsf {v}_{\ell },\mathsf {v}_{j}}$.

Lemma 45

One has

$$\begin{aligned} \mathsf {K}_{\mathsf {v}_{j},\mathsf {v}_{\ell }}\mathsf {K}^{-1}_{\mathsf {v}_{\ell },\mathsf {v}_{j}}=\frac{1}{2}-\frac{1-2H(2\theta _e)}{2{{\mathrm{cn }}}\theta _e}. \end{aligned}$$

Proof

Instead of $\mathsf {K}_{\mathsf {v}_{j},\mathsf {v}_{\ell }}\mathsf {K}^{-1}_{\mathsf {v}_{\ell },\mathsf {v}_{j}}$ we compute $\mathsf {K}_{\mathsf {v}_{\ell },\mathsf {v}_{j}}\mathsf {K}^{-1}_{\mathsf {v}_{j},\mathsf {v}_{\ell }}$. Both quantities are obviously equal, but the second one happens to be more convenient when applying the results of Sect. 3.

We start from the expression of $\mathsf {K}^{-1}$ given in (20) of Theorem 11 (note that by (19), $C_{\mathsf {v}_{j},\mathsf {v}_{\ell }}=0$):

$$\begin{aligned} \mathsf {K}^{-1}_{\mathsf {v}_{j},\mathsf {v}_{\ell }}= & {} \frac{i k'}{8\pi }\oint _{\mathscr { C}_{\mathsf {v}_{j},\mathsf {v}_{\ell }}}\mathsf {g}_{(\mathsf {v}_{j},\mathsf {v}_{\ell })}(u)H(u)\mathrm {d}u\\= & {} \frac{i k'}{8\pi }\oint _{\mathscr { C}_{\mathsf {v}_{j},\mathsf {v}_{\ell }}}\mathsf {f}_{\mathsf {v}_{j}}(u+2K)\mathsf {f}_{\mathsf {v}_{\ell }}(u){{\mathrm{\mathsf {e}}}}_{({\mathbf x},{\mathbf y})}(u)H(u)\mathrm {d}u. \end{aligned}$$

Using the harmonicity property (16) of $\mathsf {g}_{(\mathsf {z},\cdot )}(u)$ enables us to rewrite

$$\begin{aligned} \mathsf {K}_{\mathsf {v}_j,\mathsf {v}_\ell }\mathsf {f}_{\mathsf {v}_{\ell }}(u){{\mathrm{\mathsf {e}}}}_{({\mathbf x},{\mathbf y})}(u)= -[\mathsf {K}_{\mathsf {v}_j,{\mathsf {w}_j}}\mathsf {f}_{\mathsf {w}_j}(u)+\mathsf {K}_{\mathsf {v}_j,{\mathsf {w}_{j+1}}}\mathsf {f}_{\mathsf {w}_{j+1}}(u)]. \end{aligned}$$

By definition of the function $\mathsf {f}_{\mathsf {v}_j}$, see (13), we thus have

$$\begin{aligned}&\mathsf {K}_{\mathsf {v}_j,\mathsf {v}_\ell }\mathsf {f}_{\mathsf {v}_{j}}(u+2K)\mathsf {f}_{\mathsf {v}_{\ell }}(u){{\mathrm{\mathsf {e}}}}_{({\mathbf x},{\mathbf y})}(u)\\&\quad =-\,[\mathsf {K}_{\mathsf {v}_j,\mathsf {w}_j}\mathsf {f}_{\mathsf {w}_j}(u+2K)+\mathsf {K}_{\mathsf {w}_{j+1},\mathsf {v}_j}\mathsf {f}_{\mathsf {w}_{j+1}}(u+2K)] \\&\qquad \times [\mathsf {K}_{\mathsf {v}_j,{\mathsf {w}_j}}\mathsf {f}_{\mathsf {w}_j}(u)+\,\mathsf {K}_{\mathsf {v}_j,{\mathsf {w}_{j+1}}}\mathsf {f}_{\mathsf {w}_{j+1}}(u)]. \end{aligned}$$

Recalling that the orientation of the triangle $(\mathsf {w}_j,\mathsf {v}_j,\mathsf {w}_{j+1})$ is admissible, we moreover have $\mathsf {K}_{\mathsf {v}_j,{\mathsf {w}_j}}\mathsf {K}_{\mathsf {v}_j,\mathsf {w}_{j+1}}=-\mathsf {K}_{\mathsf {w}_j,\mathsf {w}_{j+1}}$, and since $\mathsf {K}_{\mathsf {v}_j,\mathsf {v}_\ell }=-\mathsf {K}_{\mathsf {v}_\ell ,\mathsf {v}_j}$, we have

$$\begin{aligned}&\mathsf {K}_{\mathsf {v}_\ell ,\mathsf {v}_j}\mathsf {f}_{\mathsf {v}_{j}}(u+2K)\mathsf {f}_{\mathsf {v}_{\ell }}(u){{\mathrm{\mathsf {e}}}}_{({\mathbf x},{\mathbf y})}(u)\nonumber \\&\quad = \mathsf {f}_{\mathsf {w}_j}(u+2K)\mathsf {f}_{\mathsf {w}_j}(u)-\mathsf {f}_{\mathsf {w}_{j+1}}(u+2K)\mathsf {f}_{\mathsf {w}_{j+1}}(u)\nonumber \\&\qquad +\,\mathsf {K}_{\mathsf {w}_j,\mathsf {w}_{j+1}}(\mathsf {f}_{\mathsf {w}_{j+1}}(u+2K)\mathsf {f}_{\mathsf {w}_{j}}(u)-\mathsf {f}_{\mathsf {w}_j}(u+2K)\mathsf {f}_{\mathsf {w}_{j+1}}(u)). \end{aligned}$$

(75)

With (75) the quantity $\mathsf {K}_{\mathsf {v}_{\ell },\mathsf {v}_{j}}\mathsf {K}^{-1}_{\mathsf {v}_{j},\mathsf {v}_{\ell }}$ is a sum of four terms. The first two ones are computed thanks to Lemma 44: with the choice of contour $\mathscr { C}_{\mathsf {v}_j,\mathsf {v}_\ell }$, we have

$$\begin{aligned} \frac{i k'}{8\pi }\oint _{\mathscr { C}_{\mathsf {v}_j,\mathsf {v}_\ell }}[\mathsf {f}_{\mathsf {w}_j}(u+2K)\mathsf {f}_{\mathsf {w}_j}(u)-\mathsf {f}_{\mathsf {w}_{j+1}}(u+2K)\mathsf {f}_{\mathsf {w}_{j+1}}(u)]H(u)\mathrm {d}u= 2\frac{i k'}{8\pi }\frac{-2\pi i}{k'}=\frac{1}{2}. \end{aligned}$$

To compute the last two terms in (75), namely

$$\begin{aligned} \frac{i k'}{8\pi }\oint _{\mathscr { C}_{\mathsf {v}_j,\mathsf {v}_\ell }}\mathsf {K}_{\mathsf {w}_j,\mathsf {w}_{j+1}}[\mathsf {f}_{\mathsf {w}_{j+1}}(u+2K)\mathsf {f}_{\mathsf {w}_{j}}(u)-\mathsf {f}_{\mathsf {w}_{j}}(u+2K)\mathsf {f}_{\mathsf {w}_{j+1}}(u)]H(u)\mathrm {d}u, \end{aligned}$$

(76)

recall that by (11)

$$\begin{aligned} \mathsf {f}_{\mathsf {w}_{j+1}}(u)= \textstyle {{\mathrm{nc}}}\Big (\frac{u-\alpha _{j+1}}{2}\Big )= \mathsf {K}_{\mathsf {w}_j,\mathsf {w}_{j+1}}{{\mathrm{nc}}}\Big (\frac{u-\alpha _j-2\theta _e}{2}\Big ), \end{aligned}$$

so that

$$\begin{aligned} \mathsf {K}_{\mathsf {w}_j,\mathsf {w}_{j+1}}\mathsf {f}_{\mathsf {w}_{j+1}}(u+2K)\mathsf {f}_{\mathsf {w}_{j}}(u)= {{\mathrm{nc}}}\Big (\frac{u+2K-\alpha _j-2\theta _e}{2}\Big ){{\mathrm{nc}}}\Big (\frac{u-\alpha _j}{2}\Big ). \end{aligned}$$

We therefore focus on the term

$$\begin{aligned} \int _{\mathscr { C}_{\mathsf {v}_{j},\mathsf {v}_{\ell }}}{{\mathrm{nc}}}\Bigl (\frac{u+2K-\alpha _j-2\theta _e}{2} \Bigr ){{\mathrm{nc}}}\Bigl (\frac{u-\alpha _j}{2} \Bigr )H(u)\frac{\mathrm {d}u}{2\pi i}, \end{aligned}$$

in which we set, without loss of generality, $\alpha _j=0$. This is equivalent to replace the function $H(u)$ by $H(u-\alpha _j)$, which is possible because both functions satisfy the same jump conditions stated in Lemma 42. There are three residues, at $2\theta _e$, 2K and $2i\mathfrak {R}K'$. We thus have

$$\begin{aligned} \int _{\mathscr { C}_{\mathsf {v}_{j},\mathsf {v}_{\ell }}}&{{\mathrm{nc}}}\Bigl (\frac{u+2K-2\theta _e}{2} \Bigr ){{\mathrm{nc}}}\Bigl (\frac{u}{2} \Bigr )H(u)\frac{\mathrm {d}u}{2\pi i}\nonumber \\&\qquad =\frac{-2}{k'}\frac{H(2\theta _e)}{{{\mathrm{cn }}}\theta _e}+\frac{-2}{k'}\frac{H(2K)}{{{\mathrm{cn }}}(2K-\theta _e)}+\frac{2\mathfrak {R}K'}{\pi }{{\mathrm{nc}}}(i\mathfrak {R}K'+K-\theta _e){{\mathrm{nc}}}(i\mathfrak {R}K')\nonumber \\&\qquad =\frac{2}{k'}\frac{H(2K)-H(2\theta _e)}{{{\mathrm{cn }}}\theta _e}, \end{aligned}$$

(77)

since ${{\mathrm{nc}}}(i\mathfrak {R}K')=0$. The same reasoning as above gives

$$\begin{aligned}&\int _{\mathscr { C}_{\mathsf {v}_{j},\mathsf {v}_{\ell }}}{{\mathrm{nc}}}\Bigl (\frac{u-2\theta _e}{2} \Bigr ){{\mathrm{nc}}}\Bigl (\frac{u+2K}{2} \Bigr )H(u)\frac{\mathrm {d}u}{2\pi i}\nonumber \\&\qquad =\frac{2}{k'}\frac{H(2\theta _e-2K)-H(0)}{{{\mathrm{cn }}}\theta _e}+\frac{2\mathfrak {R}K'}{\pi }{{\mathrm{nc}}}(i\mathfrak {R}K'+K){{\mathrm{nc}}}(i\mathfrak {R}K'-\theta _e), \end{aligned}$$

(78)

which may be slightly simplified, using that $H(0)=0$. Thanks to (77), (78) and Table 2, we obtain that (76) equals

$$\begin{aligned} \mathsf {K}_{\mathsf {v}_{\ell },\mathsf {v}_{j}}\mathsf {K}^{-1}_{\mathsf {v}_{j},\mathsf {v}_{\ell }}=\frac{1}{2}+\frac{H(2\theta _e)-H(2K)+H(2\theta _e-2K)}{2{{\mathrm{cn }}}\theta _e}+\frac{\mathfrak {R}K' k^2}{2\pi }{{\mathrm{sd}}}\theta _e. \end{aligned}$$

The addition formula (69) results in

$$\begin{aligned} H(2\theta _e-2K)-&H(2K)\\&=H(2\theta _e-4K)-\frac{\mathfrak {R}K'k^2}{\pi }\left\{ \begin{array}{ll} {{\mathrm{sn }}}K {{\mathrm{sn }}}(\theta _e-K){{\mathrm{sn }}}(\theta _e-2K) &{} \text {if } k^2>0\\ (-k'^2){{\mathrm{sd}}}K {{\mathrm{sd}}}(\theta _e-K){{\mathrm{sd}}}(\theta _e-2K)&{} \text {if } k^2<0\end{array}\right. \\&=H(2\theta _e)-1-\frac{\mathfrak {R}K'k^2}{\pi }{{\mathrm{cn }}}\theta _e{{\mathrm{sd}}}\theta _e. \end{aligned}$$

The proof is complete. $\square$

Fig. 20

Compatibility around a rhombus

Using that $\lim _{k\rightarrow 0}H(u)=\frac{u}{2\pi }$, see Lemma 42, we find that

$$\begin{aligned} \lim _{k\rightarrow 0}\mathsf {K}_{\mathsf {v}_{j},\mathsf {v}_{\ell }}\mathsf {K}^{-1}_{\mathsf {v}_{\ell },\mathsf {v}_{j}}=\frac{1}{2}-\frac{\pi -2\theta _e}{2\pi \cos \theta _e}. \end{aligned}$$

This is in accordance with the computation of [8, Appendix A] for the case $k=0$. Indeed, the dimer model considered in the latter paper corresponds to complementary polygon configurations meaning that the probability (30) is 1 minus the probability of the paper [8].

Appendix C: Proof of Lemma 7

This section consists in the proof of Lemma 7 stating that the angles $(\overline{\alpha }_j({\mathbf x}))$ defined in Eqs. (11) and (12) are indeed well defined mod $4\pi$.

Proof

We first need to check that angles around a rhombus corresponding to two neighboring decorations are well defined, see Fig. 20.

We want to prove that the following is equal to 0 mod $4\pi$:

$$\begin{aligned} (\overline{\alpha }_{\ell +1}-\overline{\alpha }_{\ell })- (\overline{\alpha }_{j+1}-\overline{\alpha }_{j})+(\overline{\alpha }_{\ell }-\overline{\alpha }_{j})- (\overline{\alpha }_{\ell +1}-\overline{\alpha }_{j+1}). \end{aligned}$$

By definition of the angles within a decoration we have, mod $4\pi$,

$$\begin{aligned} (\overline{\alpha }_{\ell +1}-\overline{\alpha }_{\ell })-(\overline{\alpha }_{j+1}-\overline{\alpha }_{j})= {\left\{ \begin{array}{ll} 0 &{} \text {if }{{\mathrm{co}}}(\mathsf {w}_\ell ,\mathsf {w}_{\ell +1})+{{\mathrm{co}}}(\mathsf {w}_j,\mathsf {w}_{j+1}) \text { is even,}\\ 2\pi &{} \text {if }{{\mathrm{co}}}(\mathsf {w}_\ell ,\mathsf {w}_{\ell +1})+{{\mathrm{co}}}(\mathsf {w}_j,\mathsf {w}_{j+1}) \text { is odd}. \end{array}\right. } \end{aligned}$$

By definition of the angles in neighboring decorations we have, mod $4\pi$,

$$\begin{aligned} (\overline{\alpha }_{\ell }-\overline{\alpha }_{j})-(\overline{\alpha }_{\ell +1}-\overline{\alpha }_{j+1})= {\left\{ \begin{array}{ll} 0&{}\text {if } {{\mathrm{co}}}(\mathsf {w}_{j+1},\mathsf {v}_j,\mathsf {w}_j)+{{\mathrm{co}}}(\mathsf {w}_{\ell +1},\mathsf {v}_\ell ,\mathsf {w}_\ell ) \text { is even,}\\ 2\pi &{}\text {if } {{\mathrm{co}}}(\mathsf {w}_{j+1},\mathsf {v}_j,\mathsf {w}_j)+{{\mathrm{co}}}(\mathsf {w}_{\ell +1},\mathsf {v}_\ell ,\mathsf {w}_\ell ) \text { is odd}. \end{array}\right. } \end{aligned}$$

This implies that, mod $4\pi$, we have:

$$\begin{aligned}&(\overline{\alpha }_{\ell +1}-\overline{\alpha }_{\ell })- (\overline{\alpha }_{j+1}-\overline{\alpha }_{j})+ (\overline{\alpha }_{\ell }-\overline{\alpha }_{j})- (\overline{\alpha }_{\ell +1}-\overline{\alpha }_{j+1})\\&\qquad = {\left\{ \begin{array}{ll} 0&{}\text {if }{{\mathrm{co}}}(\mathsf {w}_{j+1},\mathsf {v}_j,\mathsf {w}_j,\mathsf {w}_{j+1})+{{\mathrm{co}}}(\mathsf {w}_{\ell +1},\mathsf {v}_\ell ,\mathsf {w}_\ell ,\mathsf {w}_{\ell +1}) \text { is even,} \\ 2\pi &{} \text {if }{{\mathrm{co}}}(\mathsf {w}_{j+1},\mathsf {v}_j,\mathsf {w}_j,\mathsf {w}_{j+1})+{{\mathrm{co}}}(\mathsf {w}_{\ell +1},\mathsf {v}_\ell ,\mathsf {w}_\ell ,\mathsf {w}_{\ell +1}) \text { is odd}. \end{array}\right. } \end{aligned}$$

But since the orientation of the graph is admissible, we have that ${{\mathrm{co}}}(\mathsf {w}_{j+1},\mathsf {v}_j,\mathsf {w}_j,\mathsf {w}_{j+1})$ and ${{\mathrm{co}}}(\mathsf {w}_{\ell +1},\mathsf {v}_\ell ,\mathsf {w}_\ell ,\mathsf {w}_{\ell +1})$ are odd, implying that the sum is even, thus concluding the proof for angles around a rhombus.

Fig. 21

Notation for a cycle C arising from the boundary of a face of $\mathsf {G}$

We now need to prove that when doing the inductive procedure around a cycle of ${\mathsf {G}}^{\mathrm F}$, we recover the same angle mod $4\pi$. There are two types of cycles to consider: inner cycles of decorations and cycles arising from boundary of faces of $\mathsf {G}$.

Consider a cycle of a decoration corresponding to a vertex ${\mathbf x}$ of $\mathsf {G}$ of degree d, and let n be the number of edges of the inner cycle oriented clockwise. By definition of the angles, we have:

$$\begin{aligned} \overline{\alpha }_{d+1}- \overline{\alpha }_{1}=\sum _{j=1}^d (\overline{\alpha }_{j+1}-\overline{\alpha }_{j}) =\sum _{j=1}^d 2\overline{\theta }_{j}+2\pi n=2\pi (n+1). \end{aligned}$$

Since the orientation of the edges is admissible, n is odd, thus proving that $\overline{\alpha }_{d+1}-\overline{\alpha }_{1}=0\ [4\pi ]$.

Now consider a cycle C arising from the boundary ${\mathbf x}_1,\dots ,{\mathbf x}_m$ of a face of $\mathsf {G}$, with vertices labeled in clockwise order, see also Fig. 21. Up to a relabeling of vertices of the decorations, this cycle can be written as

$$\begin{aligned} C=(\mathsf {w}_{2}({\mathbf x}_1),\mathsf {v}_2({\mathbf x}_1),\mathsf {v}_{1}({\mathbf x}_2),\mathsf {w}_{2}({\mathbf x}_2),\dots ,\mathsf {v}_{1}({\mathbf x}_1)). \end{aligned}$$

Using the definition of the angles within a decoration and in neighboring ones we deduce that, mod $4\pi$, we have:

$$\begin{aligned} \overline{\alpha }_{2}({\mathbf x}_j)-\overline{\alpha }_{2}({\mathbf x}_{j+1})= {\left\{ \begin{array}{ll} -\pi -2\theta _1({\mathbf x}_j) &{}\text {if } {{\mathrm{co}}}(\mathsf {w}_2({\mathbf x}_j),\mathsf {v}_2({\mathbf x}_j),\mathsf {v}_1({\mathbf x}_{j+1}),\mathsf {w}_2({\mathbf x}_{j+1})) \text { is odd,}\\ \pi -2\theta _1({\mathbf x}_j) &{}\text {if } {{\mathrm{co}}}(\mathsf {w}_2({\mathbf x}_j),\mathsf {v}_2({\mathbf x}_j),\mathsf {v}_1({\mathbf x}_{j+1}),\mathsf {w}_2({\mathbf x}_{j+1})) \text { is even}. \end{array}\right. } \end{aligned}$$

Let n(C) denote the number of portions of the cycle C where

$$\begin{aligned} {{\mathrm{co}}}(\mathsf {w}_2({\mathbf x}_j),\mathsf {v}_2({\mathbf x}_j),\mathsf {v}_1({\mathbf x}_{j+1}),\mathsf {w}_2({\mathbf x}_{j+1})) \end{aligned}$$

is odd. Then, writing ${\mathbf x}_{m+1}={\mathbf x}_1$, we have:

$$\begin{aligned} \alpha _2({\mathbf x}_{1})-\alpha _2({\mathbf x}_{m+1})=\sum _{j=1}^m \alpha _2({\mathbf x}_j)-\alpha _2({\mathbf x}_{j+1}) =\sum _{j=1}^m(\pi -2\theta _1({\mathbf x}_j))-2\pi n(C). \end{aligned}$$

Since $\sum _{j=1}^m(\pi -2\theta _1({\mathbf x}_j))$ is the sum of angles at the center of the cycle, it is equal to $2\pi$. The orientation of the cycle being admissible, n(C) is odd, thus concluding the proof. $\square$