## 3.2 Z-massive Laplacian operators

In the infinite case, we consider the Z-massive Laplacian associated to the isoradial graph G introduced in [11], whose definition is recalled in Section 2.5.6. We also consider the dual Z-massive Laplacian $\Delta^{m,*}$ of the dual isoradial graph G*.

The purpose of this section is to define the finite versions of the Z-massive Laplacian and dual Z-massive Laplacian that are used in this paper; we introduce two operators. The first is the finite version of the dual Z-massive Laplacian, denoted $\Delta^{m,*}$ as in the infinite case. It acts on $C^{V*}$, where recall that $V^*$ is the vertex set of the restricted dual G*, whose boundary vertices are defined to be those adjacent to the vertex o in $\bar{G}^*$. For every vertex f of G* of degree d in $\bar{G}^*$, denote its neighbors by $f_1, \dots, f_d$. Let $\underline{d}$ be the degree of f in G*, then if f is a boundary vertex of G* we have $\underline{d} \neq d$, and we label the vertices so that the first $\underline{d}$ ones are common to G* and $\bar{G}^*$. For every $j \in \{1, \dots, \underline{d}\}$, $(f, f_j)$ is an edge of G* and we let $\bar{\theta}_j^*$ denote its half-angle in $G^\circ$; for every $j \in \{\underline{d}+1, \dots, d\}$, $(f, f_j) = (f, \underline{o})$ corresponds to an edge $fw_j$ of the double graph $G^D$, where $w_j$ is the unique vertex of $G^D$ which belongs to the edge $\underline{o}$, we let $\bar{\theta}_j^*$ denote the half-angle of the edge $(f, w_j)$ in $G^{\circ/2}$. Then, non-zero coefficients of the finite version of the dual Z-massive Laplacian $\Delta^{m,*}$ are given by,

$$ \Delta_{f,f'}^{m,*} = \begin{cases} -\mathrm{sc}(\theta^*) & \text{if } (f, f') \text{ is an edge of } G^* \\ \sum_{j=1}^{d} A(\theta_j^*) & \text{if } f' = f \text{ is a vertex of } G^*. \end{cases} $$

The corresponding conductances $\rho^*$ and masses $m^*$ are as defined in Section 2.4.

The second operator acts on $C^{V'}$; it is denoted $\Delta^{m,\partial}(u)$ and we restrict to $u \in C'$. It is not the natural finite version of the operator $\Delta^m$ but has boundary conditions inherited from those of the $Z^u$-Dirac operator $K^\partial(u)$, see the proof of Theorem 3.6 below. We use the notation of the infinite Z-massive Laplacian operator $\Delta^m$, see Section 2.5.6, and also the notation of Figure 10 for the boundary rhombus pairs of $\mathbb{R}^\partial$.

In the following, the two boundary vertices $v^r, v^\ell$ of $\mathbb{R}^\partial$ enter the same framework, and we denote such a vertex by $v^\partial$. For every $v^\partial$ of $\mathbb{R}^\partial$ of degree d in G, let $v_1, \dots, v_d$ denote its neighbors in cclw order with $v_1$ on the boundary of G on the left of $v^\partial$, see Figure 12 (right). Note that if $v^\partial = v^{\partial,r}$ belongs to the root pair, then $v^c = r$ is considered as a neighbor of $v^{\partial,r}$. The coefficients of $\Delta^{m,\partial}(u)$ differing from those of $\Delta^m$ are those corresponding to edges $(v^c, v^\ell)$ of $\mathbb{R}^{\partial,r}$, and to vertices $v^r, v^\ell, v^c$ of $\mathbb{R}^\partial$ such that $v^c \neq r$:

$$ \Delta^{m,\partial}(u)_{v,v'} = \begin{cases} -\mathrm{sc}(\theta^\partial) \frac{\mathrm{cd}(u_{\beta^r})}{\mathrm{cd}(u_{\alpha^\ell})} & \text{if } (v,v') \in \{(v^c, v^\ell) \in \mathbb{R}^{\partial,r}\} \\ k' \mathrm{sc}(\theta^\partial) \frac{\mathrm{nd}(u_{\beta^\ell}) \mathrm{nd}(u_{\beta^r})}{\mathrm{cn}(u_{\alpha^\ell}) \mathrm{cn}(u_{\alpha^\ell})} (\mathrm{cn}(u_{\beta^r}) + \mathrm{cn}(u_{\alpha^\ell})) & \text{if } v' = v = v^c \in \{v^c \in \mathbb{R}^{\partial,r}\} \\ k' \sum_{j=1}^d \mathrm{sc}(\theta_j) \mathrm{nd}(u_{\alpha_j}) \mathrm{nd}(u_{\alpha_{j+1}}) & \text{if } v' = v = v^\partial \in \{v^r, v^\ell \in \mathbb{R}^\partial\}. \end{cases} \quad (3.6) $$

The corresponding conductances and masses are denoted $\rho^\partial(u)$ and $m^\partial(u)$. Note that the diagonal term $\Delta^{m,\partial}(u)_{v^\partial, v^\partial}$ has the same form as that of the natural finite version of the Z-massive Laplacian.

In order to state Theorem 3.6 relating the $Z^u$-Dirac and the Z-massive Laplacian operators, in the finite case we need to introduce the matrix $Q(u)$. It has rows indexed by black vertices of $G^r$ and columns by those of the restricted dual G*. The only non-zero coefficients of $Q(u)$ are for rows corresponding to vertices $v^c$ of boundary rhombus pairs of $\mathbb{R}^{\partial,r}$. For such a vertex $v^c$, there is one non-zero coefficient, corresponding to the column $f^c$, given by

$$ Q(u)_{v^c, f^c} = -i \frac{\mathrm{nd}(u_{\beta^\ell})}{\mathrm{cd}(u_{\alpha^\ell})} (\mathrm{cd}(u_{\beta^r}) - \mathrm{cd}(u_{\alpha^\ell})). $$
### 3.3 Relating the $Z^u$-Dirac and the Z-massive Laplacian operators

Theorem 3.6 below relates the $Z^u$-Dirac operator, the Z-massive Laplacian and the dual Z-massive Laplacian in the infinite and finite cases. In the infinite case, this extends to the full Z-invariant case the results of Section 6 of [45] which correspond to $k = 0$.

**Theorem 3.6.**

* Infinite case. Let $u \in C$, then the $Z^u$-Dirac operator $K(u)$, the Z-massive Laplacian $\Delta^m$ and the dual $\Delta^{m,*}$ satisfy the following identity:

$$ \overline{K(u)}^t K(u) = k' \begin{pmatrix} \Delta^m & 0 \\ 0 & \Delta^{m,*} \end{pmatrix}. $$

* Finite case. Let $u \in C'$, then the $Z^u$-Dirac operators $K(u)$, $K^\partial(u)$, the Z-massive Laplacian $\Delta^{m,\partial}(u)$ and the dual $\Delta^{m,*}$ satisfy the following identity:

$$ \overline{K^\partial(u)}^t K(u) = k' \begin{pmatrix} \Delta^{m,\partial}(u) & Q(u) \\ 0 & \Delta^{m,*} \end{pmatrix}. $$

**Remark 3.7.**

* As a consequence of Theorem 3.6 in the infinite case, we have that if $F \in C^B$ is a $Z^u$-holomorphic function, then $F|_V$ is Z-massive harmonic on $G$ and $F|_{V^*}$ is Z-massive harmonic on $G^*$.

* From the proof of Theorem 3.6 it follows that, in the finite case, we also have the following matrix relation:

$$ \overline{K(u)}^t K(u) = k' \begin{pmatrix} \Delta^m(u) & 0 \\ 0 & \Delta^{m,*} \end{pmatrix}, $$

where $\Delta^m(u)$ is the natural finite version of the Z-massive Laplacian, i.e., where diagonal coefficients of boundary vertices of $\Delta^m(u)$ are defined as in the last line of (3.6).

*Proof.* As in the critical case [45], the proof consists in showing that matrix coefficients on both sides are equal. We separate the infinite case together with the part of the finite case which enters the infinite framework, from the part of the finite case which is specific. To simplify notation, we omit the argument $u$ from the operators.

When two black vertices $x, y$ of $G^D$, resp. $G^{D,r}$, are at distance more than two, the corresponding coefficient $[\overline{K^t K}]_{x,y}$, resp. $[\overline{K^{\partial^t} K}]_{x,y}$, is trivially equal to 0, we thus suppose that $x, y$ are at distance 2 or 0.

Infinite case and part of the finite case. Let $(x, y) = (v, f)$ or $(f, v)$, with $v$ a vertex of $G$ and $f$ a vertex of $G^*$, at distance two. Denote by $w, w'$ the two white vertices of the quadrangle of $G^D/G^{D,r}$ containing $v$ and $f$; let $e^{i\bar{\alpha}}, e^{i\bar{\beta}}$, resp. $e^{i\bar{\alpha}'}, e^{i\bar{\beta}'}$, be the rhombus vectors of $G^{\partial/2}$ associated to the edge $(w, v)$, resp. $(w', v)$, and let $\bar{\theta}$, resp. $\bar{\theta}'$, be the corresponding half-angle, see Figure 11. In the finite case, we moreover suppose that $(v, f) \neq (v^c, f^c)$ for all boundary rhombus pairs of $\mathcal{R}^{\partial,r}$, then we have $K^\partial = K$ for all coefficients involved. Let us prove that $[\overline{K^t K}]_{v,f} = \overline{K}_{w,v}K_{w,f} + \overline{K}_{w',v}K_{w',f} = 0$. By definition of $K$, we have

$$ \begin{aligned} \overline{K}_{w,v} K_{w,f} &= e^{-i \frac{\bar{\alpha}+\bar{\beta}}{2}} [\mathrm{sc}(\theta) \mathrm{dn}(u_{\alpha}) \mathrm{dn}(u_{\beta})]^{\frac{1}{2}} e^{i \frac{\bar{\beta}-\pi+\bar{\alpha}}{2}} [k' \mathrm{cs}(\theta) \mathrm{nd}(u_{\beta-2K}) \mathrm{nd}(u_{\alpha})]^{\frac{1}{2}} \\ &= -i \mathrm{dn}(u_{\beta}) \end{aligned} $$

$$ \begin{aligned} \overline{K}_{w',v} K_{w',f} &= e^{-i \frac{\bar{\alpha}'+\bar{\beta}'}{2}} [\mathrm{sc}(\theta') \mathrm{dn}(u_{\alpha'}) \mathrm{dn}(u_{\beta'})]^{\frac{1}{2}} e^{i \frac{\bar{\beta}'+\bar{\alpha}'+\pi}{2}} [k' \mathrm{cs}(\theta') \mathrm{nd}(u_{\beta'}) \mathrm{nd}(u_{\alpha'+2K})]^{\frac{1}{2}} \\ &= i \mathrm{dn}(u_{\alpha'}). \end{aligned} $$
### 3.3 Relating the $Z^u$-Dirac and the Z-massive Laplacian operators

Theorem 3.6 below relates the $Z^u$-Dirac operator, the Z-massive Laplacian and the dual Z-massive Laplacian in the infinite and finite cases. In the infinite case, this extends to the full Z-invariant case the results of Section 6 of [45] which correspond to $k = 0$.

**Theorem 3.6.**

* Infinite case. Let $u \in C$, then the $Z^u$-Dirac operator $K(u)$, the Z-massive Laplacian $\Delta^m$ and the dual $\Delta^{m,*}$ satisfy the following identity:

$$ \overline{K(u)}^t K(u) = k' \begin{pmatrix} \Delta^m & 0 \\ 0 & \Delta^{m,*} \end{pmatrix}. $$

* Finite case. Let $u \in C'$, then the $Z^u$-Dirac operators $K(u)$, $K^\partial(u)$, the Z-massive Laplacian $\Delta^{m,\partial}(u)$ and the dual $\Delta^{m,*}$ satisfy the following identity:

$$ \overline{K^\partial(u)}^t K(u) = k' \begin{pmatrix} \Delta^{m,\partial}(u) & Q(u) \\ 0 & \Delta^{m,*} \end{pmatrix}. $$

**Remark 3.7.**

* As a consequence of Theorem 3.6 in the infinite case, we have that if $F \in C^B$ is a $Z^u$-holomorphic function, then $F|_V$ is Z-massive harmonic on $G$ and $F|_{V^*}$ is Z-massive harmonic on $G^*$.

* From the proof of Theorem 3.6 it follows that, in the finite case, we also have the following matrix relation:

$$ \overline{K(u)}^t K(u) = k' \begin{pmatrix} \Delta^m(u) & 0 \\ 0 & \Delta^{m,*} \end{pmatrix}, $$

where $\Delta^m(u)$ is the natural finite version of the Z-massive Laplacian, i.e., where diagonal coefficients of boundary vertices of $\Delta^m(u)$ are defined as in the last line of (3.6).

*Proof.* As in the critical case [45], the proof consists in showing that matrix coefficients on both sides are equal. We separate the infinite case together with the part of the finite case which enters the infinite framework, from the part of the finite case which is specific. To simplify notation, we omit the argument $u$ from the operators.

When two black vertices $x, y$ of $G^D$, resp. $G^{D,r}$, are at distance more than two, the corresponding coefficient $[\overline{K^t K}]_{x,y}$, resp. $[\overline{K^{\partial^t} K}]_{x,y}$, is trivially equal to 0, we thus suppose that $x, y$ are at distance 2 or 0.

Infinite case and part of the finite case. Let $(x, y) = (v, f)$ or $(f, v)$, with $v$ a vertex of $G$ and $f$ a vertex of $G^*$, at distance two. Denote by $w, w'$ the two white vertices of the quadrangle of $G^D/G^{D,r}$ containing $v$ and $f$; let $e^{i\bar{\alpha}}, e^{i\bar{\beta}}$, resp. $e^{i\bar{\alpha}'}, e^{i\bar{\beta}'}$, be the rhombus vectors of $G^{\partial/2}$ associated to the edge $(w, v)$, resp. $(w', v)$, and let $\bar{\theta}$, resp. $\bar{\theta}'$, be the corresponding half-angle, see Figure 11. In the finite case, we moreover suppose that $(v, f) \neq (v^c, f^c)$ for all boundary rhombus pairs of $\mathcal{R}^{\partial,r}$, then we have $K^\partial = K$ for all coefficients involved. Let us prove that $[\overline{K^t K}]_{v,f} = \overline{K}_{w,v}K_{w,f} + \overline{K}_{w',v}K_{w',f} = 0$. By definition of $K$, we have

$$ \begin{aligned} \overline{K}_{w,v} K_{w,f} &= e^{-i \frac{\bar{\alpha} + \bar{\beta}}{2}} [\mathrm{sc}(\theta) \mathrm{dn}(u_{\alpha}) \mathrm{dn}(u_{\beta})]^{\frac{1}{2}} e^{i \frac{\bar{\beta} - \pi + \bar{\alpha}}{2}} [k' \mathrm{cs}(\theta) \mathrm{nd}(u_{\beta-2K}) \mathrm{nd}(u_{\alpha})]^{\frac{1}{2}} \\ &= -i \mathrm{dn}(u_{\beta}) \end{aligned} $$

$$ \begin{aligned} \overline{K}_{w',v} K_{w',f} &= e^{-i \frac{\bar{\alpha}' + \bar{\beta}'}{2}} [\mathrm{sc}(\theta') \mathrm{dn}(u_{\alpha'}) \mathrm{dn}(u_{\beta'})]^{\frac{1}{2}} e^{i \frac{\bar{\beta}' + \alpha' + \pi}{2}} [k' \mathrm{cs}(\theta') \mathrm{nd}(u_{\beta'}) \mathrm{nd}(u_{\alpha'+2K})]^{\frac{1}{2}} \\ &= i \mathrm{dn}(u_{\alpha'}). \end{aligned} $$
<figure><img src="image_13.png" /><figcaption>Figure 11: Notation for computing $[\bar{K}^t K]_{v,f}$.</figcaption></figure>

Since $\bar{\beta} = \bar{\alpha}'[2\pi]$, we deduce that $[\bar{K}^t K]_{v,f} = 0$. We also have $[\bar{K}^t K]_{f,v} = 0$, because $[\bar{K}^t K]_{f,v} = [\bar{K}^t K]_{v,f}$.

Next, let $(x, y) = (v_1, v_2)$ be an edge of the graph $G/G^r$ corresponding to a path $v_1, w, v_2$ of $G^p/G^{D,r}$. In the finite case, we moreover suppose that $(v_1, v_2) \neq (v^c, v^l)$ for all boundary rhombus pairs of $\mathcal{R}^{\theta,r}$, then we have $K^{\theta} = K$ for all coefficients involved. Using the notation of Figure 9, we have

$$ \begin{aligned} [\bar{K}^t K]_{v_1, v_2} &= \bar{K}_{w_1, v_1} K_{w_2, v_2} \\ &= e^{-i \frac{\bar{\alpha} + \bar{\beta} + 2\pi}{2}} [\mathrm{sc}(\theta) \mathrm{dn}(u_{\alpha+2K}) \mathrm{dn}(u_{\beta+2K})]^{\frac{1}{2}} e^{i \frac{\bar{\alpha} + \bar{\beta}}{2}} [\mathrm{sc}(\theta) \mathrm{dn}(u_{\alpha}) \mathrm{dn}(u_{\beta})]^{\frac{1}{2}} \\ &= -k' \mathrm{sc}(\theta) = k' \Delta_{v_1, v_2}^m. \end{aligned} $$

We now handle the case where $x, y$ are at distance 0, i.e., when $x = y$ and we suppose that $x$ is a vertex $v$ of $G$ of degree $d$. In the finite case, we moreover suppose that $v \neq v^c$ for all boundary rhombus pairs of $\mathcal{R}^{\theta,r}$, then we have $K^{\theta} = K$ for all coefficients involved. Using the notation of Figure 12 we have:

$$ \begin{aligned} [\bar{K}^t K]_{v,v} &= \sum_{j=1}^{d} \bar{K}_{w_j, v} K_{w_j, v} = \sum_{j=1}^{d} |K_{w_j, v}|^2 \\ &= \sum_{j=1}^{d} \mathrm{sc}(\theta_j) \mathrm{dn}(u_{\alpha_j+2K}) \mathrm{dn}(u_{\alpha_{j+1}+2K}) = (k')^2 \sum_{j=1}^{d} \mathrm{sc}(\theta_j) \mathrm{nd}(u_{\alpha_j}) \mathrm{nd}(u_{\alpha_{j+1}}). \end{aligned} $$

<figure><img src="image_14.png" /><figcaption>Figure 12: Notation for computing $[\bar{K}^t K]_{v,v}$, when $v$ is not a boundary vertex (left), and when $v$ is a boundary vertex $v^\theta \in \{v^\ell, v^r \in \mathcal{R}^\theta\}$ (the additional half-rhombi along the boundary are pictures in grey).</figcaption></figure>

In the finite case, when $v \in \{v^\ell, v^r \in \mathcal{R}^\theta\}$, both vertices enter the same framework and we denote $v$ by $v^\theta$. We stop the computation here, and returning to (3.6), we have that
<figure><img src="image_13.png" /><figcaption>Figure 11: Notation for computing $[\bar{K}^t K]_{v,f}$.</figcaption></figure>

Since $\bar{\beta} = \bar{\alpha}'[2\pi]$, we deduce that $[\bar{K}^t K]_{v,f} = 0$. We also have $[\bar{K}^t K]_{f,v} = 0$, because $[\bar{K}^t K]_{f,v} = [\bar{K}^t K]_{v,f}$.

Next, let $(x, y) = (v_1, v_2)$ be an edge of the graph $G/G^r$ corresponding to a path $v_1, w, v_2$ of $G^r/G^{D,r}$. In the finite case, we moreover suppose that $(v_1, v_2) \neq (v^c, v^l)$ for all boundary rhombus pairs of $\mathcal{R}^{\theta;r}$, then we have $K^{\theta} = K$ for all coefficients involved. Using the notation of Figure 9, we have

$$ \begin{aligned} [\bar{K}^t K]_{v_1, v_2} &= \bar{K}_{w_1, v_1} K_{w_2, v_2} \\ &= e^{-i \frac{\bar{\alpha} + \bar{\beta} + 2\pi}{2}} [\mathrm{sc}(\theta) \mathrm{dn}(u_{\alpha+2K}) \mathrm{dn}(u_{\beta+2K})]^{\frac{1}{2}} e^{i \frac{\bar{\alpha} + \bar{\beta}}{2}} [\mathrm{sc}(\theta) \mathrm{dn}(u_{\alpha}) \mathrm{dn}(u_{\beta})]^{\frac{1}{2}} \\ &= -k' \mathrm{sc}(\theta) = k' \Delta_{v_1, v_2}^m. \end{aligned} $$

We now handle the case where $x, y$ are at distance 0, i.e., when $x = y$ and we suppose that $x$ is a vertex $v$ of $G$ of degree $d$. In the finite case, we moreover suppose that $v \neq v^c$ for all boundary rhombus pairs of $\mathcal{R}^{\theta;r}$, then we have $K^{\theta} = K$ for all coefficients involved. Using the notation of Figure 12 we have:

$$ \begin{aligned} [\bar{K}^t K]_{v,v} &= \sum_{j=1}^{d} \bar{K}_{w_j, v} K_{w_j, v} = \sum_{j=1}^{d} |K_{w_j, v}|^2 \\ &= \sum_{j=1}^{d} \mathrm{sc}(\theta_j) \mathrm{dn}(u_{\alpha_j+2K}) \mathrm{dn}(u_{\alpha_{j+1}+2K}) = (k')^2 \sum_{j=1}^{d} \mathrm{sc}(\theta_j) \mathrm{nd}(u_{\alpha_j}) \mathrm{nd}(u_{\alpha_{j+1}}). \end{aligned} $$

<figure><img src="image_14.png" /><figcaption>Figure 12: Notation for computing $[\bar{K}^t K]_{v,v}$, when $v$ is not a boundary vertex (left), and when $v$ is a boundary vertex $v^\theta \in \{v^\ell, v^r \in \mathcal{R}^\theta\}$ (the additional half-rhombi along the boundary are pictures in grey).</figcaption></figure>

In the finite case, when $v \in \{v^\ell, v^r \in \mathcal{R}^\theta\}$, both vertices enter the same framework and we denote $v$ by $v^\theta$. We stop the computation here, and returning to (3.6), we have that
$[\bar{K}^t K]_{v\partial, v\partial} = k' \Delta_{v\partial}^{m,\partial}$. In the infinite case or in the finite case when $v$ is not a boundary vertex we write, $\mathrm{nd}(u_\alpha) = \mathrm{nd}((u - 2iK')_\alpha + iK') = i \mathrm{sc}((u - 2iK')_\alpha)$ and obtain

$$[\bar{K}^t K]_{v,v} = -(k')^2 \sum_{j=1}^{d} \mathrm{sc}(\theta_j) \mathrm{sc}((u - 2iK')_{\alpha_j}) \mathrm{sc}((u - 2iK')_{\alpha_{j+1}}).$$

Since in this case we have $\bar{\alpha}_{d+1} = \bar{\alpha}_1 + 2\pi$, we can use Proposition 11 of [11] and obtain that, for every $u \in \mathbb{T}(k)$,

$$-(k')^2 \sum_{j=1}^{d} \mathrm{sc}(\theta_j) \mathrm{sc}(u_{\alpha_j}) \mathrm{sc}(u_{\alpha_{j+1}}) = k' [\sum_{j=1}^{d} A(\theta_j)].$$

Evaluating this identity at $u - 2iK'$, and recalling the definition of $\Delta^m$, see (2.9), we deduce that $[\bar{K}^t K]_{v,v} = k' [\sum_{j=1}^{d} A(\theta_j)] = k' \Delta_{v,v}^m$.

We are left with handling the cases where $(x, y) = (f_1, f_2)$ is an edge of $G^*$, and where $x = y = f$ is a vertex of $G^*$. First note that coefficients of $K, K^\partial$ involved in these computations arise from dual edges, and that for these edges we have $K^\partial = K$. Next, observe that coefficients of the operator $K$ on edges arising from the primal and from the dual differ in that the weight function $c$ is evaluated at $u$ or $u^* = K - u$, see (3.1). Given that the value of $[\bar{K}^t K]_{v_1, v_2}$ is independent of $u$, we immediately deduce the corresponding result for $[\bar{K}^t K]_{f_1, f_2}$. In the infinite case and in the finite case if $v$ is an inner vertex of $G$, $[\bar{K}^t K]_{v,v}$ is also independent of $u$ thus giving the corresponding result for $[\bar{K}^t K]_{f,f}$. The computation of $[\bar{K}^t K]_{v,v}$ when $v$ is a boundary vertex $v^\partial \in \{v^\ell, v^r\}$, yields a result which depends on $u$. Recall that what prevents us from proceeding with the computation as in the non-boundary case is the fact that $\bar{\alpha}_{d+1} \neq \bar{\alpha}_1 + 2\pi$. Since for boundary vertices of the restricted dual the condition $\bar{\alpha}_{d+1} = \bar{\alpha}_1 + 2\pi$ is satisfied, we proceed with the computation as for inner vertices and obtain the corresponding result for $[\bar{K}^t K]_{f,f}$. This ends the proof of Theorem 3.6 in the infinite case and part of the finite case.

Remaining part of the finite case. When $(x, y) = (v, f)$ or $(f, v)$, with $v, f$ at distance 2, we are left with the case where $(v, f) = (v^c, f^c)$ for some boundary rhombus pair of $\mathbb{R}^{\partial,r}$. Since the operator $[\bar{K}^\partial^t K]$ is not skew-symmetric, we have to consider the coefficients $(v^c, f)$ and $(f, v^c)$ separately. For the coefficient $(f^c, v^c)$, using the notation of Figure 10, we have $[\bar{K}^\partial^t K]_{f^c, v^c} = \bar{K}^\partial w^\ell, f^c K_{w^\ell, v^c} + \bar{K}^\partial w^r, f^c K_{w^r, v^c}$. This is equal to 0 as in the finite non-boundary case, because $K^\partial$ is equal to $K$ on the edges $(w^\ell, f^c)$ and $(w^r, f^c)$. For the coefficient $(v^c, f^c)$, we have

$$\begin{aligned}
[\bar{K}^\partial^t K]_{v^c, f^c} &= \bar{K}^\partial w^\ell, v^c K_{w^\ell, f^c} + \bar{K}^\partial w^r, v^c K_{w^r, f^c} \\
&= \frac{\mathrm{cd}(u_{\beta^r})}{\mathrm{cd}(u_{\alpha^\ell})} \bar{K}_{w^\ell, v^c} K_{w^\ell, f^c} + \bar{K}_{w^r, v^c} K_{w^r, f^c}, \quad \text{by definition of } K^\partial, \text{ see (3.5)} \\
&= -i \mathrm{dn}(u_{\beta^\ell+2K}) \left( \frac{\mathrm{cd}(u_{\beta^r})}{\mathrm{cd}(u_{\alpha^\ell})} - 1 \right), \quad \text{by the finite non-boundary computation} \\
&= -ik' \frac{\mathrm{nd}(u_{\beta^\ell})}{\mathrm{cd}(u_{\alpha^\ell})} (\mathrm{cd}(u_{\beta^r}) - \mathrm{cd}(u_{\alpha^\ell})) = k' Q_{v^c, f^c}.
\end{aligned}$$

Now consider an edge $(v_1, v_2)$ of $G^r$. Then, we are left with the case where $(v_1, v_2) = (v^c, v^\ell)$ for some boundary rhombus pair of $\mathbb{R}^{\partial,r}$. We have,

$$\begin{aligned}
[\bar{K}^\partial^t K]_{v^c, v^\ell} &= \bar{K}^\partial w^\ell, v^c K_{w^\ell, v^\ell} = \frac{\mathrm{cd}(u_{\beta^r})}{\mathrm{cd}(u_{\alpha^\ell})} \bar{K}_{w^\ell, v^c} K_{w^\ell, v^\ell}, \quad \text{by definition of } K^\partial, \text{ see (3.5)} \\
&= -k' \frac{\mathrm{cd}(u_{\beta^r})}{\mathrm{cd}(u_{\alpha^\ell})} \mathrm{sc}(\theta^\partial) = k' \Delta_{v^c, v^\ell}^{m,\partial}, \quad \text{by the finite non-boundary computation.}
\end{aligned}$$
$[\bar{K}^t K]_{v\partial, v\partial} = k' \Delta_{v\partial}^{m,\partial}$. In the infinite case or in the finite case when $v$ is not a boundary vertex we write, $\mathrm{nd}(u_\alpha) = \mathrm{nd}((u - 2iK')_\alpha + iK') = i \mathrm{sc}((u - 2iK')_\alpha)$ and obtain

$$[\bar{K}^t K]_{v,v} = -(k')^2 \sum_{j=1}^{d} \mathrm{sc}(\theta_j) \mathrm{sc}((u - 2iK')_{\alpha_j}) \mathrm{sc}((u - 2iK')_{\alpha_{j+1}}).$$

Since in this case we have $\bar{\alpha}_{d+1} = \bar{\alpha}_1 + 2\pi$, we can use Proposition 11 of [11] and obtain that, for every $u \in \mathbb{T}(k)$,

$$-(k')^2 \sum_{j=1}^{d} \mathrm{sc}(\theta_j) \mathrm{sc}(u_{\alpha_j}) \mathrm{sc}(u_{\alpha_{j+1}}) = k' [\sum_{j=1}^{d} A(\theta_j)].$$

Evaluating this identity at $u - 2iK'$, and recalling the definition of $\Delta^m$, see (2.9), we deduce that $[\bar{K}^t K]_{v,v} = k' [\sum_{j=1}^d A(\theta_j)] = k' \Delta_{v,v}^m$.

We are left with handling the cases where $(x, y) = (f_1, f_2)$ is an edge of $G^*$, and where $x = y = f$ is a vertex of $G^*$. First note that coefficients of $K, K^\partial$ involved in these computations arise from dual edges, and that for these edges we have $K^\partial = K$. Next, observe that coefficients of the operator $K$ on edges arising from the primal and from the dual differ in that the weight function $c$ is evaluated at $u$ or $u^* = K - u$, see (3.1). Given that the value of $[\bar{K}^t K]_{v_1, v_2}$ is independent of $u$, we immediately deduce the corresponding result for $[\bar{K}^t K]_{f_1, f_2}$. In the infinite case and in the finite case if $v$ is an inner vertex of $G$, $[\bar{K}^t K]_{v,v}$ is also independent of $u$ thus giving the corresponding result for $[\bar{K}^t K]_{f,f}$. The computation of $[\bar{K}^t K]_{v,v}$ when $v$ is a boundary vertex $v^\partial \in \{v^\ell, v^r\}$, yields a result which depends on $u$. Recall that what prevents us from proceeding with the computation as in the non-boundary case is the fact that $\bar{\alpha}_{d+1} \neq \bar{\alpha}_1 + 2\pi$. Since for boundary vertices of the restricted dual the condition $\bar{\alpha}_{d+1} = \bar{\alpha}_1 + 2\pi$ is satisfied, we proceed with the computation as for inner vertices and obtain the corresponding result for $[\bar{K}^t K]_{f,f}$. This ends the proof of Theorem 3.6 in the infinite case and part of the finite case.

Remaining part of the finite case. When $(x, y) = (v, f)$ or $(f, v)$, with $v, f$ at distance 2, we are left with the case where $(v, f) = (v^c, f^c)$ for some boundary rhombus pair of $\mathbb{R}^{\partial,r}$. Since the operator $[\bar{K}^\partial^t K]$ is not skew-symmetric, we have to consider the coefficients $(v^c, f)$ and $(f, v^c)$ separately. For the coefficient $(f^c, v^c)$, using the notation of Figure 10, we have $[\bar{K}^\partial^t K]_{f^c, v^c} = \bar{K}^\partial_{w^\ell, v^c} K_{w^\ell, v^c} + \bar{K}^\partial_{w^r, f^c} K_{w^r, v^c}$. This is equal to 0 as in the finite non-boundary case, because $K^\partial$ is equal to $K$ on the edges $(w^\ell, f^c)$ and $(w^r, f^c)$. For the coefficient $(v^c, f^c)$, we have

$$\begin{aligned}
[\bar{K}^\partial^t K]_{v^c, f^c} &= \bar{K}^\partial_{w^\ell, v^c} K_{w^\ell, f^c} + \bar{K}^\partial_{w^r, v^c} K_{w^r, f^c} \\
&= \frac{\mathrm{cd}(u_{\beta^r})}{\mathrm{cd}(u_{\alpha^\ell})} \bar{K}_{w^\ell, v^c} K_{w^\ell, f^c} + \bar{K}_{w^r, v^c} K_{w^r, f^c}, \quad \text{by definition of } K^\partial, \text{ see (3.5)} \\
&= -i \mathrm{dn}(u_{\beta^\ell+2K}) \left( \frac{\mathrm{cd}(u_{\beta^r})}{\mathrm{cd}(u_{\alpha^\ell})} - 1 \right), \quad \text{by the finite non-boundary computation} \\
&= -ik' \frac{\mathrm{nd}(u_{\beta^\ell})}{\mathrm{cd}(u_{\alpha^\ell})} (\mathrm{cd}(u_{\beta^r}) - \mathrm{cd}(u_{\alpha^\ell})) = k' Q_{v^c, f^c}.
\end{aligned}$$

Now consider an edge $(v_1, v_2)$ of $G^r$. Then, we are left with the case where $(v_1, v_2) = (v^c, v^\ell)$ for some boundary rhombus pair of $\mathbb{R}^{\partial,r}$. We have,

$$\begin{aligned}
[\bar{K}^\partial^t K]_{v^c, v^\ell} &= \bar{K}^\partial_{w^\ell, v^c} K_{w^\ell, v^\ell} = \frac{\mathrm{cd}(u_{\beta^r})}{\mathrm{cd}(u_{\alpha^\ell})} \bar{K}_{w^\ell, v^c} K_{w^\ell, v^\ell}, \quad \text{by definition of } K^\partial, \text{ see (3.5)} \\
&= -k' \frac{\mathrm{cd}(u_{\beta^r})}{\mathrm{cd}(u_{\alpha^\ell})} \mathrm{sc}(\theta^\partial) = k' \Delta_{v^c, v^\ell}^{m,\partial}, \quad \text{by the finite non-boundary computation.}
\end{aligned}$$
When $x, y$ are at distance 0, and $x = y = v$ is a vertex of $G$, we are left with the case where $v = v^c$ for some boundary rhombus pair of $\mathcal{R}^{\partial,r}$. We have,

$$ 
\begin{aligned}
[\bar{K}^{\partial} K]_{v^c, v^c} &= \bar{K}^{\partial}_{w^l, v^c} K_{w^l, v^c} + \bar{K}^{\partial}_{w^r, v^c} K_{w^r, v^c} \\
&= |K_{w^l, v^c}|^2 \frac{\mathrm{cd}(u_{\beta^r})}{\mathrm{cd}(u_{\alpha^l})} + |K_{w^r, v^c}|^2, \text{ by definition of } \bar{K}^{\partial}, \text{ see (3.5)} \\
&= (k')^2 \mathrm{sc}(\theta^{\partial}) \left( \mathrm{nd}(u_{\alpha^l}) \mathrm{nd}(u_{\beta^l}) \frac{\mathrm{cd}(u_{\beta^r})}{\mathrm{cd}(u_{\alpha^l})} + \mathrm{nd}(u_{\alpha^r}) \mathrm{nd}(u_{\beta^r}) \right), \\
& \qquad \text{by the finite non-boundary computation} \\
&= (k')^2 \mathrm{sc}(\theta^{\partial}) \frac{\mathrm{nd}(u_{\beta^l}) \mathrm{nd}(u_{\beta^r})}{\mathrm{cn}(u_{\alpha^l})} (\mathrm{cn}(u_{\beta^r}) + \mathrm{cn}(u_{\alpha^l})) = k' \Delta_{v^c, v^c}^{m, \partial}, \text{ since } \bar{\beta}_l = \bar{\alpha}_r [2\pi]. \quad \Box
\end{aligned}
$$

## 3.4 Determinants of the Z$^u$-Dirac and Z-massive Laplacian operators

We restrict to the finite case and consider the Z$^u$-Dirac operators $K(u)$ and $K^{\partial}(u)$ of Section 3.1 (finite case), the Z-massive Laplacian $\Delta^{m,\partial}(u)$, and the dual Z-massive Laplacian $\Delta^{m,*}$ of Section 3.2 (finite case). Theorem 3.8 below proves that the determinants of $K(u)$ and $\Delta^{m,*}$ are equal up to an explicit constant depending on $u$. Corollary 3.10 establishes a similar result for the determinants of $K^{\partial}(u)$ and $\Delta^{m,\partial}(u)$, thus implying identities between partition functions.

### 3.4.1 Determinants as partition functions

Returning to Section 2.3.1 on Temperley's bijection, in particular to Equation (2.3), let $\tilde{c}(u)$, be the weight function on directed edges of $G, \bar{G}^*$ corresponding to the weight function $c(u)$ of Equation (3.1). Then, the partition function of pairs of dual r-rooted and o-rooted directed spanning trees of $G, \bar{G}^*$ with weight function $\tilde{c}(u)$, is equal to the partition function of the dimer model on the graph $G^{D,r}$ with weight function $c(u)$ [71, 49], which is equal to $|\det K(u)|$ [72, 41]. We proceed in a similar way with dimers weighted by $c^{\partial}(u)$ and thus have,

$$ Z_{\text{dST-pairs}}^{r,o}((G, \bar{G}^*), \tilde{c}(u)) = |\det K(u)|, \quad Z_{\text{dST-pairs}}^{r,o}((G, \bar{G}^*), \tilde{c}^{\partial}(u)) = |\det K^{\partial}(u)|. \quad (3.7) $$

Returning to Section 2.4, we have that $\det \Delta^{m,*}$ counts weighted rooted spanning forests of $G^*$ with conductances $\rho^*$ and masses $m^*$, where conductances are symmetric. In a similar way, $\det \Delta^{m,\partial}(u)$ counts weighted rooted directed spanning forests of $G^r$ with conductances $\rho^{\partial}(u), m^{\partial}(u)$, where the dependence in $u$ is along the boundary only, and the conductances are symmetric away from the boundary. That is,

$$ Z_{\text{rSF}}(G^*, \rho^*, m^*) = \det \Delta^{m,*}, \quad Z_{\text{rdSF}}(G^r, \rho^\partial(u), m^\partial(u)) = \det \Delta^{m,\partial}(u). \quad (3.8) $$

### 3.4.2 Statements

Recall that every edge $e = (w, x)$ of $G^{D,r}$ is assigned two rhombus vectors $e^{i\tilde{\alpha}_e}, e^{i\tilde{\beta}_e}$ of the diamond graph $G^{\diamond/2}$ and a half-angle $\tilde{\theta}_e$. Moreover, every white vertex $w$ of $G^{D,r}$ is in the center of a rhombus of the diamond graph $G^{\diamond}$; we let $\tilde{\theta}_w$ be the half-angle of this rhombus at one of the two *primal* vertices.

**Theorem 3.8.** Let $M_0$ be a dimer configuration of $G^{D,r}$. Then, for every $u \in C$, we have

$$ |\det K(u)| = (k')^{\frac{|V^*|}{2}} \left( \prod_{w \in W} \mathrm{sc}(\theta_w)^{\frac{1}{2}} \right) \left( \prod_{e=wx \in M_0} [\mathrm{dn}(u_{\alpha_e}) \mathrm{dn}(u_{\beta_e})]^{\frac{1}{2}} \right) \det \Delta^{m,*}, \quad (3.9) $$

where, $|\det K(u)| = Z_{\text{dST-pairs}}^{r,o}((G, \bar{G}^*), \tilde{c}(u))$, $\det \Delta^{m,*} = Z_{\text{rSF}}(G^*, \rho^*, m^*)$.
When $x, y$ are at distance 0, and $x = y = v$ is a vertex of $G$, we are left with the case where $v = v^c$ for some boundary rhombus pair of $\mathcal{R}^{\partial,r}$. We have,

$$
\begin{align*}
[\bar{K}^{\partial}{}^{t} K]_{v^c, v^c} &= \bar{K}^{\partial}_{w^\ell, v^c} K_{w^\ell, v^c} + \bar{K}^{\partial}_{w^r, v^c} K_{w^r, v^c} \\
&= |K_{w^\ell, v^c}|^2 \frac{\mathrm{cd}(u_{\beta^r})}{\mathrm{cd}(u_{\alpha^\ell})} + |K_{w^r, v^c}|^2, \text{ by definition of } K^{\partial}, \text{ see (3.5)} \\
&= (k')^2 \mathrm{sc}(\theta^{\partial}) \left( \mathrm{nd}(u_{\alpha^\ell}) \mathrm{nd}(u_{\beta^r}) \frac{\mathrm{cd}(u_{\beta^r})}{\mathrm{cd}(u_{\alpha^\ell})} + \mathrm{nd}(u_{\alpha^r}) \mathrm{nd}(u_{\beta^r}) \right), \\
&\qquad \text{by the finite non-boundary computation} \\
&= (k')^2 \mathrm{sc}(\theta^{\partial}) \frac{\mathrm{nd}(u_{\beta^\ell}) \mathrm{nd}(u_{\beta^r})}{\mathrm{cn}(u_{\alpha^\ell})} (\mathrm{cn}(u_{\beta^r}) + \mathrm{cn}(u_{\alpha^\ell})) = k' \Delta^{m, \partial}_{v^c, v^c}, \text{ since } \bar{\beta}_\ell = \bar{\alpha}_r [2\pi]. \quad \Box
\end{align*}
$$

## 3.4 Determinants of the Z$^u$-Dirac and Z-massive Laplacian operators

We restrict to the finite case and consider the Z$^u$-Dirac operators $K(u)$ and $K^\partial(u)$ of Section 3.1 (finite case), the Z-massive Laplacian $\Delta^{m,\partial}(u)$, and the dual Z-massive Laplacian $\Delta^{m,*}$ of Section 3.2 (finite case). Theorem 3.8 below proves that the determinants of $K(u)$ and $\Delta^{m,*}$ are equal up to an explicit constant depending on $u$. Corollary 3.10 establishes a similar result for the determinants of $K^\partial(u)$ and $\Delta^{m,\partial}(u)$, thus implying identities between partition functions.

### 3.4.1 Determinants as partition functions

Returning to Section 2.3.1 on Temperley's bijection, in particular to Equation (2.3), let $\tilde{c}(u)$, be the weight function on directed edges of $G, \bar{G}^*$ corresponding to the weight function $c(u)$ of Equation (3.1). Then, the partition function of pairs of dual r-rooted and o-rooted directed spanning trees of $G, \bar{G}^*$ with weight function $\tilde{c}(u)$, is equal to the partition function of the dimer model on the graph $G^{D,r}$ with weight function $c(u)$ [71, 49], which is equal to $|\det K(u)|$ [72, 41]. We proceed in a similar way with dimers weighted by $c^\partial(u)$ and thus have,

$$Z_{\text{dST-pairs}}^{r,o}((G, \bar{G}^*), \tilde{c}(u)) = |\det K(u)|, \quad Z_{\text{dST-pairs}}^{r,o}((G, \bar{G}^*), \tilde{c}^\partial(u)) = |\det K^\partial(u)|. \quad (3.7)$$

Returning to Section 2.4, we have that $\det \Delta^{m,*}$ counts weighted rooted spanning forests of $G^*$ with conductances $\rho^*$ and masses $m^*$, where conductances are symmetric. In a similar way, $\det \Delta^{m,\partial}(u)$ counts weighted rooted directed spanning forests of $G^r$ with conductances $\rho^\partial(u), m^\partial(u)$, where the dependence in $u$ is along the boundary only, and the conductances are symmetric away from the boundary. That is,

$$Z_{\text{rSF}}(G^*, \rho^*, m^*) = \det \Delta^{m,*}, \quad Z_{\text{rdSF}}(G^r, \rho^\partial(u), m^\partial(u)) = \det \Delta^{m,\partial}(u). \quad (3.8)$$

### 3.4.2 Statements

Recall that every edge $e = (w, x)$ of $G^{D,r}$ is assigned two rhombus vectors $e^{i\tilde{\alpha}_e}, e^{i\tilde{\beta}_e}$ of the diamond graph $G^{\diamond/2}$ and a half-angle $\tilde{\theta}_e$. Moreover, every white vertex $w$ of $G^{D,r}$ is in the center of a rhombus of the diamond graph $G^\diamond$; we let $\tilde{\theta}_w$ be the half-angle of this rhombus at one of the two *primal* vertices.

**Theorem 3.8.** Let $M_0$ be a dimer configuration of $G^{D,r}$. Then, for every $u \in C$, we have

$$|\det K(u)| = (k')^{\frac{|v^*|}{2}} \left( \prod_{w \in W} \mathrm{sc}(\theta_w)^{\frac{1}{2}} \right) \left( \prod_{e=wx \in M_0} [\mathrm{dn}(u_{\alpha_e}) \mathrm{dn}(u_{\beta_e})]^{\frac{1}{2}} \right) \det \Delta^{m,*}, \quad (3.9)$$

where, $|\det K(u)| = Z_{\text{dST-pairs}}^{r,o}((G, \bar{G}^*), \tilde{c}(u))$, $\det \Delta^{m,*} = Z_{\text{rSF}}(G^*, \rho^*, m^*)$.
**Remark 3.9.** Comments on this theorem are given in the introduction to Section 3 especially how, in the critical case $k = 0$, it is an easy consequence of Temperley's bijection and the matrix-tree theorem, and how the argument does not directly extend to the non-critical case.

As one expects, the quantity $\prod_{e=bx \in M_0} [\mathrm{dn}(u_{\alpha_e}) \mathrm{dn}(u_{\beta_e})]^{\frac{1}{2}}$ is independent of the choice of perfect matching $M_0$. To see this, it suffices to check that the alternating product around every inner quadrangle face of $G^{D,r}$ is equal to 1; this is proved in Lemma 3.11 below.

The following is an immediate corollary of Theorem 3.8.

**Corollary 3.10.** Let $M_0$ be a dimer configuration of $G^{D,r}$. Then, for every $u \in C'$ we have,

$$|\det K^\theta(u)| = (k')^{\frac{|V|-1}{2}} \left( \prod_{w \in W} \mathrm{cs}(\theta_w)^{\frac{1}{2}} \right) \left( \prod_{e=bx \in M_0} [k' \mathrm{nd}(u_{\alpha_e}) \mathrm{nd}(u_{\beta_e})]^{\frac{1}{2}} \right) \det \Delta^{m,\theta}(u),$$

$$\text{where } |\det K^\theta(u)| = Z_{\text{dST-pairs}}^{r,0}((G, \bar{G}^*), \tilde{c}^\theta(u)), \det \Delta^{m,\theta}(u) = Z_{\text{rdSF}}(G^r, \rho^\theta(u), m^\theta(u)).$$

*Proof.* From Theorem 3.6, we have:

$$\det K^\theta(u) \det K(u) = (k')^{|V^r|+|V^*|} \det \Delta^{m,\theta}(u) \det \Delta^{m,*}.$$

Replacing $\det K(u)$ in the above by Equation (3.9) and recalling that $|W^r| = |B^r| = |V^r| + |V^*|$ and that $|V^r| = |V| - 1$, yields the result. Note that an argument similar to that used for proving Theorem 3.8 would give an alternative direct proof of Corollary 3.10. $\square$

### 3.4.3 Proof of Theorem 3.8

The proof of Theorem 3.8 is postponed until the end of this section. It is a consequence of four intermediate results, see Equations (3.10), (3.13), (3.15), (3.16) below, which we now establish. We will use Section 2.3.1 on Temperley's bijection and Appendix A on gauge transformations.

**From pairs of directed spanning trees to directed spanning trees** We start by defining a gauge transformation $K^g(u)$ of the matrix $K(u)$. Being gauge equivalent, the determinants of $K^g(u)$ and $K(u)$ are equal up to an explicit constant. We then use Temperley's bijection to deduce that the determinant of $K^g(u)$ counts weighted o-directed spanning trees of the dual graph $\bar{G}^*$.

Let $g(u)$ be the weight function on white-to-black edges of $G^{D,r}$ defined by, for every edge $wx$ of $G^{D,r}$,

$$g(u)_{wx} = [\mathrm{cs}(\theta_w) \mathrm{nd}(u_{\alpha_e}) \mathrm{nd}(u_{\beta_e})]^{\frac{1}{2}}.$$

Consider the matrix $K^g(u)$ obtained from $K(u)$ by multiplying edge-weights by $g(u)$. Returning to the definition of $K(u)$, see (3.1) and (3.2), we obtain that non-zero coefficients of $K^g(u)$ are given by,

$$K^g(u)_{w,x} = \begin{cases} e^{i \frac{\bar{\alpha}_e + \bar{\beta}_e}{2}} & \text{if } x \in V^r \\ e^{i \frac{\bar{\alpha}_e + \bar{\beta}_e}{2}} (k')^{\frac{1}{2}} \mathrm{cs}(\theta_w) \mathrm{nd}(u_{\alpha_e}) \mathrm{nd}(u_{\beta_e}) & \text{if } x \in V^* \end{cases}.$$

**Lemma 3.11.** Consider a perfect matching $M_0$ of $G^{D,r}$. The bipartite, weighted adjacency matrices $K^g(u)$ and $K(u)$ are gauge equivalent and we have,

$$|\det K(u)| = \left( \prod_{w \in W} \mathrm{sc}(\theta_w)^{\frac{1}{2}} \right) \left( \prod_{e=bx \in M_0} [\mathrm{dn}(u_{\alpha_e}) \mathrm{dn}(u_{\beta_e})]^{\frac{1}{2}} \right) |\det K^g(u)|. \quad (3.10)$$
**Remark 3.9.** Comments on this theorem are given in the introduction to Section 3 especially how, in the critical case $k = 0$, it is an easy consequence of Temperley's bijection and the matrix-tree theorem, and how the argument does not directly extend to the non-critical case.

As one expects, the quantity $\prod_{e=zx \in M_0} [\mathrm{dn}(u_{\alpha_e}) \mathrm{dn}(u_{\beta_e})]^{\frac{1}{2}}$ is independent of the choice of perfect matching $M_0$. To see this, it suffices to check that the alternating product around every inner quadrangle face of $G^{D,r}$ is equal to 1; this is proved in Lemma 3.11 below.

The following is an immediate corollary of Theorem 3.8.

**Corollary 3.10.** Let $M_0$ be a dimer configuration of $G^{D,r}$. Then, for every $u \in C'$ we have,

$$|\det K^\theta(u)| = (k')^{\frac{|V|-1}{2}} \left( \prod_{w \in W} \mathrm{cs}(\theta_w)^{\frac{1}{2}} \right) \left( \prod_{e=zx \in M_0} [k' \mathrm{nd}(u_{\alpha_e}) \mathrm{nd}(u_{\beta_e})]^{\frac{1}{2}} \right) \det \Delta^{m,\theta}(u),$$

$$\text{where } |\det K^\theta(u)| = Z_{\text{dST-pairs}}^{r,o}((G, \bar{G}^*), \tilde{c}^\theta(u)), \det \Delta^{m,\theta}(u) = Z_{\text{rdSF}}(G^r, \rho^\theta(u), m^\theta(u)).$$

*Proof.* From Theorem 3.6, we have:

$$\det K^\theta(u) \det K(u) = (k')^{|V^r|+|V^*|} \det \Delta^{m,\theta}(u) \det \Delta^{m,*}.$$

Replacing $\det K(u)$ in the above by Equation (3.9) and recalling that $|W^r| = |B^r| = |V^r| + |V^*|$ and that $|V^r| = |V| - 1$, yields the result. Note that an argument similar to that used for proving Theorem 3.8 would give an alternative direct proof of Corollary 3.10. $\square$

### 3.4.3 Proof of Theorem 3.8

The proof of Theorem 3.8 is postponed until the end of this section. It is a consequence of four intermediate results, see Equations (3.10), (3.13), (3.15), (3.16) below, which we now establish. We will use Section 2.3.1 on Temperley's bijection and Appendix A on gauge transformations.

**From pairs of directed spanning trees to directed spanning trees** We start by defining a gauge transformation $K^g(u)$ of the matrix $K(u)$. Being gauge equivalent, the determinants of $K^g(u)$ and $K(u)$ are equal up to an explicit constant. We then use Temperley's bijection to deduce that the determinant of $K^g(u)$ counts weighted o-directed spanning trees of the dual graph $\bar{G}^*$.

Let $g(u)$ be the weight function on white-to-black edges of $G^{D,r}$ defined by, for every edge $wx$ of $G^{D,r}$,

$$g(u)_{wx} = [\mathrm{cs}(\theta_w) \mathrm{nd}(u_{\alpha_e}) \mathrm{nd}(u_{\beta_e})]^{\frac{1}{2}}.$$

Consider the matrix $K^g(u)$ obtained from $K(u)$ by multiplying edge-weights by $g(u)$. Returning to the definition of $K(u)$, see (3.1) and (3.2), we obtain that non-zero coefficients of $K^g(u)$ are given by,

$$K^g(u)_{w,x} = \begin{cases} e^{i \frac{\bar{\alpha}_e + \bar{\beta}_e}{2}} & \text{if } x \in V^r \\ e^{i \frac{\bar{\alpha}_e + \bar{\beta}_e}{2}} (k')^{\frac{1}{2}} \mathrm{cs}(\theta_w) \mathrm{nd}(u_{\alpha_e}) \mathrm{nd}(u_{\beta_e}) & \text{if } x \in V^* \end{cases}.$$

**Lemma 3.11.** Consider a perfect matching $M_0$ of $G^{D,r}$. The bipartite, weighted adjacency matrices $K^g(u)$ and $K(u)$ are gauge equivalent and we have,

$$|\det K(u)| = \left( \prod_{w \in W} \mathrm{sc}(\theta_w)^{\frac{1}{2}} \right) \left( \prod_{e=zx \in M_0} [\mathrm{dn}(u_{\alpha_e}) \mathrm{dn}(u_{\beta_e})]^{\frac{1}{2}} \right) |\det K^g(u)|. \quad (3.10)$$
**Proof.** Details on gauge equivalences for bipartite, weighted adjacency matrices are given in Section A.2 of Appendix A. To prove that $K$ and $K^g$ are gauge equivalent, it suffices to show that the alternating products of $K$ and $K^g$ around inner face-cycles of $G^{D,r}$ are equal. Inner face-cycles of $G^{D,r}$ are quadrangles; using the notation of Figure 11 we have,

$$ \frac{K_{w,v}^g K_{w',f}^g}{K_{w,f}^g K_{w',v}^g} = \frac{K_{w,v} K_{w',f}}{K_{w,f} K_{w',v}} \Leftrightarrow \frac{g_{w,v}}{g_{w',f}} \frac{g_{w,f}}{g_{w',v}} = 1. $$

By definition of $g$ we have,

$$ \frac{g_{w,v} g_{w',f}}{g_{w,f} g_{w',v}} = \frac{\mathrm{cs}(\theta) \mathrm{nd}(u_{\alpha}) \mathrm{nd}(u_{\beta})}{\mathrm{cs}(\theta) \mathrm{nd}(u_{\beta-2K}) \mathrm{nd}(u_{\alpha})} \frac{\mathrm{cs}(\theta') \mathrm{nd}(u_{\beta'}) \mathrm{nd}(u_{\alpha'+2K})}{\mathrm{cs}(\theta') \mathrm{nd}(u_{\alpha'}) \mathrm{nd}(u_{\beta'})} = 1, $$

since $\bar{\alpha}' = \bar{\beta}[2\pi]$. The equality between determinants comes from [53], see also Corollary A.8, and from the fact that, $\prod_{w \in M_0} \mathrm{sc}(\theta_w) = \prod_{w \in W} \mathrm{sc}(\theta_w)$, since a dimer configuration covers all white vertices $W^r$ of $G^{D,r}$, and $W^r = W$. $\square$

Now, the matrix $K^g(u)$ is a complex, bipartite Kasteleyn matrix of the graph $G^{D,r}$, where edges are assigned the positive weight function $c^g(u)$ given by:

$$ c^g(u)_{w,x} = \begin{cases} 1 & \text{if } x \in V^r \\ (k')^{\frac{1}{2}} \mathrm{cs}(\theta_w) \mathrm{nd}(u_{\alpha_e}) \mathrm{nd}(u_{\beta_e}) & \text{if } x \in V^*. \end{cases} $$

Returning to Section 2.3.1, the weight function $c^g(u)$ determines a weight function $\tilde{c}^g(u)$ on directed edges of $G, \bar{G}^*$, see (2.3). Then, $\tilde{c}^g(u)_{v,v'} = 1$, for all directed edges $(v, v')$ of $G$ such that $v$ is a vertex of $G^r$. Let us now express the weight function $\tilde{c}^g(u)$ on directed edges of $\bar{G}^*$ using the rhombus vectors and half-angles assigned to those edges. Recall that an edge $(f, f')$ of the restricted dual $G^*$ is assigned two rhombus vectors $2e^{i\bar{\alpha}}, 2e^{i\bar{\beta}}$ and a half-angle $\bar{\theta}^*$ of $G^\circ$. An edge $(f, o)$ of $\bar{G}^*$ corresponds to an edge $fw$ of $G^D$, and one assigns to $(f, o)$ the rhombus vectors $e^{i\bar{\alpha}}, e^{i\bar{\beta}}$ and half-angle $\bar{\theta}^*$ of $G^{\circ/2}$ associated to the edge $(f, w)$. Then, for every directed edge $(f, f')$ of $\bar{G}^*$ such that $f$ is a vertex of $G^*$ we have, using the notation of (2.3):

$$ \begin{aligned} \tilde{c}^g(u)_{f,f'} &= c^g(u)_{w,f} = (k')^{\frac{1}{2}} \mathrm{cs}(\theta_w) \mathrm{nd}(u_{\alpha_e}) \mathrm{nd}(u_{\beta_e}) \\ &= (k')^{-\frac{1}{2}} \mathrm{sc}(\theta^*) \mathrm{dn}(u_{\alpha}) \mathrm{dn}(u_{\beta}), \end{aligned} \quad (3.11) $$

using that $\bar{\theta}^* = \frac{\pi}{2} - \bar{\theta}_w, \bar{\alpha}_e = \bar{\alpha} \pm \pi, \bar{\beta}_e = \bar{\beta} \pm \pi$; and $\tilde{c}^g(u)_{o,f'} = 0$, for every edge $(o, f')$ of $\bar{G}^*$.

Since edges of the primal graph have weight 1, by the KPW-Temperley bijection [71, 49], the determinant of $K^g(u)$ counts weighted o-directed spanning trees of $\mathcal{T}^\circ(\bar{G}^*)$, where directed edges of $\bar{G}^*$ are assigned the weight function $\gamma^*(u)$ given by, for every directed edge $(f, f')$ of $\bar{G}^*$,

$$ \gamma^*(u)_{f,f'} := \tilde{c}^g(u)_{f,f'} = \begin{cases} (k')^{-\frac{1}{2}} \mathrm{sc}(\theta^*) \mathrm{dn}(u_\alpha) \mathrm{dn}(u_\beta) & \text{if } f \text{ is a vertex of } \bar{G}^* \\ 0 & \text{if } f = o. \end{cases} \quad (3.12) $$

That is,

$$ |\det K^g(u)| = Z_{\mathrm{dST}}^\circ(\bar{G}^*, \gamma^*(u)). \quad (3.13) $$

**Matrix-tree theorem** An alternative way of computing the partition function (3.13) is to use the directed version of the matrix-tree theorem [50, 73]. Let $\Delta^*(u)$ be the
**Proof.** Details on gauge equivalences for bipartite, weighted adjacency matrices are given in Section A.2 of Appendix A. To prove that $K$ and $K^g$ are gauge equivalent, it suffices to show that the alternating products of $K$ and $K^g$ around inner face-cycles of $G^{D,r}$ are equal. Inner face-cycles of $G^{D,r}$ are quadrangles; using the notation of Figure 11 we have,

$$ \frac{K_{w,v}^g K_{w',f}^g}{K_{w,f}^g K_{w',v}^g} = \frac{K_{w,v} K_{w',f}}{K_{w,f} K_{w',v}} \Leftrightarrow \frac{g_{w,v}}{g_{w',f}} \frac{g_{w,f}}{g_{w',v}} = 1. $$

By definition of $g$ we have,

$$ \frac{g_{w,v} g_{w',f}}{g_{w,f} g_{w',v}} = \frac{\mathrm{cs}(\theta) \mathrm{nd}(u_\alpha) \mathrm{nd}(u_\beta)}{\mathrm{cs}(\theta) \mathrm{nd}(u_{\beta-2K}) \mathrm{nd}(u_\alpha)} \frac{\mathrm{cs}(\theta') \mathrm{nd}(u_{\beta'}) \mathrm{nd}(u_{\alpha'+2K})}{\mathrm{cs}(\theta') \mathrm{nd}(u_{\alpha'}) \mathrm{nd}(u_{\beta'})} = 1, $$

since $\bar{\alpha}' = \bar{\beta}[2\pi]$. The equality between determinants comes from [53], see also Corollary A.8, and from the fact that, $\prod_{w \in M_0} \mathrm{sc}(\theta_w) = \prod_{w \in W} \mathrm{sc}(\theta_w)$, since a dimer configuration covers all white vertices $W^r$ of $G^{D,r}$, and $W^r = W$. $\square$

Now, the matrix $K^g(u)$ is a complex, bipartite Kasteleyn matrix of the graph $G^{D,r}$, where edges are assigned the positive weight function $c^g(u)$ given by:

$$ c^g(u)_{w,x} = \begin{cases} 1 & \text{if } x \in V^r \\ (k')^{\frac{1}{2}} \mathrm{cs}(\theta_w) \mathrm{nd}(u_{\alpha_e}) \mathrm{nd}(u_{\beta_e}) & \text{if } x \in V^*. \end{cases} $$

Returning to Section 2.3.1, the weight function $c^g(u)$ determines a weight function $\tilde{c}^g(u)$ on directed edges of $G, \bar{G}^*$, see (2.3). Then, $\tilde{c}^g(u)_{v,v'} = 1$, for all directed edges $(v, v')$ of $G$ such that $v$ is a vertex of $G^r$. Let us now express the weight function $\tilde{c}^g(u)$ on directed edges of $\bar{G}^*$ using the rhombus vectors and half-angles assigned to those edges. Recall that an edge $(f, f')$ of the restricted dual $G^*$ is assigned two rhombus vectors $2e^{i\bar{\alpha}}, 2e^{i\bar{\beta}}$ and a half-angle $\bar{\theta}^*$ of $G^\circ$. An edge $(f, o)$ of $\bar{G}^*$ corresponds to an edge $fw$ of $G^D$, and one assigns to $(f, o)$ the rhombus vectors $e^{i\bar{\alpha}}, e^{i\bar{\beta}}$ and half-angle $\bar{\theta}^*$ of $G^{\circ/2}$ associated to the edge $(f, w)$. Then, for every directed edge $(f, f')$ of $\bar{G}^*$ such that $f$ is a vertex of $G^*$ we have, using the notation of (2.3):

$$ \begin{aligned} \tilde{c}^g(u)_{f,f'} &= c^g(u)_{w,f} = (k')^{\frac{1}{2}} \mathrm{cs}(\theta_w) \mathrm{nd}(u_{\alpha_e}) \mathrm{nd}(u_{\beta_e}) \\ &= (k')^{-\frac{1}{2}} \mathrm{sc}(\theta^*) \mathrm{dn}(u_\alpha) \mathrm{dn}(u_\beta), \end{aligned} \quad (3.11) $$

using that $\bar{\theta}^* = \frac{\pi}{2} - \bar{\theta}_w, \bar{\alpha}_e = \bar{\alpha} \pm \pi, \bar{\beta}_e = \bar{\beta} \pm \pi$; and $\tilde{c}^g(u)_{o,f'} = 0$, for every edge $(o, f')$ of $\bar{G}^*$.

Since edges of the primal graph have weight 1, by the KPW-Temperley bijection [71, 49], the determinant of $K^g(u)$ counts weighted o-directed spanning trees of $\mathcal{T}^\circ(\bar{G}^*)$, where directed edges of $\bar{G}^*$ are assigned the weight function $\gamma^*(u)$ given by, for every directed edge $(f, f')$ of $\bar{G}^*$,

$$ \gamma^*(u)_{f,f'} := \tilde{c}^g(u)_{f,f'} = \begin{cases} (k')^{-\frac{1}{2}} \mathrm{sc}(\theta^*) \mathrm{dn}(u_\alpha) \mathrm{dn}(u_\beta) & \text{if } f \text{ is a vertex of } \bar{G}^* \\ 0 & \text{if } f = o. \end{cases} \quad (3.12) $$

That is,

$$ |\det K^g(u)| = Z_{\mathrm{dST}}^\circ(\bar{G}^*, \gamma^*(u)). \quad (3.13) $$

**Matrix-tree theorem** An alternative way of computing the partition function (3.13) is to use the directed version of the matrix-tree theorem [50, 73]. Let $\Delta^*(u)$ be the
(non-massive) Laplacian matrix of the graph $\bar{G}^*$ with conductances $\gamma^*(u)$ on directed edges. The non-zero coefficients of the matrix $\Delta^*(u)$ are given by:

$$ \Delta^*(u)_{f,f'} = \begin{cases} -\gamma^*(u)_{f,f'} & \text{if } (f, f') \text{ is an edge of } \bar{G}^* \\ \sum_{j=1}^d \gamma^*(u)_{f,f_j} & \text{if } f' = f \text{ is a vertex of } \bar{G}^*. \end{cases} \quad (3.14) $$

Let $\Delta^{*,\circ}(u)$ be the matrix obtained from $\Delta^*(u)$ by removing the row and column corresponding to the vertex $\circ$. Then,

$$ Z_{\mathrm{dST}}^{\circ}(\bar{G}^{*}, \gamma^{*}(u)) = \det \Delta^{*,\circ}(u). \quad (3.15) $$

**From o-directed spanning trees to rooted spanning forests** The last step consists in going from o-directed spanning trees of $\bar{G}^*$ counted by $\det \Delta^{*,\circ}(u)$ to rooted spanning forests of $\bar{G}^*$ counted by $\det \Delta^{m,*}$ using gauge equivalences on weighted adjacency matrices of digraphs, see Section A.1 of Appendix A.

**Lemma 3.12.** The weighted adjacency matrices $(k')^{-\frac{1}{2}}\Delta^{*,\circ}(u)$ and $\Delta^{m,*}$ are gauge equivalent and we have,

$$ \det \Delta^{*,\circ}(u) = (k')^{\frac{|\nu^*|}{2}} \det \Delta^{m,*}. \quad (3.16) $$

*Proof.* Both matrices $(k')^{-\frac{1}{2}}\Delta^{*,\circ}(u)$ and $\Delta^{m,*}$ have the same associated digraph which is the restricted dual $\bar{G}^*$ where a loop is added at every vertex, and each undirected edge is replaced by the two possible directed edges. The graph $\bar{G}^*$ being connected, the associated digraph is strongly connected. We use Lemma A.5 to prove gauge equivalence of the matrices.

For every vertex $f$ of $\bar{G}^*$, define $q_{f,f} = 1$. Next, consider two distinct vertices $f_1, f_2$ of $\bar{G}^*$ and a simple di-path $\gamma$ from $f_1$ to $f_2$. Set,

$$ q_{f_1, f_2} = \prod_{e \in \gamma} [(k')^{-1} \mathrm{dn}(u_{\alpha}) \mathrm{dn}(u_{\beta})]. $$

The function $q$ is well defined because it is the exponential function of [11] evaluated at $u - 2K - 2iK'$, see also Equation (2.11). Indeed,

$$ \begin{aligned} (k')^{-1} \mathrm{dn}(u_{\alpha}) \mathrm{dn}(u_{\beta}) &= (k')^{-1} \mathrm{dn}((u - 2K - 2iK')_{\alpha} + K + iK') \times \\ & \quad \times \mathrm{dn}((u - 2K - 2iK')_{\beta} + K + iK') \\ &= [i(k')^{\frac{1}{2}} \mathrm{sc}((u - 2K - 2iK')_{\alpha})][i(k')^{\frac{1}{2}} \mathrm{sc}((u - 2K - 2iK')_{\beta})]. \end{aligned} $$

Fix a vertex $f_0$ of $\bar{G}^*$, and define $D^{f_0}$ to be the diagonal matrix whose diagonal coefficient $D_{f,f_0}^{f_0}$ corresponding to the vertex $f$ of $\bar{G}^*$ is $q_{f_0,f}$. Let us prove that

$$ \Delta^{m,*} = (k')^{-\frac{1}{2}} D^{f_0} \Delta^{*,\circ}(u) (D^{f_0})^{-1}. \quad (3.17) $$

Recall the definition of the Laplacian matrix $\Delta^{*,\circ}(u)$, see (3.14) and (3.12). For the diagonal coefficient corresponding to a vertex $f$ of $\bar{G}^*$, we have

$$ \begin{aligned} [(k')^{-\frac{1}{2}} D^{f_0} \Delta^{*,\circ}(u) (D^{f_0})^{-1}]_{f,f} &= (k')^{-\frac{1}{2}} \Delta^{*,\circ}(u)_{f,f} = (k')^{-1} \sum_{j=1}^{d} \mathrm{sc}(\theta_j^*) \mathrm{dn}(u_{\alpha_j}) \mathrm{dn}(u_{\beta_j}), \\ &= \sum_{j=1}^{d} A(\theta_j^*) = \Delta_{f,f}^{m,*}, \quad \text{by Proposition 11 of [11].} \end{aligned} $$
(non-massive) Laplacian matrix of the graph $\bar{G}^*$ with conductances $\gamma^*(u)$ on directed edges. The non-zero coefficients of the matrix $\Delta^*(u)$ are given by:

$$ \Delta^*(u)_{f,f'} = \begin{cases} -\gamma^*(u)_{f,f'} & \text{if } (f, f') \text{ is an edge of } \bar{G}^* \\ \sum_{j=1}^d \gamma^*(u)_{f,f_j} & \text{if } f' = f \text{ is a vertex of } \bar{G}^*. \end{cases} \qquad (3.14) $$

Let $\Delta^{*,\circ}(u)$ be the matrix obtained from $\Delta^*(u)$ by removing the row and column corresponding to the vertex o. Then,

$$ Z_{\mathrm{dST}}^{\circ}(\bar{G}^{*}, \gamma^{*}(u)) = \det \Delta^{*,\circ}(u). \qquad (3.15) $$

**From o-directed spanning trees to rooted spanning forests** The last step consists in going from o-directed spanning trees of $\bar{G}^*$ counted by $\det \Delta^{*,\circ}(u)$ to rooted spanning forests of $\bar{G}^*$ counted by $\det \Delta^{m,*}$ using gauge equivalences on weighted adjacency matrices of digraphs, see Section A.1 of Appendix A.

**Lemma 3.12.** The weighted adjacency matrices $(k')^{-\frac{1}{2}} \Delta^{*,\circ}(u)$ and $\Delta^{m,*}$ are gauge equivalent and we have,

$$ \det \Delta^{*,\circ}(u) = (k')^{\frac{|\nu^*|}{2}} \det \Delta^{m,*}. \qquad (3.16) $$

*Proof.* Both matrices $(k')^{-\frac{1}{2}} \Delta^{*,\circ}(u)$ and $\Delta^{m,*}$ have the same associated digraph which is the restricted dual $\bar{G}^*$ where a loop is added at every vertex, and each undirected edge is replaced by the two possible directed edges. The graph $\bar{G}^*$ being connected, the associated digraph is strongly connected. We use Lemma A.5 to prove gauge equivalence of the matrices.

For every vertex $f$ of $\bar{G}^*$, define $q_{f,f} = 1$. Next, consider two distinct vertices $f_1, f_2$ of $\bar{G}^*$ and a simple di-path $\gamma$ from $f_1$ to $f_2$. Set,

$$ q_{f_1, f_2} = \prod_{e \in \gamma} [(k')^{-1} \mathrm{dn}(u_{\alpha}) \mathrm{dn}(u_{\beta})]. $$

The function $q$ is well defined because it is the exponential function of [11] evaluated at $u - 2K - 2iK'$, see also Equation (2.11). Indeed,

$$ \begin{aligned} (k')^{-1} \mathrm{dn}(u_{\alpha}) \mathrm{dn}(u_{\beta}) &= (k')^{-1} \mathrm{dn}((u - 2K - 2iK')_{\alpha} + K + iK') \times \\ & \qquad \times \mathrm{dn}((u - 2K - 2iK')_{\beta} + K + iK') \\ &= [i(k')^{\frac{1}{2}} \mathrm{sc}((u - 2K - 2iK')_{\alpha})][i(k')^{\frac{1}{2}} \mathrm{sc}((u - 2K - 2iK')_{\beta})]. \end{aligned} $$

Fix a vertex $f_0$ of $\bar{G}^*$, and define $D^{f_0}$ to be the diagonal matrix whose diagonal coefficient $D_{f,f_0}^{f_0}$ corresponding to the vertex $f$ of $\bar{G}^*$ is $q_{f_0,f}$. Let us prove that

$$ \Delta^{m,*} = (k')^{-\frac{1}{2}} D^{f_0} \Delta^{*,\circ}(u) (D^{f_0})^{-1}. \qquad (3.17) $$

Recall the definition of the Laplacian matrix $\Delta^{*,\circ}(u)$, see (3.14) and (3.12). For the diagonal coefficient corresponding to a vertex $f$ of $\bar{G}^*$, we have

$$ \begin{aligned} [(k')^{-\frac{1}{2}} D^{f_0} \Delta^{*,\circ}(u) (D^{f_0})^{-1}]_{f,f} &= (k')^{-\frac{1}{2}} \Delta^{*,\circ}(u)_{f,f} = (k')^{-1} \sum_{j=1}^{d} \mathrm{sc}(\theta_j^*) \mathrm{dn}(u_{\alpha_j}) \mathrm{dn}(u_{\beta_j}), \\ &= \sum_{j=1}^{d} A(\theta_j^*) = \Delta_{f,f}^{m,*}, \text{ by Proposition 11 of [11].} \end{aligned} $$
For an edge $(f, f')$ of $G^*$, we have

$$ \begin{aligned} [(k')^{-\frac{1}{2}} D^{f_0} \Delta^{*,\circ}(u) (D^{f_0})^{-1}]_{f,f'} &= (k')^{-\frac{1}{2}} \frac{q_{f_0,f}}{q_{f_0,f'}} \Delta^{*,\circ}(u)_{f,f'} = (k')^{-\frac{1}{2}} \frac{1}{q_{f,f'}} \Delta^{*,\circ}(u)_{f,f'} \\ &= -(k')^{-\frac{1}{2}}(k') \mathrm{nd}(u_\alpha) \mathrm{nd}(u_\beta)(k')^{-\frac{1}{2}} \mathrm{sc}(\theta^*) \mathrm{dn}(u_\alpha) \mathrm{dn}(u_\beta) \\ &= -\mathrm{sc}(\theta^*) = \Delta^{m,*}_{f,f'}. \end{aligned} $$

By Lemma A.5, Equation (3.17) implies that the matrices $(k')^{-\frac{1}{2}} \Delta^{*,\circ}(u)$ and $\Delta^{m,*}$ are gauge equivalent, and we have equality of the determinants: $\det \Delta^{m,*} = \det[(k')^{-\frac{1}{2}} \Delta^{*,\circ}(u)]$. $\square$

*Proof of Theorem 3.8.* Combining Equation (3.10), (3.13), (3.15) and (3.16), we obtain,

$$ \begin{aligned} |\det K(u)| &= \left( \prod_{w \in W} \mathrm{sc}(\theta_w)^{\frac{1}{2}} \right) \left( \prod_{e=w x \in M_0} \left[ \mathrm{dn}(u_{\alpha_e}) \mathrm{dn}(u_{\beta_e}) \right]^{\frac{1}{2}} \right) |\det K^g(u)| \\ &= \left( \prod_{w \in W} \mathrm{sc}(\theta_w)^{\frac{1}{2}} \right) \left( \prod_{e=w x \in M_0} \left[ \mathrm{dn}(u_{\alpha_e}) \mathrm{dn}(u_{\beta_e}) \right]^{\frac{1}{2}} \right) \det \Delta^{*,\circ}(u) \\ &= (k')^{\frac{|V^*|}{2}} \left( \prod_{w \in W} \mathrm{sc}(\theta_w)^{\frac{1}{2}} \right) \left( \prod_{e=w x \in M_0} \left[ \mathrm{dn}(u_{\alpha_e}) \mathrm{dn}(u_{\beta_e}) \right]^{\frac{1}{2}} \right) \det \Delta^{m,*}. \end{aligned} \square $$

### 3.5 Z-invariance of the $Z^u$-Dirac operator

According to Baxter [4, 5, 6] a model of statistical mechanics is Z-invariant if, when decomposing the partition function according to the possible configurations outside of the hexagon of the diamond graph $G^\diamond$ defining the star/triangle, it only changes by a constant independent of the outer configurations when performing a $Y-\Delta$ move.

Suppose that the isoradial graph $G$ is finite. The $Z^u$-Dirac operator $K(u)$ is the bipartite Kasteleyn matrix of the dimer model on the double graph $G^{D,r}$ with weight function $c(u)$ on the edges given by (3.1). By Section 3.4.1 this dimer model is in bijection with pairs of dual directed spanning trees of $T^{r,\circ}(G, \bar{G}^*)$, with weight function $\tilde{c}(u)$ on directed edges of $G, \bar{G}^*$:

$$ Z_{\mathrm{dimer}}(G^{D,r}, c(u)) = |\det K(u)| = Z_{\mathrm{dST-pairs}}^{r,\circ}((G, \bar{G}^*), \tilde{c}(u)). $$

By Lemma 3.11 and Equation (3.13), $|\det K(u)|$ is equal up to a constant to $|\det K^g(u)|$ which counts o-directed spanning trees of $T^\circ(\bar{G}^*)$ with conductances $\gamma^*(u)$ on directed edges of $\bar{G}^*$ given by (3.12); that is:

$$ Z_{\mathrm{dimer}}(G^{D,r}, c(u)) = C(u) \cdot Z_{\mathrm{dST}}^{\circ}(\bar{G}^*, \gamma^*(u)), $$

where $C(u)$ is given in Lemma 3.11. By Temperley's bijection again, there is a one-to-one correspondence between dimer configurations of $G^{D,r}$ and o-directed spanning trees of $T^\circ(\bar{G}^*)$ (such that the primal tree is r-rooted). We prove Z-invariance of this o-rooted directed spanning tree model on $\bar{G}^*$; using the above, the decomposition of the partition function has a direct interpretation in terms of the dimer model on $G^{D,r}$.

Since duality preserves isoradiality, we actually show Z-invariance of the r-directed spanning tree model on $G$, where directed edges of $G$ are assigned conductances $\gamma(u)$ given by, for every directed edge $(v, v')$ of $G$,

$$ \gamma(u)_{v,v'} = (k')^{-\frac{1}{2}} \mathrm{sc}(\theta) \mathrm{dn}(u_\alpha) \mathrm{dn}(u_\beta), $$

where $2e^{i\bar{\alpha}}, 2e^{i\bar{\beta}}, \bar{\theta}$ are the rhombus vectors and half-angle of $G^\diamond$ associated to the edge $(v, v')$; $\gamma(u)$ is the primal version of the conductances $\gamma^*(u)$ of (3.12).
For an edge $(f, f')$ of $G^*$, we have

$$ \begin{aligned} [(k')^{-\frac{1}{2}} D^{f_0} \Delta^{*,\circ}(u) (D^{f_0})^{-1}]_{f,f'} &= (k')^{-\frac{1}{2}} \frac{q_{f_0,f}}{q_{f_0,f'}} \Delta^{*,\circ}(u)_{f,f'} = (k')^{-\frac{1}{2}} \frac{1}{q_{f,f'}} \Delta^{*,\circ}(u)_{f,f'} \\ &= -(k')^{-\frac{1}{2}}(k') \mathrm{nd}(u_\alpha) \mathrm{nd}(u_\beta)(k')^{-\frac{1}{2}} \mathrm{sc}(\theta^*) \mathrm{dn}(u_\alpha) \mathrm{dn}(u_\beta) \\ &= -\mathrm{sc}(\theta^*) = \Delta_{f,f'}^{m,*}. \end{aligned} $$

By Lemma A.5, Equation (3.17) implies that the matrices $(k')^{-\frac{1}{2}} \Delta^{*,\circ}(u)$ and $\Delta^{m,*}$ are gauge equivalent, and we have equality of the determinants: $\det \Delta^{m,*} = \det[(k')^{-\frac{1}{2}} \Delta^{*,\circ}(u)]$. $\square$

*Proof of Theorem 3.8.* Combining Equation (3.10), (3.13), (3.15) and (3.16), we obtain,

$$ \begin{aligned} |\det K(u)| &= \left( \prod_{w \in W} \mathrm{sc}(\theta_w)^{\frac{1}{2}} \right) \left( \prod_{e=w x \in M_0} \left[ \mathrm{dn}(u_{\alpha_e}) \mathrm{dn}(u_{\beta_e}) \right]^{\frac{1}{2}} \right) |\det K^g(u)| \\ &= \left( \prod_{w \in W} \mathrm{sc}(\theta_w)^{\frac{1}{2}} \right) \left( \prod_{e=w x \in M_0} \left[ \mathrm{dn}(u_{\alpha_e}) \mathrm{dn}(u_{\beta_e}) \right]^{\frac{1}{2}} \right) \det \Delta^{*,\circ}(u) \\ &= (k')^{\frac{|V^*|}{2}} \left( \prod_{w \in W} \mathrm{sc}(\theta_w)^{\frac{1}{2}} \right) \left( \prod_{e=w x \in M_0} \left[ \mathrm{dn}(u_{\alpha_e}) \mathrm{dn}(u_{\beta_e}) \right]^{\frac{1}{2}} \right) \det \Delta^{m,*}. \end{aligned} \square $$

### 3.5 Z-invariance of the $Z^u$-Dirac operator

According to Baxter [4, 5, 6] a model of statistical mechanics is Z-invariant if, when decomposing the partition function according to the possible configurations outside of the hexagon of the diamond graph $G^\diamond$ defining the star/triangle, it only changes by a constant independent of the outer configurations when performing a $Y-\Delta$ move.

Suppose that the isoradial graph $G$ is finite. The $Z^u$-Dirac operator $K(u)$ is the bipartite Kasteleyn matrix of the dimer model on the double graph $G^{D,r}$ with weight function $c(u)$ on the edges given by (3.1). By Section 3.4.1 this dimer model is in bijection with pairs of dual directed spanning trees of $T^{r,\circ}(G, \bar{G}^*)$, with weight function $\tilde{c}(u)$ on directed edges of $G, \bar{G}^*$:

$$ Z_{\mathrm{dimer}}(G^{D,r}, c(u)) = |\det K(u)| = Z_{\mathrm{dST-pairs}}^{r,\circ}((G, \bar{G}^*), \tilde{c}(u)). $$

By Lemma 3.11 and Equation (3.13), $|\det K(u)|$ is equal up to a constant to $|\det K^g(u)|$ which counts o-directed spanning trees of $T^\circ(\bar{G}^*)$ with conductances $\gamma^*(u)$ on directed edges of $\bar{G}^*$ given by (3.12); that is:

$$ Z_{\mathrm{dimer}}(G^{D,r}, c(u)) = C(u) \cdot Z_{\mathrm{dST}}^{\circ}(\bar{G}^*, \gamma^*(u)), $$

where $C(u)$ is given in Lemma 3.11. By Temperley's bijection again, there is a one-to-one correspondence between dimer configurations of $G^{D,r}$ and o-directed spanning trees of $T^\circ(\bar{G}^*)$ (such that the primal tree is r-rooted). We prove Z-invariance of this o-rooted directed spanning tree model on $\bar{G}^*$; using the above, the decomposition of the partition function has a direct interpretation in terms of the dimer model on $G^{D,r}$.

Since duality preserves isoradiality, we actually show Z-invariance of the r-directed spanning tree model on $G$, where directed edges of $G$ are assigned conductances $\gamma(u)$ given by, for every directed edge $(v, v')$ of $G$,

$$ \gamma(u)_{v,v'} = (k')^{-\frac{1}{2}} \mathrm{sc}(\theta) \mathrm{dn}(u_\alpha) \mathrm{dn}(u_\beta), $$

where $2e^{i\bar{\alpha}}, 2e^{i\bar{\beta}}, \bar{\theta}$ are the rhombus vectors and half-angle of $G^\diamond$ associated to the edge $(v, v')$; $\gamma(u)$ is the primal version of the conductances $\gamma^*(u)$ of (3.12).
**Proposition 3.13.** Consider a finite isoradial graph G, and let $u \in C$. Then, the model of r-directed spanning trees on G, with weight function $\gamma(u)$ on the edges is Z-invariant.

**Remark 3.14.**

* By Theorem 3.8, we have:

$$Z_{\text{dimer}}(G^{\text{D},r}, c(u)) = C'(u) \cdot Z_{\text{rSF}}(G^*, \rho^*, m^*),$$

and in the paper [11], the model of rooted spanning forests with these weights is shown to be Z-invariant. But, since the proof of Theorem 3.8 does not provide a bijection between directed spanning trees and rooted spanning forests, the two decompositions of the partition functions are not directly comparable; they should nevertheless be compatible. Note that the computations of the proof of Proposition 3.13 are reminiscent but much simpler than those of [11] (the latter have been removed from the published version).

* The critical spanning tree model of [45] with conductances $\tan(\theta)$ is Z-invariant [43]. The result of [11] extends this to rooted spanning forest while Proposition 3.13 extends it to directed spanning trees.

*Proof.* Let $G^\gamma$ and $G^\Delta$ be two finite isoradial graphs differing by a star-triangle transformation, and let $r$ be a fixed root on the boundary of the graph, outside of the hexagon defining the star/triangle. Let $\gamma^\Delta(u)$, resp. $\gamma^\gamma(u)$, be the weight function on r-dST of $G^\Delta$, resp. $G^\gamma$. We use the notation of Figure 13, and write $\gamma_{i,j}^\Delta/\gamma_{i,j}^\gamma$ for the weight of the edge $(v_i, v_j)$. Using the identities $\text{dn}(u+K) = k' \text{nd}(u)$, $\text{sc}(\theta^*) = \text{sc}(K-\theta) = (k')^{-1} \text{cs}(\theta)$, we have the following: for every $j \in \{1, 2, 3\}$, with cyclic notation for indices:

$$\begin{aligned}
\gamma_{0,j}^{\gamma} &= (k')^{-\frac{1}{2}} \text{sc}(\theta_j) \text{dn}(u_{\alpha_j}) \text{dn}(u_{\alpha_{j+1}}), & \gamma_{j,j+1}^{\Delta} &= (k')^{-\frac{1}{2}} \text{cs}(\theta_{j-1}) \text{nd}(u_{\alpha_j}) \text{dn}(u_{\alpha_{j-1}}), \\
\gamma_{j,0}^{\gamma} &= (k')^{\frac{3}{2}} \text{sc}(\theta_j) \text{nd}(u_{\alpha_j}) \text{nd}(u_{\alpha_{j+1}}), & \gamma_{j+1,j}^{\Delta} &= (k')^{-\frac{1}{2}} \text{cs}(\theta_{j-1}) \text{dn}(u_{\alpha_j}) \text{nd}(u_{\alpha_{j-1}}).
\end{aligned}$$

<figure><img src="image_15.png" /><figcaption>Figure 13: Notation for vertices, angles and rhombus vectors in a star-triangle transformation.</figcaption></figure>

Consider a directed edge configuration T outside of the hexagon which can be extended to an r-dST of $G^\gamma/G^\Delta$. Then given T, the configurations inside the Y-Δ only depend on connection properties of T outside. As in the non-directed case [43], we thus have three kinds of configurations to consider.

I. $\{T : \{v_1, v_2, v_3\} \text{ are connected to } r\}$.

II. $\{T : \{v_j, v_{j+1}\} \text{ are connected to } \{r\}, j \in \{1, 2, 3\}\}$.

III. $\{T : \{v_j\} \text{ is connected to } \{r\}, j \in \{1, 2, 3\}\}$.
**Proposition 3.13.** Consider a finite isoradial graph G, and let $u \in C$. Then, the model of r-directed spanning trees on G, with weight function $\gamma(u)$ on the edges is Z-invariant.

**Remark 3.14.**

* By Theorem 3.8, we have:

$$Z_{\text{dimer}}(G^{\text{D},r}, c(u)) = C'(u) \cdot Z_{\text{rSF}}(G^*, \rho^*, m^*),$$

and in the paper [11], the model of rooted spanning forests with these weights is shown to be Z-invariant. But, since the proof of Theorem 3.8 does not provide a bijection between directed spanning trees and rooted spanning forests, the two decompositions of the partition functions are not directly comparable; they should nevertheless be compatible. Note that the computations of the proof of Proposition 3.13 are reminiscent but much simpler than those of [11] (the latter have been removed from the published version).

* The critical spanning tree model of [45] with conductances $\tan(\theta)$ is Z-invariant [43]. The result of [11] extends this to rooted spanning forest while Proposition 3.13 extends it to directed spanning trees.

*Proof.* Let $G^\gamma$ and $G^\Delta$ be two finite isoradial graphs differing by a star-triangle transformation, and let r be a fixed root on the boundary of the graph, outside of the hexagon defining the star/triangle. Let $\gamma^\Delta(u)$, resp. $\gamma^\gamma(u)$, be the weight function on r-dST of $G^\Delta$, resp. $G^\gamma$. We use the notation of Figure 13, and write $\gamma_{i,j}^\Delta/\gamma_{i,j}^\gamma$ for the weight of the edge $(v_i, v_j)$. Using the identities $\text{dn}(u+K) = k' \text{nd}(u)$, $\text{sc}(\theta^*) = \text{sc}(K-\theta) = (k')^{-1} \text{cs}(\theta)$, we have the following: for every $j \in \{1, 2, 3\}$, with cyclic notation for indices:

$$\begin{align*}
\gamma_{0,j}^{\gamma} &= (k')^{-\frac{1}{2}} \text{sc}(\theta_j) \text{dn}(u_{\alpha_j}) \text{dn}(u_{\alpha_{j+1}}), & \gamma_{j,j+1}^{\Delta} &= (k')^{-\frac{1}{2}} \text{cs}(\theta_{j-1}) \text{nd}(u_{\alpha_j}) \text{dn}(u_{\alpha_{j-1}}), \\
\gamma_{j,0}^{\gamma} &= (k')^{\frac{3}{2}} \text{sc}(\theta_j) \text{nd}(u_{\alpha_j}) \text{nd}(u_{\alpha_{j+1}}), & \gamma_{j+1,j}^{\Delta} &= (k')^{-\frac{1}{2}} \text{cs}(\theta_{j-1}) \text{dn}(u_{\alpha_j}) \text{nd}(u_{\alpha_{j-1}}).
\end{align*}$$

<figure><img src="image_15.png" /><figcaption>Figure 13: Notation for vertices, angles and rhombus vectors in a star-triangle transformation.</figcaption></figure>

Consider a directed edge configuration T outside of the hexagon which can be extended to an r-dST of $G^\gamma/G^\Delta$. Then given T, the configurations inside the Y-Δ only depend on connection properties of T outside. As in the non-directed case [43], we thus have three kinds of configurations to consider.

I. $\{T : \{v_1, v_2, v_3\} \text{ are connected to } r\}$.

II. $\{T : \{v_j, v_{j+1}\} \text{ are connected to } \{r\}, j \in \{1, 2, 3\}\}$.

III. $\{T : \{v_j\} \text{ is connected to } \{r\}, j \in \{1, 2, 3\}\}$.
Let $Z(G^\gamma | i)$, resp. $Z(G^\Delta | i)$, be the partition function of $G^\gamma$, resp. $G^\Delta$, restricted to an outer configuration T belonging to the set defined in $i$, $i \in \{\text{I, II, III}\}$, divided by the contribution of the configuration T. Then, proving Z-invariance amounts to showing that there exists a constant $\mathcal{C}$, such that

$$ \forall i \in \{\text{I, II, III}\}, \quad Z(G^\gamma | i) = \mathcal{C} \cdot Z(G^\Delta | i). $$

Let us prove that this is indeed the case, with $\mathcal{C} = (k')^{\frac{3}{2}} \prod_{j=1}^{3} \text{sc}(\theta_j)$.

**Restriction I** We have:

$$ Z(G^\gamma | I) = (k')^{-\frac{1}{2}} \sum_{j=1}^{3} \text{sc}(\theta_j) \text{dn}(u_{\alpha_j}) \text{dn}(u_{\alpha_{j+1}}) = (k')^{\frac{1}{2}} \sum_{j=1}^{3} A(\theta_j) $$

$$ Z(G^\Delta | I) = 1, $$

$$ \text{implying that } \mathcal{C} = (k')^{\frac{1}{2}} \sum_{j=1}^{3} A(\theta_j) = (k')^{\frac{3}{2}} \prod_{j=1}^{3} \text{sc}(\theta_j), \qquad (3.18) $$

by Equation (71) of Lemma 47 of [11].

**Restriction II** We have, $\forall j \in \{1, 2, 3\}$,

$$ \begin{aligned} Z(G^\gamma | \text{II}) &= \gamma_{j+2,0}^\gamma (\gamma_{0,j}^\gamma + \gamma_{0,j+1}^\gamma) \\ &= k' \text{sc}(\theta_{j+2}) \text{nd}(u_{\alpha_{j+2}}) \text{nd}(u_{\alpha_j}) [\text{sc}(\theta_j) \text{dn}(u_{\alpha_j}) \text{dn}(u_{\alpha_{j+1}}) + \\ & \qquad \qquad + \text{sc}(\theta_{j+1}) \text{dn}(u_{\alpha_{j+1}}) \text{dn}(u_{\alpha_{j+2}})] \\ &= k' \text{sc}(\theta_{j+2}) [\text{sc}(\theta_j) \text{nd}(u_{\alpha_{j+2}}) \text{dn}(u_{\alpha_{j+1}}) + \text{sc}(\theta_{j+1}) \text{nd}(u_{\alpha_j}) \text{dn}(u_{\alpha_{j+1}})], \end{aligned} $$

$$ \begin{aligned} Z(G^\Delta | \text{II}) &= \gamma_{j+2,j}^\Delta + \gamma_{j+2,j+1}^\Delta \\ &= (k')^{-\frac{1}{2}} [\text{cs}(\theta_{j+1}) \text{nd}(u_{\alpha_{j+2}}) \text{dn}(u_{\alpha_{j+1}}) + \text{cs}(\theta_j) \text{dn}(u_{\alpha_{j+1}}) \text{nd}(u_{\alpha_j})]. \end{aligned} $$

Using Equation (3.18), it is straightforward that $Z(G^\gamma | \text{II}) = \mathcal{C} Z(G^\Delta | \text{II})$.

**Restriction III** We have, $\forall j \in \{1, 2, 3\}$,

$$ \begin{aligned} Z(G^\gamma | \text{III}) &= \gamma_{j+1,0}^\gamma \gamma_{j+2,0}^\gamma \gamma_{0,j}^\gamma \\ &= (k')^{\frac{5}{2}} \text{sc}(\theta_{j+1}) \text{nd}(u_{\alpha_{j+1}}) \text{nd}(u_{\alpha_{j+2}}) \times \text{sc}(\theta_{j+2}) \text{nd}(u_{\alpha_{j+2}}) \text{nd}(u_{\alpha_j}) \times \\ & \qquad \times \text{sc}(\theta_j) \text{dn}(u_{\alpha_j}) \text{dn}(u_{\alpha_{j+1}}) \\ &= (k')^{\frac{5}{2}} \text{sc}(\theta_j) \text{sc}(\theta_{j+1}) \text{sc}(\theta_{j+2}) \text{nd}^2(u_{\alpha_{j+2}}), \end{aligned} $$

$$ \begin{aligned} Z(G^\Delta | \text{III}) &= \gamma_{j+1,j+2}^\Delta \gamma_{j+2,j}^\Delta + \gamma_{j+2,j+1}^\Delta \gamma_{j+1,j}^\Delta + \gamma_{j+1,j}^\Delta \gamma_{j+2,j}^\Delta \\ &= (k')^{-1} [\text{cs}(\theta_j) \text{nd}(u_{\alpha_{j+1}}) \text{dn}(u_{\alpha_j}) \text{cs}(\theta_{j+1}) \text{nd}(u_{\alpha_{j+2}}) \text{dn}(u_{\alpha_{j+1}}) + \\ & \qquad + \text{cs}(\theta_j) \text{dn}(u_{\alpha_{j+1}}) \text{nd}(u_{\alpha_j}) \text{cs}(\theta_{j+2}) \text{dn}(u_{\alpha_j}) \text{nd}(u_{\alpha_{j+2}}) + \\ & \qquad + \text{cs}(\theta_{j+2}) \text{dn}(u_{\alpha_j}) \text{nd}(u_{\alpha_{j+2}}) \text{cs}(\theta_{j+1}) \text{nd}(u_{\alpha_{j+2}}) \text{dn}(u_{\alpha_{j+1}})] \\ &= \frac{(k')^{-1}}{\prod_{j=1}^{3} \text{sc}(\theta_j)} \text{nd}^2(u_{\alpha_{j+2}}) [\text{sc}(\theta_{j+2}) \text{dn}(u_{\alpha_{j+2}}) \text{dn}(u_{\alpha_j}) + \\ & \qquad + \text{sc}(\theta_{j+1}) \text{dn}(u_{\alpha_{j+1}}) \text{dn}(u_{\alpha_{j+2}}) + \text{sc}(\theta_j) \text{dn}(u_{\alpha_j}) \text{dn}(u_{\alpha_{j+1}})] \\ &= k' \text{nd}^2(u_{\alpha_{j+2}}), \text{ by the proof of Restriction I.} \end{aligned} $$

The proof that $Z(G^\gamma | \text{III}) = \mathcal{C} Z(G^\Delta | \text{III})$ is concluded by using Equation (3.18) again. $\square$
Let $Z(G^\gamma | i)$, resp. $Z(G^\Delta | i)$, be the partition function of $G^\gamma$, resp. $G^\Delta$, restricted to an outer configuration T belonging to the set defined in $i$, $i \in \{\text{I, II, III}\}$, divided by the contribution of the configuration T. Then, proving Z-invariance amounts to showing that there exists a constant $\mathcal{C}$, such that

$$ \forall i \in \{\text{I, II, III}\}, \quad Z(G^\gamma | i) = \mathcal{C} \cdot Z(G^\Delta | i). $$

Let us prove that this is indeed the case, with $\mathcal{C} = (k')^{\frac{3}{2}} \prod_{j=1}^{3} \text{sc}(\theta_j)$.

**Restriction I** We have:

$$ Z(G^\gamma | I) = (k')^{-\frac{1}{2}} \sum_{j=1}^{3} \text{sc}(\theta_j) \text{dn}(u_{\alpha_j}) \text{dn}(u_{\alpha_{j+1}}) = (k')^{\frac{1}{2}} \sum_{j=1}^{3} A(\theta_j) $$

$$ Z(G^\Delta | I) = 1, $$

$$ \text{implying that } \mathcal{C} = (k')^{\frac{1}{2}} \sum_{j=1}^{3} A(\theta_j) = (k')^{\frac{3}{2}} \prod_{j=1}^{3} \text{sc}(\theta_j), \qquad (3.18) $$

by Equation (71) of Lemma 47 of [11].

**Restriction II** We have, $\forall j \in \{1, 2, 3\}$,

$$ \begin{aligned} Z(G^\gamma | \text{II}) &= \gamma_{j+2,0}^\gamma (\gamma_{0,j}^\gamma + \gamma_{0,j+1}^\gamma) \\ &= k' \text{sc}(\theta_{j+2}) \text{nd}(u_{\alpha_{j+2}}) \text{nd}(u_{\alpha_j}) [\text{sc}(\theta_j) \text{dn}(u_{\alpha_j}) \text{dn}(u_{\alpha_{j+1}}) + \\ &\quad + \text{sc}(\theta_{j+1}) \text{dn}(u_{\alpha_{j+1}}) \text{dn}(u_{\alpha_{j+2}})] \\ &= k' \text{sc}(\theta_{j+2}) [\text{sc}(\theta_j) \text{nd}(u_{\alpha_{j+2}}) \text{dn}(u_{\alpha_{j+1}}) + \text{sc}(\theta_{j+1}) \text{nd}(u_{\alpha_j}) \text{dn}(u_{\alpha_{j+1}})], \end{aligned} $$

$$ \begin{aligned} Z(G^\Delta | \text{II}) &= \gamma_{j+2,j}^\Delta + \gamma_{j+2,j+1}^\Delta \\ &= (k')^{-\frac{1}{2}} [\text{cs}(\theta_{j+1}) \text{nd}(u_{\alpha_{j+2}}) \text{dn}(u_{\alpha_{j+1}}) + \text{cs}(\theta_j) \text{dn}(u_{\alpha_{j+1}}) \text{nd}(u_{\alpha_j})]. \end{aligned} $$

Using Equation (3.18), it is straightforward that $Z(G^\gamma | \text{II}) = \mathcal{C} Z(G^\Delta | \text{II})$.

**Restriction III** We have, $\forall j \in \{1, 2, 3\}$,

$$ \begin{aligned} Z(G^\gamma | \text{III}) &= \gamma_{j+1,0}^\gamma \gamma_{j+2,0}^\gamma \gamma_{0,j}^\gamma \\ &= (k')^{\frac{5}{2}} \text{sc}(\theta_{j+1}) \text{nd}(u_{\alpha_{j+1}}) \text{nd}(u_{\alpha_{j+2}}) \times \text{sc}(\theta_{j+2}) \text{nd}(u_{\alpha_{j+2}}) \text{nd}(u_{\alpha_j}) \times \\ &\quad \times \text{sc}(\theta_j) \text{dn}(u_{\alpha_j}) \text{dn}(u_{\alpha_{j+1}}) \\ &= (k')^{\frac{5}{2}} \text{sc}(\theta_j) \text{sc}(\theta_{j+1}) \text{sc}(\theta_{j+2}) \text{nd}^2(u_{\alpha_{j+2}}), \end{aligned} $$

$$ \begin{aligned} Z(G^\Delta | \text{III}) &= \gamma_{j+1,j+2}^\Delta \gamma_{j+2,j}^\Delta + \gamma_{j+2,j+1}^\Delta \gamma_{j+1,j}^\Delta + \gamma_{j+1,j}^\Delta \gamma_{j+2,j}^\Delta \\ &= (k')^{-1} [\text{cs}(\theta_j) \text{nd}(u_{\alpha_{j+1}}) \text{dn}(u_{\alpha_j}) \text{cs}(\theta_{j+1}) \text{nd}(u_{\alpha_{j+2}}) \text{dn}(u_{\alpha_{j+1}}) + \\ &\quad + \text{cs}(\theta_j) \text{dn}(u_{\alpha_{j+1}}) \text{nd}(u_{\alpha_j}) \text{cs}(\theta_{j+2}) \text{dn}(u_{\alpha_j}) \text{nd}(u_{\alpha_{j+2}}) + \\ &\quad + \text{cs}(\theta_{j+2}) \text{dn}(u_{\alpha_j}) \text{nd}(u_{\alpha_{j+2}}) \text{cs}(\theta_{j+1}) \text{nd}(u_{\alpha_{j+2}}) \text{dn}(u_{\alpha_{j+1}})] \\ &= \frac{(k')^{-1}}{\prod_{j=1}^3 \text{sc}(\theta_j)} \text{nd}^2(u_{\alpha_{j+2}}) [\text{sc}(\theta_{j+2}) \text{dn}(u_{\alpha_{j+2}}) \text{dn}(u_{\alpha_j}) + \\ &\quad + \text{sc}(\theta_{j+1}) \text{dn}(u_{\alpha_{j+1}}) \text{dn}(u_{\alpha_{j+2}}) + \text{sc}(\theta_j) \text{dn}(u_{\alpha_j}) \text{dn}(u_{\alpha_{j+1}})] \\ &= k' \text{nd}^2(u_{\alpha_{j+2}}), \quad \text{by the proof of Restriction I.} \end{aligned} $$

The proof that $Z(G^\gamma | \text{III}) = \mathcal{C} Z(G^\Delta | \text{III})$ is concluded by using Equation (3.18) again. $\square$
## 3.6 Inverse $Z^u$-Dirac operator and dimer model on the double graph

Using Theorem 3.6, in Corollaries 3.15 and 3.17, we express the inverse $Z^u$-Dirac operators $K(u)^{-1}$ and $K^{\partial}(u)^{-1}$ using the Z-massive and dual Z-massive Green functions. In the infinite case, this allows to prove in Theorem 3.19 an explicit local expression for a Gibbs measure for the dimer model on the double graph with weight function $c(u)$.

### 3.6.1 Inverse $Z^u$-Dirac operator and Z-massive Green functions

Infinite case. Consider an infinite isoradial graph G, and the $Z^u$-Dirac operator $K(u)$. Consider also the Z-massive Laplacian $\Delta^m$ on G, the dual Z-massive Laplacian $\Delta^{m,*}$ of the dual $G^*$ [11]. When $k \neq 0$, let $G^m$ and $G^{m,*}$ be the Z-massive and dual massive Green functions of G and $G^*$ of [11], whose definition is recalled in Section 2.5.6. When $k = 0$, the mass is 0 and we let $G^0$ and $G^{0,*}$ be the Green and dual Green functions of [45].

**Notation for coefficients of Corollary 3.15** Let $\nu$, resp. $f$, be a vertex of G, resp. $G^*$. Let $w$ be a white vertex of $G^D$, its neighbors in $G^D$ are $v_1, f_1, v_2, f_2$, and let $e^{i\bar{\alpha}_f}, e^{i\bar{\beta}_f}, \bar{\theta}_f$ be the rhombus vectors and half-angle of $G^{\diamond/2}$ associated to the edge $(w, v_2)$, where the subscript "f" stands for "final", see Figure 14.

<figure><img src="image_16.png" /><figcaption>Figure 14: Notation for coefficients of Corollary 3.15 and 3.17.</figcaption></figure>

As a consequence of Theorem 3.6 (infinite case) we obtain,

**Corollary 3.15.** For every $u \in C$, consider the operator $K(u)^{-1}$ mapping $C^W$ to $C^B$ defined by:

* Matrix form. $K(u)^{-1} = (k')^{-1} \begin{pmatrix} G^m & 0 \\ 0 & G^{m,*} \end{pmatrix} \overline{K(u)}^t$.

* Coefficients. For every $\nu, f, w$ as in the notation above,

$$K(u)_{\nu,w}^{-1} = e^{-i \frac{\bar{\alpha}_f + \bar{\beta}_f}{2}} (k')^{-1} \text{sc}(\theta_f) \frac{1}{2} \left( \left[ \text{dn}(u_{\alpha_f}) \text{dn}(u_{\beta_f}) \right]^{\frac{1}{2}} G_{\nu,v_2}^m - \left[ \text{dn}(u_{\alpha_f+2K}) \text{dn}(u_{\beta_f+2K}) \right]^{\frac{1}{2}} G_{\nu,v_1}^m \right)$$

$$K(u)_{f,w}^{-1} = -ie^{-i \frac{\bar{\alpha}_f + \bar{\beta}_f}{2}} (k')^{-1} \text{sc}(\theta_f^*) \frac{1}{2} \left( \left[ \text{dn}((u_{\beta_f})^*) \text{dn}((u_{\alpha_f+2K})^*) \right]^{\frac{1}{2}} G_{f,f_2}^{m,*} + \right. \\ \left. - \left[ \text{dn}((u_{\beta_f-2K})^*) \text{dn}((u_{\alpha_f})^*) \right]^{\frac{1}{2}} G_{f,f_1}^{m,*} \right).$$

Then $K(u)^{-1}$ is the unique inverse of the operator $K(u)$ decreasing to zero at infinity.

**Remark 3.16.**

* When $k = 0$, then $dn \equiv 1$ and we recover Corollary 7.2 of [45]. When the graph is $Z^2$ and the weights are specific, Chhita [20] has a result in the same flavor, relating the inverse Kasteleyn operator of pairs of dual directed spanning trees to a massive Green function.

* The operator $K(u)^{-1}$ is local. Indeed the expression for $K(u)_{\bar{x},w}^{-1}$, with $\bar{x} = \nu \in V$ or $\bar{x} = f \in V^*$, only depends on: two paths of the diamond graph $G^{\diamond}$ from $\bar{x}$ to
## 3.6 Inverse $Z^u$-Dirac operator and dimer model on the double graph

Using Theorem 3.6, in Corollaries 3.15 and 3.17, we express the inverse $Z^u$-Dirac operators $K(u)^{-1}$ and $K^{\partial}(u)^{-1}$ using the Z-massive and dual Z-massive Green functions. In the infinite case, this allows to prove in Theorem 3.19 an explicit local expression for a Gibbs measure for the dimer model on the double graph with weight function $c(u)$.

### 3.6.1 Inverse $Z^u$-Dirac operator and Z-massive Green functions

Infinite case. Consider an infinite isoradial graph G, and the $Z^u$-Dirac operator $K(u)$. Consider also the Z-massive Laplacian $\Delta^m$ on G, the dual Z-massive Laplacian $\Delta^{m,*}$ of the dual $G^*$ [11]. When $k \neq 0$, let $G^m$ and $G^{m,*}$ be the Z-massive and dual massive Green functions of G and $G^*$ of [11], whose definition is recalled in Section 2.5.6. When $k = 0$, the mass is 0 and we let $G^0$ and $G^{0,*}$ be the Green and dual Green functions of [45].

**Notation for coefficients of Corollary 3.15** Let $v$, resp. $f$, be a vertex of G, resp. $G^*$. Let $w$ be a white vertex of $G^D$, its neighbors in $G^D$ are $v_1, f_1, v_2, f_2$, and let $e^{i\bar{\alpha}_f}, e^{i\bar{\beta}_f}, \bar{\theta}_f$ be the rhombus vectors and half-angle of $G^{\diamond/2}$ associated to the edge $(w, v_2)$, where the subscript "f" stands for "final", see Figure 14.

<figure><img src="image_16.png" /><figcaption>Figure 14: Notation for coefficients of Corollary 3.15 and 3.17.</figcaption></figure>

As a consequence of Theorem 3.6 (infinite case) we obtain,

**Corollary 3.15.** For every $u \in C$, consider the operator $K(u)^{-1}$ mapping $C^W$ to $C^B$ defined by:

* Matrix form. $K(u)^{-1} = (k')^{-1} \begin{pmatrix} G^m & 0 \\ 0 & G^{m,*} \end{pmatrix} \overline{K(u)}^t$.

* Coefficients. For every $v, f, w$ as in the notation above,

$$K(u)_{v,w}^{-1} = e^{-i \frac{\bar{\alpha}_f + \bar{\beta}_f}{2}} (k')^{-1} \text{sc}(\theta_f) \frac{1}{2} \left( \left[ \text{dn}(u_{\alpha_f}) \text{dn}(u_{\beta_f}) \right]^{\frac{1}{2}} G_{v,v_2}^m - \left[ \text{dn}(u_{\alpha_f+2K}) \text{dn}(u_{\beta_f+2K}) \right]^{\frac{1}{2}} G_{v,v_1}^m \right)$$

$$K(u)_{f,w}^{-1} = -ie^{-i \frac{\bar{\alpha}_f + \bar{\beta}_f}{2}} (k')^{-1} \text{sc}(\theta_f^*) \frac{1}{2} \left( \left[ \text{dn}((u_{\beta_f})^*) \text{dn}((u_{\alpha_f+2K})^*) \right]^{\frac{1}{2}} G_{f,f_2}^{m,*} + \right. \\ \left. - \left[ \text{dn}((u_{\beta_f-2K})^*) \text{dn}((u_{\alpha_f})^*) \right]^{\frac{1}{2}} G_{f,f_1}^{m,*} \right).$$

Then $K(u)^{-1}$ is the unique inverse of the operator $K(u)$ decreasing to zero at infinity.

**Remark 3.16.**

* When $k = 0$, then $dn \equiv 1$ and we recover Corollary 7.2 of [45]. When the graph is $Z^2$ and the weights are specific, Chhita [20] has a result in the same flavor, relating the inverse Kasteleyn operator of pairs of dual directed spanning trees to a massive Green function.

* The operator $K(u)^{-1}$ is local. Indeed the expression for $K(u)_{x,w}^{-1}$, with $x = v \in V$ or $x = f \in V^*$, only depends on: two paths of the diamond graph $G^{\diamond}$ from $x$ to
the two neighbors $x_1, x_2$ of $w$ in $G$ or $G^*$, where the two paths are those used in computing the massive exponential function of coefficients of the massive Green function of [11]; and from the rhombus vectors and half-angles of $G^{\diamond/2}$ associated to the edges $(w, x_1), (w, x_2)$.

**Proof.** By definition, see Section 2.2.2, to show that $K(u)^{-1}$ is an inverse, we need to prove that $K(u)^{-1}K(u) = \text{Id}$, and that $K(u)_{x,w}^{-1} \to 0$ as $|w - x| \to \infty$. For the second part, when $k \neq 0$ we prove in [11] that the Green function $G^m$ and $G^{m,*}$ decrease exponentially fast to 0 at infinity; since the function $dn$ is uniformly bounded, the same holds for $K(u)^{-1}$. When $k = 0$, the Green function explodes like $-\log|x - x'|$ at infinity, but the difference converges polynomially fast to 0 as is proved in [45]. For the first part, we apply $K(u)$ to the right of the definition of $K(u)^{-1}$ in matrix form and use Theorem 3.6. Note that this step also uses associativity of the infinite matrix product, which holds in this case. Uniqueness when $k = 0$ is proved in [45, Theorem 4.1]; when $k \neq 0$, it follows from the fact that 0 is the unique function in the kernel of $K(u)$ decreasing to zero at infinity, see [8, Proposition 47] (note that when $k \neq 0$, subexponential growth suffices for this argument to work). $\square$

**Finite case.** Consider a finite isoradial graph $G$ and the $Z^u$-Dirac operators $K(u)$ and $K^{\partial}(u)$ of Section 3.1 (finite case). Consider also the $Z$-massive Laplacian $\Delta^{m,\partial}(u)$ and dual Laplacian $\Delta^{m,*}$ of Section 3.2 (finite case). For the purpose of handling the Ising model, our goal is to obtain an explicit expression for the inverse of the operator $K^{\partial}(u)$. In order to ensure that $K^{\partial}(u)$ and $\Delta^{m,\partial}(u)$ are invertible, we restrict $u$ to:

$$C'' := C \setminus \{\alpha^{\ell}[2K], \beta^{\ell}[2K] : e^{i\bar{\alpha}^{\ell}}, e^{i\bar{\beta}^{\ell}} \in \mathbb{R}^{\partial}\}. \quad (3.19)$$

Indeed, by the forthcoming Remark 4.4 this ensures that $\det K^{\partial}(u) \neq 0$ and by Corollary 3.10 that $\det \Delta^{m,\partial}(u) \neq 0$ (since $nd$ is positive when $u$ is real). Denote by $G^{m,*}$ the inverse of the (finite) matrix $\Delta^{m,*}$ and by $G^{m,\partial}(u)$ the inverse of $\Delta^{m,\partial}(u)$. Note that when some of the conductances $(\rho_{v^c, v^\ell}^{\partial}(u))$ of the boundary rhombus pairs of $\mathbb{R}^{\partial,\tau}$ are negative, there is a twist in defining the associated random walk and in giving the random walk interpretation of the Green function, but this might at most happen for half of the boundary edges.

**Notation for coefficients of Corollary 3.17** Let $w$ be a white vertex of $G^{\mathbb{D},\tau}$; if $w \notin \{w^\ell, w^r \in \mathbb{R}^\partial\}$, then we use the notation of the infinite case, see Figure 14; if $w \in \{w^\ell, w^r \in \mathbb{R}^\partial\}$, then the vertex $f_2$ is absent. As an immediate consequence of Theorem 3.6 (finite case) we obtain

**Corollary 3.17.** For every $u \in C''$, the inverse matrix $K^{\partial}(u)^{-1}$ has the following explicit expression.

* Matrix form. $K^{\partial}(u)^{-1} = (k')^{-1} [\overline{K(u)} \begin{pmatrix} G^{m,\partial}(u) & -G^{m,\partial}(u) \overline{Q(u)} G^{m,*} \\ 0 & G^{m,*} \end{pmatrix}]^t$.

* Coefficients. For every vertex $v$ of $G^r$, $f$ of $G^*$, and every white vertex $w$ of $G^{\mathbb{D},\tau}$, using the notation of Figure 14,

$$K^{\partial}(u)_{v,w}^{-1} = e^{-i \frac{\bar{\alpha}_{\ell} + \bar{\beta}_{\ell}}{2}} (k')^{-1} \text{sc}(\theta_{\ell})^{\frac{1}{2}} \left( [\text{dn}(u_{\alpha_{\ell}}) \text{dn}(u_{\beta_{\ell}})]^{\frac{1}{2}} G^{m,\partial}(u)_{v_2,v} + \right. \\ \left. - [\text{dn}(u_{\alpha_{\ell}+2K}) \text{dn}(u_{\beta_{\ell}+2K})]^{\frac{1}{2}} G^{m,\partial}(u)_{v_1,v} \right) \\ K^{\partial}(u)_{f,w}^{-1} = -i e^{-i \frac{\bar{\alpha}_{f} + \bar{\beta}_{f}}{2}} (k')^{-1} \text{sc}(\theta_{f}^{*})^{\frac{1}{2}} \left( \mathbb{I}_{\{w \notin \{w^{\ell}, w^{r} \in \mathbb{R}^{\partial}\}\}} [\text{dn}((u_{\beta_{f}})^{*}) \text{dn}((u_{\alpha_{f}+2K})^{*})]^{\frac{1}{2}} G^{m,*}_{f_2,f} + \right. \\ \left. - [\text{dn}((u_{\beta_{f}-2K})^{*}) \text{dn}((u_{\alpha_{f}})^{*})]^{\frac{1}{2}} G^{m,*}_{f_1,f} \right) + \\ + i \sum_{(v^c, f^c) \in \mathbb{R}^{\partial,\tau}} \frac{\text{nd}(u_{\beta^c \ell})}{\text{cd}(u_{\alpha^c \ell})} (\text{cd}(u_{\beta^r}) - \text{cd}(u_{\alpha^r})) K^{\partial}(u)_{v^c,w}^{-1} \cdot G^{m,*}_{f^c,f^c},$$
the two neighbors $x_1, x_2$ of $w$ in $G$ or $G^*$, where the two paths are those used in computing the massive exponential function of coefficients of the massive Green function of [11]; and from the rhombus vectors and half-angles of $G^{\diamond/2}$ associated to the edges $(w, x_1), (w, x_2)$.

**Proof.** By definition, see Section 2.2.2, to show that $K(u)^{-1}$ is an inverse, we need to prove that $K(u)^{-1}K(u) = \text{Id}$, and that $K(u)_{x,w}^{-1} \to 0$ as $|w - x| \to \infty$. For the second part, when $k \neq 0$ we prove in [11] that the Green function $G^m$ and $G^{m,*}$ decrease exponentially fast to 0 at infinity; since the function $d\nu$ is uniformly bounded, the same holds for $K(u)^{-1}$. When $k = 0$, the Green function explodes like $-\log|x - x'|$ at infinity, but the difference converges polynomially fast to 0 as is proved in [45]. For the first part, we apply $K(u)$ to the right of the definition of $K(u)^{-1}$ in matrix form and use Theorem 3.6. Note that this step also uses associativity of the infinite matrix product, which holds in this case. Uniqueness when $k = 0$ is proved in [45, Theorem 4.1]; when $k \neq 0$, it follows from the fact that 0 is the unique function in the kernel of $K(u)$ decreasing to zero at infinity, see [8, Proposition 47] (note that when $k \neq 0$, subexponential growth suffices for this argument to work). $\square$

**Finite case.** Consider a finite isoradial graph $G$ and the $Z^u$-Dirac operators $K(u)$ and $K^{\partial}(u)$ of Section 3.1 (finite case). Consider also the $Z$-massive Laplacian $\Delta^{m,\partial}(u)$ and dual Laplacian $\Delta^{m,*}(u)$ of Section 3.2 (finite case). For the purpose of handling the Ising model, our goal is to obtain an explicit expression for the inverse of the operator $K^{\partial}(u)$. In order to ensure that $K^{\partial}(u)$ and $\Delta^{m,\partial}(u)$ are invertible, we restrict $u$ to:

$$ C'' := C \setminus \{\alpha^{\ell}[2K], \beta^{\ell}[2K] : e^{i\bar{\alpha}^{\ell}}, e^{i\bar{\beta}^{\ell}} \in \mathbb{R}^{\partial}\}. \quad (3.19) $$

Indeed, by the forthcoming Remark 4.4 this ensures that $\det K^{\partial}(u) \neq 0$ and by Corollary 3.10 that $\det \Delta^{m,\partial}(u) \neq 0$ (since $nd$ is positive when $u$ is real). Denote by $G^{m,*}$ the inverse of the (finite) matrix $\Delta^{m,*}$ and by $G^{m,\partial}(u)$ the inverse of $\Delta^{m,\partial}(u)$. Note that when some of the conductances $(\rho_{v^c, v^\ell}^{\partial}(u))$ of the boundary rhombus pairs of $\mathbb{R}^{\partial,\tau}$ are negative, there is a twist in defining the associated random walk and in giving the random walk interpretation of the Green function, but this might at most happen for half of the boundary edges.

**Notation for coefficients of Corollary 3.17** Let $w$ be a white vertex of $G^{\mathbb{D},\tau}$; if $w \notin \{w^\ell, w^r \in \mathbb{R}^\partial\}$, then we use the notation of the infinite case, see Figure 14; if $w \in \{w^\ell, w^r \in \mathbb{R}^\partial\}$, then the vertex $f_2$ is absent. As an immediate consequence of Theorem 3.6 (finite case) we obtain

**Corollary 3.17.** For every $u \in C''$, the inverse matrix $K^{\partial}(u)^{-1}$ has the following explicit expression.

* Matrix form. $K^{\partial}(u)^{-1} = (k')^{-1} [\overline{K(u)} \begin{pmatrix} G^{m,\partial}(u) & -G^{m,\partial}(u) \overline{Q(u)} G^{m,*} \\ 0 & G^{m,*} \end{pmatrix}]^t$.

* Coefficients. For every vertex $v$ of $G^r$, $f$ of $G^*$, and every white vertex $w$ of $G^{\mathbb{D},\tau}$, using the notation of Figure 14,

$$ \begin{align*} K^{\partial}(u)_{v,w}^{-1} &= e^{-i \frac{\bar{\alpha}_{\ell} + \bar{\beta}_{\ell}}{2}} (k')^{-1} \text{sc}(\theta_{\ell})^{\frac{1}{2}} \left( [\text{dn}(u_{\alpha_{\ell}}) \text{dn}(u_{\beta_{\ell}})]^{\frac{1}{2}} G^{m,\partial}(u)_{v_2,v} + \right. \\ & \qquad \left. - [\text{dn}(u_{\alpha_{\ell}+2K}) \text{dn}(u_{\beta_{\ell}+2K})]^{\frac{1}{2}} G^{m,\partial}(u)_{v_1,v} \right) \\ K^{\partial}(u)_{f,w}^{-1} &= -i e^{-i \frac{\bar{\alpha}_{f} + \bar{\beta}_{f}}{2}} (k')^{-1} \text{sc}(\theta_{f}^{*})^{\frac{1}{2}} \left( \mathbb{I}_{\{w \notin \{w^{\ell}, w^{r} \in \mathbb{R}^{\partial}\}\}} [\text{dn}((u_{\beta_{f}})^{*}) \text{dn}((u_{\alpha_{f}+2K})^{*})]^{\frac{1}{2}} G_{f_2,f}^{m,*} + \right. \\ & \qquad \left. - [\text{dn}((u_{\beta_{f}-2K})^{*}) \text{dn}((u_{\alpha_{f}})^{*})]^{\frac{1}{2}} G_{f_1,f}^{m,*} \right) + \\ & \qquad + i \sum_{(v^c, f^c) \in \mathbb{R}^{\partial,\tau}} \frac{\text{nd}(u_{\beta^c \ell})}{\text{cd}(u_{\alpha^c \ell})} (\text{cd}(u_{\beta^r}) - \text{cd}(u_{\alpha^r})) K^{\partial}(u)_{v^c,w}^{-1} \cdot G_{f^c,f}^{m,*} \end{align*} $$
where $\mathrm{I}_{\{w \notin \{w^l, w^r \in \mathbb{R}^\partial\}\}}$ is equal to 0 if $w$ is a boundary vertex of $G^{\mathrm{D},r}$ and 1 otherwise. In the sum over $(v^c, f^c) \in \mathbb{R}^\partial, r$, we use the notation of Figure 10 for vertices and rhombus vectors of the boundary rhombus pairs of $\mathbb{R}^\partial, r$; $K^\partial(u)_{v^c, w}^{-1}$ in the formula for $K^\partial(u)_{f, w}^{-1}$ is given by the first formula with $v = v^c$.

**Example 3.18.** Let us compute the explicit values of Corollaries 3.15 and 3.17 in the case where $u = \mathbf{u}^f := \frac{\alpha_f + \beta_f}{2} + K$. This will be used again in Sections 4.3 and 5.3. In the finite case, we will also need to evaluate it at $u = \mathbf{v}^f := \frac{\alpha_f + \beta_f}{2} - K$. We have,

$$ 
\begin{align} 
\mathbf{v}^f &= \mathbf{u}^f - 2K, \quad \mathbf{u}_{\alpha_f}^f = \frac{K+\theta_f}{2}, \quad \mathbf{u}_{\beta_f}^f = \frac{K-\theta_f}{2}, \quad \mathbf{v}_{\alpha_f}^f = -\mathbf{u}_{\beta_f}^f, \quad \mathbf{v}_{\beta_f}^f = -\mathbf{u}_{\alpha_f}^f \nonumber \\
\mathbf{u}_{\alpha_f}^f + \mathbf{u}_{\beta_f}^f &= K, \quad \mathbf{u}_{\alpha_f+2K}^f + \mathbf{u}_{\beta_f+2K}^f = -K, \quad \mathbf{u}_{\alpha_f+2K}^f + \mathbf{u}_{\beta_f}^f = 0, \quad \mathbf{u}_{\alpha_f}^f + \mathbf{u}_{\beta_f-2K}^f = 2K \tag{3.20} 
\end{align} 
$$

$$ \mathbf{v}_{\alpha_f}^f + \mathbf{v}_{\beta_f}^f = -K, \quad \mathbf{v}_{\alpha_f+2K}^f + \mathbf{v}_{\beta_f+2K}^f = -3K, \quad \mathbf{v}_{\alpha_f+2K}^f + \mathbf{v}_{\beta_f}^f = -2K, \quad \mathbf{v}_{\alpha_f}^f + \mathbf{v}_{\beta_f-2K}^f = 0. $$

Using that $\mathrm{dn}(u \pm K) = k' \mathrm{nd}(u)$, $\mathrm{dn}(u \pm 2K) = \mathrm{dn}(u)$, $\mathrm{dn}(-u) = \mathrm{dn}(u)$ and the above relations, we obtain, for $u \in \{\mathbf{u}^f, \mathbf{v}^f\}$,

$$ 
\begin{align*} 
[\mathrm{dn}(u_{\alpha_f}) \mathrm{dn}(u_{\beta_f})]^{\frac{1}{2}} &= [\mathrm{dn}(u_{\alpha_f+2K}) \mathrm{dn}(u_{\beta_f+2K})]^{\frac{1}{2}} = (k')^{\frac{1}{2}} \\
[\mathrm{dn}((u_{\beta_f})^*) \mathrm{dn}((u_{\alpha_f+2K})^*)]^{\frac{1}{2}} &= k' \mathrm{nd}(u_{\beta_f}) = k' \begin{cases} \mathrm{nd}(\frac{K-\theta_f}{2}) & \text{if } u = \mathbf{u}^f \\ \mathrm{nd}(\frac{K+\theta_f}{2}) & \text{if } u = \mathbf{v}^f \end{cases} \\
[\mathrm{dn}((u_{\beta_f-2K})^*) \mathrm{dn}((u_{\alpha_f})^*)]^{\frac{1}{2}} &= k' \mathrm{nd}(u_{\alpha_f}) = k' \begin{cases} \mathrm{nd}(\frac{K+\theta_f}{2}) & \text{if } u = \mathbf{u}^f \\ \mathrm{nd}(\frac{K-\theta_f}{2}) & \text{if } u = \mathbf{v}^f, \end{cases} 
\end{align*} 
$$

an thus, with the notation of Corollaries 3.15 and 3.17:

* Infinite case.

$$ 
\begin{align*} 
K(\mathbf{u}^f)^{-1}_{v,w} &= e^{-i \frac{\bar{\alpha}_f + \bar{\beta}_f}{2}} \mathrm{sc}(\theta_f)^{\frac{1}{2}} (k')^{-\frac{1}{2}} (G^m_{v,v_2} - G^m_{v,v_1}) \\ 
K(\mathbf{u}^f)^{-1}_{f,w} &= -ie^{-i \frac{\bar{\alpha}_f + \bar{\beta}_f}{2}} \mathrm{sc}(\theta_f^*)^{\frac{1}{2}} \left( \mathrm{nd}\left(\frac{K-\theta_f}{2}\right) G^{m,*}_{f,f_2} - \mathrm{nd}\left(\frac{K+\theta_f}{2}\right) G^{m,*}_{f,f_1} \right). 
\end{align*} 
$$

* Finite case.

$$ 
\begin{align*} 
K^\partial(\mathbf{u}^f)^{-1}_{v,w} &= e^{-i \frac{\bar{\alpha}_f + \bar{\beta}_f}{2}} (k')^{-\frac{1}{2}} \mathrm{sc}(\theta_f)^{\frac{1}{2}} \left( G^{m,\partial}(\mathbf{u}^f)_{v_2,v} - G^{m,\partial}(\mathbf{u}^f)_{v_1,v} \right) \\ 
K^\partial(\mathbf{u}^f)^{-1}_{f,w} &= -ie^{-i \frac{\bar{\alpha}_f + \bar{\beta}_f}{2}} \mathrm{sc}(\theta_f^*)^{\frac{1}{2}} \left( \mathrm{I}_{\{w \notin \{w^l, w^r \in \mathbb{R}^\partial\}\}} \mathrm{nd}\left(\frac{K-\theta_f}{2}\right) G^{m,*}_{f_2,f} - \mathrm{nd}\left(\frac{K+\theta_f}{2}\right) G^{m,*}_{f_1,f} \right) + \\ 
& \qquad + i \sum_{(v^c, f^c) \in \mathbb{R}^\partial, r} \frac{\mathrm{nd}(\mathbf{u}^f_{\beta^c})}{\mathrm{cd}(\mathbf{u}^f_{\alpha^c})} (\mathrm{cd}(\mathbf{u}^f_{\beta^r}) - \mathrm{cd}(\mathbf{u}^f_{\alpha^r})) K^\partial(\mathbf{u}^f)^{-1}_{v^c,w} \cdot G^{m,*}_{f^c,f^c}, 
\end{align*} 
$$

$$ K^\partial(\mathbf{v}^f)^{-1}_{v,w} = e^{-i \frac{\bar{\alpha}_f + \bar{\beta}_f}{2}} (k')^{-\frac{1}{2}} \mathrm{sc}(\theta_f)^{\frac{1}{2}} \left( G^{m,\partial}(\mathbf{v}^f)_{v_2,v} - G^{m,\partial}(\mathbf{v}^f)_{v_1,v} \right) $$

$$ 
\begin{align*} 
K^\partial(\mathbf{v}^f)^{-1}_{f,w} &= -ie^{-i \frac{\bar{\alpha}_f + \bar{\beta}_f}{2}} \mathrm{sc}(\theta_f^*)^{\frac{1}{2}} \left( \mathrm{I}_{\{w \notin \{w^l, w^r \in \mathbb{R}^\partial\}\}} \mathrm{nd}\left(\frac{K+\theta_f}{2}\right) G^{m,*}_{f_2,f} - \mathrm{nd}\left(\frac{K-\theta_f}{2}\right) G^{m,*}_{f_1,f} \right) + \\ 
& \qquad + i \sum_{(v^c, f^c) \in \mathbb{R}^\partial, r} \frac{\mathrm{nd}(\mathbf{v}^f_{\beta^c})}{\mathrm{cd}(\mathbf{v}^f_{\alpha^c})} (\mathrm{cd}(\mathbf{v}^f_{\beta^r}) - \mathrm{cd}(\mathbf{v}^f_{\alpha^r})) K^\partial(\mathbf{v}^f)^{-1}_{v^c,w} \cdot G^{m,*}_{f^c,f^c}. 
\end{align*} 
$$

### 3.6.2 Dimer model on an infinite isoradial double graph $G^D$

Suppose that the isoradial graph $G$ is infinite; when it is moreover $\mathbb{Z}^2$-periodic, we consider the natural exhaustion $(G^D_n)_{n \ge 1}$ of $G^D$ by toroidal graphs, where $G^D_n = G^D/n\mathbb{Z}^2$. Let $\mathcal{F}$ denote the $\sigma$-field generated by cylinder sets of $G^D$. Using arguments of [23, 47, 28], we obtain an explicit, *local* expression for a Gibbs measure of the dimer model on $G^D$ with weight function $c(u)$. Since the proof closely follows that done in the papers [28, 10, 12], we do not repeat it here. The key requirements are that the operator $K(u)^{-1}$ is local and unique in the $\mathbb{Z}^2$-periodic case.
where $\mathbb{I}_{\{w \notin \{w^l, w^r \in \mathbb{R}^\partial\}\}}$ is equal to 0 if $w$ is a boundary vertex of $G^{\mathbb{D},r}$ and 1 otherwise. In the sum over $(v^c, f^c) \in \mathbb{R}^{\partial,r}$, we use the notation of Figure 10 for vertices and rhombus vectors of the boundary rhombus pairs of $\mathbb{R}^{\partial,r}$; $K^\partial(u)_{v^c,w}^{-1}$ in the formula for $K^\partial(u)_{f,w}^{-1}$ is given by the first formula with $v = v^c$.

**Example 3.18.** Let us compute the explicit values of Corollaries 3.15 and 3.17 in the case where $u = \mathbf{u}^f := \frac{\alpha_f + \beta_f}{2} + K$. This will be used again in Sections 4.3 and 5.3. In the finite case, we will also need to evaluate it at $u = \mathbf{v}^f := \frac{\alpha_f + \beta_f}{2} - K$. We have,

$$ 
\begin{align} 
\mathbf{v}^f &= \mathbf{u}^f - 2K, \quad \mathbf{u}_{\alpha_f}^f = \frac{K+\theta_f}{2}, \quad \mathbf{u}_{\beta_f}^f = \frac{K-\theta_f}{2}, \quad \mathbf{v}_{\alpha_f}^f = -\mathbf{u}_{\beta_f}^f, \quad \mathbf{v}_{\beta_f}^f = -\mathbf{u}_{\alpha_f}^f \nonumber \\
\mathbf{u}_{\alpha_f}^f + \mathbf{u}_{\beta_f}^f &= K, \quad \mathbf{u}_{\alpha_f+2K}^f + \mathbf{u}_{\beta_f+2K}^f = -K, \quad \mathbf{u}_{\alpha_f+2K}^f + \mathbf{u}_{\beta_f}^f = 0, \quad \mathbf{u}_{\alpha_f}^f + \mathbf{u}_{\beta_f-2K}^f = 2K \tag{3.20} 
\end{align} 
$$

$$ \mathbf{v}_{\alpha_f}^f + \mathbf{v}_{\beta_f}^f = -K, \quad \mathbf{v}_{\alpha_f+2K}^f + \mathbf{v}_{\beta_f+2K}^f = -3K, \quad \mathbf{v}_{\alpha_f+2K}^f + \mathbf{v}_{\beta_f}^f = -2K, \quad \mathbf{v}_{\alpha_f}^f + \mathbf{v}_{\beta_f-2K}^f = 0. $$

Using that $dn(u \pm K) = k' \text{nd}(u)$, $dn(u \pm 2K) = \text{dn}(u)$, $dn(-u) = \text{dn}(u)$ and the above relations, we obtain, for $u \in \{\mathbf{u}^f, \mathbf{v}^f\}$,

$$ 
\begin{align*} 
[\text{dn}(u_{\alpha_f}) \text{dn}(u_{\beta_f})]^{\frac{1}{2}} &= [\text{dn}(u_{\alpha_f+2K}) \text{dn}(u_{\beta_f+2K})]^{\frac{1}{2}} = (k')^{\frac{1}{2}} \\
[\text{dn}((u_{\beta_f})^*) \text{dn}((u_{\alpha_f+2K})^*)]^{\frac{1}{2}} &= k' \text{nd}(u_{\beta_f}) = k' \begin{cases} \text{nd}(\frac{K-\theta_f}{2}) & \text{if } u = \mathbf{u}^f \\ \text{nd}(\frac{K+\theta_f}{2}) & \text{if } u = \mathbf{v}^f \end{cases} \\
[\text{dn}((u_{\beta_f-2K})^*) \text{dn}((u_{\alpha_f})^*)]^{\frac{1}{2}} &= k' \text{nd}(u_{\alpha_f}) = k' \begin{cases} \text{nd}(\frac{K+\theta_f}{2}) & \text{if } u = \mathbf{u}^f \\ \text{nd}(\frac{K-\theta_f}{2}) & \text{if } u = \mathbf{v}^f, \end{cases} 
\end{align*} 
$$

an thus, with the notation of Corollaries 3.15 and 3.17:

* Infinite case.

$$ 
\begin{align*} 
K(\mathbf{u}^f)^{-1}_{v,w} &= e^{-i \frac{\bar{\alpha}_f + \bar{\beta}_f}{2}} \text{sc}(\theta_f)^{\frac{1}{2}} (k')^{-\frac{1}{2}} (G^{m,\partial}_{v,v_2} - G^{m,\partial}_{v,v_1}) \\
K(\mathbf{u}^f)^{-1}_{f,w} &= -ie^{-i \frac{\bar{\alpha}_f + \bar{\beta}_f}{2}} \text{sc}(\theta_f^*)^{\frac{1}{2}} \left( \text{nd}(\frac{K-\theta_f}{2}) G^{m,*}_{f,f_2} - \text{nd}(\frac{K+\theta_f}{2}) G^{m,*}_{f,f_1} \right). 
\end{align*} 
$$

* Finite case.

$$ 
\begin{align*} 
K^\partial(\mathbf{u}^f)^{-1}_{v,w} &= e^{-i \frac{\bar{\alpha}_f + \bar{\beta}_f}{2}} (k')^{-\frac{1}{2}} \text{sc}(\theta_f)^{\frac{1}{2}} (G^{m,\partial}(\mathbf{u}^f)_{v_2,v} - G^{m,\partial}(\mathbf{u}^f)_{v_1,v}) \\
K^\partial(\mathbf{u}^f)^{-1}_{f,w} &= -ie^{-i \frac{\bar{\alpha}_f + \bar{\beta}_f}{2}} \text{sc}(\theta_f^*)^{\frac{1}{2}} \left( \mathbb{I}_{\{w \notin \{w^l, w^r \in \mathbb{R}^\partial\}\}} \text{nd}(\frac{K-\theta_f}{2}) G^{m,*}_{f_2,f} - \text{nd}(\frac{K+\theta_f}{2}) G^{m,*}_{f_1,f} \right) + \\
&\quad + i \sum_{(v^c, f^c) \in \mathbb{R}^{\partial,r}} \frac{\text{nd}(\mathbf{u}^f_{\beta^c})}{\text{cd}(\mathbf{u}^f_{\alpha^c})} (\text{cd}(\mathbf{u}^f_{\beta^r}) - \text{cd}(\mathbf{u}^f_{\alpha^c})) K^\partial(\mathbf{u}^f)^{-1}_{v^c,w} \cdot G^{m,*}_{f^c,f}, \\
K^\partial(\mathbf{v}^f)^{-1}_{v,w} &= e^{-i \frac{\bar{\alpha}_f + \bar{\beta}_f}{2}} (k')^{-\frac{1}{2}} \text{sc}(\theta_f)^{\frac{1}{2}} (G^{m,\partial}(\mathbf{v}^f)_{v_2,v} - G^{m,\partial}(\mathbf{v}^f)_{v_1,v}) \\
K^\partial(\mathbf{v}^f)^{-1}_{f,w} &= -ie^{-i \frac{\bar{\alpha}_f + \bar{\beta}_f}{2}} \text{sc}(\theta_f^*)^{\frac{1}{2}} \left( \mathbb{I}_{\{w \notin \{w^l, w^r \in \mathbb{R}^\partial\}\}} \text{nd}(\frac{K+\theta_f}{2}) G^{m,*}_{f_2,f} - \text{nd}(\frac{K-\theta_f}{2}) G^{m,*}_{f_1,f} \right) + \\
&\quad + i \sum_{(v^c, f^c) \in \mathbb{R}^{\partial,r}} \frac{\text{nd}(\mathbf{v}^f_{\beta^c})}{\text{cd}(\mathbf{v}^f_{\alpha^c})} (\text{cd}(\mathbf{v}^f_{\beta^r}) - \text{cd}(\mathbf{v}^f_{\alpha^c})) K^\partial(\mathbf{v}^f)^{-1}_{v^c,w} \cdot G^{m,*}_{f^c,f}. 
\end{align*} 
$$

### 3.6.2 Dimer model on an infinite isoradial double graph $G^\mathbb{D}$

Suppose that the isoradial graph $G$ is infinite; when it is moreover $\mathbb{Z}^2$-periodic, we consider the natural exhaustion $(G_n^\mathbb{D})_{n \ge 1}$ of $G^\mathbb{D}$ by toroidal graphs, where $G_n^\mathbb{D} = G^\mathbb{D}/n\mathbb{Z}^2$. Let $\mathcal{F}$ denote the $\sigma$-field generated by cylinder sets of $G^\mathbb{D}$. Using arguments of [23, 47, 28], we obtain an explicit, *local* expression for a Gibbs measure of the dimer model on $G^\mathbb{D}$ with weight function $c(u)$. Since the proof closely follows that done in the papers [28, 10, 12], we do not repeat it here. The key requirements are that the operator $K(u)^{-1}$ is local and unique in the $\mathbb{Z}^2$-periodic case.
**Theorem 3.19.** For every $u \in C$, there exists a unique probability measure on $(M(G^D), \mathcal{F})$, denoted $P_{\text{dimer}}^{D,u}$, such that the probability of occurrence of a subset of edges $\mathcal{E} = \{w_1x_1, \dots, w_lx_l\}$ in a dimer configuration of $G^D$ is given by:

$$P_{\text{dimer}}^{D,u}(w_1x_1, \dots, w_lx_l) = \left( \prod_{j=1}^{l} K(u)_{w_j, x_j} \right) \det(K(u)^{-1})_{\mathcal{E}},$$

where $(K(u)^{-1})_{\mathcal{E}}$ is the sub-matrix of $K(u)^{-1}$ given by Corollary 3.15, whose rows are indexed by $x_1, \dots, x_l$ and columns by $w_1, \dots, w_l$. The measure $P_{\text{dimer}}^{D,u}$ is a Gibbs measure. Moreover, when the graph $G^D$ is $\mathbb{Z}^2$-periodic, the probability measure $P_{\text{dimer}}^{D,u}$ is obtained as weak limit of the Boltzmann measures on the toroidal exhaustion $(G_n^D)_{n \ge 1}$.

**Remark 3.20.**

* As mentioned in the introduction to Section 3, this theorem is a directed version of the transfer impedance theorem of [14]. A result in the same flavor, i.e., computing probabilities of pairs of directed spanning trees using the massive Green functions of massive non-directed random walks, is obtained by [20] in the case where $G = \mathbb{Z}^2$ with specific weights.

* A version of this theorem in the finite case can be obtained using Remark 3.7.

**Example 3.21.** As an example of application we express the probability of single edges occurring in dimer configurations of $G^D$ chosen with respect to the measure $P_{\text{dimer}}^{D,u}$, using the Z-massive and dual Z-massive Green functions of [11]. We use the notation of Figure 14 and omit the subscript “f” since there is no confusion possible between the initial and final vertices. Details of computations are given in Appendix B.1.

$$P_{\text{dimer}}^{D,u}(wv_2) = \frac{\text{sc}(\theta)}{k'} \left[ \text{dn}(u_\alpha) \text{dn}(u_\beta) G_{v_2,v_2}^m - k' G_{v_2,v_1}^m \right] \\ = H(2u_\alpha) - H(2u_\beta),$$

$$P_{\text{dimer}}^{D,u}(wf_2) = \frac{\text{sc}(\theta^*)}{k'} \left[ \text{dn}((u_\beta)^*) \text{dn}((u_{\alpha+2K})^*) G_{f_2,f_2}^{m,*} - k' G_{f_2,f_1}^{m,*} \right] \\ = H(2(u_{\alpha+2K})^*) - H(2(u_\beta)^*),$$

where $H(u|k) = -\frac{ik'}{\pi} A(\frac{iu}{2}|k')$, and $A(u|k) = \frac{1}{k'} \left( \int_0^u \text{dc}^2(v|k) \text{dv} + \frac{E-K}{K} u \right)$, see [11, (9)].

# 4 Kasteleyn operator of the graph $G^Q$ and $Z^u$-Dirac operator

Let $G$ be an isoradial graph, infinite or finite. We consider the isoradial embedding of the bipartite graph $G^Q = (V^Q, E^Q)$ given in Section 2.5.2, and the weight function $\nu^j$ of Equation (2.8) arising from the Z-invariant Ising model. Let $K^Q$ be the associated complex, bipartite Kasteleyn matrix defined in Section 2.5.7, with rows indexed by black vertices. In the finite case we moreover consider the diagonal matrix $D^{Q,B}$, resp. $D^{Q,W}$, whose rows/columns are indexed by black/white vertices of $G^Q$, and whose diagonal coefficients are:

$$\forall b \in B, \quad D_{b,b}^{Q,B} = \begin{cases} \text{sn}(\theta^b) & \text{if } b \in \{b^l, b^r \in \mathbb{R}^b\} \\ 1 & \text{otherwise,} \end{cases}$$

$$\forall w \in W, \quad D_{w,w}^{Q,W} = \begin{cases} \text{sn}(\theta^w)^{-1} & \text{if } w \in \{w^c \in \mathbb{R}^c\} \\ 1 & \text{otherwise.} \end{cases}$$

Let $\underline{K}^Q$ be the modified, complex, bipartite Kasteleyn matrix defined by

$$\underline{K}^Q = D^{Q,B} \underline{K}^Q D^{Q,W}, \qquad (4.1)$$
**Theorem 3.19.** For every $u \in C$, there exists a unique probability measure on $(M(G^D), \mathcal{F})$, denoted $P_{\text{dimer}}^{\text{D,u}}$, such that the probability of occurrence of a subset of edges $\mathcal{E} = \{w_1x_1, \dots, w_lx_l\}$ in a dimer configuration of $G^D$ is given by:

$$P_{\text{dimer}}^{\text{D,u}}(w_1x_1, \dots, w_lx_l) = \left( \prod_{j=1}^{l} K(u)_{w_j, x_j} \right) \det(K(u)^{-1})_{\mathcal{E}},$$

where $(K(u)^{-1})_{\mathcal{E}}$ is the sub-matrix of $K(u)^{-1}$ given by Corollary 3.15, whose rows are indexed by $x_1, \dots, x_l$ and columns by $w_1, \dots, w_l$. The measure $P_{\text{dimer}}^{\text{D,u}}$ is a Gibbs measure. Moreover, when the graph $G^D$ is $\mathbb{Z}^2$-periodic, the probability measure $P_{\text{dimer}}^{\text{D,u}}$ is obtained as weak limit of the Boltzmann measures on the toroidal exhaustion $(G_n^D)_{n \ge 1}$.

**Remark 3.20.**

* As mentioned in the introduction to Section 3, this theorem is a directed version of the transfer impedance theorem of [14]. A result in the same flavor, i.e., computing probabilities of pairs of directed spanning trees using the massive Green functions of massive non-directed random walks, is obtained by [20] in the case where $G = \mathbb{Z}^2$ with specific weights.

* A version of this theorem in the finite case can be obtained using Remark 3.7.

**Example 3.21.** As an example of application we express the probability of single edges occurring in dimer configurations of $G^D$ chosen with respect to the measure $P_{\text{dimer}}^{\text{D,u}}$, using the Z-massive and dual Z-massive Green functions of [11]. We use the notation of Figure 14 and omit the subscript “f” since there is no confusion possible between the initial and final vertices. Details of computations are given in Appendix B.1.

$$P_{\text{dimer}}^{\text{D,u}}(wv_2) = \frac{\text{sc}(\theta)}{k'} \left[ \text{dn}(u_\alpha) \text{dn}(u_\beta) G_{v_2,v_2}^m - k' G_{v_2,v_1}^m \right] \\ = H(2u_\alpha) - H(2u_\beta),$$

$$P_{\text{dimer}}^{\text{D,u}}(wf_2) = \frac{\text{sc}(\theta^*)}{k'} \left[ \text{dn}((u_\beta)^*) \text{dn}((u_{\alpha+2K})^*) G_{f_2,f_2}^{m,*} - k' G_{f_2,f_1}^{m,*} \right] \\ = H(2(u_{\alpha+2K})^*) - H(2(u_\beta)^*),$$

where $H(u|k) = -\frac{ik'}{\pi} A(\frac{iu}{2}|k')$, and $A(u|k) = \frac{1}{k'} \left( \int_0^u \text{dc}^2(v|k) \text{dv} + \frac{E-K}{K} u \right)$, see [11, (9)].

# 4 Kasteleyn operator of the graph $G^Q$ and $Z^u$-Dirac operator

Let $G$ be an isoradial graph, infinite or finite. We consider the isoradial embedding of the bipartite graph $G^Q = (V^Q, E^Q)$ given in Section 2.5.2, and the weight function $\nu^j$ of Equation (2.8) arising from the Z-invariant Ising model. Let $K^Q$ be the associated complex, bipartite Kasteleyn matrix defined in Section 2.5.7, with rows indexed by black vertices. In the finite case we moreover consider the diagonal matrix $D^{Q,B}$, resp. $D^{Q,W}$, whose rows/columns are indexed by black/white vertices of $G^Q$, and whose diagonal coefficients are:

$$\forall b \in B, \quad D_{b,b}^{Q,B} = \begin{cases} \text{sn}(\theta^b) & \text{if } b \in \{b^\ell, b^r \in \mathbb{R}^b\} \\ 1 & \text{otherwise,} \end{cases}$$

$$\forall w \in W, \quad D_{w,w}^{Q,W} = \begin{cases} \text{sn}(\theta^w)^{-1} & \text{if } w \in \{w^c \in \mathbb{R}^c\} \\ 1 & \text{otherwise.} \end{cases}$$

Let $\underline{K}^Q$ be the modified, complex, bipartite Kasteleyn matrix defined by

$$\underline{K}^Q = D^{Q,B} \underline{K}^Q D^{Q,W}, \qquad (4.1)$$
that is, $\underline{K}^Q$ is obtained from $K^Q$ by multiplying the weight of the edges $b^\ell w^\ell, b^r w^r$ of all boundary rhombus pairs of $\mathcal{R}^\theta$ by $\mathrm{sn}(\theta^\theta)$.

Consider also the isoradial embedding of the double graph $G^D$ of Section 2.5.2. For every $u \in C$, let $K(u)$ be the $Z^u$-Dirac operator and, in the finite case, for every $u \in C'$, let $K^\theta(u)$ be the $Z^u$-Dirac operator with specific boundary conditions, as defined in Equation (3.1).

The main result of this section, and actually the key result of this paper, is Theorem 4.2 of Section 4.1 proving, for every $u \in C$, explicit linear relations between the matrices $K^Q$ and $K(u)$ in the infinite case, and between $\underline{K}^Q$ and $K^\theta(u)$ in the finite case.

In Section 4.2 we restrict to the finite case; using Theorem 4.2 and a combinatorial argument, we prove in Theorem 4.3 that the determinant of the Kasteleyn matrix $K^Q$ is equal, up to an explicit multiplicative constant depending on $u$, to the determinant of the $Z^u$-Dirac operator $K^\theta(u)$. Combining this with Theorem 3.8 proves that the determinant of $K^Q$ is equal, up to an explicit constant, to the determinant of the Z-massive Laplacian $\Delta^{m,\theta}(u)$. Interpreting these determinants as partition functions, Theorem 4.3 proves that the squared partition function of the Z-invariant Ising model with + boundary conditions is equal, up to an explicit constant, to the partition function of weighted rooted directed spanning forests counted by $\Delta^{m,\theta}(u)$, where the dependence in $u$ is along the boundary only. This generalizes to the full Z-invariant case the results of [29, 30] proved in the Z-invariant critical case, and to the case of simply connected domains the result of [12] proved in the toroidal case. The proof we provide here has a slight combinatorial flavor but is mainly based on matrix relations, so quite different from [29, 30]. Note that the combinatorics argument of [30] can be generalized to the full Z-invariant case and would give an alternative proof. Note also that the boundary trick of Chelkak and Smirnov [18] allows us to remove dual trees along the boundary which we could not do in [30].

Using Theorem 4.2, Corollaries 4.12 and 4.14 of Section 4.3 prove linear relations between the inverse operator $(\underline{K}^Q)^{-1}$ and the inverse $Z^u$-Dirac operator. Choosing specific values of $u$ allows us to express the dimer measure of the graph $G^Q$ using the inverse $Z^u$-Dirac operator and the Z-massive Green functions, see Corollaries 4.16 and 4.18. This also provides an alternative direct way of finding a local formula for $(\underline{K}^Q)^{-1}$ [12], explicitly relating it to the Z-massive Green functions.

## 4.1 Relating the Kasteleyn operator $\underline{K}^Q$ and the $Z^u$-Dirac operator

The main result of this section is Theorem 4.2 proving an explicit relation between the matrices $K^Q$ and $K(u)$ in the infinite case, and between $\underline{K}^Q$ and $K^\theta(u)$ in the finite case. In order to state this theorem we need to introduce two additional matrices $S(u)$ and $T(u)$. Both of them are “rectangular” with “twice” more rows than columns.

The matrix $S(u)$ has rows indexed by black vertices of $G^Q$ and columns by white vertices of $G^D$, resp. of $G^{D,r}$, in the infinite case, resp. finite case. If $G^Q$ is infinite, let $b$ be a black vertex; if $G^Q$ is finite, let $b$ be a black vertex of an inner quadrangle. Let $w$ be the white vertex of $G^D$ corresponding to the quadrangle to which $b$ belongs. Then, the only non-zero coefficient of the row corresponding to $b$ is:

$$s(u)_{b,w} = e^{-i \frac{\beta}{2}} \mathrm{cn}(u_{\beta}) [\mathrm{sn}(\theta) \mathrm{cn}(\theta) \mathrm{nd}(u_{\alpha}) \mathrm{nd}(u_{\beta})]^{\frac{1}{2}}, \quad (4.2)$$

with the following notation, see Figure 15 (left): $w$ is the white vertex of $G^Q$ such that $bw$ is parallel to an edge of $G$; $e^{i\bar{\alpha}}$, $e^{i\bar{\beta}}$, $\bar{\theta}$ are the rhombus vectors and half-angle of $G^{\diamond/2}$ assigned to the edge (b, w).

Suppose that $G^Q$ is finite, and let $b$ be a black vertex of a boundary quadrangle of $G^Q$. Then $b \in \{b^r, b^\ell\}$ for some boundary rhombus pair of $\mathcal{R}^\theta$, see Figure 15 (right), and the
that is, $\underline{K}^Q$ is obtained from $K^Q$ by multiplying the weight of the edges $b^\ell w^\ell, b^r w^r$ of all boundary rhombus pairs of $\mathcal{R}^\theta$ by $\mathrm{sn}(\theta^\theta)$.

Consider also the isoradial embedding of the double graph $G^D$ of Section 2.5.2. For every $u \in C$, let $K(u)$ be the $Z^u$-Dirac operator and, in the finite case, for every $u \in C'$, let $K^\theta(u)$ be the $Z^u$-Dirac operator with specific boundary conditions, as defined in Equation (3.1).

The main result of this section, and actually the key result of this paper, is Theorem 4.2 of Section 4.1 proving, for every $u \in C$, explicit linear relations between the matrices $K^Q$ and $K(u)$ in the infinite case, and between $\underline{K}^Q$ and $K^\theta(u)$ in the finite case.

In Section 4.2 we restrict to the finite case; using Theorem 4.2 and a combinatorial argument, we prove in Theorem 4.3 that the determinant of the Kasteleyn matrix $K^Q$ is equal, up to an explicit multiplicative constant depending on $u$, to the determinant of the $Z^u$-Dirac operator $K^\theta(u)$. Combining this with Theorem 3.8 proves that the determinant of $K^Q$ is equal, up to an explicit constant, to the determinant of the Z-massive Laplacian $\Delta^{m,\theta}(u)$. Interpreting these determinants as partition functions, Theorem 4.3 proves that the squared partition function of the Z-invariant Ising model with + boundary conditions is equal, up to an explicit constant, to the partition function of weighted rooted directed spanning forests counted by $\Delta^{m,\theta}(u)$, where the dependence in $u$ is along the boundary only. This generalizes to the full Z-invariant case the results of [29, 30] proved in the Z-invariant critical case, and to the case of simply connected domains the result of [12] proved in the toroidal case. The proof we provide here has a slight combinatorial flavor but is mainly based on matrix relations, so quite different from [29, 30]. Note that the combinatorics argument of [30] can be generalized to the full Z-invariant case and would give an alternative proof. Note also that the boundary trick of Chelkak and Smirnov [18] allows us to remove dual trees along the boundary which we could not do in [30].

Using Theorem 4.2, Corollaries 4.12 and 4.14 of Section 4.3 prove linear relations between the inverse operator $(\underline{K}^Q)^{-1}$ and the inverse $Z^u$-Dirac operator. Choosing specific values of $u$ allows us to express the dimer measure of the graph $G^Q$ using the inverse $Z^u$-Dirac operator and the Z-massive Green functions, see Corollaries 4.16 and 4.18. This also provides an alternative direct way of finding a local formula for $(\underline{K}^Q)^{-1}$ [12], explicitly relating it to the Z-massive Green functions.

## 4.1 Relating the Kasteleyn operator $\underline{K}^Q$ and the $Z^u$-Dirac operator

The main result of this section is Theorem 4.2 proving an explicit relation between the matrices $K^Q$ and $K(u)$ in the infinite case, and between $\underline{K}^Q$ and $K^\theta(u)$ in the finite case. In order to state this theorem we need to introduce two additional matrices $S(u)$ and $T(u)$. Both of them are “rectangular” with “twice” more rows than columns.

The matrix $S(u)$ has rows indexed by black vertices of $G^Q$ and columns by white vertices of $G^D$, resp. of $G^{D,r}$, in the infinite case, resp. finite case. If $G^Q$ is infinite, let $b$ be a black vertex; if $G^Q$ is finite, let $b$ be a black vertex of an inner quadrangle. Let $w$ be the white vertex of $G^D$ corresponding to the quadrangle to which $b$ belongs. Then, the only non-zero coefficient of the row corresponding to $b$ is:

$$s(u)_{b,w} = e^{-i \frac{\beta}{2}} \mathrm{cn}(u_{\beta}) [\mathrm{sn}(\theta) \mathrm{cn}(\theta) \mathrm{nd}(u_{\alpha}) \mathrm{nd}(u_{\beta})]^{\frac{1}{2}}, \quad (4.2)$$

with the following notation, see Figure 15 (left): $w$ is the white vertex of $G^Q$ such that $bw$ is parallel to an edge of $G$; $e^{i\bar{\alpha}}$, $e^{i\bar{\beta}}$, $\bar{\theta}$ are the rhombus vectors and half-angle of $G^{\diamond/2}$ assigned to the edge (b, w).

Suppose that $G^Q$ is finite, and let $b$ be a black vertex of a boundary quadrangle of $G^Q$. Then $b \in \{b^r, b^\ell\}$ for some boundary rhombus pair of $\mathcal{R}^\theta$, see Figure 15 (right), and the
non-zero coefficient of the row corresponding to b is:

$$ 
\begin{align} 
s(u)_{b^r, w^r} &= e^{-i \frac{\bar{\beta}^r}{2}} \operatorname{cn}(u_{\beta^r}) [\operatorname{sn}(\theta^\theta) \operatorname{cn}(\theta^\theta) \operatorname{nd}(u_{\alpha^r}) \operatorname{nd}(u_{\beta^r})]^{\frac{1}{2}} \tag{4.3} \\ 
s(u)_{b^\ell, w^\ell} &= e^{-i \frac{\bar{\alpha}^\ell}{2}} \operatorname{cn}(u_{\alpha^\ell}) [\operatorname{sn}(\theta^\theta) \operatorname{cn}(\theta^\theta) \operatorname{nd}(u_{\alpha^\ell}) \operatorname{nd}(u_{\beta^\ell})]^{\frac{1}{2}}, \nonumber 
\end{align} 
$$

where for $b^r$ the definition is coherent with that of the non-boundary case, since the rhombus vectors of $G^{\diamond/2}$ assigned to the edge ($b^r, w^r$) are $e^{i\bar{\alpha}^r}$, $e^{i\bar{\beta}^r}$. For $b^\ell$, the rhombus vectors assigned to the edge ($b^\ell, w^\ell$) are $e^{i\bar{\alpha}^\ell}$, $e^{i\bar{\beta}^\ell}$ so that there is a change of definition; this comes from our choice of embedding of $G^Q$ which exchanges the bipartite coloring of vertices in the left rhombus of the pair.

<figure><img src="image_17.png" /><figcaption>Figure 15: Notation around a black vertex b and a white vertex w of $G^Q$: non-boundary case (left), boundary case (right).</figcaption></figure>

The matrix $T(u)$ has rows indexed by white vertices of $G^Q$ and columns by black vertices of $G^D$, resp. of $G^{D,r}$, in the infinite case, resp. finite case. If $G^Q$ is infinite, let w be a white vertex; if $G^Q$ is finite, let w be a white vertex such that $w \neq w^c$ for all boundary rhombus pairs of $\mathcal{R}^\theta$. The vertex w is on a rhombus edge $vf$ of the diamond graph $G^\theta$, where v is a vertex of G and f a vertex of $G^*$, see Figure 15 (left). Then the row of $T(u)$ corresponding to w has two non-zero coefficients defined by,

$$ 
\begin{align} 
t(u)_{w,v} &= e^{-i \frac{\bar{\beta}}{2}} \operatorname{cn}(u_{\beta}) \tag{4.4} \\ 
t(u)_{w,f} &= e^{-i \frac{\bar{\beta}+\pi}{2}} \operatorname{sn}((u_{\beta+2K})^*) = e^{-i \frac{\bar{\beta}+\pi}{2}} \operatorname{cd}(u_{\beta+2K}), \nonumber 
\end{align} 
$$

where $e^{i\bar{\beta}}$ is the rhombus vector $(w, v)$ and $e^{i(\bar{\beta}+\pi)}$ is the rhombus vector $(w, f)$.

Suppose that $G^Q$ is finite, and let $w = w^c$ for some boundary rhombus pair of $\mathcal{R}^\theta$, then $w^c$ is on a rhombus edge $vf^c$ of $G^\theta$, see Figure 15 (right). As long as the rhombus pair is not the root one, i.e., the one where $v^c = r$, the row of $T(u)$ corresponding to $w^c$ has non-zero coefficients given by,

$$ 
\begin{align} 
t(u)_{w^c, v^c} &= -ik' e^{-i \frac{\bar{\beta}^r}{2}} \operatorname{sn}(\theta^\theta) \operatorname{nd}(u_{\alpha^r}) \operatorname{cd}(u_{\beta^r}), \tag{4.5} \\ 
t(u)_{w^c, f^c} &= e^{-i \frac{\bar{\beta}^\ell}{2}} \operatorname{sn}((u_{\beta^\ell})^*) = e^{-i \frac{\bar{\beta}^\ell}{2}} \operatorname{cd}(u_{\beta^\ell}). \nonumber 
\end{align} 
$$

When the boundary rhombus pair is the root pair, then only the term $t(u)_{w^c, f^c}$ is defined. Note that the definition is specific for $v^c$, and coherent with the non-boundary case for $f^c$ since the rhombus vector $(w^c, f^c)$ is $e^{i\bar{\beta}^\ell}$.

**Remark 4.1.** Rhombus angles are well defined mod $2\pi$ implying that half-angles are well defined mod $\pi$, but the coefficients of $S(u)$ and $T(u)$ are nevertheless well defined. Indeed, keeping in mind that by definition $\frac{\bar{\beta}-\bar{\alpha}}{2} \in (\varepsilon, \frac{\pi}{2}-\varepsilon)$, one has for example, see
non-zero coefficient of the row corresponding to b is:

$$ 
\begin{align} 
s(u)_{b^r, w^r} &= e^{-i \frac{\bar{\beta}^r}{2}} \operatorname{cn}(u_{\beta^r}) [\operatorname{sn}(\theta^\theta) \operatorname{cn}(\theta^\theta) \operatorname{nd}(u_{\alpha^r}) \operatorname{nd}(u_{\beta^r})]^{\frac{1}{2}} \tag{4.3} \\ 
s(u)_{b^\ell, w^\ell} &= e^{-i \frac{\bar{\alpha}^\ell}{2}} \operatorname{cn}(u_{\alpha^\ell}) [\operatorname{sn}(\theta^\theta) \operatorname{cn}(\theta^\theta) \operatorname{nd}(u_{\alpha^\ell}) \operatorname{nd}(u_{\beta^\ell})]^{\frac{1}{2}}, \nonumber 
\end{align} 
$$

where for $b^r$ the definition is coherent with that of the non-boundary case, since the rhombus vectors of $G^{\diamond/2}$ assigned to the edge ($b^r, w^r$) are $e^{i\bar{\alpha}^r}$, $e^{i\bar{\beta}^r}$. For $b^\ell$, the rhombus vectors assigned to the edge ($b^\ell, w^\ell$) are $e^{i\bar{\alpha}^\ell}$, $e^{i\bar{\beta}^\ell}$ so that there is a change of definition; this comes from our choice of embedding of $G^Q$ which exchanges the bipartite coloring of vertices in the left rhombus of the pair.

<figure><img src="image_17.png" /><figcaption>Figure 15: Notation around a black vertex b and a white vertex w of $G^Q$: non-boundary case (left), boundary case (right).</figcaption></figure>

The matrix $T(u)$ has rows indexed by white vertices of $G^Q$ and columns by black vertices of $G^D$, resp. of $G^{D,r}$, in the infinite case, resp. finite case. If $G^Q$ is infinite, let w be a white vertex; if $G^Q$ is finite, let w be a white vertex such that $w \neq w^c$ for all boundary rhombus pairs of $\mathcal{R}^\theta$. The vertex w is on a rhombus edge $vf$ of the diamond graph $G^\theta$, where v is a vertex of G and f a vertex of $G^*$, see Figure 15 (left). Then the row of $T(u)$ corresponding to w has two non-zero coefficients defined by,

$$ 
\begin{align} 
t(u)_{w,v} &= e^{-i \frac{\bar{\beta}}{2}} \operatorname{cn}(u_{\beta}) \tag{4.4} \\ 
t(u)_{w,f} &= e^{-i \frac{\bar{\beta}+\pi}{2}} \operatorname{sn}((u_{\beta+2K})^*) = e^{-i \frac{\bar{\beta}+\pi}{2}} \operatorname{cd}(u_{\beta+2K}), \nonumber 
\end{align} 
$$

where $e^{i\bar{\beta}}$ is the rhombus vector $(w, v)$ and $e^{i(\bar{\beta}+\pi)}$ is the rhombus vector $(w, f)$.

Suppose that $G^Q$ is finite, and let $w = w^c$ for some boundary rhombus pair of $\mathcal{R}^\theta$, then $w^c$ is on a rhombus edge $vf^c$ of $G^\theta$, see Figure 15 (right). As long as the rhombus pair is not the root one, i.e., the one where $v^c = r$, the row of $T(u)$ corresponding to $w^c$ has non-zero coefficients given by,

$$ 
\begin{align} 
t(u)_{w^c, v^c} &= -ik' e^{-i \frac{\bar{\beta}^r}{2}} \operatorname{sn}(\theta^\theta) \operatorname{nd}(u_{\alpha^r}) \operatorname{cd}(u_{\beta^r}), \tag{4.5} \\ 
t(u)_{w^c, f^c} &= e^{-i \frac{\bar{\beta}^\ell}{2}} \operatorname{sn}((u_{\beta^\ell})^*) = e^{-i \frac{\bar{\beta}^\ell}{2}} \operatorname{cd}(u_{\beta^\ell}). \nonumber 
\end{align} 
$$

When the boundary rhombus pair is the root pair, then only the term $t(u)_{w^c, f^c}$ is defined. Note that the definition is specific for $v^c$, and coherent with the non-boundary case for $f^c$ since the rhombus vector $(w^c, f^c)$ is $e^{i\bar{\beta}^\ell}$.

**Remark 4.1.** Rhombus angles are well defined mod $2\pi$ implying that half-angles are well defined mod $\pi$, but the coefficients of $S(u)$ and $T(u)$ are nevertheless well defined. Indeed, keeping in mind that by definition $\frac{\bar{\beta}-\bar{\alpha}}{2} \in (\varepsilon, \frac{\pi}{2}-\varepsilon)$, one has for example, see
Equation (4.2):

$$e^{-i \frac{\beta+2\pi}{2}} \operatorname{cn}(u_{\beta+4K})[\operatorname{sn}(\theta) \operatorname{cn}(\theta) \operatorname{nd}(u_{\alpha+4K}) \operatorname{nd}(u_{\beta+4K})]^{\frac{1}{2}} = \\ = (-e^{-i \frac{\beta}{2}})(-\operatorname{cn}(u_{\beta})) [\operatorname{sn}(\theta) \operatorname{cn}(\theta) \operatorname{nd}(u_{\alpha}) \operatorname{nd}(u_{\beta})]^{\frac{1}{2}} = s_{b,w},$$

using that $u_{\beta+4K} = u_{\beta} - 2K$ and $\operatorname{cn}(u - 2K) = -\operatorname{cn}(u)$, $\operatorname{nd}(u - 2K) = \operatorname{nd}(u)$. Similar arguments hold for other coefficients.

We are now ready to state our main result.

**Theorem 4.2.**

* Infinite case. Let $u \in C$, then the Kasteleyn matrix $K^Q$, the $Z^u$-Dirac operator $K(u)$ and the matrices $S(u)$, $T(u)$ are related by the following identity:

$$K^Q T(u) = S(u) K(u). \qquad (4.6)$$

* Finite case. Let $u \in C'$, then the Kasteleyn matrix $\underline{K}^Q$, the $Z^u$-Dirac operator $K^\theta(u)$ and the matrices $S(u)$, $T(u)$ are related by the following identity:

$$\underline{K}^Q T(u) = S(u) \underline{K}^\theta(u). \qquad (4.7)$$

*Proof.* In the whole of the proof, we omit the argument $u$ from the matrices.

Infinite case and finite non-boundary case. Figure 16 below sets the notation. Let $b$ be a black vertex of $G^Q$, then $b$ belongs to a quadrangle corresponding to a vertex $w$ of $G^D$. If $G^Q$ is finite, suppose further that the quadrangle is not a boundary one, or equivalently that $w$ is not a boundary vertex of $G^D$. Let $v_1, f_1, v_2, f_2$ be the four black vertices of $G^D$ incident to $w$. Denote by $w_1, w_2, w_3$ the three white vertices of $G^Q$ incident to $b$, and let $e^{i\hat{\alpha}}, e^{i\hat{\beta}}$ be the rhombus vectors of the edge $(b, w_1)$.

<figure><img src="image_18.png" /><figcaption>Figure 16: Notation around a black vertex b of $G^Q$.</figcaption></figure>

The coefficient $[\underline{K}^Q T]_{b,x}$ of the LHS of (4.6) and (4.7) is non-zero only when $x \in \{v_1, v_2, f_1, f_2\}$, and

$$[\underline{K}^Q T]_{b,v_1} = K^Q_{b,w_2} t_{w_2,v_1} + K^Q_{b,w_3} t_{w_3,v_1} \\ [\underline{K}^Q T]_{b,v_2} = K^Q_{b,w_1} t_{w_1,v_2} \\ [\underline{K}^Q T]_{b,f_1} = K^Q_{b,w_1} t_{w_1,f_1} + K^Q_{b,w_3} t_{w_3,f_1} \\ [\underline{K}^Q T]_{b,f_2} = K^Q_{b,w_2} t_{w_2,f_2}.$$

The coefficients of $T$ involved are, see definition (4.4),

$$t_{w_2,v_1} = -ie^{-i \frac{\beta}{2}} \operatorname{cn}(u_{\beta+2K}), \quad t_{w_3,v_1} = -ie^{-i \frac{\alpha}{2}} \operatorname{cn}(u_{\alpha+2K}), \quad t_{w_1,v_2} = e^{-i \frac{\beta}{2}} \operatorname{cn}(u_{\beta})$$

$$t_{w_1,f_1} = ie^{-i \frac{\beta}{2}} \operatorname{sn}((u_{\beta-2K})^*), \quad t_{w_3,f_1} = e^{-i \frac{\alpha}{2}} \operatorname{sn}((u_{\alpha})^*), \quad t_{w_2,f_2} = e^{-i \frac{\beta}{2}} \operatorname{sn}((u_{\beta})^*).$$
Equation (4.2):

$$e^{-i\frac{\beta+2\pi}{2}} \mathrm{cn}(u_{\beta+4K})[\mathrm{sn}(\theta) \mathrm{cn}(\theta) \mathrm{nd}(u_{\alpha+4K}) \mathrm{nd}(u_{\beta+4K})]^{\frac{1}{2}} = \\ = (-e^{-i\frac{\beta}{2}})(-\mathrm{cn}(u_{\beta})) [\mathrm{sn}(\theta) \mathrm{cn}(\theta) \mathrm{nd}(u_{\alpha}) \mathrm{nd}(u_{\beta})]^{\frac{1}{2}} = s_{b,w},$$

using that $u_{\beta+4K} = u_{\beta} - 2K$ and $\mathrm{cn}(u - 2K) = -\mathrm{cn}(u)$, $\mathrm{nd}(u - 2K) = \mathrm{nd}(u)$. Similar arguments hold for other coefficients.

We are now ready to state our main result.

**Theorem 4.2.**

* Infinite case. Let $u \in C$, then the Kasteleyn matrix $K^Q$, the $Z^u$-Dirac operator $K(u)$ and the matrices $S(u)$, $T(u)$ are related by the following identity:

$$K^Q T(u) = S(u) K(u). \qquad (4.6)$$

* Finite case. Let $u \in C'$, then the Kasteleyn matrix $\underline{K}^Q$, the $Z^u$-Dirac operator $K^\theta(u)$ and the matrices $S(u)$, $T(u)$ are related by the following identity:

$$\underline{K}^Q T(u) = S(u) K^\theta(u). \qquad (4.7)$$

*Proof.* In the whole of the proof, we omit the argument $u$ from the matrices.

Infinite case and finite non-boundary case. Figure 16 below sets the notation. Let $b$ be a black vertex of $G^Q$, then $b$ belongs to a quadrangle corresponding to a vertex $w$ of $G^D$. If $G^Q$ is finite, suppose further that the quadrangle is not a boundary one, or equivalently that $w$ is not a boundary vertex of $G^D$. Let $v_1, f_1, v_2, f_2$ be the four black vertices of $G^D$ incident to $w$. Denote by $w_1, w_2, w_3$ the three white vertices of $G^Q$ incident to $b$, and let $e^{i\tilde{\alpha}}, e^{i\tilde{\beta}}$ be the rhombus vectors of the edge $(b, w_1)$.

<figure><img src="image_18.png" /><figcaption>Figure 16: Notation around a black vertex b of $G^Q$.</figcaption></figure>

The coefficient $[\underline{K}^Q T]_{b,x}$ of the LHS of (4.6) and (4.7) is non-zero only when $x \in \{v_1, v_2, f_1, f_2\}$, and

$$[\underline{K}^Q T]_{b,v_1} = K^Q_{b,w_2} t_{w_2,v_1} + K^Q_{b,w_3} t_{w_3,v_1} \\ [\underline{K}^Q T]_{b,v_2} = K^Q_{b,w_1} t_{w_1,v_2} \\ [\underline{K}^Q T]_{b,f_1} = K^Q_{b,w_1} t_{w_1,f_1} + K^Q_{b,w_3} t_{w_3,f_1} \\ [\underline{K}^Q T]_{b,f_2} = K^Q_{b,w_2} t_{w_2,f_2}.$$

The coefficients of $T$ involved are, see definition (4.4),

$$t_{w_2,v_1} = -ie^{-i\frac{\beta}{2}} \mathrm{cn}(u_{\beta+2K}), \quad t_{w_3,v_1} = -ie^{-i\frac{\alpha}{2}} \mathrm{cn}(u_{\alpha+2K}), \quad t_{w_1,v_2} = e^{-i\frac{\beta}{2}} \mathrm{cn}(u_{\beta})$$

$$t_{w_1,f_1} = ie^{-i\frac{\beta}{2}} \mathrm{sn}((u_{\beta-2K})^*), \quad t_{w_3,f_1} = e^{-i\frac{\alpha}{2}} \mathrm{sn}((u_{\alpha})^*), \quad t_{w_2,f_2} = e^{-i\frac{\beta}{2}} \mathrm{sn}((u_{\beta})^*).$$
Replacing coefficients of K$^Q$ yields for the LHS:

$$[\underline{K}^Q T]_{b,v_1} = e^{i \frac{\beta+\bar{\alpha}+\pi}{2}} \mathrm{cn}(\theta) t_{w_2,v_1} + e^{i \frac{\bar{\alpha}-\pi+\bar{\alpha}}{2}} t_{w_3,v_1} = e^{i \frac{\bar{\alpha}}{2}} [\mathrm{cn}(\theta) \mathrm{cn}(u_{\beta+2K}) - \mathrm{cn}(u_{\alpha+2K})]$$

$$[\underline{K}^Q T]_{b,v_2} = e^{i \frac{\beta+\bar{\alpha}}{2}} \mathrm{sn}(\theta) t_{w_1,v_2} = e^{i \frac{\bar{\alpha}}{2}} \mathrm{sn}(\theta) \mathrm{cn}(u_{\beta})$$

$$[\underline{K}^Q T]_{b,f_1} = e^{i \frac{\beta+\bar{\alpha}}{2}} \mathrm{sn}(\theta) t_{w_1,f_1} + e^{i \frac{\bar{\alpha}-\pi+\bar{\alpha}}{2}} t_{w_3,f_1} = i e^{i \frac{\bar{\alpha}}{2}} [\mathrm{sn}(\theta) \mathrm{sn}((u_{\beta-2K})^*) - \mathrm{sn}((u_{\alpha})^*)]$$

$$[\underline{K}^Q T]_{b,f_2} = e^{i \frac{\beta+\bar{\alpha}+\pi}{2}} \mathrm{cn}(\theta) t_{w_2,f_2} = i e^{i \frac{\bar{\alpha}}{2}} \mathrm{cn}(\theta) \mathrm{sn}((u_{\beta})^*)$$

Note that in the finite case, we have $K^\theta = K$ for coefficients involved. The coefficient $[SK]_{b,x}$ of the RHS of (4.6) and (4.7) is also non-zero when $x \in \{v_1, v_2, f_1, f_2\}$ and we have $[SK]_{b,x} = s_{b,w}K_{w,x}$. To compute these terms, we first express the part $\mathrm{cn}(u_\beta)[\mathrm{nd}(u_\alpha) \mathrm{nd}(u_\beta)]^{\frac{1}{2}} := (\diamond) \text{ in } s_{b,w}, \text{ see (4.2), using the angles and parameters involved in the four coefficients of K.}$

$$\begin{aligned} (\diamond) &= -\mathrm{sn}(u_{\beta+2K})[\mathrm{dn}(u_{\alpha+2K}) \mathrm{nd}(u_{\beta+2K})]^{\frac{1}{2}} \\ &= \mathrm{cn}(u_\beta)[\mathrm{nd}(u_\alpha) \mathrm{nd}(u_\beta)]^{\frac{1}{2}} \\ &= (k')^{-\frac{1}{2}} \mathrm{cn}((u_{\beta-2K})^*)[\mathrm{nd}((u_{\beta-2K})^*) \mathrm{dn}((u_\alpha)^*)]^{\frac{1}{2}} \\ &= (k')^{\frac{1}{2}} \mathrm{sn}((u_\beta)^*)[\mathrm{nd}((u_\beta)^*) \mathrm{nd}((u_{\alpha+2K})^*)]^{\frac{1}{2}}, \end{aligned}$$

where in the first line we used that $u_\alpha = K + u_{\alpha+2K}$, in the third that $u_\beta = -(u_{\beta-2K})^*$, $u_\alpha = K - (u_\alpha)^*$, and in the fourth that $u_\beta = K - (u_\beta)^*$, $u_\alpha = 2K - (u_{\alpha+2K})^*$. Replacing coefficients of K by their definition gives for the RHS of (4.6),

$$[\underline{S K}]_{b,v_1} = s_{b,w} K_{w,v_1} = e^{i \frac{\bar{\alpha}}{2}} \mathrm{sn}(\theta) \mathrm{sn}(u_{\beta+2K}) \mathrm{dn}(u_{\alpha+2K})$$

$$[\underline{S K}]_{b,v_2} = s_{b,w} K_{w,v_2} = e^{i \frac{\bar{\alpha}}{2}} \mathrm{sn}(\theta) \mathrm{cn}(u_\beta)$$

$$[\underline{S K}]_{b,f_1} = s_{b,w} K_{w,f_1} = -i (k')^{-1} e^{i \frac{\bar{\alpha}}{2}} \mathrm{cn}(\theta) \mathrm{cn}((u_{\beta-2K})^*) \mathrm{dn}((u_\alpha)^*)$$

$$[\underline{S K}]_{b,f_2} = s_{b,w} K_{w,f_2} = i e^{i \frac{\bar{\alpha}}{2}} \mathrm{cn}(\theta) \mathrm{sn}((u_\beta)^*)$$

The equality $[K^Q T]_{b,x} = [S K]_{b,x}$ is then straightforward when $x = v_2$ and $x = f_2$. When $x = v_1$, this is a consequence of the identity, see [54, chap.2, ex.32 (i)],

$$\mathrm{cn} \, u \, \mathrm{cn} \, v - \mathrm{cn}(u + v) = \mathrm{sn} \, u \, \mathrm{sn} \, v \, \mathrm{dn}(u + v), \quad (4.8)$$

evaluated at $u = \theta$, $v = u_{\beta+2K}$, $u + v = u_{\alpha+2K}$.

We are left with proving the case $x = f_1$. Multiplying the identity (4.8) by $\mathrm{nd}(u - v)$, using that $\mathrm{nd}(u - v) = (k')^{-1} \mathrm{dn}(K - (u - v))$, $\mathrm{cd}(u - v) = \mathrm{sn}(K - (u - v))$ we obtain,

$$(k')^{-1} \mathrm{cn} \, u \, \mathrm{cn} \, v \, \mathrm{dn}(K - (u - v)) = \mathrm{sn}(K - (u - v)) - \mathrm{sn} \, u \, \mathrm{sn} \, v.$$

The proof is concluded by evaluating the above at $u = \theta$, $v = (u_{\beta-2K})^*$, $K - (u - v) = (u_\alpha)^*$.

Finite boundary case. The notation used are those of Figure 15 (right). Let $b \in \{b^\ell, b^r\}$ be a black vertex of some boundary rhombus pair of $\mathbb{R}^\theta$. Suppose first that this rhombus pair is not the root one. Then, if $b = b^r$, resp. $b = b^\ell$, the coefficients of the LHS and RHS of (4.7) are non-zero when $x \in \{v^c, v^r, f^c\}$, resp. $x \in \{v^c, v^\ell, f^c\}$. We need to prove:

$$\underline{K}^Q_{b^r, w^c} t_{w^c, v^c} = s_{b^r, w^r} K^\partial_{w^r, v^c}$$

$$\underline{K}^Q_{b^r, w^r} t_{w^r, v^r} = s_{b^r, w^r} K^\partial_{w^r, v^r}$$

$$\underline{K}^Q_{b^r, w^r} t_{w^r, f^c} + \underline{K}^Q_{b^r, w^c} t_{w^c, f^c} = s_{b^r, w^r} K^\partial_{w^r, f^c}$$

and

$$\underline{K}^Q_{b^\ell, w^c} t_{w^c, v^c} = s_{b^\ell, w^\ell} K^\partial_{w^\ell, v^c}$$

$$\underline{K}^Q_{b^\ell, w^\ell} t_{w^\ell, v^\ell} = s_{b^\ell, w^\ell} K^\partial_{w^\ell, v^\ell}$$

$$\underline{K}^Q_{b^\ell, w^\ell} t_{w^\ell, f^c} + \underline{K}^Q_{b^\ell, w^c} t_{w^c, f^c} = s_{b^\ell, w^\ell} K^\partial_{w^\ell, f^c}$$

In both cases, the last two equalities are as in the full plane with the appropriate change of notation. We thus need to check the first equality of each case. Returning to the
Replacing coefficients of $K^Q$ yields for the LHS:

$$[\underline{K}^Q T]_{b,v_1} = e^{i \frac{\beta+\bar{\alpha}+\pi}{2}} \mathrm{cn}(\theta) t_{w_2,v_1} + e^{i \frac{\bar{\alpha}-\pi+\bar{\alpha}}{2}} t_{w_3,v_1} = e^{i \frac{\bar{\alpha}}{2}} [\mathrm{cn}(\theta) \mathrm{cn}(u_{\beta+2K}) - \mathrm{cn}(u_{\alpha+2K})]$$

$$[\underline{K}^Q T]_{b,v_2} = e^{i \frac{\beta+\bar{\alpha}}{2}} \mathrm{sn}(\theta) t_{w_1,v_2} = e^{i \frac{\bar{\alpha}}{2}} \mathrm{sn}(\theta) \mathrm{cn}(u_{\beta})$$

$$[\underline{K}^Q T]_{b,f_1} = e^{i \frac{\beta+\bar{\alpha}}{2}} \mathrm{sn}(\theta) t_{w_1,f_1} + e^{i \frac{\bar{\alpha}-\pi+\bar{\alpha}}{2}} t_{w_3,f_1} = i e^{i \frac{\bar{\alpha}}{2}} [\mathrm{sn}(\theta) \mathrm{sn}((u_{\beta-2K})^*) - \mathrm{sn}((u_{\alpha})^*)]$$

$$[\underline{K}^Q T]_{b,f_2} = e^{i \frac{\beta+\bar{\alpha}+\pi}{2}} \mathrm{cn}(\theta) t_{w_2,f_2} = i e^{i \frac{\bar{\alpha}}{2}} \mathrm{cn}(\theta) \mathrm{sn}((u_{\beta})^*)$$

Note that in the finite case, we have $K^\theta = K$ for coefficients involved. The coefficient $[SK]_{b,x}$ of the RHS of (4.6) and (4.7) is also non-zero when $x \in \{v_1, v_2, f_1, f_2\}$ and we have $[SK]_{b,x} = s_{b,w}K_{w,x}$. To compute these terms, we first express the part $\mathrm{cn}(u_\beta)[\mathrm{nd}(u_\alpha) \mathrm{nd}(u_\beta)]^{\frac{1}{2}} := (\diamond) \text{ in } s_{b,w}, \text{ see (4.2), using the angles and parameters involved in the four coefficients of K.}$

$$\begin{aligned}
(\diamond) &= -\mathrm{sn}(u_{\beta+2K})[\mathrm{dn}(u_{\alpha+2K}) \mathrm{nd}(u_{\beta+2K})]^{\frac{1}{2}} \\
&= \mathrm{cn}(u_\beta)[\mathrm{nd}(u_\alpha) \mathrm{nd}(u_\beta)]^{\frac{1}{2}} \\
&= (k')^{-\frac{1}{2}} \mathrm{cn}((u_{\beta-2K})^*)[\mathrm{nd}((u_{\beta-2K})^*) \mathrm{dn}((u_\alpha)^*)]^{\frac{1}{2}} \\
&= (k')^{\frac{1}{2}} \mathrm{sn}((u_\beta)^*)[\mathrm{nd}((u_\beta)^*) \mathrm{nd}((u_{\alpha+2K})^*)]^{\frac{1}{2}},
\end{aligned}$$

where in the first line we used that $u_\alpha = K + u_{\alpha+2K}$, in the third that $u_\beta = -(u_{\beta-2K})^*$, $u_\alpha = K - (u_\alpha)^*$, and in the fourth that $u_\beta = K - (u_\beta)^*$, $u_\alpha = 2K - (u_{\alpha+2K})^*$. Replacing coefficients of K by their definition gives for the RHS of (4.6),

$$[\underline{S K}]_{b,v_1} = s_{b,w} K_{w,v_1} = e^{i \frac{\bar{\alpha}}{2}} \mathrm{sn}(\theta) \mathrm{sn}(u_{\beta+2K}) \mathrm{dn}(u_{\alpha+2K})$$

$$[\underline{S K}]_{b,v_2} = s_{b,w} K_{w,v_2} = e^{i \frac{\bar{\alpha}}{2}} \mathrm{sn}(\theta) \mathrm{cn}(u_\beta)$$

$$[\underline{S K}]_{b,f_1} = s_{b,w} K_{w,f_1} = -i (k')^{-1} e^{i \frac{\bar{\alpha}}{2}} \mathrm{cn}(\theta) \mathrm{cn}((u_{\beta-2K})^*) \mathrm{dn}((u_\alpha)^*)$$

$$[\underline{S K}]_{b,f_2} = s_{b,w} K_{w,f_2} = i e^{i \frac{\bar{\alpha}}{2}} \mathrm{cn}(\theta) \mathrm{sn}((u_\beta)^*)$$

The equality $[K^Q T]_{b,x} = [S K]_{b,x}$ is then straightforward when $x = v_2$ and $x = f_2$. When $x = v_1$, this is a consequence of the identity, see [54, chap.2, ex.32 (i)],

$$\mathrm{cn} \, u \, \mathrm{cn} \, v - \mathrm{cn}(u + v) = \mathrm{sn} \, u \, \mathrm{sn} \, v \, \mathrm{dn}(u + v), \quad (4.8)$$

evaluated at $u = \theta$, $v = u_{\beta+2K}$, $u + v = u_{\alpha+2K}$.

We are left with proving the case $x = f_1$. Multiplying the identity (4.8) by $\mathrm{nd}(u - v)$, using that $\mathrm{nd}(u - v) = (k')^{-1} \mathrm{dn}(K - (u - v))$, $\mathrm{cd}(u - v) = \mathrm{sn}(K - (u - v))$ we obtain,

$$(k')^{-1} \mathrm{cn} \, u \, \mathrm{cn} \, v \, \mathrm{dn}(K - (u - v)) = \mathrm{sn}(K - (u - v)) - \mathrm{sn} \, u \, \mathrm{sn} \, v.$$

The proof is concluded by evaluating the above at $u = \theta$, $v = (u_{\beta-2K})^*$, $K - (u - v) = (u_\alpha)^*$.

Finite boundary case. The notation used are those of Figure 15 (right). Let $b \in \{b^\ell, b^r\}$ be a black vertex of some boundary rhombus pair of $\mathbb{R}^\theta$. Suppose first that this rhombus pair is not the root one. Then, if $b = b^r$, resp. $b = b^\ell$, the coefficients of the LHS and RHS of (4.7) are non-zero when $x \in \{v^c, v^r, f^c\}$, resp. $x \in \{v^c, v^\ell, f^c\}$. We need to prove:

$$\underline{K}^Q_{b^r, w^c} t_{w^c, v^c} = s_{b^r, w^r} K^\partial_{w^r, v^c}$$

$$\underline{K}^Q_{b^r, w^r} t_{w^r, v^r} = s_{b^r, w^r} K^\partial_{w^r, v^r}$$

$$\underline{K}^Q_{b^r, w^r} t_{w^r, f^c} + \underline{K}^Q_{b^r, w^c} t_{w^c, f^c} = s_{b^r, w^r} K^\partial_{w^r, f^c}$$

and

$$\underline{K}^Q_{b^\ell, w^c} t_{w^c, v^c} = s_{b^\ell, w^\ell} K^\partial_{w^\ell, v^c}$$

$$\underline{K}^Q_{b^\ell, w^\ell} t_{w^\ell, v^\ell} = s_{b^\ell, w^\ell} K^\partial_{w^\ell, v^\ell}$$

$$\underline{K}^Q_{b^\ell, w^\ell} t_{w^\ell, f^c} + \underline{K}^Q_{b^\ell, w^c} t_{w^c, f^c} = s_{b^\ell, w^\ell} K^\partial_{w^\ell, f^c}$$

In both cases, the last two equalities are as in the full plane with the appropriate change of notation. We thus need to check the first equality of each case. Returning to the
definition of $\underline{K}^Q$ and boundary values of the matrices $S$ and $T$ defined in Equations (4.3) and (4.5) we have,

$$ 
\begin{aligned}
\frac{\underline{K}^Q_{b^r, w^c} t_{w^c, v^c}}{s_{b^r, w^r}} &= \frac{e^{i \frac{\bar{\alpha}^r - \pi + \bar{\alpha}^r}{2}} \times (-ik' e^{-i \frac{\bar{\alpha}^r}{2}} \operatorname{sn}(\theta^\partial) \operatorname{nd}(u_{\alpha^r}) \operatorname{cd}(u_{\beta^r}))}{e^{-i \frac{\bar{\beta}^r}{2}} \operatorname{cn}(u_{\beta^r}) [\operatorname{sn}(\theta^\partial) \operatorname{cn}(\theta^\partial) \operatorname{nd}(u_{\alpha^r}) \operatorname{nd}(u_{\beta^r})]^{\frac{1}{2}}} \\
&= -k' e^{i \frac{\bar{\alpha}^r + \bar{\beta}^r}{2}} [\operatorname{sc}(\theta^\partial) \operatorname{nd}(u_{\alpha^r}) \operatorname{nd}(u_{\beta^r})]^{\frac{1}{2}} \\
&= e^{i \frac{\bar{\alpha}^r + \pi + \bar{\beta}^r + \pi}{2}} [\operatorname{sc}(\theta^\partial) \operatorname{nd}(u_{\alpha^r + 2K}) \operatorname{nd}(u_{\beta^r + 2K})]^{\frac{1}{2}} = \underline{K}_{w^r, v^c} = \underline{K}_{w^r, v^c}^\partial.
\end{aligned}
$$

In a similar way,

$$ 
\begin{aligned}
\frac{\underline{K}^Q_{b^\ell, w^c} t_{w^c, v^c}}{s_{b^\ell, w^\ell}} &= \frac{e^{i \frac{\bar{\beta}^\ell + \bar{\beta}^\ell + \pi}{2}} \times (-ik' e^{-i \frac{\bar{\alpha}^r}{2}} \operatorname{sn}(\theta^\partial) \operatorname{nd}(u_{\alpha^r}) \operatorname{cd}(u_{\beta^r}))}{e^{-i \frac{\bar{\alpha}^\ell}{2}} \operatorname{cn}(u_{\alpha^\ell}) [\operatorname{sn}(\theta^\partial) \operatorname{cn}(\theta^\partial) \operatorname{nd}(u_{\alpha^\ell}) \operatorname{nd}(u_{\beta^\ell})]^{\frac{1}{2}}} \\
&= k' e^{i \frac{\bar{\alpha}^\ell + \bar{\beta}^\ell + (\bar{\beta}^\ell - \bar{\alpha}^r)}{2}} [\operatorname{sc}(\theta^\partial) \operatorname{dn}(u_{\alpha^\ell}) \operatorname{dn}(u_{\beta^\ell})]^{\frac{1}{2}} \operatorname{nd}(u_{\alpha^r}) \operatorname{cd}(u_{\beta^r}) \operatorname{nc}(u_{\alpha^\ell}), \\
&= -k' e^{i \frac{\bar{\alpha}^\ell + \bar{\beta}^\ell}{2}} [\operatorname{sc}(\theta^\partial) \operatorname{nd}(u_{\alpha^\ell}) \operatorname{nd}(u_{\beta^\ell})]^{\frac{1}{2}} \operatorname{cd}(u_{\beta^r}) \operatorname{dc}(u_{\alpha^\ell}), \quad \text{since } \bar{\beta}^\ell - \bar{\alpha}^r = 2\pi \\
&= e^{i \frac{\bar{\alpha}^\ell + \pi + \bar{\beta}^\ell + \pi}{2}} [\operatorname{sc}(\theta^\partial) \operatorname{dn}(u_{\alpha^\ell + 2K}) \operatorname{nd}(u_{\beta^\ell + 2K})]^{\frac{1}{2}} \operatorname{cd}(u_{\beta^r}) \operatorname{dc}(u_{\alpha^\ell}) \\
&= \underline{K}_{w^\ell, v^c} \operatorname{cd}(u_{\beta^r}) \operatorname{dc}(u_{\alpha^\ell}) = \underline{K}_{w^\ell, v^c}^\partial.
\end{aligned}
$$

Note that we always have $s_{b^r, w^r}$, $s_{b^\ell, w^\ell} \neq 0$ on $C'$, see (3.4), so that it makes sense to divide by these quantities.

The last case we need to consider is if $b$ belongs to the root boundary rhombus pair of $\mathcal{R}^\partial$. But then, of the three equations above, the first one is absent since we have $v^c = r$, so we are left with the last two which are as in the full plane case. This ends the proof of (4.7) and thus finishes the proof of Theorem 4.2. $\square$

## 4.2 Determinants of the Kasteleyn matrix $K^Q$ and of the $Z^u$-Dirac operator

We restrict to the finite case. Theorem 4.3 proves that the determinants of the matrices $K^Q$ and $K^\partial(u)$ are equal up to an explicit multiplicative constant depending on $u$. By [32], see also Equation (2.5), the determinant of $K^Q$ is equal, up to a constant, to the squared partition function of the Ising model with + boundary conditions. Using Corollary 3.10, the determinant of $K^\partial(u)$ is equal, up to an explicit constant, to the determinant of $\Delta^{m,\partial}(u)$ which counts weighted rooted directed spanning forests. Theorem 4.3 thus implies identities between partition functions made explicit in Corollary 4.5.

In the whole of this section, we restrict the domain of $u$ to

$$ 
\begin{aligned}
C'' &= C \setminus \{\alpha^\ell[2K], \beta^\ell[2K] : e^{i\bar{\alpha}^\ell}, e^{i\bar{\beta}^\ell} \in \mathcal{R}^\partial\} \\
&= C \setminus \{\alpha^r[2K], \beta^r[2K] : e^{i\bar{\alpha}^r}, e^{i\bar{\beta}^r} \in \mathcal{R}^\partial\}.
\end{aligned}
\quad (4.9) $$

Note that by Section 2.5.3 on train-tracks, this amounts to removing all the parallel directions of the train-tracks of the isoradial graph $G$.

### 4.2.1 Results

Every edge $e = (w, x)$ of $G^{D,r}$ is assigned two rhombus vectors $e^{i\bar{\alpha}_e}, e^{i\bar{\beta}_e}$ of the diamond graph $G^{\diamond/2}$ and a half-angle $\bar{\theta}_e$. We partition the set of white vertices $W^r = W$ of $G^{D,r}$ as $W^\partial \cup W^\circ$, where $W^\partial$ consists of boundary vertices of $W$ and $W^\circ$ of inner ones. Every white vertex $w$ of $G^{D,r}$ is in the center of a rhombus of the diamond graph $G^{\diamond}$; we let $\bar{\theta}_w$ be the half-angle of this rhombus at one of the two *primal* vertices.

In the statement below, a specific role is played by the boundary rhombus pair of $\mathcal{R}^\partial$ containing the root $r$. We use the notation of Figure 15 (right) and add a superscript $r$ to specify vertices/angles of this root pair.
definition of $K^Q$ and boundary values of the matrices $S$ and $T$ defined in Equations (4.3) and (4.5) we have,

$$ 
\begin{aligned}
\frac{K_{b^r, w^c}^Q t_{w^c, v^c}}{s_{b^r, w^r}} &= \frac{e^{i \frac{\bar{\alpha}^r - \pi + \bar{\alpha}^r}{2}} \times (-ik' e^{-i \frac{\bar{\alpha}^r}{2}} \operatorname{sn}(\theta^\partial) \operatorname{nd}(u_{\alpha^r}) \operatorname{cd}(u_{\beta^r}))}{e^{-i \frac{\bar{\beta}^r}{2}} \operatorname{cn}(u_{\beta^r}) [\operatorname{sn}(\theta^\partial) \operatorname{cn}(\theta^\partial) \operatorname{nd}(u_{\alpha^r}) \operatorname{nd}(u_{\beta^r})]^{\frac{1}{2}}} \\
&= -k' e^{i \frac{\bar{\alpha}^r + \bar{\beta}^r}{2}} [\operatorname{sc}(\theta^\partial) \operatorname{nd}(u_{\alpha^r}) \operatorname{nd}(u_{\beta^r})]^{\frac{1}{2}} \\
&= e^{i \frac{\bar{\alpha}^r + \pi + \bar{\beta}^r + \pi}{2}} [\operatorname{sc}(\theta^\partial) \operatorname{nd}(u_{\alpha^r + 2K}) \operatorname{nd}(u_{\beta^r + 2K})]^{\frac{1}{2}} = K_{w^r, v^c} = K_{w^r, v^c}^\partial.
\end{aligned}
$$

In a similar way,

$$ 
\begin{aligned}
\frac{K_{b^\ell, w^c}^Q t_{w^c, v^c}}{s_{b^\ell, w^\ell}} &= \frac{e^{i \frac{\bar{\beta}^\ell + \bar{\beta}^\ell + \pi}{2}} \times (-ik' e^{-i \frac{\bar{\alpha}^\ell}{2}} \operatorname{sn}(\theta^\partial) \operatorname{nd}(u_{\alpha^\ell}) \operatorname{cd}(u_{\beta^\ell}))}{e^{-i \frac{\bar{\alpha}^\ell}{2}} \operatorname{cn}(u_{\alpha^\ell}) [\operatorname{sn}(\theta^\partial) \operatorname{cn}(\theta^\partial) \operatorname{nd}(u_{\alpha^\ell}) \operatorname{nd}(u_{\beta^\ell})]^{\frac{1}{2}}} \\
&= k' e^{i \frac{\bar{\alpha}^\ell + \bar{\beta}^\ell + (\bar{\beta}^\ell - \bar{\alpha}^\ell)}{2}} [\operatorname{sc}(\theta^\partial) \operatorname{dn}(u_{\alpha^\ell}) \operatorname{dn}(u_{\beta^\ell})]^{\frac{1}{2}} \operatorname{nd}(u_{\alpha^\ell}) \operatorname{cd}(u_{\beta^\ell}) \operatorname{nc}(u_{\alpha^\ell}), \\
&= -k' e^{i \frac{\bar{\alpha}^\ell + \bar{\beta}^\ell}{2}} [\operatorname{sc}(\theta^\partial) \operatorname{nd}(u_{\alpha^\ell}) \operatorname{nd}(u_{\beta^\ell})]^{\frac{1}{2}} \operatorname{cd}(u_{\beta^\ell}) \operatorname{dc}(u_{\alpha^\ell}), \text{ since } \bar{\beta}^\ell - \bar{\alpha}^\ell = 2\pi \\
&= e^{i \frac{\bar{\alpha}^\ell + \pi + \bar{\beta}^\ell + \pi}{2}} [\operatorname{sc}(\theta^\partial) \operatorname{dn}(u_{\alpha^\ell + 2K}) \operatorname{nd}(u_{\beta^\ell + 2K})]^{\frac{1}{2}} \operatorname{cd}(u_{\beta^\ell}) \operatorname{dc}(u_{\alpha^\ell}) \\
&= K_{w^\ell, v^c} \operatorname{cd}(u_{\beta^\ell}) \operatorname{dc}(u_{\alpha^\ell}) = K_{w^\ell, v^c}^\partial.
\end{aligned}
$$

Note that we always have $s_{b^r, w^r}$, $s_{b^\ell, w^\ell} \neq 0$ on $C'$, see (3.4), so that it makes sense to divide by these quantities.

The last case we need to consider is if $b$ belongs to the root boundary rhombus pair of $\mathcal{R}^\partial$. But then, of the three equations above, the first one is absent since we have $v^c = r$, so we are left with the last two which are as in the full plane case. This ends the proof of (4.7) and thus finishes the proof of Theorem 4.2. $\square$

## 4.2 Determinants of the Kasteleyn matrix $K^Q$ and of the $Z^u$-Dirac operator

We restrict to the finite case. Theorem 4.3 proves that the determinants of the matrices $K^Q$ and $K^\partial(u)$ are equal up to an explicit multiplicative constant depending on $u$. By [32], see also Equation (2.5), the determinant of $K^Q$ is equal, up to a constant, to the squared partition function of the Ising model with + boundary conditions. Using Corollary 3.10, the determinant of $K^\partial(u)$ is equal, up to an explicit constant, to the determinant of $\Delta^{m,\partial}(u)$ which counts weighted rooted directed spanning forests. Theorem 4.3 thus implies identities between partition functions made explicit in Corollary 4.5.

In the whole of this section, we restrict the domain of $u$ to

$$ 
\begin{aligned}
C'' &= C \setminus \{\alpha^\ell[2K], \beta^\ell[2K] : e^{i\bar{\alpha}^\ell}, e^{i\bar{\beta}^\ell} \in \mathcal{R}^\partial\} \\
&= C \setminus \{\alpha^r[2K], \beta^r[2K] : e^{i\bar{\alpha}^r}, e^{i\bar{\beta}^r} \in \mathcal{R}^\partial\}.
\end{aligned}
\quad (4.9)
$$

Note that by Section 2.5.3 on train-tracks, this amounts to removing all the parallel directions of the train-tracks of the isoradial graph $G$.

### 4.2.1 Results

Every edge $e = (w, x)$ of $G^{D,r}$ is assigned two rhombus vectors $e^{i\bar{\alpha}_e}, e^{i\bar{\beta}_e}$ of the diamond graph $G^{\diamond/2}$ and a half-angle $\bar{\theta}_e$. We partition the set of white vertices $W^r = W$ of $G^{D,r}$ as $W^\partial \cup W^\circ$, where $W^\partial$ consists of boundary vertices of $W$ and $W^\circ$ of inner ones. Every white vertex $w$ of $G^{D,r}$ is in the center of a rhombus of the diamond graph $G^{\diamond}$; we let $\bar{\theta}_w$ be the half-angle of this rhombus at one of the two *primal* vertices.

In the statement below, a specific role is played by the boundary rhombus pair of $\mathcal{R}^\partial$ containing the root $r$. We use the notation of Figure 15 (right) and add a superscript $r$ to specify vertices/angles of this root pair.
**Theorem 4.3.** Let $M_1$ be a dimer configuration of $G^{D,r}$. Then, for every $u \in C''$, we have

$$|\det K^Q| = \tilde{C}(u) \cdot |\det K^\theta(u)|,$$

where $\tilde{C}(u)$ is equal to:

$$(k')^{\frac{|V|-1}{2}} \left( \prod_{w \in W^{\theta}} \mathrm{sn}(\theta_w)^{-1} \right) \left( \prod_{w \in W} [\mathrm{cn}(\theta_w) \mathrm{sn}(\theta_w)]^{\frac{1}{2}} \right) \times \\ \times \left( \prod_{wx \in M_1: w \in W} |\mathrm{sc}(u_{\alpha_e}) \mathrm{sc}(u_{\beta_e})|^{\frac{1}{2}} \right) \mathrm{sn}(\theta^{\theta,r}) \left| \frac{\mathrm{cd}(u_{\beta,r,r})}{\mathrm{sn}(u_{\alpha,r,r})} \right|.$$

**Remark 4.4.**

* Since for $u \in C''$, $\mathrm{sc}(u_{\alpha_e}) \mathrm{sc}(u_{\beta_e}) \neq 0$, and since $\det K^Q \neq 0$, we have that the matrix $K^\theta(u)$ is invertible for these values of $u$.

* The quantity $\prod_{wx \in M_1: w \in W} |\mathrm{sc}(u_{\alpha_e}) \mathrm{sc}(u_{\beta_e})|^{\frac{1}{2}}$ is independent of the choice of perfect matching $M_1$. Similarly to Remark 3.9, this consists in showing that the alternating product around every inner face of $G^{D,r}$ is equal to 1; the proof is similar to that of Lemma 3.11.

* A surprising fact is that the LHS is independent of $u$ while the RHS does not seem to be, also the RHS seems to depend on the choice of root vertex $r$. It is not straightforward to see why this indeed not the case. An alternative way of proving this theorem is to extend to the full Z-invariant case the combinatorial argument of [30]; one then better sees the parameter $u$ and the choice of root vertex $r$ appearing.

Combining Theorems 4.3 and 3.8, we deduce that the squared partition function of the Z-invariant Ising model with + boundary conditions is equal, up to an explicit constant, to the determinant of the massive Laplacian $\Delta^{m,\theta}(u)$, i.e., to the partition function of rooted directed spanning forests. This generalizes to the full Z-invariant case the result of [29, 30] and to simply connected domains the result of [12] proved for the characteristic polynomial in the toroidal case in an abstract way.

**Corollary 4.5.** For every $u \in C''$, we have

$$1. \quad [Z_{\text{Ising}}^{+}(G, J)]^2 = 2^{|V|-1} (k')^{|E|} \left( \prod_{w \in W^{\theta}} \frac{1+\mathrm{sn}(\theta_w)}{2 \mathrm{sn}(\theta_w)} \right) \times \\ \times \left( \prod_{wx \in M_1: w \in W} |\mathrm{sc}(u_{\alpha_e}) \mathrm{sc}(u_{\beta_e}) \mathrm{nd}(u_{\alpha_e}) \mathrm{nd}(u_{\beta_e})|^{\frac{1}{2}} \right) \mathrm{sn}(\theta^{\theta,r}) \left| \frac{\mathrm{cd}(u_{\beta,r,r})}{\mathrm{sn}(u_{\alpha,r,r})} \right| |\det \Delta^{m,\theta}(u)|.$$

$$2. \quad [Z_{\text{Ising}}^{+}(G, J)]^2 = 2^{|V|-1} (k')^{-|E|} \left( \prod_{w \in W^{\theta}} \frac{1+\mathrm{sn}(\theta_w)}{2 \mathrm{sn}(\theta_w)} \right) \mathrm{sn}(\theta^{\theta,r}) \left| \frac{\mathrm{cd}(u_{\beta,r,r}) \mathrm{sn}(u_{\beta,r,r})}{\mathrm{cd}(u_{\alpha,r,r}) \mathrm{sn}(u_{\alpha,r,r})} \right|^{\frac{1}{2}} \times \\ \times |\det \Delta^{m,\theta}(u) \det \Delta^{m,\theta}(u+2K)|^{\frac{1}{2}}.$$

*Proof.* Equality 2. is obtained by writing $[Z_{\text{Ising}}^{+}(G, J)]^2 = ([Z_{\text{Ising}}^{+}(G, J)]^2[Z_{\text{Ising}}^{+}(G, J)]^2)^{\frac{1}{2}}$, using Equality 1. evaluated once at $u$ and once at $u+2K$, and using the identities $|\mathrm{sc}(u-K)| = |(k')^{-1} \mathrm{cs}(u)|$, $|\mathrm{nd}(u-K)| = |(k')^{-1} \mathrm{dn}(u)|$, $|\mathrm{cd}(u-K)| = |\mathrm{sn}(u)|$, $|\mathrm{sn}(u-K)| = |\mathrm{cd}(u)|$.

Let us prove Equality 1. From Equation (2.5) we have, for all coupling constants J,

$$[Z_{\text{Ising}}^{+}(G, J)]^2 = 2^{|V|-1} \left( \prod_{e \in E^{\theta}} \frac{e^{2J_e}}{2} \right) \left( \prod_{e^* \in E^*} \cosh(2J_e) \right) Z_{\text{dimer}}(G^Q, \nu^J).$$

Returning to the definition of the Z-invariant weights (2.6) and (2.8) gives,

$$\left( \prod_{e \in E^{\theta}} \frac{e^{2J_e}}{2} \right) \left( \prod_{e^* \in E^*} \cosh(2J_e) \right) = \left( \prod_{e \in E^{\theta}} \frac{1+\mathrm{sn}(\theta_e)}{2 \mathrm{cn}(\theta_e)} \right) \left( \prod_{e^* \in E^*} \frac{1}{\mathrm{cn}(\theta_e)} \right) \\ = \left( \prod_{w \in W^{\theta}} \frac{1+\mathrm{sn}(\theta_w)}{2} \right) \left( \prod_{w \in W} \frac{1}{\mathrm{cn}(\theta_w)} \right),$$
**Theorem 4.3.** Let $M_1$ be a dimer configuration of $G^{D,r}$. Then, for every $u \in C''$, we have

$$|\det K^Q| = \tilde{C}(u) \cdot |\det K^\theta(u)|,$$

where $\tilde{C}(u)$ is equal to:

$$(k')^{\frac{|V|-1}{2}} \left( \prod_{w \in W^{\theta}} \mathrm{sn}(\theta_w)^{-1} \right) \left( \prod_{w \in W} [\mathrm{cn}(\theta_w) \mathrm{sn}(\theta_w)]^{\frac{1}{2}} \right) \times \\ \times \left( \prod_{wx \in M_1: w \in W} |\mathrm{sc}(u_{\alpha_e}) \mathrm{sc}(u_{\beta_e})|^{\frac{1}{2}} \right) \mathrm{sn}(\theta^{\theta,r}) \left| \frac{\mathrm{cd}(u_{\beta,r,r})}{\mathrm{sn}(u_{\alpha,r,r})} \right|.$$

**Remark 4.4.**

* Since for $u \in C''$, $\mathrm{sc}(u_{\alpha_e}) \mathrm{sc}(u_{\beta_e}) \neq 0$, and since $\det K^Q \neq 0$, we have that the matrix $K^\theta(u)$ is invertible for these values of $u$.

* The quantity $\prod_{wx \in M_1: w \in W} |\mathrm{sc}(u_{\alpha_e}) \mathrm{sc}(u_{\beta_e})|^{\frac{1}{2}}$ is independent of the choice of perfect matching $M_1$. Similarly to Remark 3.9, this consists in showing that the alternating product around every inner face of $G^{D,r}$ is equal to 1; the proof is similar to that of Lemma 3.11.

* A surprising fact is that the LHS is independent of $u$ while the RHS does not seem to be, also the RHS seems to depend on the choice of root vertex $r$. It is not straightforward to see why this indeed not the case. An alternative way of proving this theorem is to extend to the full Z-invariant case the combinatorial argument of [30]; one then better sees the parameter $u$ and the choice of root vertex $r$ appearing.

Combining Theorems 4.3 and 3.8, we deduce that the squared partition function of the Z-invariant Ising model with + boundary conditions is equal, up to an explicit constant, to the determinant of the massive Laplacian $\Delta^{m,\theta}(u)$, i.e., to the partition function of rooted directed spanning forests. This generalizes to the full Z-invariant case the result of [29, 30] and to simply connected domains the result of [12] proved for the characteristic polynomial in the toroidal case in an abstract way.

**Corollary 4.5.** For every $u \in C''$, we have

$$1. \quad [Z_{\text{Ising}}^{+}(G, J)]^2 = 2^{|V|-1} (k')^{|E|} \left( \prod_{w \in W^{\theta}} \frac{1+\mathrm{sn}(\theta_w)}{2 \mathrm{sn}(\theta_w)} \right) \times \\ \times \left( \prod_{wx \in M_1: w \in W} |\mathrm{sc}(u_{\alpha_e}) \mathrm{sc}(u_{\beta_e}) \mathrm{nd}(u_{\alpha_e}) \mathrm{nd}(u_{\beta_e})|^{\frac{1}{2}} \right) \mathrm{sn}(\theta^{\theta,r}) \left| \frac{\mathrm{cd}(u_{\beta,r,r})}{\mathrm{sn}(u_{\alpha,r,r})} \right| |\det \Delta^{m,\theta}(u)|.$$

$$2. \quad [Z_{\text{Ising}}^{+}(G, J)]^2 = 2^{|V|-1} (k')^{-|E|} \left( \prod_{w \in W^{\theta}} \frac{1+\mathrm{sn}(\theta_w)}{2 \mathrm{sn}(\theta_w)} \right) \mathrm{sn}(\theta^{\theta,r}) \left| \frac{\mathrm{cd}(u_{\beta,r,r}) \mathrm{sn}(u_{\beta,r,r})}{\mathrm{cd}(u_{\alpha,r,r}) \mathrm{sn}(u_{\alpha,r,r})} \right|^{\frac{1}{2}} \times \\ \times |\det \Delta^{m,\theta}(u) \det \Delta^{m,\theta}(u+2K)|^{\frac{1}{2}}.$$

*Proof.* Equality 2. is obtained by writing $[Z_{\text{Ising}}^{+}(G, J)]^2 = ([Z_{\text{Ising}}^{+}(G, J)]^2[Z_{\text{Ising}}^{+}(G, J)]^2)^{\frac{1}{2}}$, using Equality 1. evaluated once at $u$ and once at $u+2K$, and using the identities $|\mathrm{sc}(u-K)| = |(k')^{-1} \mathrm{cs}(u)|$, $|\mathrm{nd}(u-K)| = |(k')^{-1} \mathrm{dn}(u)|$, $|\mathrm{cd}(u-K)| = |\mathrm{sn}(u)|$, $|\mathrm{sn}(u-K)| = |\mathrm{cd}(u)|$.

Let us prove Equality 1. From Equation (2.5) we have, for all coupling constants J,

$$[Z_{\text{Ising}}^{+}(G, J)]^2 = 2^{|V|-1} \left( \prod_{e \in E^{\theta}} \frac{e^{2J_e}}{2} \right) \left( \prod_{e^* \in E^*} \cosh(2J_e) \right) Z_{\text{dimer}}(G^Q, \nu^J).$$

Returning to the definition of the Z-invariant weights (2.6) and (2.8) gives,

$$\left( \prod_{e \in E^{\theta}} \frac{e^{2J_e}}{2} \right) \left( \prod_{e^* \in E^*} \cosh(2J_e) \right) = \left( \prod_{e \in E^{\theta}} \frac{1+\mathrm{sn}(\theta_e)}{2 \mathrm{cn}(\theta_e)} \right) \left( \prod_{e^* \in E^*} \frac{1}{\mathrm{cn}(\theta_e)} \right) \\ = \left( \prod_{w \in W^{\theta}} \frac{1+\mathrm{sn}(\theta_w)}{2} \right) \left( \prod_{w \in W} \frac{1}{\mathrm{cn}(\theta_w)} \right),$$
where in the second equality we use the notation introduced before the statement of
Theorem 4.3. Since $Z_{\text{dimer}}(G^Q, \nu^j) = |\det K^Q|$, Theorem 4.3 yields

$$
[Z_{\text{Ising}}^{+}(G, J)]^2 = 2^{|V|-1}(k')^{\frac{|V^*|}{2}} \prod_{w \in W} [\text{sc}(\theta_w)]^{\frac{1}{2}} \left( \prod_{w \in W^{\partial}} \frac{1+\text{sn}(\theta_w)}{2\text{sn}(\theta_w)} \right) \times \\
\left( \prod_{wx \in M_1: w \in W} |\text{sc}(u_{\alpha_e}) \text{sc}(u_{\beta_e})|^{\frac{1}{2}} \right) \times \text{sn}(\theta^{\partial,r}) \left| \frac{\text{cd}(u_{\beta r,r})}{\text{sn}(u_{\alpha r,r})} \right| |\det K^{\partial}(u)|.
$$

The proof is concluded by using Corollary 3.10, choosing $M_1$ as perfect matching of $G^{D,r}$ and using that $|V| - 1 = -|V^*| + |E|$. $\square$

**4.2.2 Proof of Theorem 4.3**

Let us prove that for the modified Kasteleyn matrix $\underline{K}^Q$, we have

$$
\det \underline{K}^Q = (k')^{\frac{|V^*|}{2}} \left( \prod_{w \in W} [\text{sn}(\theta_w) \text{cn}(\theta_w)]^{\frac{1}{2}} \right) \left( \prod_{w \in W^{\partial}} \text{sn}(\theta_w)^{-\frac{1}{2}} \right) \times \\
\left( \prod_{wx \in M_1: w \in W} |\text{sc}(u_{\alpha_e}) \text{sc}(u_{\beta_e})|^{\frac{1}{2}} \right) \times \text{sn}(\theta^{\partial,r}) \left| \frac{\text{cd}(u_{\beta r,r})}{\text{sn}(u_{\alpha r,r})} \right| |\det \underline{K}^{\partial}(u)|.
$$

To obtain the result for $K^Q$, we use Relation (4.1) which implies that

$$
\det \underline{K}^Q = \left( \prod_{w \in W^{\partial}} \text{sn}(\theta_w)^{-\frac{1}{2}} \right) \det \underline{K}^Q,
$$

because $|\{b_\ell, b_r \in \mathbb{R}^\partial\}| = 2|\{w^c \in \mathbb{R}^\partial\}| = 2|W^\partial|$. The proof has three main steps: the first consists in using a partition of black/white vertices of $G^Q$ and Theorem 4.2 for comparing $\det \underline{K}^Q$ and $\det \underline{K}^\partial(u)$. After this first step, there remains auxiliary determinant computations that are hard to perform for general choices of partitions. In the second step, we identify specific choices of partitions that make these computations actually doable. The latter are the subject of the third step.

**Partition of the vertices of G**<sup>Q</sup> **and Theorem 4.2** We partition black vertices of G<sup>Q</sup>
into two subsets: B = B<sub>1</sub> ∪ B<sub>2</sub>, where B<sub>1</sub> has one black vertex per quadrangle of G<sup>Q</sup>
and B<sub>2</sub> = B \ B<sub>1</sub>. Since boundary quadrangles of G<sup>Q</sup> are reduced to edges, B<sub>1</sub> contains
all black vertices of boundary quadrangles, and half of the black vertices of inner
quadrangles. As a consequence, B<sub>1</sub> has a natural partition as B<sub>1</sub><sup>q</sup> ∪ B<sub>1</sub><sup>r</sup>, where B<sub>1</sub><sup>q</sup>, resp.
B<sub>1</sub><sup>r</sup>, consists of boundary quadrangle, resp. inner quadrangles, black vertices. Then B<sub>2</sub>
has no boundary quadrangle black vertices. In a similar way, we partition white vertices
of G<sup>Q</sup>: W = W<sub>1</sub> ∪ W<sub>2</sub>, where W<sub>1</sub> = W<sub>1</sub><sup>q</sup> ∪ W<sub>1</sub><sup>r</sup>.

Recalling that there is a natural bijection between quadrangles of GQ and white vertices of GD,r, we have the following natural bijections:

$$
W_1^\partial \leftrightarrow W^\partial, \quad W_1^\circ \leftrightarrow W^\circ, \quad W_2 \leftrightarrow W^\circ.
$$

Here are some notation for sub-matrices of the matrices $S(u)$ and $T(u)$ defined in
Section 4.1, and for sub-matrices of the matrix $\underline{K}^Q$.

$$
\begin{align*}
S_1(u) &= S(u)_{B_1}^{W^r}, & S_2(u) &= S(u)_{B_2}^{W^\circ} \\
S_1^\circ(u) &= S(u)_{B_1^\circ}^{W^\partial}, & S_1^\circ(u) &= S(u)_{B_1^\circ}^{W^\circ} \\
T_1(u) &= T(u)_{W_1}^{B^r}, & T_2(u) &= T(u)_{W_2}^{B^r} \\
\forall i,j \in \{1,2\}, & \underline{K}_{ij}^Q &= (\underline{K}^Q)_{B_i}^{W_j}, & \underline{K}_{1\circ,j}^Q &= (\underline{K}^Q)_{B_1^\circ}^{W_j}.
\end{align*}
$$

Using Theorem 4.2, we obtain the following lemma.
where in the second equality we use the notation introduced before the statement of Theorem 4.3. Since $Z_{\text{dimer}}(G^Q, \nu^j) = |\det K^Q|$, Theorem 4.3 yields

$$ [Z_{\text{Ising}}^{+}(G, J)]^2 = 2^{|V|-1}(k')^{\frac{|V^*|}{2}} \prod_{w \in W} [\text{sc}(\theta_w)]^{\frac{1}{2}} \left( \prod_{w \in W^{\partial}} \frac{1+\text{sn}(\theta_w)}{2\text{sn}(\theta_w)} \right) \times \\ \left( \prod_{wx \in M_1: w \in W} |\text{sc}(u_{\alpha_e}) \text{sc}(u_{\beta_e})|^{\frac{1}{2}} \right) \times \text{sn}(\theta^{\partial,r}) \left| \frac{\text{cd}(u_{\beta r,r})}{\text{sn}(u_{\alpha r,r})} \right| |\det K^{\partial}(u)|. $$

The proof is concluded by using Corollary 3.10, choosing $M_1$ as perfect matching of $G^{D,r}$ and using that $|V| - 1 = -|V^*| + |E|$. $\square$

### 4.2.2 Proof of Theorem 4.3

Let us prove that for the modified Kasteleyn matrix $K^Q$, we have

$$ \det K^Q = (k')^{\frac{|V^*|}{2}} \left( \prod_{w \in W} [\text{sn}(\theta_w) \text{cn}(\theta_w)]^{\frac{1}{2}} \right) \left( \prod_{w \in W^{\partial}} \text{sn}(\theta_w)^{-\frac{1}{2}} \right) \times \\ \left( \prod_{wx \in M_1: w \in W} |\text{sc}(u_{\alpha_e}) \text{sc}(u_{\beta_e})|^{\frac{1}{2}} \right) \times \text{sn}(\theta^{\partial,r}) \left| \frac{\text{cd}(u_{\beta r,r})}{\text{sn}(u_{\alpha r,r})} \right| |\det K^{\partial}(u)|. $$

To obtain the result for $K^Q$, we use Relation (4.1) which implies that

$$ \det K^Q = \left( \prod_{w \in W^{\partial}} \text{sn}(\theta_w)^{-\frac{1}{2}} \right) \det K^Q, $$

because $|\{b_l, b_r \in \mathbb{R}^\partial\}| = 2|\{w^c \in \mathbb{R}^\partial\}| = 2|W^\partial|$. The proof has three main steps: the first consists in using a partition of black/white vertices of $G^Q$ and Theorem 4.2 for comparing $\det K^Q$ and $\det K^\partial(u)$. After this first step, there remains auxiliary determinant computations that are hard to perform for general choices of partitions. In the second step, we identify specific choices of partitions that make these computations actually doable. The latter are the subject of the third step.

**Partition of the vertices of $G^Q$ and Theorem 4.2** We partition black vertices of $G^Q$ into two subsets: $B = B_1 \cup B_2$, where $B_1$ has one black vertex per quadrangle of $G^Q$ and $B_2 = B \setminus B_1$. Since boundary quadrangles of $G^Q$ are reduced to edges, $B_1$ contains all black vertices of boundary quadrangles, and half of the black vertices of inner quadrangles. As a consequence, $B_1$ has a natural partition as $B_1^\circ \cup B_1^\circ$, where $B_1^\circ$, resp. $B_1^r$, consists of boundary quadrangle, resp. inner quadrangles, black vertices. Then $B_2$ has no boundary quadrangle black vertices. In a similar way, we partition white vertices of $G^Q$: $W = W_1 \cup W_2$, where $W_1 = W_1^\circ \cup W_1^r$.

Recalling that there is a natural bijection between quadrangles of $G^Q$ and white vertices of $G^{D,r}$, we have the following natural bijections:

$$ W_1^\circ \leftrightarrow W^\circ, \quad W_1 \leftrightarrow W^\circ, \quad W_2 \leftrightarrow W^\circ. $$

Here are some notation for sub-matrices of the matrices $S(u)$ and $T(u)$ defined in Section 4.1, and for sub-matrices of the matrix $K^Q$.

$$ \begin{align*} S_1(u) &= S(u)_{B_1}^{W^r}, & S_2(u) &= S(u)_{B_2}^{W^\circ} \\ S_1^\circ(u) &= S(u)_{B_1^\circ}^{W^\circ}, & S_1^\circ(u) &= S(u)_{B_1^\circ}^{W^\circ} \\ T_1(u) &= T(u)_{W_1}^{B^r}, & T_2(u) &= T(u)_{W_2}^{B^r} \\ \forall i, j \in \{1, 2\}, & K_{ij}^Q &= (K^Q)_{B_i}^{W_j}, & K_{1\circ j}^Q &= (K^Q)_{B_1^\circ}^{W_j}. \end{align*} $$

Using Theorem 4.2, we obtain the following lemma.
**Lemma 4.6.** Consider a partition of the black and white vertices of $G^Q$ as above. Then, for every $u \in C''$,

$$ \det \underline{K}^Q \cdot \det T_1(u) = \det S_1^\theta(u) \cdot \det S_1^\circ(u) \cdot \det S_2(u) \cdot \det R(u) \cdot \det \underline{K}^\theta(u), $$

$$ \text{where } R(u) := S_2(u)^{-1} \underline{K}_{22}^Q - S_1^\circ(u)^{-1} \underline{K}_{12}^Q. $$

**Proof.** Note that $S_2(u)$ and $S_1^\circ(u)$ are invertible because we choose $u \in C''$. By Theorem 4.2, we have the following identity:

$$ B_1^\circ \left( \begin{array}{c|c} \underline{K}_{11}^Q & \underline{K}_{12}^Q \\ \hline \underline{K}_{21}^Q & \underline{K}_{22}^Q \end{array} \right) \quad B_2 \left( \begin{array}{c|c} \underline{K}_{12}^Q & \underline{K}_{21}^Q \\ \hline \underline{K}_{22}^Q & \underline{K}_{22}^Q \end{array} \right) $$

$$ B_1^\circ \left( \begin{array}{c|c} \underline{K}_{11}^Q & \underline{K}_{12}^Q \\ \hline \underline{K}_{21}^Q & \underline{K}_{22}^Q \end{array} \right) \quad B_2 \left( \begin{array}{c|c} \underline{K}_{12}^Q & \underline{K}_{21}^Q \\ \hline \underline{K}_{22}^Q & \underline{K}_{22}^Q \end{array} \right) $$

$$ = \begin{array}{c} \begin{array}{c|c} W^\theta & W^\circ \\ \hline W^\circ & W^\theta \end{array} \quad \begin{array}{c|c} W^\theta & W^\circ \\ \hline W^\circ & W^\theta \end{array} \quad \begin{array}{c|c} B^r & W_2 \\ \hline W^\theta & W^\circ \end{array} \\ \begin{array}{c} B_1^\circ \\ B_2 \end{array} \left( \begin{array}{c|c} S_1^\theta(u) & 0 \\ \hline 0 & S_1^\circ(u) \end{array} \right) \quad \begin{array}{c|c} W^\theta & W^\circ \\ \hline W^\circ & W^\theta \end{array} \quad \begin{array}{c|c} \begin{array}{c|c} K^\theta(u) & \tilde{R}(u) \\ \hline 0 & R(u) \end{array} & W_2 \end{array} \end{array} $$

$$ \text{with} \quad \begin{cases} \underline{K}_{12}^Q = S_1^\circ(u) \tilde{R}(u) \underline{W}_2^\circ \\ \underline{K}_{22}^Q = S_2(u) \tilde{R}(u) \underline{W}_2^\circ + S_2(u) R(u). \end{cases} $$

We extract $\tilde{R}(u)\underline{W}_2^\circ$ from the first equation. Plugging $\tilde{R}(u)\underline{W}_2^\circ$ in the second equation gives $R(u)$; taking the determinant ends the proof. $\square$

**Combinatorial partition of the vertices of $G^Q$** Let us specify the partition of the black/white vertices of $G^Q$. It is constructed from a well chosen perfect matching $M_1$ of $G^{D,r}$, which we now define. Recall that by Temperley's bijection, perfect matchings of $G^{D,r}$ are in bijection with pairs of dual directed spanning trees of $\mathcal{T}^{r,\circ}(G, \bar{G}^*)$, see Section 2.3.1. Consider a spanning tree of the restricted dual $G^*$, and root it at the vertex $f^{c,r}$ incident to the root vertex $v^c = r$ in the diamond graph $G^\circ$. To this spanning tree, add the edge $(f^{c,r}, o)$ which is the dual of the edge $rv^{\ell,r}$ of $G$. This defines an o-directed spanning tree $T^*$ of $\bar{G}^*$. Consider the r-directed spanning tree $T$ of $G$ which is the dual of $T^*$, rooted at the vertex $r$. Then $(T, T^*)$ is a pair of dual spanning trees of $\mathcal{T}^{r,\circ}(G, \bar{G}^*)$, and we let $M_1$ be the corresponding perfect matching of $G^{D,r}$, see Figure 17.

We now define the partition of black/white vertices of $G^Q$ arising from $M_1$; this amounts to specifying the partition for vertices of inner quadrangle since those of boundary quadrangles define $B_1^\circ/W_1^\circ$. For $j \in \{1, 2\}$, $b_j/w_j$ denotes a black/white vertex of $B_j/W_j$. Consider an inner quadrangle of $G^Q$ corresponding to a white vertex $w$ of $G^{D,r}$. Then, exactly one edge of the quadrangle is crossed by an edge $wx$ of the perfect matching $M_1$. The partition is defined as follows: if $x = v \in V^r$, then $w_1$ is the white vertex on the right of the edge $(w, v)$, and $b_1$ is the black vertex on the left; if $x = f \in V^*$, then $w_1$ is the white vertex on the left of the edge $(w, f)$, and $b_1$ is the black vertex on the right. An example is provided in Figure 17 (right) and Figure 19.

Let us prove a combinatorial lemma. Note that because $u \in C''$, the fact that a coefficient of $T_1(u)$ or $R(u)$ is non-zero is independent of $u$. As in Section A.2 of Appendix A, to the matrix $T_1(u)$ corresponds a bipartite graph $G(T_1) = (W_1 \cup B^r, E(T_1))$, where there is an edge $w_1x$ iff $t(u)_{w_1,x} \neq 0$, with $w_1 \in W_1$, $x \in B^r = V^r \cup V^*$. In a similar way, to the matrix $R(u)$ of Lemma 4.6 corresponds a bipartite graph $G(R) =$
**Lemma 4.6.** Consider a partition of the black and white vertices of $G^Q$ as above. Then, for every $u \in C''$,

$$ \det \underline{K}^Q \cdot \det T_1(u) = \det S_1^\theta(u) \cdot \det S_1^\circ(u) \cdot \det S_2(u) \cdot \det R(u) \cdot \det \underline{K}^\theta(u), $$

$$ \text{where } R(u) := S_2(u)^{-1} \underline{K}_{22}^Q - S_1^\circ(u)^{-1} \underline{K}_{1\circ 2}^Q. $$

**Proof.** Note that $S_2(u)$ and $S_1^\circ(u)$ are invertible because we choose $u \in C''$. By Theorem 4.2, we have the following identity:

$$ \begin{array}{c} \begin{array}{c} \underline{K}_{11}^Q \\ \underline{K}_{21}^Q \end{array} \\ \underline{K}_{12}^Q \\ \underline{K}_{22}^Q \end{array} \begin{array}{c} \underline{K}_{12}^Q \\ \underline{K}_{22}^Q \end{array} \begin{array}{c} W_1^\theta \\ W_1^\circ \\ W_2^\circ \end{array} \begin{array}{c} W_2^\theta \\ W_2^\circ \\ W_2^\circ \end{array} \begin{array}{c} T_1(u) \\ T_2(u) \\ I \end{array} = \begin{array}{c} W_1^\theta \\ W_1^\circ \\ W_2^\circ \end{array} \begin{array}{c} T_1(u) \\ T_2(u) \\ I \end{array} $$

$$ = \begin{array}{c} \begin{array}{c} \underline{K}_{11}^\theta \\ \underline{K}_{12}^\theta \end{array} \begin{array}{c} S_1^\theta(u) \\ S_1^\circ(u) \end{array} \begin{array}{c} 0 \\ S_2(u) \end{array} \begin{array}{c} W_1^\theta \\ W_1^\circ \end{array} \begin{array}{c} 0 \\ S_2(u) \end{array} \begin{array}{c} W_2^\theta \\ W_2^\circ \end{array} \begin{array}{c} \underline{K}^\theta(u) \\ \underline{R}(u) \end{array} \\ \underline{K}_{12}^\theta \\ \underline{K}_{22}^\theta \end{array} \begin{array}{c} W_1^\theta \\ W_1^\circ \\ W_2^\circ \end{array} \begin{array}{c} 0 \\ S_2(u) \\ S_2(u) \end{array} \begin{array}{c} W_2^\theta \\ W_2^\circ \end{array} \begin{array}{c} \underline{R}(u) \\ \underline{R}(u) \end{array} $$

$$ \text{with} \begin{cases} \underline{K}_{1\circ 2}^Q = S_1^\circ(u) \tilde{R}(u) \underline{W}_2^\circ \\ \underline{K}_{22}^Q = S_2(u) \tilde{R}(u) \underline{W}_2^\circ + S_2(u) R(u). \end{cases} $$

We extract $\tilde{R}(u)\underline{W}_2^\circ$ from the first equation. Plugging $\tilde{R}(u)\underline{W}_2^\circ$ in the second equation gives $R(u)$; taking the determinant ends the proof. $\square$

**Combinatorial partition of the vertices of $G^Q$** Let us specify the partition of the black/white vertices of $G^Q$. It is constructed from a well chosen perfect matching $M_1$ of $G^{D,r}$, which we now define. Recall that by Temperley's bijection, perfect matchings of $G^{D,r}$ are in bijection with pairs of dual directed spanning trees of $\mathcal{T}^{r,\circ}(G, \bar{G}^*)$, see Section 2.3.1. Consider a spanning tree of the restricted dual $G^*$, and root it at the vertex $f^{c,r}$ incident to the root vertex $v^c = r$ in the diamond graph $G^\circ$. To this spanning tree, add the edge $(f^{c,r}, o)$ which is the dual of the edge $rv^{\ell,r}$ of $G$. This defines an o-directed spanning tree $T^*$ of $\bar{G}^*$. Consider the r-directed spanning tree $T$ of $G$ which is the dual of $T^*$, rooted at the vertex $r$. Then $(T, T^*)$ is a pair of dual spanning trees of $\mathcal{T}^{r,\circ}(G, \bar{G}^*)$, and we let $M_1$ be the corresponding perfect matching of $G^{D,r}$, see Figure 17.

We now define the partition of black/white vertices of $G^Q$ arising from $M_1$; this amounts to specifying the partition for vertices of inner quadrangle since those of boundary quadrangles define $B_1^\theta/W_1^\theta$. For $j \in \{1, 2\}$, $b_j/w_j$ denotes a black/white vertex of $B_j/W_j$. Consider an inner quadrangle of $G^Q$ corresponding to a white vertex $w$ of $G^{D,r}$. Then, exactly one edge of the quadrangle is crossed by an edge $wx$ of the perfect matching $M_1$. The partition is defined as follows: if $x = v \in V^r$, then $w_1$ is the white vertex on the right of the edge $(w, v)$, and $b_1$ is the black vertex on the left; if $x = f \in V^*$, then $w_1$ is the white vertex on the left of the edge $(w, f)$, and $b_1$ is the black vertex on the right. An example is provided in Figure 17 (right) and Figure 19.

Let us prove a combinatorial lemma. Note that because $u \in C''$, the fact that a coefficient of $T_1(u)$ or $R(u)$ is non-zero is independent of $u$. As in Section A.2 of Appendix A, to the matrix $T_1(u)$ corresponds a bipartite graph $G(T_1) = (W_1 \cup B^r, E(T_1))$, where there is an edge $w_1x$ iff $t(u)_{w_1,x} \neq 0$, with $w_1 \in W_1$, $x \in B^r = V^r \cup V^*$. In a similar way, to the matrix $R(u)$ of Lemma 4.6 corresponds a bipartite graph $G(R) =$
<figure><img src="image_19.png" /><figcaption>Figure 17: Left: a pair (T, T*) of dual directed spanning trees of $\mathcal{T}^{r,0}(G, \tilde{G}^*)$ such that T* arises from a spanning tree of the restricted dual G*. Right: corresponding dimer configuration $M_1$ of $G^{D,r}$ (red); associated partition of black/white vertices of $G^Q$: vertices of $B_1$, $W_1$ are in green, those of $B_2$, $W_2$ are in blue.</figcaption></figure>

$(W^\circ \cup W_2, E(R))$ where there is an edge $ww_2$ iff $r(u)_{w,w_2} \neq 0$, with $w \in W^\circ$, $w_2 \in W_2$. The matrix $T_1(u)$, resp. $R(u)$, is then a bipartite, weighted adjacency matrix of the graph $G(T_1)$, resp. $G(R)$.

**Lemma 4.7.** Consider a partition of the black/white vertices of $G^Q$ arising from the perfect matching $M_1$ of $G^{D,r}$. Then, the graph $G(T_1)$ is a spanning tree on the vertex set $W_1 \cup B^r$, and the graph $G(R)$ is a union of trees on the vertex set $W^\circ \cup W_2$, spanning all vertices of $W^\circ \cup W_2$.

*Proof.* Let us prove that $G(T_1)$ is a spanning tree. Every vertex $w_1$ of $W_1$ has degree 2 in $G(T_1)$: it is adjacent to a vertex $x$ of $B^r$ such that $wx$ is an edge of the perfect matching $M_1$, and to a vertex $x'$ such that $xx'$ is an edge of the diamond graph $G^\circ$ with $w_1$ in its middle. Using the natural bijection $W_1 \leftrightarrow W$, that is, identifying every vertex $w_1$ with the corresponding vertex $w$ of $W^r$ does not change the combinatorics of the graph $G(T_1)$. With this identification, the graph $G(T_1)$ contains all edges of the perfect matching $M_1$, and the second edge $wx'$ incident to $w$ is such that $(w, x')$ is on the right, resp. left, of $(x, w)$ if $x \in V^*$, resp. $x \in V^r$. Then, by Proposition 7.3. of [30], the graph $G(T_1)$ is a spanning tree. An example of $G(T_1)$ is pictured with red edges in Figure 18.

Let us prove that $G(R)$ is a union of trees spanning all vertices of $W^\circ \cup W_2$. By definition, a white vertex $w$ of $W^\circ$ is adjacent to all white vertices of $W_2$ that are adjacent to the black vertices $b_1$ and $b_2$ of the quadrangle $w$ corresponds to; that is, $w$ is adjacent to the vertex $w_2$ of the quadrangle of $w$ and maybe to another vertex of $W_2$. As a consequence $G(R)$ is spanning all vertices of $W^\circ \cup W_2$. It cannot contain a cycle for otherwise it would mean that there is a vertex of $B^r$ which does not belong to $M_1$, which is a contradiction with it being a perfect matching. An example of $G(R)$ is pictured with blue edges in Figure 18. $\square$

**Corollary 4.8.** Consider a partition of the black/white vertices of $G^Q$ arising from the perfect matching $M_1$ of $G^{D,r}$.

* Using the identification $W_1 \leftrightarrow W^r = W$, we have,

$$|\det T_1(u)| = \prod_{wx \in M_1: w \in W} |t(u)_{w_1,x}| = \prod_{wx \in M: w \in W^\circ} |t(u)_{w_1,x}| \prod_{wx \in M: w \in W^\circ} |t(u)_{w_1,x}|$$
<figure><img src="image_19.png" /><figcaption>Figure 17: Left: a pair (T, T*) of dual directed spanning trees of $\mathcal{T}^{r,0}(G, \tilde{G}^*)$ such that $T^*$ arises from a spanning tree of the restricted dual $G^*$. Right: corresponding dimer configuration $M_1$ of $G^{D,r}$ (red); associated partition of black/white vertices of $G^Q$: vertices of $B_1, W_1$ are in green, those of $B_2, W_2$ are in blue.</figcaption></figure>

$(W^\circ \cup W_2, E(R))$ where there is an edge $ww_2$ iff $r(u)_{w,w_2} \neq 0$, with $w \in W^\circ$, $w_2 \in W_2$. The matrix $T_1(u)$, resp. $R(u)$, is then a bipartite, weighted adjacency matrix of the graph $G(T_1)$, resp. $G(R)$.

**Lemma 4.7.** Consider a partition of the black/white vertices of $G^Q$ arising from the perfect matching $M_1$ of $G^{D,r}$. Then, the graph $G(T_1)$ is a spanning tree on the vertex set $W_1 \cup B^r$, and the graph $G(R)$ is a union of trees on the vertex set $W^\circ \cup W_2$, spanning all vertices of $W^\circ \cup W_2$.

*Proof.* Let us prove that $G(T_1)$ is a spanning tree. Every vertex $w_1$ of $W_1$ has degree 2 in $G(T_1)$: it is adjacent to a vertex $x$ of $B^r$ such that $wx$ is an edge of the perfect matching $M_1$, and to a vertex $x'$ such that $xx'$ is an edge of the diamond graph $G^\circ$ with $w_1$ in its middle. Using the natural bijection $W_1 \leftrightarrow W$, that is, identifying every vertex $w_1$ with the corresponding vertex $w$ of $W^r$ does not change the combinatorics of the graph $G(T_1)$. With this identification, the graph $G(T_1)$ contains all edges of the perfect matching $M_1$, and the second edge $wx'$ incident to $w$ is such that $(w, x')$ is on the right, resp. left, of $(x, w)$ if $x \in V^*$, resp. $x \in V^r$. Then, by Proposition 7.3. of [30], the graph $G(T_1)$ is a spanning tree. An example of $G(T_1)$ is pictured with red edges in Figure 18.

Let us prove that $G(R)$ is a union of trees spanning all vertices of $W^\circ \cup W_2$. By definition, a white vertex $w$ of $W^\circ$ is adjacent to all white vertices of $W_2$ that are adjacent to the black vertices $b_1$ and $b_2$ of the quadrangle $w$ corresponds to; that is, $w$ is adjacent to the vertex $w_2$ of the quadrangle of $w$ and maybe to another vertex of $W_2$. As a consequence $G(R)$ is spanning all vertices of $W^\circ \cup W_2$. It cannot contain a cycle for otherwise it would mean that there is a vertex of $B^r$ which does not belong to $M_1$, which is a contradiction with it being a perfect matching. An example of $G(R)$ is pictured with blue edges in Figure 18. $\square$

**Corollary 4.8.** Consider a partition of the black/white vertices of $G^Q$ arising from the perfect matching $M_1$ of $G^{D,r}$.

* Using the identification $W_1 \leftrightarrow W^r = W$, we have,

$$|\det T_1(u)| = \prod_{wx \in M_1: w \in W} |t(u)_{w_1,x}| = \prod_{wx \in M: w \in W^\circ} |t(u)_{w_1,x}| \prod_{wx \in M: w \in W^\circ} |t(u)_{w_1,x}|.$$
<figure><img src="image_20.png" /><figcaption>Figure 18: Example of graph G(T₁) (red edges) and of graph G(R) (blue edges).</figcaption></figure>

• Using the identification $W_2 \leftrightarrow W^\circ$, we have,

$$|\det R(u)| = \prod_{w \in W^\circ} |r(u)_{w, w_2}|,$$

where for every $w \in W^\circ$, $r(u)_{w, w_2} = s(u)_{b_2, w}^{-1} \underline{K}_{b_2, w_2}^Q - s(u)_{b_1, w}^{-1} \underline{K}_{b_1, w_2}^Q$.

**Proof.** Writing the determinant as a sum over permutations, we have that non-zero terms in the expansion of $\det T_1(u)$, resp. $\det R(u)$, correspond to perfect matchings of the bipartite graph $G(T_1)$, resp. $G(R)$. Since these two graphs are trees or union of trees spanning all vertices, they have at most one perfect matching; indeed if they had more, the union of two different ones would yield a cycle which is in contradiction with being a tree. Using the identification $W_1 \leftrightarrow W$, The graph $G(T_1)$ has one perfect matching given by edges of $M_1$ (pictured in thick red lines in Figure 18), while the graph $G(R)$ has one perfect matching given by the natural identification $W^\circ \leftrightarrow W_2$ (pictured in thick blue lines in Figure 18). Since, the contribution of a perfect matching to the determinant is the product of the edge-weights (up to a sign), this ends the proof of the corollary. □

Since $S_1^\beta(u), S_1^\alpha(u), S_2(u)$ are diagonal matrices, their determinant is the product of the diagonal terms. Combining Lemma 4.6 and Corollary 4.8 we thus obtain the following.

**Corollary 4.9.** Consider a partition of the black/white vertices of $G^Q$ arising from the perfect matching $M_1$ of $G^{D,r}$. Then, for every $u \in C''$,

$$|\det \underline{K}^Q| = \underbrace{\left| \frac{\prod_{w \in W^\circ} s(u)_{b_1, w}}{\prod_{wx \in M_1: w \in W^\circ} t(u)_{w_1, x}} \right|}_{(I)} \underbrace{\left| \frac{\prod_{w \in W^\circ} r'(u)_{w, w_2}}{\prod_{wx \in M_1: w \in W^\circ} t(u)_{w_1, x}} \right|}_{(II)} |\det \underline{K}^\beta(u)|, \quad (4.10)$$

where $r'(u)_{w, w_2} = s(u)_{b_1, w} \underline{K}_{b_2, w_2}^Q - s(u)_{b_2, w} \underline{K}_{b_1, w_2}^Q$.
<figure><img src="image_20.png" /><figcaption>Figure 18: Example of graph G(T₁) (red edges) and of graph G(R) (blue edges).</figcaption></figure>

• Using the identification $W_2 \leftrightarrow W^\circ$, we have,

$$|\det R(u)| = \prod_{w \in W^\circ} |r(u)_{w,w_2}|,$$

where for every $w \in W^\circ$, $r(u)_{w,w_2} = s(u)_{b_2,w}^{-1} \underline{K}_{b_2,w_2}^Q - s(u)_{b_1,w}^{-1} \underline{K}_{b_1,w_2}^Q$.

**Proof.** Writing the determinant as a sum over permutations, we have that non-zero terms in the expansion of $\det T_1(u)$, resp. $\det R(u)$, correspond to perfect matchings of the bipartite graph $G(T_1)$, resp. $G(R)$. Since these two graphs are trees or union of trees spanning all vertices, they have at most one perfect matching; indeed if they had more, the union of two different ones would yield a cycle which is in contradiction with being a tree. Using the identification $W_1 \leftrightarrow W$, The graph $G(T_1)$ has one perfect matching given by edges of $M_1$ (pictured in thick red lines in Figure 18), while the graph $G(R)$ has one perfect matching given by the natural identification $W^\circ \leftrightarrow W_2$ (pictured in thick blue lines in Figure 18). Since, the contribution of a perfect matching to the determinant is the product of the edge-weights (up to a sign), this ends the proof of the corollary. □

Since $S_1^\beta(u), S_1^\alpha(u), S_2(u)$ are diagonal matrices, their determinant is the product of the diagonal terms. Combining Lemma 4.6 and Corollary 4.8 we thus obtain the following.

**Corollary 4.9.** Consider a partition of the black/white vertices of $G^Q$ arising from the perfect matching $M_1$ of $G^{D,r}$. Then, for every $u \in C''$,

$$|\det \underline{K}^Q| = \underbrace{\left| \frac{\prod_{w \in W^\circ} s(u)_{b_1,w}}{\prod_{wx \in M_1: w \in W^\circ} t(u)_{w_1,x}} \right|}_{(I)} \underbrace{\left| \frac{\prod_{w \in W^\circ} r'(u)_{w,w_2}}{\prod_{wx \in M_1: w \in W^\circ} t(u)_{w_1,x}} \right|}_{(II)} |\det \underline{K}^\beta(u)|, \quad (4.10)$$

where $r'(u)_{w,w_2} = s(u)_{b_1,w} \underline{K}_{b_2,w_2}^Q - s(u)_{b_2,w} \underline{K}_{b_1,w_2}^Q$.
**Computation of (I)(II) in Identity (4.10)** For every $u \in C''$, define the weight function $\eta(u)$ on edges of $G^{D,r}$ as follows.

$$ \eta(u)_{wx} = \begin{cases} \left| \frac{\mathrm{sn}(u_{\alpha_e})}{\mathrm{cn}(u_{\beta_e})} \right| [\mathrm{nd}(u_{\alpha_e}) \mathrm{dn}(u_{\beta_e})]^{\frac{1}{2}} & \text{if } x = v \\ (k')^{\frac{1}{2}} \left| \frac{\mathrm{sn}(u_{\beta_e})}{\mathrm{cn}(u_{\alpha_e})} \right| [\mathrm{dn}(u_{\alpha_e}) \mathrm{nd}(u_{\beta_e})]^{\frac{1}{2}} & \text{if } x = f. \end{cases} \quad (4.11) $$

**Lemma 4.10.** The product (I)(II) is equal to:

$$ \mathrm{sn}(\theta^{\beta,r}) \left| \frac{\mathrm{cd}(u_{\beta,r})}{\mathrm{sn}(u_{\alpha,r})} \right| \left( \prod_{w \in W^{\beta}} \mathrm{sn}(\theta_w)^{-\frac{1}{2}} \right) \left( \prod_{w \in W} |\mathrm{sn}(\theta_w) \mathrm{cn}(\theta_w)|^{\frac{1}{2}} \right) \prod_{wx \in M_1: w \in W} \eta(u)_{wx}. $$

*Proof.* In the whole of the proof, we simply denote $\bar{\theta}_w$ by $\bar{\theta}$, and omit the argument $u$ from matrix coefficients.

We first handle Part (II) involving inner vertices. Let $w \in W^\circ$ and $wx$ be an edge of the perfect matching $M_1$, with $x = v$ or $f$. By definition of the partition of black/white vertices arising from $M_1$, we have $b_1, b_2, w_1, w_2$ as in Figure 19. Let $v'$ be the primal vertex such that the edge $v'v$ crosses the quadrangle, and let $2e^{i\bar{\alpha}}$, $2e^{i\bar{\beta}}$ be the two rhombus vectors of $G^\circ$ associated to the edge $(v', v)$.

<figure><img src="image_21.png" /><figcaption>Figure 19: Notation for computing $\frac{r'_{w,w_2}}{t_{w_1,x}}$, when $w \in W^\circ$ and $x = v$ (left), $x = f$ (right).</figcaption></figure>

We compute the term $r'_{w,w_2}$ when $x = v$. Using the notation of Figure 19, the rhombus vectors of $G^{\circ/2}$ assigned to the edge $(w, v)$ are $e^{i\bar{\alpha}_e} = e^{i\bar{\alpha}}$, $e^{i\bar{\beta}_e} = e^{i\bar{\beta}}$. Returning to the definition of the matrices $K^Q$ and $S$, see (4.2), we have

$$ s_{b_1,w} \frac{K^Q_{b_2,w_2}}{K^Q_{b_1,w_1}} = e^{-i \frac{\bar{\beta}+\pi}{2}} \mathrm{cn}(u_{\beta+2K}) [\mathrm{sn}(\theta) \mathrm{cn}(\theta) \mathrm{nd}(u_{\alpha+2K}) \mathrm{nd}(u_{\beta+2K})]^{\frac{1}{2}} \times e^{i \frac{\bar{\beta}+\bar{\alpha}+\pi}{2}} \mathrm{cn}(\theta) \\ = e^{i \frac{\bar{\alpha}}{2}} \mathrm{cn}(\theta) \mathrm{sn}(u_{\beta}) \mathrm{dn}(u_{\alpha}) [\mathrm{sn}(\theta) \mathrm{cn}(\theta) \mathrm{nd}(u_{\alpha}) \mathrm{nd}(u_{\beta})]^{\frac{1}{2}} $$

$$ s_{b_2,w} \frac{K^Q_{b_1,w_2}}{K^Q_{b_2,w_1}} = e^{-i \frac{\bar{\beta}}{2}} \mathrm{cn}(u_{\beta}) [\mathrm{sn}(\theta) \mathrm{cn}(\theta) \mathrm{nd}(u_{\alpha}) \mathrm{nd}(u_{\beta})]^{\frac{1}{2}} \times e^{i \frac{\bar{\alpha}+\pi+\bar{\beta}+\pi}{2}} \mathrm{sn}(\theta) \\ = -e^{i \frac{\bar{\alpha}}{2}} \mathrm{sn}(\theta) \mathrm{cn}(u_{\beta}) [\mathrm{sn}(\theta) \mathrm{cn}(\theta) \mathrm{nd}(u_{\alpha}) \mathrm{nd}(u_{\beta})]^{\frac{1}{2}}. $$

As a consequence,

$$ \begin{aligned} r'_{w,w_2} &= e^{i \frac{\bar{\alpha}}{2}} [\mathrm{sn}(\theta) \mathrm{cn}(\theta) \mathrm{nd}(u_{\alpha}) \mathrm{nd}(u_{\beta})]^{\frac{1}{2}} (\mathrm{cn}(\theta) \mathrm{sn}(u_{\beta}) \mathrm{dn}(u_{\alpha}) + \mathrm{sn}(\theta) \mathrm{cn}(u_{\beta})) \\ &= e^{i \frac{\bar{\alpha}}{2}} [\mathrm{sn}(\theta) \mathrm{cn}(\theta) \mathrm{nd}(u_{\alpha}) \mathrm{nd}(u_{\beta})]^{\frac{1}{2}} \mathrm{dn}(u_{\beta}) \mathrm{sn}(u_{\alpha}) \\ &= e^{i \frac{\bar{\alpha}}{2}} \mathrm{sn}(u_{\alpha}) [\mathrm{sn}(\theta) \mathrm{cn}(\theta) \mathrm{nd}(u_{\alpha}) \mathrm{dn}(u_{\beta})]^{\frac{1}{2}}, \end{aligned} $$

using the identity $\mathrm{dn}(u) \mathrm{sn}(u+v) = \mathrm{cn}(u) \mathrm{sn}(v) + \mathrm{sn}(u) \mathrm{cn}(v) \mathrm{dn}(u+v)$ [54, chap.2, ex.32 (ii)], evaluated at $u = u_{\beta}, v = \theta, u+v = u_{\alpha}$, in the penultimate line. Since $e^{i\bar{\alpha}_e} = e^{i\bar{\alpha}}$, $e^{i\bar{\beta}_e} = e^{i\bar{\beta}}$, we have

$$ r'_{w,w_2} = e^{i \frac{\bar{\alpha}_e}{2}} \mathrm{sn}(u_{\alpha_e}) [\mathrm{sn}(\theta) \mathrm{cn}(\theta) \mathrm{nd}(u_{\alpha_e}) \mathrm{dn}(u_{\beta_e})]^{\frac{1}{2}}. \quad (4.12) $$
**Computation of (I)(II) in Identity (4.10)** For every $u \in C''$, define the weight function $\eta(u)$ on edges of $G^{D,r}$ as follows.

$$ \eta(u)_{wx} = \begin{cases} \left| \frac{\mathrm{sn}(u_{\alpha_e})}{\mathrm{cn}(u_{\beta_e})} \right| [\mathrm{nd}(u_{\alpha_e}) \mathrm{dn}(u_{\beta_e})]^{\frac{1}{2}} & \text{if } x = v \\ (k')^{\frac{1}{2}} \left| \frac{\mathrm{sn}(u_{\beta_e})}{\mathrm{cn}(u_{\alpha_e})} \right| [\mathrm{dn}(u_{\alpha_e}) \mathrm{nd}(u_{\beta_e})]^{\frac{1}{2}} & \text{if } x = f. \end{cases} \quad (4.11) $$

**Lemma 4.10.** The product (I)(II) is equal to:

$$ \mathrm{sn}(\theta^{\beta,r}) \left| \frac{\mathrm{cd}(u_{\beta,r})}{\mathrm{sn}(u_{\alpha,r})} \right| \left( \prod_{w \in W^{\beta}} \mathrm{sn}(\theta_w)^{-\frac{1}{2}} \right) \left( \prod_{w \in W} |\mathrm{sn}(\theta_w) \mathrm{cn}(\theta_w)|^{\frac{1}{2}} \right) \prod_{wx \in M_1: w \in W} \eta(u)_{wx}. $$

*Proof.* In the whole of the proof, we simply denote $\bar{\theta}_w$ by $\bar{\theta}$, and omit the argument $u$ from matrix coefficients.

We first handle Part (II) involving inner vertices. Let $w \in W^\circ$ and $wx$ be an edge of the perfect matching $M_1$, with $x = v$ or $f$. By definition of the partition of black/white vertices arising from $M_1$, we have $b_1, b_2, w_1, w_2$ as in Figure 19. Let $v'$ be the primal vertex such that the edge $v'v$ crosses the quadrangle, and let $2e^{i\bar{\alpha}}$, $2e^{i\bar{\beta}}$ be the two rhombus vectors of $G^\circ$ associated to the edge $(v', v)$.

<figure><img src="image_21.png" /><figcaption>Figure 19: Notation for computing $\frac{r'_{w,w_2}}{t_{w_1,x}}$, when $w \in W^\circ$ and $x = v$ (left), $x = f$ (right).</figcaption></figure>

We compute the term $r'_{w,w_2}$ when $x = v$. Using the notation of Figure 19, the rhombus vectors of $G^{\circ/2}$ assigned to the edge $(w, v)$ are $e^{i\bar{\alpha}_e} = e^{i\bar{\alpha}}$, $e^{i\bar{\beta}_e} = e^{i\bar{\beta}}$. Returning to the definition of the matrices $K^Q$ and $S$, see (4.2), we have

$$ s_{b_1,w} \frac{K^Q_{b_2,w_2}}{K^Q_{b_1,w_1}} = e^{-i \frac{\bar{\beta}+\pi}{2}} \mathrm{cn}(u_{\beta+2K}) [\mathrm{sn}(\theta) \mathrm{cn}(\theta) \mathrm{nd}(u_{\alpha+2K}) \mathrm{nd}(u_{\beta+2K})]^{\frac{1}{2}} \times e^{i \frac{\bar{\beta}+\bar{\alpha}+\pi}{2}} \mathrm{cn}(\theta) \\ = e^{i \frac{\bar{\alpha}}{2}} \mathrm{cn}(\theta) \mathrm{sn}(u_{\beta}) \mathrm{dn}(u_{\alpha}) [\mathrm{sn}(\theta) \mathrm{cn}(\theta) \mathrm{nd}(u_{\alpha}) \mathrm{nd}(u_{\beta})]^{\frac{1}{2}} $$

$$ s_{b_2,w} \frac{K^Q_{b_1,w_2}}{K^Q_{b_2,w_1}} = e^{-i \frac{\bar{\beta}}{2}} \mathrm{cn}(u_{\beta}) [\mathrm{sn}(\theta) \mathrm{cn}(\theta) \mathrm{nd}(u_{\alpha}) \mathrm{nd}(u_{\beta})]^{\frac{1}{2}} \times e^{i \frac{\bar{\alpha}+\pi+\bar{\beta}+\pi}{2}} \mathrm{sn}(\theta) \\ = -e^{i \frac{\bar{\alpha}}{2}} \mathrm{sn}(\theta) \mathrm{cn}(u_{\beta}) [\mathrm{sn}(\theta) \mathrm{cn}(\theta) \mathrm{nd}(u_{\alpha}) \mathrm{nd}(u_{\beta})]^{\frac{1}{2}}. $$

As a consequence,

$$ \begin{aligned} r'_{w,w_2} &= e^{i \frac{\bar{\alpha}}{2}} [\mathrm{sn}(\theta) \mathrm{cn}(\theta) \mathrm{nd}(u_{\alpha}) \mathrm{nd}(u_{\beta})]^{\frac{1}{2}} (\mathrm{cn}(\theta) \mathrm{sn}(u_{\beta}) \mathrm{dn}(u_{\alpha}) + \mathrm{sn}(\theta) \mathrm{cn}(u_{\beta})) \\ &= e^{i \frac{\bar{\alpha}}{2}} [\mathrm{sn}(\theta) \mathrm{cn}(\theta) \mathrm{nd}(u_{\alpha}) \mathrm{nd}(u_{\beta})]^{\frac{1}{2}} \mathrm{dn}(u_{\beta}) \mathrm{sn}(u_{\alpha}) \\ &= e^{i \frac{\bar{\alpha}}{2}} \mathrm{sn}(u_{\alpha}) [\mathrm{sn}(\theta) \mathrm{cn}(\theta) \mathrm{nd}(u_{\alpha}) \mathrm{dn}(u_{\beta})]^{\frac{1}{2}}, \end{aligned} $$

using the identity $\mathrm{dn}(u) \mathrm{sn}(u+v) = \mathrm{cn}(u) \mathrm{sn}(v) + \mathrm{sn}(u) \mathrm{cn}(v) \mathrm{dn}(u+v)$ [54, chap.2, ex.32 (ii)], evaluated at $u = u_{\beta}, v = \theta, u+v = u_{\alpha}$, in the penultimate line. Since $e^{i\bar{\alpha}_e} = e^{i\bar{\alpha}}$, $e^{i\bar{\beta}_e} = e^{i\bar{\beta}}$, we have

$$ r'_{w,w_2} = e^{i \frac{\bar{\alpha}_e}{2}} \mathrm{sn}(u_{\alpha_e}) [\mathrm{sn}(\theta) \mathrm{cn}(\theta) \mathrm{nd}(u_{\alpha_e}) \mathrm{dn}(u_{\beta_e})]^{\frac{1}{2}}. \quad (4.12) $$