# Orthodiagonal Maps, Tilings of Rectangles, and their Convergence to
Conformal Maps

Ilia Binder and David Pechersky

###### Abstract

A classic result of Brooks, Smith, Stone and Tutte [6] associates to any
finite planar network with distinguished source and sink vertices, a tiling of
a rectangle by smaller subrectangles whose aspect ratios are given by the
conductances of corresponding edges in the network. This tiling can be viewed
as a discrete analogue of the uniformizing conformal map that maps a simply
connected domain with four distinguished prime ends to a rectangle, so that
the four prime ends are mapped to the four corners of the rectangle.

We make this intuition precise by showing that if $\Omega$ is a simply
connected domain with four distinguished prime ends $A,B,C,D$ in
counterclockwise order and $(\Omega_{n})_{n\geq{1}}$ is a sequence of
orthodiagonal maps with distinguished boundary vertices
$A_{n},B_{n},C_{n},D_{n}$ in counterclockwise order, that are finer and finer
approximations of $\Omega$ with its distinguished boundary points $A,B,C,D$,
then the corresponding “rectangle tiling maps" converge uniformly on compacts
to the aforementioned conformal map on $\Omega$.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
    1. 2.1 The Theory of Electrical Networks
    2. 2.2 Extremal Length and Planar Networks
    3. 2.3 Tilings of Rectangles
    4. 2.4 Orthodiagonal Approximations of Planar Domains
  3. 3 Precompactness of the Tiling Maps
    1. 3.1 Modulus of Continuity for the Limiting Conformal Map
    2. 3.2 Modulus of Continuity for Tiling Maps
    3. 3.3 Two- Sided Estimates for Extremal Length on Orthodiagonal Maps
  4. 4 Limits of Discrete Holomorphic Functions are Holomorphic
  5. 5 Convergence of Tilings to Conformal Maps
    1. 5.1 Proof of Theorem 2.15
  6. A Appendix
    1. A.1 Elementary Results in Topology and Complex Analysis

##  1 Introduction

In the past two decades, discrete complex analysis has proven to be a powerful
tool for solving problems in critical 2D statistical physics. For instance, it
was used to prove Cardy’s formula for crossing probabilities in critical
percolation [26], the convergence of interfaces in the critical Ising model to
$SLE_{3}$ [8], and the convergence of percolation interfaces to $SLE_{6}$ [7].
Here $SLE_{\kappa}$ refers to Schramm Loewner evolution, a $1$\- parameter
family of random curves parametrized by $\kappa>0$ that is characterized by
conformal invariance and a certain Markov property. All of these arguments go
as follows:

  1. 1.

You start off with a statistical physics model defined on a planar lattice,
embedded in the plane.

  2. 2.

You identify a discrete holomorphic observable in this model. That is, you
find some functional of your statistical physics model that satisfies a
discrete version of the Cauchy- Riemann equations.

  3. 3.

A priori regularity estimates for discrete holomorphic functions are used to
show that as the lattice mesh tends to $0$, discrete holomorphic observables
are precompact with respect to the topology of local uniform convergence. This
gives us subsequential limits. Discrete holomorphicity of the observable gives
us holomorphicity for the limiting function.

  4. 4.

Since the limiting observable is holomorphic, we can uniquely identify it by
its boundary behavior. Since all subsequential limits agree, we conclude that
our discrete holomorphic observables converge to the relevant continuum limit.

  5. 5.

Convergence of the observable is then leveraged to prove convergence of the
discrete random objects of interest to their continuum limits (i.e.
percolation interfaces to $SLE_{6}$, interfaces in the Ising model to
$SLE_{3}$, etc.).

A recurring theme in thermodynamics is the notion of universality.
Universality is a phenomenon where, due to some overarching principle,
physical phenomenon with radically different microscopic details (i.e.
magnets, porous media, growing polymer chains), give rise to the same
macroscopic behavior when you zoom out (i.e. $SLE$ curves). In the context of
$2D$ statistical physics, universality suggests that for the same model, it
shouldn’t matter how we chop up space- i.e. it doesn’t matter if we define our
model on a triangular or a square lattice- the resulting large- scale behavior
should be the same.

Orthodiagonal maps [17] are a very general class of discretizations of $2D$
space that accommodate a notion of discrete complex analysis. Thus, at least
in principle, the same techniques that were used previously to prove
convergence of observables and interfaces in $2D$ critical statistical physics
models for isoradial and more restrictive lattices, could be generalized to
demonstrate the same convergence when our statistical physics happens on an
orthodiagonal map. At present, doing this for any model is an open problem.

In this work, we take a step in this direction by using discrete complex
analysis to solve a purely deterministic problem in this very general setting.
A classic paper of Brooks, Smith, Stone and Tutte describes how planar
electrical networks give rise to tilings of rectangles by smaller
subrectangles [6]. Each subrectangle in the tiling corresponds to an edge of
the network and its aspect ratio is precisely the conductance of this
corresponding edge. These tilings can be thought of as discrete analogues of
the uniformizing conformal map that maps a simply connected domain to a
rectangle so that four distinguished points on the boundary of our simply
connected domain are mapped to the four corners of the rectangle. We make this
idea rigorous by showing that for any simply connected domain, if we have an
increasingly fine sequence of orthodiagonal approximations, the associated
tilings converge to the corresponding uniformizing conformal map. This
significantly improves on a previous result of Georgakopoulos and Panagiotis
who prove this convergence in the case where the approximating orthodiagonal
map is just a chunk of the square grid [16]. Furthermore, our approach is
significantly different from the one in [16] which relies heavily on the fact
that reflected random walks on $\delta\mathbb{Z}^{2}$ converge in law to
reflected Brownian motion as $\delta\rightarrow{0}$. To our knowledge, this
result is not known for any other lattices.

In recent work, Albin, Lind, and Poggi- Corradini provide an explicit rate of
convergence for these tilings to the limiting conformal map in the general
orthodiagonal setting (this is effectively Theorem 3 of [3]), subject to
certain assumptions on the smoothness of the boundary of the simply connected
domain that is being approximated. They then use this to prove convergence of
the probabilistic interpretation of modulus as well as convergence of discrete
extremal length to continuous extremal length in this setting (see Theorems 2
and 4 of [3]). By modifying their approach, we manage to avoid making any
assumptions about the smoothness of the boundary of our simply connected
domain, at the expense of providing an explicit rate of convergence.

Our result can also be interpreted as the rectangle tiling analogue of similar
results that are known for circle packings. The Koebe-Andreev-Thurston theorem
tells us that any finite triangulation can be realized as the tangency graph
of a circle packing in the plane. With this in mind, Bill Thurston made the
observation that if you fill a simply connected domain with circles packed
together, the Koebe-Andreev-Thurston theorem gives you a natural way to repack
these circles in the unit disk in a way that preserves tangency. Since this
“repacking map" sends circles to circles, if we fill our simply connected
domain with smaller and smaller circles, the corresponding circles in the
images should also get smaller and smaller. In the limit, these repacking maps
should converge to a function that sends infinitisimal circles to
infinitisimal circles. In other words, a conformal map. Thus, Thurston
conjectured that circle packings should give us a way to approximate the
uniformizing conformal map from a simply connected domain to the unit disk
[28]. This was proven by Rodin and Sullivan when the circle packings in the
simply connected domain consist of circles, all having the same radii, packed
together in a honeycomb pattern [25]. This was later generalized to circle
packings with arbitrary combinatorics by He and Schramm [18]. For more on
circle packings and their connection to complex analysis, see [27].

Finally, it is worth noting that the closely related tilings of cylinders have
been the object of recent study by Benjamini and Schramm [4], Georgakopoulos
[15], and Hutchcroft and Peres [19] in connection with the Poisson boundary of
infinite planar graphs.

## Acknowledgements

Many thanks to Dmitry Chelkak for discussing the results of [9] with us and
commenting on the state-of-the-art in discrete complex analysis more broadly.

##  2 Preliminaries

###  2.1 The Theory of Electrical Networks

Following [17], a finite network is a finite graph $G=(V,E)$ along with a
weight function $c:E\rightarrow{\mathbb{R}_{>0}}$. For any edge $e\in{E}$ we
say that $c(e)$ is the conductance of that edge. The reciprocal
$r(e)=\frac{1}{c(e)}$ is the resistance of that edge.

A function $\theta:\vec{E}\rightarrow{\mathbb{R}}$ is said to be antisymmetric
if $\theta(-\vec{e})=-\theta(\vec{e})$ for all $\vec{e}\in{\vec{E}}$.
Intuitively, antisymmetric functions on a network $G$ are the discrete
analogues of vector fields. Let $\ell_{-}^{2}(\vec{E})$ denote the space of
antisymmetric functions on $\vec{E}$ with the inner product:

$\langle\theta,\psi\rangle_{r}:=\frac{1}{2}\sum_{\vec{e}\in{\vec{E}}}r(e)\theta(\vec{e})\psi(\vec{e})$

The energy of $\theta\in{\ell_{-}^{2}(\vec{E})}$ is:

$\mathcal{E}(\theta)=\|\theta\|_{r}^{2}=\langle\theta,\theta\rangle_{r}$

Given $f:V\rightarrow{\mathbb{R}}$, its gradient
$cdf:\vec{E}\rightarrow{\mathbb{R}}$ is given by:

$(cdf)(\vec{e})=c(e)(f(e^{+})-f(e^{-}))$

for any directed edge $\vec{e}=(e^{-},e^{+})$. For any function
$f:V\rightarrow{\mathbb{R}}$, the gradient is antisymmetric. Thus, we can
define the energy of a function $f:V\rightarrow{\mathbb{R}}$ as the energy of
its gradient:

$\mathcal{E}(f)=\mathcal{E}(cdf)=\frac{1}{2}\sum_{\vec{e}\in{\vec{E}}}c(e)\big{(}f(e^{+})-f(e^{-})\big{)}^{2}$

Given a function in $\ell^{2}_{-}(\vec{E})$, we are often interested in the
energy of its restriction to some subgraph of $G$. To make it clear where it
is we are computing the energy, if $\theta\in{\ell_{-}^{2}(\vec{E})}$ and
$G^{\prime}=(V^{\prime},E^{\prime})$ is a subgraph of $G=(V,E)$ with the same
edge weights, then:

$\mathcal{E}(\theta;G^{\prime})=\frac{1}{2}\sum_{\vec{e}\in{\vec{E^{\prime}}}}r(e)\theta(\vec{e})^{2}$

Similarly, if $f:V\rightarrow{\mathbb{R}}$,

$\mathcal{E}(f;G^{\prime})=\frac{1}{2}\sum_{\vec{e}\in{\vec{E^{\prime}}}}c(e)\big{(}f(e^{+})-f(e^{-})\big{)}^{2}$

A function $\theta\in{\ell_{-}^{2}(\vec{E})}$ satisfies the cycle law if for
any directed cycle $\gamma=(\vec{e}_{1},\vec{e}_{2},...,\vec{e}_{m})$ in $G$,

$\sum_{i=1}^{m}r(e_{i})\theta(\vec{e}_{i})=0$

It is not hard to see that $\theta\in{\ell_{-}^{2}(\vec{E})}$ satisfies the
cycle law if and only if $\theta=cdf$ for some function
$f:V\rightarrow{\mathbb{R}}$. Given $\theta\in{\ell_{-}^{2}(\vec{E})}$, its
divergence $\text{div}(\theta):V\rightarrow{\mathbb{R}}$ is given by:

$\big{(}\text{div}(\theta)\big{)}(x)=\sum_{e^{-}=x}\theta(\vec{e})$

Similar to the continuous setting, the divergence of $\theta$ at $x$ measures
the net flow out of $x$ by $\theta$. Given distinct vertices $a,z\in{V}$, a
function $\theta\in{\ell_{-}^{2}(\vec{E})}$ is a flow from $a$ to $z$ if:

$\big{(}\text{div}(\theta)\big{)}(x)=0\hskip 5.0pt\text{for all
$x\in{V\setminus{\\{a,z\\}}}$.}$

Given a flow $\theta$ from $a$ to $z$ its strength, denoted by $\|\theta\|$,
is defined as follows:

$\|\theta\|=\sum_{x:x\sim{a}}\theta(a,x)=\big{(}\text{div}(\theta)\big{)}(a)$

For every flow $\theta$ from $a$ to $z$,

$\|\theta\|=\sum_{y:y\sim{z}}\theta(y,z)=-\big{(}\text{div}(\theta)\big{)}(z)$

This is because:

$\sum_{\vec{e}\in{\vec{E}}}\theta(\vec{e})=0=\sum_{x\in{V}}\sum_{y:y\sim{x}}\theta(x,y)=\sum_{x\in{V}}\big{(}\text{div}(\theta)\big{)}(x)=\big{(}\text{div}(\theta)\big{)}(a)+\big{(}\text{div}(\theta)\big{)}(z)$

The first equality follows from the antisymmetry of $\theta$. Given
$f:V\rightarrow{\mathbb{R}}$, its Laplacian
$\Delta{f}:V\rightarrow{\mathbb{R}}$ is given by:

$\Delta{f}(x)=\big{(}\text{div}(cdf)\big{)}(x)=\sum_{e^{-}=x}c(e)\big{(}f(e^{+})-f(e^{-})\big{)}=\sum_{y:y\sim{x}}c(x,y)\big{(}f(y)-f(x)\big{)}$

If $\Delta{f}(x)=0$, we say that $f$ is harmonic at $x$. Equivalently, $f$ is
harmonic at $x$ if:

$f(x)=\frac{1}{\pi_{x}}\sum_{y:y\sim{x}}c(x,y)f(y)$

where:

$\pi_{x}=\sum\limits_{y:y\sim{x}}c(x,y)$

From this formula, it is immediate that harmonic functions satisfy the maximum
principle:

######  Proposition 2.1.

Suppose that $G=(V,E,c)$ is a finite network and $h:V\rightarrow{\mathbb{R}}$
is harmonic on $U\subsetneq{V}$. Define:

$\partial{U}=\\{w\in{V}\setminus{U}:w\sim{u}\hskip 5.0pt\text{for some}\hskip
5.0ptu\in{U}\\}$

Then:

$\max_{u\in{U}}h(u)\leq{\max_{v\in{\partial{U}}}}h(v)$

Following [22], a simple random walk on the network $G=(V,E,c)$ is the
discrete time Markov process $(X_{n})_{n\geq{0}}$ with transition
probabilities:

$P(x,y)=\frac{c(x,y)}{\pi_{x}}1_{(x\sim{y})}$

Given a function $f:V\rightarrow{\mathbb{R}}$ and vertices $a,z\in{V}$, it is
clear that $cdf$ is a flow from $a$ to $z$ if and only if $\Delta{f}(x)=0$ for
all $x\in{V\setminus{\\{a,z\\}}}$. We call such a function a voltage. Since
the discrete boundary value problem:

$\displaystyle h(a)$ $\displaystyle=\alpha$ $\displaystyle h(b)$
$\displaystyle=\beta$ $\displaystyle\Delta{h}(x)$ $\displaystyle=0\hskip
5.0pt\text{for all $x\in{V\setminus{\\{a,z\\}}}$}$

has a unique solution for any choice of $\alpha,\beta\in{\mathbb{R}}$,
voltages form a two-parameter family. The flow $cdh$ corresponding to any
voltage $h:V\rightarrow{\mathbb{R}}$ is known as the corresponding current
flow. Given distinct vertices $a,z\in{V}$, the effective resistance between
$a$ and $z$ in $G$, denoted by $R_{\text{eff}}(a\leftrightarrow{z};G)$, is
given by:

$R_{\text{eff}}(a\leftrightarrow{z};G)=\frac{h(z)-h(a)}{\|cdh\|}$

where $h$ is any nonconstant voltage. To see that this quantity is well-
defined, just observe that adding a constant doesn’t affect the voltage
difference between $a$ and $z$, $h(z)-h(a)$, or the current flow $cdh$.
Similarly, multiplying $h$ by a nonzero constant scales the voltage difference
and the strength of the corresponding current flow by the same factor, leaving
the effective resistance unchanged.

More generally, given disjoint sets of vertices $A,Z\subseteq{V}$, we can
define a new network by identifying the vertices of $A$ to a single vertex $a$
and identifying the vertices of $Z$ to a single vertex $z$. Then the effective
resistance $R_{\text{eff}}(A\leftrightarrow{Z};G)$ between $A$ and $Z$ in $G$,
is given by the electrical resistance between the vertices $a$ and $z$ in this
new network.

In this paper, we will frequently need to bound effective resistances from
above and below. To do this, we will use the following pair of variational
formulas. Dirichlet’s Principle allows us to bound effective resistances from
below by finding functions with small discrete Dirichlet energy:

######  Proposition 2.2 (Dirichlet’s Principle).

If $G=(V,E,c)$ is a finite network with distinct vertices $a,z\in{V}$ then:

$R_{\text{eff}}(a\leftrightarrow{z};G)=\sup\left\\{\frac{1}{\mathcal{E}(h)}:h:V\rightarrow{\mathbb{R}},\hskip
2.0pth(a)=0,\hskip 2.0pth(z)=1\right\\}$

Thomson’s Principle allows us to bound effective resistances from above, by
finding low- energy flows:

######  Proposition 2.3 (Thomson’s Principle).

If $G=(V,E,c)$ is a finite network with distinct vertices $a,z\in{V}$ then:

$R_{\text{eff}}(a\leftrightarrow{z};G)=\inf\left\\{\mathcal{E}(\theta):\|\theta\|=1,\hskip
2.0pt\text{$\theta$ is a flow from $a$ to $z$}\right\\}$

Given $f:V\rightarrow{\mathbb{R}}$ and $A,B\subseteq{V}$ nonempty, disjoint
sets of vertices, we define the quantity $\text{gap}_{A,B}(f)$ as follows:

$\text{gap}_{A,B}(f)=\min\limits_{b\in{B}}f(b)-\max\limits_{a\in{A}}f(a)$

Additionally, recall Proposition 4.11 of [17] which tells us that:

######  Proposition 2.4.

If $G=(V,E,c)$ is a finite network, $A,B\subseteq{V}$ are disjoint, nonempty
sets of vertices then for any flow $\theta$ on $G$ and any function
$f:V\rightarrow{\mathbb{R}}$ such that $\text{gap}_{A,B}(f)\geq{0}$,

$\|\theta\|\cdot{\text{gap}_{A,B}(f)}\leq{\mathcal{E}(\theta)^{1/2}\mathcal{E}(f)^{1/2}}$

This inequality follows almost immediately from the Cauchy- Schwarz inequality
on $\ell^{2}_{-}(\vec{E})$. Furthermore, Dirichlet’s Principle and Thomson’s
Principle can both be recovered cheaply as corollaries of this inequality.

###  2.2 Extremal Length and Planar Networks

Suppose $G=(V,E,c)$ is a finite network and $\Gamma$ is a nonempty collection
of paths in $G$. Then the extremal length of the collection of paths $\Gamma$
in $G$ is given by the following variational formula:

$\lambda(\Gamma,G):=\sup_{\rho}\frac{\ell^{2}(\rho,\Gamma)}{A(\rho)}$ (2.1)

where our supremum is taken over all nonzero metrics
$\rho:E\rightarrow{\mathbb{R}_{\geq{0}}}$, and:

$\displaystyle\ell(\rho,\Gamma)$
$\displaystyle:=\min\\{\sum_{e\in{\gamma}}\rho(e):\gamma\in{\Gamma}\\},$
$\displaystyle A(\rho)$ $\displaystyle:=\sum_{e\in{E}}c(e)\rho(e)^{2}$

Note that the quantity $\ell^{2}(\rho,\Gamma)/A(\rho)$ doesn’t change if we
replace $\rho$ by some scalar multiple $\lambda\rho$ where $\lambda>0$. Thus:

$\lambda(\Gamma,G)=\sup_{\rho}\frac{l^{2}(\rho,\Gamma)}{A(\rho)}=\sup_{A(\rho)=1}l^{2}(\rho,\Gamma)=\sup_{l(\rho,\Gamma)=1}\frac{1}{A(\rho)}=\big{(}\inf_{l(\rho,\Gamma)=1}A(\rho)\big{)}^{-1}$

The set of metrics $\rho$ on $G$ such that $l(\rho,\Gamma)=1$ is referred to
as the set of admissible metrics and is denoted by $\mathcal{A}(\Gamma)$. We
say that a metric $\rho$ on $G$ is extremal for $\lambda(\Gamma,G)$ if
$\lambda(\Gamma,G)=l^{2}(\Gamma,G)/A(\rho)$. Looking at the second equality
above, we see that when we compute the extremal length of the path family
$\Gamma$, we are optimizing a continuous function,
$\rho\mapsto{l^{2}(\rho,\Gamma)}$, over the set of
$\rho\in{\mathbb{R}_{\geq{0}}^{E}}$ such that $A(\rho)=1$. This is a compact
subset of $\mathbb{R}^{E}$ with respect to the standard topology on
$\mathbb{R}^{E}$. Thus, in contrast to the continuous setting (see Exercise
IV.9 of [14]), for a finite network we always have an extremal metric.

This extremal metric is unique up to multiplication by a scalar. This follows
by the same argument as the in the continuous setting: suppose that $\rho_{1}$
and $\rho_{2}$ are both extremal for $\lambda(\Gamma,G)$. First we rescale so
that $A(\rho_{1})=A(\rho_{2})=1$. It follows that
$l^{2}(\Gamma,\rho_{1})=l^{2}(\Gamma,\rho_{2})=\lambda(\Gamma,G)$. Consider
the metric $\nu:=\frac{1}{2}(\rho_{1}+\rho_{2})$. Trivially,

$l(\nu,\Gamma)\geq{\frac{1}{2}\big{(}l(\rho_{1},\Gamma)+l(\rho_{2},\Gamma)\big{)}}=\sqrt{\lambda(\Gamma,G)}\hskip
5.0pt\implies\hskip 5.0ptl^{2}(\Gamma,\nu)\geq{\lambda(\Gamma,G)}$ (2.2)

On the other hand, by Cauchy- Schwartz,
$A(\nu)\leq{\frac{1}{2}\big{(}A(\rho_{1})+A(\rho_{2})\big{)}}=1$ with equality
iff $\rho_{1}$ is a scalar multiple of $\rho_{2}$. If $A(\nu)<1$ then
$l^{2}(\nu,\Gamma)/A(\nu)>\lambda(\Gamma,G)$. This is not possible since
$\lambda(\Gamma,G)$ is the supremum of $l^{2}(\rho,\Gamma)/A(\rho)$ over all
metrics $\rho$. Thus, $\rho_{2}$ is a scalar multiple of $\rho_{1}$. Since
$A(\rho_{1})=A(\rho_{2})=1$ we actually have that $\rho_{1}=\rho_{2}$.

In all of the cases we’re interested in, the path family $\Gamma$ will be the
set of paths $\gamma$ in $G$ that start at a vertex of $S$ and end at a vertex
of $T$ for $S$, $T$ nonempty disjoint subsets of $V$. We denote the extremal
length of this path family by $\lambda(S\leftrightarrow{T};G)$. It turns out
that the quantity $\lambda(S\leftrightarrow{T};G)$ is precisely the effective
resistance between $S$ and $T$ from the theory of electrical networks:

######  Proposition 2.5.

(Theorem 2 of [12]) Suppose $G=(V,E,c)$ is a finite network and $S,T$ are
nonempty, disjoint subsets of $V$. Then:

$\lambda(S\leftrightarrow{T};G)=R_{\text{eff}}(S\leftrightarrow{T};G)$

One nice property of extremal length is blocking duality. Given, $S,T$
nonempty, disjoint sets of vertices in $G$, we say that a set $F\subset{E}$ is
an $S$\- $T$ cut if $F$ separates $S$ from $T$ in $G$. That is, if we remove
the edges of $F$ from $G$, there is no nearest- neighbor path in $G$ starting
at a vertex of $S$ and ending at a vertex of $T$.

Let $B(S,T;G)$ denote the set of $S$\- $T$ cuts in $G$. Analogous to how we
defined the extremal length of a path family, we can talk about the extremal
length of the set of $S$\- $T$ cuts in $G$. This is denoted by
$\lambda(S\not\leftrightarrow{T};G)$ and defined as follows:

$\lambda(S\not\leftrightarrow{T};G)=\sup_{\rho}\frac{\ell^{2}(\rho,B(S,T;G))}{A(\rho)}$

where our supremum is taken over all nonzero metrics
$\rho:E\rightarrow{\mathbb{R}_{\geq{0}}}$, and:

$\displaystyle\ell(\rho,B(S,T;G))$
$\displaystyle=\min\\{\sum_{e\in{F}}\rho(e):F\in{B(S,T;G)}\\},$ $\displaystyle
A(\rho)$ $\displaystyle=\sum_{e\in{E}}c(e)\rho(e)^{2}$ (2.3)

More generally, while we initially restricted our attention to path families
so as to draw parallels with the continuous theory, it is clear that if we let
$\Gamma$ be any family of multisets of edges in $G$, Definition 2.1 still
makes sense. Thus, we can actually talk about the extremal length of any
family of multisets of edges of $G$. For instance, the modulus of the set of
spanning trees of a network has been the subject of recent study in [2].

A classic result of Ford and Fulkerson relates the extremal length of paths
from $S$ to $T$ to the extremal length of the set of $S$\- $T$ cuts:

######  Proposition 2.6.

(Theorem 1 of [13]) If $G=(V,E,c)$ and $H=(V,E,r)$ are finite networks so that
$r:E\rightarrow{\mathbb{R}_{>0}}$ is the resistance function corresponding to
the conductance function $c:E\rightarrow{\mathbb{R}_{>0}}$ and $S,T$ are
nonempty, disjoint sets of vertices in $G$, then:

$\lambda(S\leftrightarrow{T};G)\cdot{\lambda(S\not\leftrightarrow{T};H)}=1$

For a more modern treatment of this result as well as a generalization to the
case of $p$-extremal length, see [1]. This result is particularly useful in
the case where our graph $G$ is planar, in which case we can identify the set
of $S$\- $T$ cuts with path families in the dual graph.

A finite planar map is a finite planar graph $(V,E)$ along with a proper
embedding of this graph into the Riemann sphere, viewed up to homeomorphism of
the Riemann sphere. Specifying a proper embedding of a graph in the Riemann
sphere up to orientation- preserving homeomorphism is equivalent to assigning
a coherent system of orientations to the edges about each vertex (for details,
see Section 1.1.2 of [11]). Thus, despite the topology present in our initial
definition, planar maps can be viewed as purely combinatorial objects.
Equivalently, we can think of finite planar maps as gluings of polygons along
edges so that the resulting topological manifold is a sphere.

A quadrangulation with boundary is a bipartite planar map all of whose faces
are quadrilaterals, with the possible exception of some finite number of
distinguished faces which we think of as “holes" in our planar map. Notice
that requiring our planar map to be bipartite is equivalent to asking that all
of these “hole" faces have an even number of sides. Given a quadrangulation
with boundary $G=(V,E)$, we refer to these distinguished faces as the exterior
faces of $G$. The remaining faces are called the interior faces of $G$. The
edges and vertices tangent to the exterior faces of $G$ are known as the
boundary vertices and edges of $G$. We denote these by $\partial{V}$ and
$\partial{E}$. A quadrangulation with boundary $G=(V,E)$ is simply- connected
if $G$ has a unique exterior face whose boundary is a simple closed curve.

Since our quadrangulations are bipartite, we have a natural bipartition of the
vertices $V=V^{\bullet}\sqcup{V^{\circ}}$. The vertices of $V^{\bullet}$ are
known as the primal vertices of $G$ and are typically colored black. The
vertices of $V^{\circ}$ are known as the dual vertices and are typically
colored white. These give rise to the primal and dual graphs
$G^{\bullet}=(V^{\bullet},E^{\bullet})$ and $G^{\circ}=(V^{\circ},E^{\circ})$.
$G^{\bullet}$ is formed by connecting any pair of primal vertices that share
an interior face in $G$. Similarly, $G^{\circ}$ is formed by connecting any
pair of dual vertices that share an interior face in $G$. Since the interior
faces of $G$ are all quadrilaterals, each interior face corresponds to one
primal and one dual edge. In this way, there is a natural correspondence
between primal and dual edges.

Based on the paragraph above, it might seem that the setting we are working in
is very restrictive. On the contrary, observe that this procedure of
recovering a graph $G^{\bullet}$ (and its dual $G^{\circ}$) from a
quadrangulation $G$ gives us a one- to- one correspondence between the set of
quadrangulations with $n$ faces and the set of planar maps with $n$ edges (see
Section 2.2.1 of [11]). In other words, restricting our attention to bipartite
quadrangulations with $k$ holes is equivalent to restricting our attention to
embeddings of a graph and its dual in the Riemann sphere, up to orientation-
preserving homeomorphism, so that the resulting discrete object has the
topology of the Riemann sphere with k discs removed.

Given a quadrangulation with boundary $G=(V^{\bullet}\sqcup{V^{\circ}},E)$, a
conformal metric on $G$ is a function
$c:E^{\bullet}\sqcup{E^{\circ}}\rightarrow{(0,\infty))}$ such that:

$c(e^{\circ})=\frac{1}{c(e^{\bullet})}$

for $e^{\circ}\in{E^{\circ}}$, $e^{\bullet}\in{E^{\bullet}}$ so that
$e^{\circ}$ is the dual edge corresponding to the primal edge $e^{\bullet}$.
Let $c^{\bullet}:E^{\bullet}\rightarrow{(0,\infty)}$ and
$c^{\circ}:E^{\circ}\rightarrow{(0,\infty)}$ denote the conductances on
$G^{\bullet}$ and $G^{\circ}$ produced by restricting $c$ to $E^{\bullet}$ and
$E^{\circ}$ respectively. If $\theta\in{\ell_{-}^{2}(\vec{E^{\bullet}})}$ and
$f:V^{\bullet}\rightarrow{\mathbb{R}}$, we write:

$\displaystyle\mathcal{E}^{\bullet}(\theta)$
$\displaystyle=\mathcal{E}(\theta;G^{\bullet}),$
$\displaystyle\mathcal{E}^{\bullet}(f)$
$\displaystyle=\mathcal{E}(f;G^{\bullet})$

to emphasize that these energies are being computed on the primal graph
$G^{\bullet}$. Similarly, given a subgraph $H$ of $G^{\bullet}$, we write:

$\displaystyle\mathcal{E}^{\bullet}(\theta;H)$
$\displaystyle=\mathcal{E}(\theta;H),$
$\displaystyle\mathcal{E}^{\bullet}(f;H)$ $\displaystyle=\mathcal{E}(f;H)$

Given $\omega\in{l^{2}_{-}(\vec{E^{\circ}})}$,
$g:V^{\circ}\rightarrow{\mathbb{R}}$, and a subgraph $H$ of $G^{\circ}$, the
quantities $\mathcal{E}^{\circ}(\omega)$, $\mathcal{E}^{\circ}(g)$,
$\mathcal{E}^{\circ}(\omega;H)$, $\mathcal{E}^{\circ}(g;H)$ are defined
analogously.

A discrete conformal rectangle is a simply connected, bipartite
quadrangulation with boundary endowed with a conformal metric
$c:E^{\bullet}\sqcup{E^{\circ}}\rightarrow{(0,\infty)}$ and four distinguished
boundary points
$A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet}\in{\partial{V^{\bullet}}}$,
listed in counterclockwise order. Since our quadrangulation with boundary is
simply connected, it has a unique exterior face $f$, so
$A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet}$ must all lie along $f$.
Furthermore, having embedded our quadrangulation in the Riemann sphere, we can
talk about orientation.

The distinguished boundary points
$A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet}\in{\partial{V^{\bullet}}}$ of
our discrete conformal rectangle give rise to primal boundary arcs
$[A^{\bullet},B^{\bullet}],[C^{\bullet},D^{\bullet}]\subseteq{\partial{V^{\bullet}}}$.
$[A^{\bullet},B^{\bullet}]$ refers to the set of primal vertices that lie
along the counterclockwise path from $A^{\bullet}$ to $B^{\bullet}$ along the
boundary of $f$. Similarly, $[C^{\bullet},D^{\bullet}]$ is the set of primal
vertices that lie along the counterclockwise path from $C^{\bullet}$ to
$D^{\bullet}$ along the boundary of $f$. These primal boundary arcs have
corresponding dual arcs
$[B^{\circ},C^{\circ}],[D^{\circ},A^{\circ}]\subseteq{\partial{V^{\circ}}}$
where $[B^{\circ},C^{\circ}]$ consists of the set of dual vertices that lie
along the counterclockwise path from $B^{\bullet}$ to $C^{\bullet}$ along the
boundary of $f$. Similarly, $[D^{\circ},A^{\circ}]$ is the set of dual
vertices that lie along the counterclockwise path from $D^{\bullet}$ to
$A^{\bullet}$ along the boundary of $f$.

Figure 1: A discrete conformal rectangle with its four distinguished boundary
arcs.

We say that an $S$\- $T$ cut, $F\subseteq{E}$, is minimal if for any edge
$e\in{F}$, $F\setminus{\\{e\\}}$ is no longer an $S$\- $T$ cut.

######  Lemma 2.7.

(Lemma VIII.1 of [12]) If $(G,c)$ is a discrete conformal rectangle with
distinguished boundary points
$A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet}\in{\partial{V^{\bullet}}}$
giving rise to primal boundary arcs $[A^{\bullet},B^{\bullet}]$,
$[C^{\bullet},D^{\bullet}]\subseteq{\partial{V^{\bullet}}}$ with corresponding
dual arcs
$[B^{\circ},C^{\circ}],[D^{\circ},A^{\circ}]\subseteq{\partial{V^{\circ}}}$,
then the set of minimal $[A^{\bullet},B^{\bullet}]$\-
$[C^{\bullet},D^{\bullet}]$ cuts in $G^{\bullet}$ is in one- to- one
correspondence with the set of simple paths from $[B^{\circ},C^{\circ}]$ to
$[D^{\circ},A^{\circ}]$ in $G^{\circ}$.

When computing the extremal length of the family of paths between disjoint
vertex sets $S$ and $T$ in $G$, for a fixed metric $\rho$, we are interested
in the quantity:

$\inf_{\gamma}\left\\{\sum_{e\in{\gamma}}\rho(e)\right\\}$

where our infimum is taken over all paths $\gamma$ in $G$ between $S$ and $T$.
Since any such path that isn’t simple has a simple subpath of smaller $\rho$\-
weight, when taking this infimum, it actually suffices to restrict our
attention to simple paths $\gamma$ from $S$ to $T$. Similarly, when we compute
the extremal length of the set of $S$\- $T$ cuts in a network, rather than
taking an infimum over all $S$\- $T$ cuts, it suffices to restrict our
attention only to minimal $S$\- $T$ cuts. Thus, as an immediate corollary of
Lemma 2.7, we have that:

######  Corollary 2.8.

If $(G,c)$ is a discrete rectangle with distinguished boundary points
$A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet}\in{\partial{V^{\bullet}}}$
giving rise to primal boundary arcs
$[A^{\bullet},B^{\bullet}],[C^{\bullet},D^{\bullet}]\subseteq{\partial{V^{\bullet}}}$
with corresponding dual arcs
$[B^{\circ},C^{\circ}],[D^{\circ},A^{\circ}]\subseteq{\partial{V^{\circ}}}$,
then:

$\lambda([A^{\bullet},B^{\bullet}]\not\leftrightarrow{[C^{\bullet},D^{\bullet}]};(G^{\bullet},c^{\bullet}))=\lambda([B^{\circ},C^{\circ}]\leftrightarrow{[D^{\circ},A^{\circ}]};(G^{\circ},c^{\circ}))$

Suppose $G=(V^{\bullet}\sqcup{V^{\circ}},E)$ is a bipartite quadrangulation
with boundary endowed with a conformal metric
$c:E^{\bullet}\sqcup{E^{\circ}}\rightarrow{\mathbb{R}}$. We say that
$h:V^{\bullet}\rightarrow{\mathbb{R}}$ is harmonic on $G^{\bullet}$ if $h$ is
harmonic on
$\text{Int}(V^{\bullet})=V^{\bullet}\setminus{\partial{V^{\bullet}}}$.
$\widetilde{h}:V^{\circ}\rightarrow{\mathbb{R}}$ is a harmonic conjugate of
$h$ on $G$ if for any interior face $f$ of $G$, we have that:

$\big{(}\widetilde{h}(w_{2})-\widetilde{h}(w_{1})\big{)}=c^{\bullet}(v_{1},v_{2})\big{(}h(v_{2})-h(v_{1})\big{)}$
(2.4)

where $v_{1},w_{1},v_{2},w_{2}$ are the vertices of $f$ listed in
counterclockwise order so that $v_{1},v_{2}\in{V^{\bullet}}$ and
$w_{1},w_{2}\in{V^{\circ}}$. Equation 2.4 is a discrete analogue of the
Cauchy- Riemann equations for a quadrangulation with boundary $G$, endowed
with a conformal metric. It is easy to check that the conjugate of a harmonic
function on $G^{\bullet}$ is harmonic on $G^{\circ}$. Additionally, since $c$
is a conformal metric, if $\widetilde{h}$ is a harmonic conjugate of $h$, then
$h$ is a harmonic conjugate of $\widetilde{h}$. The next two propositions are
well-known, though they are rarely stated in this generality:

######  Proposition 2.9.

Suppose $G=(V^{\bullet}\sqcup{V^{\circ}},E)$ is a bipartite quadrangulation
with boundary endowed with a conformal metric
$c:E^{\bullet}\sqcup{E^{\circ}}\rightarrow{\mathbb{R}}$. If
$h:V^{\bullet}\rightarrow{\mathbb{R}}$ is harmonic on $G^{\bullet}$, then the
harmonic conjugate of $h$, if it exists, is unique up to an additive constant.

######  Proposition 2.10.

Suppose $G=(V^{\bullet}\sqcup{V^{\circ}},E)$ is a bipartite quadrangulation
with boundary endowed with a conformal metric
$c:E^{\bullet}\sqcup{E^{\circ}}\rightarrow{\mathbb{R}}$ and
$h:V^{\bullet}\rightarrow{\mathbb{R}}$ is harmonic on $G^{\bullet}$. Let
$H=(V_{H}^{\bullet}\sqcup{V_{H}^{\circ}},E_{H})$ be submap of $G$ which is
itself a simply- connected, bipartite quadrangulation with boundary. Then $h$
has a harmonic conjugate $\widetilde{h}:V_{H}^{\circ}\rightarrow{\mathbb{R}}$
on $H$.

######  Remark 2.11.

Ford- Fulkerson duality is a general statement that holds for any finite
network. However, in the case of discrete rectangles, it has a particularly
simple proof stemming from the fact that if $h$ is the function on
$V^{\bullet}$ that is equal to $0$ on $[A^{\bullet},B^{\bullet}]$, $1$ on
$[C^{\bullet},D^{\bullet}]$ and is harmonic elsewhere so that:

$\lambda^{\bullet}=\lambda([A^{\bullet},B^{\bullet}]\leftrightarrow{[C^{\bullet},D^{\bullet}]};G)=\frac{1}{\mathcal{E}^{\bullet}(h)}$

its harmonic conjugate $\widetilde{h}$ is (up to an additive constant) equal
to $0$ on $[B^{\circ},C^{\circ}]$, $\frac{1}{\lambda^{\bullet}}$ on
$[D^{\circ},A^{\circ}]$ and is harmonic elsewhere. Furthermore, by the
discrete Cauchy-Riemann equations,
$\mathcal{E}^{\circ}(\widetilde{h})=\mathcal{E}^{\bullet}(h)$. Hence:

$\lambda^{\circ}=\lambda([B^{\circ},C^{\circ}]\leftrightarrow{[D^{\circ},A^{\circ}]};G^{\circ})=\frac{(1/\lambda^{\bullet})^{2}}{\mathcal{E}^{\circ}(\widetilde{h})}=\frac{(1/\lambda^{\bullet})^{2}}{(1/\lambda^{\bullet})}=\frac{1}{\lambda^{\bullet}}$

###  2.3 Tilings of Rectangles

Suppose $\Omega\subseteq{\mathbb{C}}$ is a Jordan domain with analytic
boundary and distinguished boundary points $A,B,C,D$ listed in
counterclockwise order. Let
$[A,B]_{\partial{\Omega}},[B,C]_{\partial{\Omega}},[C,D]_{\partial{\Omega}},[D,A]_{\partial{\Omega}}$
denote the closed boundary arcs stretching counterclockwise from $A$ to $B$,
$B$ to $C$, $C$ to $D$ and $D$ to $A$ along $\partial{\Omega}$. Let
$(A,B)_{\partial{\Omega}},(B,C)_{\partial{\Omega}},(C,D)_{\partial{\Omega}},(D,A)_{\partial{\Omega}}$
be the corresponding open boundary arcs. Let $\widetilde{h}$ be the solution
to the following boundary value problem:

$\displaystyle\widetilde{h}(x)$ $\displaystyle=0\text{ for
}x\in{[B,C]_{\partial{\Omega}}}$ $\displaystyle\widetilde{h}(x)$
$\displaystyle=1\text{ for }x\in{[A,D]_{\partial{\Omega}}}$
$\displaystyle\Delta{\widetilde{h}(x)}$ $\displaystyle=0\text{ for
}x\in{\Omega}$ $\displaystyle\partial_{n}\widetilde{h}(x)$
$\displaystyle=0\text{ for
}x\in{(B,C)_{\partial{\Omega}}\cup{(D,A)_{\partial{\Omega}}}}$

The existence and uniqueness of the solution to this problem is clear by
conformal invariance. Namely, if $L$ is the extremal length from
$[A,B]_{\partial{\Omega}}$ to $[C,D]_{\partial{\Omega}}$ in $\Omega$ and
$\mathcal{R}_{L}:=(0,L)\times{(0,1)}$ then there is a unique conformal map
$\phi:\Omega\rightarrow{\mathcal{R}_{L}}$ sending $A,B,C,D$ to the corners of
$\mathcal{R}_{L}$. Since our boundary value problem is conformally invariant,
$\widetilde{h}(x)=Im(\phi(x))$. Furthermore, the conjugate harmonic function
$h(x)=Re(\phi(x))$ satisfies:

$\displaystyle h(x)$ $\displaystyle=0\text{ for
}x\in{[A,B]_{\partial{\Omega}}}$ $\displaystyle h(x)$ $\displaystyle=L\text{
for }x\in{[C,D]_{\partial{\Omega}}}$ $\displaystyle\partial_{n}h(x)$
$\displaystyle=0\text{ for
}x\in{(B,C)_{\partial{\Omega}}\cup{(D,A)_{\partial{\Omega}}}}$

Thus, if we are presented with a Jordan domain $\Omega$ with four
distinguished boundary points $A,B,C,D$ listed in counterclockwise order and
$\phi:\Omega\rightarrow{\mathcal{R}_{L}}$ is the conformal map that maps
$A,B,C,D$ to the corners of $\mathcal{R}_{L}$, we can intuitively think of the
real and imaginary parts of this conformal map as solving the aforementioned
boundary value problems.

Suppose $(G,c)$ is a discrete rectangle with distinguished boundary points
$A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet}\in{\partial{V^{\bullet}}}$
listed in counterclockwise order, giving rise to primal boundary arcs
$[A^{\bullet},B^{\bullet}],[C^{\bullet},D^{\bullet}]\subseteq{\partial{V^{\bullet}}}$
and corresponding dual boundary arcs
$[B^{\circ},C^{\circ}],[D^{\circ},A^{\circ}]\subseteq{\partial{V^{\circ}}}$.
Let $\widetilde{h}$ be the solution to the following boundary value problem on
$(G^{\circ},c^{\circ})$:

$\displaystyle\widetilde{h}(x)$ $\displaystyle=0\hskip 2.0pt\text{ for
$x\in{[B^{\circ},C^{\circ}]}$}$ $\displaystyle\widetilde{h}(x)$
$\displaystyle=1\hskip 2.0pt\text{ for $x\in{[D^{\circ},A^{\circ}]}$}$
$\displaystyle\Delta^{\circ}\widetilde{h}(x)$ $\displaystyle=0\hskip
2.0pt\text{ for
$x\in{V^{\circ}\setminus{([B^{\circ},C^{\circ}]\cup{[D^{\circ},A^{\circ}]})}}$}$

Let $h$ be the solution to the following boundary value problem on
$(G^{\bullet},c^{\bullet})$:

$\displaystyle h(x)$ $\displaystyle=0\hskip 2.0pt\text{ for
$x\in{[A^{\bullet},B^{\bullet}]}$}$ $\displaystyle h(x)$
$\displaystyle=L\hskip 2.0pt\text{ for $x\in{[C^{\bullet},D^{\bullet}]}$}$
$\displaystyle\Delta^{\bullet}{h}(x)$ $\displaystyle=0\hskip 2.0pt\text{ for
$x\in{V^{\bullet}\setminus{([A^{\bullet},B^{\bullet}]\cup{[C^{\bullet},D^{\bullet}]})}}$}$

where $L$ is the effective resistance between $[A^{\bullet},B^{\bullet}]$ and
$[C^{\bullet},D^{\bullet}]$ in $(G^{\bullet},c^{\bullet})$. Just as in the
continuous setting, $\widetilde{h}$ is the harmonic conjugate of $h$. Since
they are defined in terms of analogous boundary value problems, the functions
$h$ and $\widetilde{h}$ are discrete analogues of the real and imaginary parts
of the uniformizing conformal map that takes a simply connected domain with
four distinguished prime ends to a rectangle so that the four distinguished
prime ends are mapped to the four corners of the rectangle.

Suppose $(G,c)$ is a discrete rectangle with distinguished boundary points
$A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet}\in{\partial{V^{\bullet}}}$
listed in counterclockwise order, giving rise to primal boundary arcs
$[A^{\bullet},B^{\bullet}],[C^{\bullet},D^{\bullet}]\subseteq{\partial{V^{\bullet}}}$
and corresponding dual boundary arcs
$[B^{\circ},C^{\circ}],[D^{\circ},A^{\circ}]\subseteq{\partial{V^{\circ}}}$.
Let $h:V^{\bullet}\rightarrow{\mathbb{R}}$,
$\widetilde{h}:V^{\circ}\rightarrow{\mathbb{R}}$ be conjugate harmonic
functions defined as above. For any interior face $f$ of $G$ with incident
vertices $x,y,u,v$ where $x,y\in{V^{\bullet}}$, $u,v\in{V^{\circ}}$, the image
of $f$ under the tiling map $\phi$ is defined as follows:

$\phi(f)=[h(x),h(y)]\times{[\widetilde{h}(u),\widetilde{h}(v)]}$

where the order of $x,y$ and $u,v$ is chosen so that:

$\displaystyle h(x)$ $\displaystyle\leq{h(y)},$
$\displaystyle\widetilde{h}(u)$ $\displaystyle\leq{\widetilde{h}(v)}$

As the name suggests, $\phi$ corresponds to a tiling of the rectangle
$\mathcal{R}_{L}$ by smaller subrectangles:

######  Theorem 2.12.

(Theorem 4.31 of [6]) Suppose $\phi$ is the tiling map associated with the
discrete rectangle $(G,c)$ with distinguished boundary points
$A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet}\in{\partial{V^{\bullet}}}$
listed in counterclockwise order. Then for any pair of distinct inner faces
$f,f^{\prime}$ of $G$, the rectangles $\phi(f)$ and $\phi(f^{\prime})$ have
disjoint interiors. Furthermore, if $F_{\text{in}}$ is the set of interior
faces of $G$,

$\bigcup\limits_{f\in{F_{\text{in}}}}\phi(f)=[0,L]\times{[0,1]}$

Since $h$ and $\widetilde{h}$ are conjugate, the aspect ratio of the rectangle
$\phi(f)$ corresponding to the face $f$ of $G$ with incident vertices
$x,y\in{V^{\bullet}}$ and $u,v\in{V^{\circ}}$ is precisely the resistance of
the primal edge $\\{x,y\\}\in{E^{\bullet}}$ or equivalently, the conductance
of the dual edge $\\{u,v\\}\in{E^{\circ}}$.

###  2.4 Orthodiagonal Approximations of Planar Domains

Our reason for working in the level of generality that we did in Sections 2.2
and 2.3 was to illustrate the scope of the theory of planar electrical
networks. That said, in order to get a sequence of discrete conformal maps
that converge to the relevant uniformizing comformal map in the limit, we need
to do two things:

  1. 1.

Fix an embedding of our graph in the complex plane.

  2. 2.

Pick a conformal metric that is tied to the geometry of this embedding.

With this in mind, an orthodiagonal map is a finite, bipartite quadrangulation
with boundary $G=(V^{\bullet}\sqcup{V^{\circ}},E)$ with a fixed, proper
embedding in the plane so that:

  * •

Each edge is a straight line segment.

  * •

Each interior face is a quadrilateral with orthogonal diagonals.

We allow non- convex quadrilaterals, whose diagonals do not intersect. We
endow this with a conformal metric
$c:E^{\bullet}\sqcup{E^{\circ}}\rightarrow{(0,\infty)}$ defined as follows:

$\displaystyle c(e^{\bullet})$
$\displaystyle=\frac{|e^{\circ}|}{|e^{\bullet}|},$ $\displaystyle
c(e^{\circ})$ $\displaystyle=\frac{|e^{\bullet}|}{|e^{\circ}|}$

for $e^{\circ}\in{E^{\circ}}$, $e^{\bullet}\in{E^{\bullet}}$ so that
$e^{\circ}$ is the dual edge corresponding to the primal edge $e^{\bullet}$.
For any edge $e\in{E^{\bullet}\sqcup{E^{\circ}}}$, $|e|$ is the length of the
edge $e$ in our embedding. Recall that $E^{\bullet}$ and $E^{\circ}$ are the
edges of $G^{\bullet}$ and $G^{\circ}$ respectively, not the edges of $G$.
That is, they correspond to the diagonals of interior faces of $G$.

To make it clear that the discussion that follows is not totally vacuous,
observe that the square lattice, the triangular lattice and the hexagonal
lattice all have this property that primal and dual edges are orthogonal. More
generally, finite subdomains of isoradial lattices, which have been widely
studied in the context of critical statistical physics in 2D (i.e. see [20]
and [10]), are precisely orthodiagonal maps whose faces are all rhombii.
Furthermore, as a consequence of the double circle packing theorem, a wide
variety of planar graphs admit an orthodiagonal embedding (see Section 2 of
[17]).

While our choice of conformal metric might seem strange at first, observe that
if $G$ is an orthodiagonal map with conformal metric $c$ as above:

  * •

simple random walk on $(G^{\bullet},c^{\bullet})$ and $(G^{\circ},c^{\circ})$
is a martingale.

  * •

as a Markov chain, simple random walk on $(G^{\bullet},c^{\bullet})$ and
$(G^{\circ},c^{\circ})$ is reversible.

The orthogonality of edges and dual edges gives us a natural way to write down
the Cauchy- Riemann equations on an orthodiagonal map. Namely, a function
$F:V^{\bullet}\sqcup{V^{\circ}}\rightarrow{\mathbb{C}}$ is said to be discrete
holomorphic if for every interior face $Q$ of $G$ with primal diagonal
$e^{\bullet}=\\{v_{1},v_{2}\\}$ and dual diagonal
$e^{\circ}=\\{w_{1},w_{2}\\}$ we have:

$\frac{F(v_{2})-F(v_{1})}{v_{2}-v_{1}}=\frac{F(w_{2})-F(w_{1})}{w_{2}-w_{1}}$
(2.5)

From this definition, it follows that:

  * •

discrete contour integrals vanish if the integrand is discrete holomorphic.
That is, if $F$ is discrete holomorphic on $G$ and $\gamma$ is a simple,
closed, directed curve in $G$ so that the faces of $G$ enclosed by $\gamma$
are all interior faces of $G$, then:

$\sum_{\begin{subarray}{c}\vec{e}\hskip 1.0pt\in{\gamma}\\\
\vec{e}=(e^{-},e^{+})\end{subarray}}\big{(}F(e^{-})+F(e^{+})\big{)}(e^{+}-e^{-})=0$

  * •

the real and imaginary parts of any discrete holomorphic function are harmonic
with respect to the edge weights $c^{\bullet}$ and $c^{\circ}$. That is,
$\text{Re}(F)|_{V^{\bullet}},\text{Im}(F)|_{V^{\bullet}}$ are harmonic on
$(G^{\bullet},c^{\bullet})$ and
$\text{Re}(F)|_{V^{\circ}},\text{Im}(F)|_{V^{\circ}}$ are harmonic on
$(G^{\circ},c^{\circ})$. Moreover, $\text{Im}(F)|_{V^{\circ}}$ is the
conjugate harmonic function of $\text{Re}(F)|_{V^{\bullet}}$ and
$\text{Re}(F)|_{V^{\circ}}$ is the conjugate harmonic function of
$\text{Im}(F)|_{V^{\bullet}}$.

In short, orthodiagonal maps provide us with a notion of discrete complex
analysis.

An orthodiagonal rectangle is an orthodiagonal map $G$ with a unique
distinguished outer face whose boundary is a simple closed curve with four
distinguished boundary points
$A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet}\subseteq\partial{V}^{\bullet}$
listed in counterclockwise order. As in the case of discrete conformal
rectangles, these give rise to primal boundary arcs
$[A^{\bullet},B^{\bullet}],[C^{\bullet},D^{\bullet}]$ and corresponding dual
arcs $[B^{\circ},C^{\circ}],[D^{\circ},A^{\circ}]$. Given an orthodiagonal map
$G$, let $\widehat{G}$ denote the subdomain of $\mathbb{C}$ formed by taking
the interior of the union of the faces of $G$.

Figure 2: An orthodiagonal rectangle with its four distinguished boundary
arcs.

Suppose $\Omega$ is a connected, proper subdomain of $\mathbb{C}$.
$\gamma\subseteq{\overline{\Omega}}$ is a crosscut of $\Omega$ if
$\gamma=\eta([0,1])$ for some injective, continuous function
$\eta:[0,1]\rightarrow{\overline{\Omega}}$ such that
$\eta(0,1)\subseteq{\Omega}$ and $\eta(0),\eta(1)\in{\partial{\Omega}}$, where
$\eta(0)\neq{\eta(1)}$. If $\gamma$ is a crosscut of $\Omega$,
$\Omega\setminus{\gamma}$ has two connected components. By the Jordan curve
theorem, the same is true of $\Omega\setminus{\gamma}$ when $\gamma$ is a
simple closed curve in $\Omega$. Given disjoint subsets $A,B\subseteq{\Omega}$
we say that a simple closed curve or crosscut $\gamma$ separates $A$ and $B$
in $\Omega$ if $\gamma\cap{A}=\gamma\cap{B}=\emptyset$ and $A$ and $B$ lie in
distinct connected components of $\Omega\setminus{\gamma}$.

Fix $z_{0}\in{\Omega}$. For any $z,w\in{\Omega\setminus{\\{z_{0}\\}}}$, their
Carathéodory distance with respect to the reference point $z_{0}$ is given by:

$d_{Cara}^{z_{0}}(z,w):=\inf\\{\text{length}(\gamma):\text{$\gamma$ is a
simple closed curve or crosscut that separates $z$ and $w$ from $z_{0}$}\\}$

$d_{Cara}^{z_{0}}$ is a metric on $\Omega\setminus{\\{z_{0}\\}}$ that is
locally equivalent to the usual Euclidean metric. The Carathéodory
compactification $\Omega^{*}$ of $\Omega$ is the completion of
$\Omega\setminus{\\{z_{0}\\}}$ with respect to $d_{Cara}^{z_{0}}$. As a
topological space, the Carathéodory compactification $\Omega^{*}$ is
independent of our choice of reference point $z_{0}\in{\Omega}$.
$\partial{\Omega^{*}}$ is known as the space of prime ends of $\Omega$. The
prime ends of $\Omega$ can be interpretted geometrically as equivalence
classes of chains of open sets in $\Omega$ converging to a point on the
boundary. For details, see Section 3.1 of [5] or Section 2.4 of [24]. Given
disjoint subsets $A,B\subseteq{\partial{\Omega^{\ast}}}$, we say that a
crosscut $\gamma$ of $\Omega$ joins $A$ and $B$ in $\Omega$ if one of the
endpoints of $\gamma$ lies in $A$ and the other lies in $B$. If
$A\subseteq{\Omega}$, $B\subseteq{\partial{\Omega^{\ast}}}$, we say that a
crosscut $\gamma$ of $\Omega$ joins $A$ and $B$ in $\Omega$ if
$A\cap{\gamma}=\emptyset$, $A$ is contained in one of the two connected
components of $\Omega^{\ast}\setminus{\gamma}$, and $B$ intersects the
connected component of $\Omega^{\ast}\setminus{\gamma}$ containing $A$.

If $\Omega_{1},\Omega_{2}$ are proper, simply connected subdomains of
$\mathbb{C}$ and $\phi:\Omega_{1}\rightarrow{\Omega_{2}}$ is conformal, then
$\phi$ extends to a homeomorphism
$\phi:\Omega_{1}^{*}\rightarrow{\Omega_{2}^{*}}$. This tells us that, from the
standpoint of complex analysis, the space of prime ends is the right notion of
boundary for a proper, simply connected subdomain of $\mathbb{C}$. In
particular, if $\phi:\Omega\rightarrow{\mathbb{D}}$ is the uniformizing
conformal map that maps a simply connected domain
$\Omega\subsetneq{\mathbb{C}}$ to the unit disk, we have the following
estimates for the modulus of continuity of $\phi$ and $\phi^{-1}$ with respect
to the Carathéodory metric:

######  Proposition 2.13.

Suppose $\Omega\subsetneq{\mathbb{C}}$ is a bounded simply connected domain,
$z_{0}\in{\Omega}$ and $\phi:\Omega\rightarrow{\mathbb{D}}$ is a uniformizing
conformal that maps $\Omega$ to the unit disk so that $\phi(z_{0})=0$. Then
there exists an absolute constant $C_{1}>0$ such that for any
$x,y\in{\Omega}$:

$|\phi(y)-\phi(x)|\leq{C_{1}\sqrt{\frac{d^{z_{0}}_{Cara}(x,y)}{|\phi^{\prime}(z_{0})|}}}$
(2.6)

If additionally we know that $\Omega$ is bounded, there exists an absolute
constant $C_{2}>0$ such that for any $x,y\in{\mathbb{D}}$, we have that:

$d_{Cara}^{z_{0}}(\phi^{-1}(x),\phi^{-1}(y))\leq{C_{2}\sqrt{\frac{\text{Area}(\Omega)}{\log\big{(}\frac{1}{|x-y|}\big{)}}}}$
(2.7)

The modulus of continuity for $\phi$ in Equation 2.6 is a consequence of
Beurling’s estimate (see Proposition 3.85 of [21]). The modulus of continuity
for $\phi^{-1}$ follows from Wolff’s lemma (see Proposition 2.2 of [24]). As a
consequence, if $\Omega\subsetneq{\mathbb{C}}$ is simply connected,
$\Omega^{*}$ is homeomorphic to the closed unit disk $\overline{\mathbb{D}}$
and the space of prime ends $\partial{\Omega^{*}}$ is homeomorphic to $S^{1}$.
If $x,y\in{\partial{\Omega^{\ast}}}$ are prime ends, let
$[x,y]_{\partial{\Omega^{\ast}}}$ denote the arc along
$\partial{\Omega^{\ast}}$ that travels from $x$ to $y$, counterclockwise.

A conformal rectangle is a bounded, simply connected domain $\Omega$ along
with four distinguished prime ends $A,B,C,D$, listed in counterclockwise
order. Recalling our discussion in Section 1, given a conformal rectangle
$(\Omega,A,B,C,D)$ and a sequence of orthodiagonal rectangles
$\big{(}(G_{n},A_{n}^{\bullet},B_{n}^{\bullet},C_{n}^{\bullet},D_{n}^{\bullet})\big{)}_{n=1}^{\infty}$
that are better and better approximations of $(\Omega,A,B,C,D)$, we want to
show that the associated tiling maps converge to the conformal map from
$\Omega$ to a rectangle $\mathcal{R}_{L}$ so that the prime ends $A,B,C,D$ are
mapped to the four corners of $\mathcal{R}_{L}$, and in particular,
$\phi(A)=i$. This of course begs the question: what does it mean for an
orthodiagonal rectangle $(G,A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet})$
to be a good approximation of $(\Omega,A,B,C,D)$? One natural requirement is
that the boundary arcs $[A^{\bullet},B^{\bullet}]$, $[B^{\circ},C^{\circ}]$,
$[C^{\bullet},D^{\bullet}]$, $[D^{\circ},A^{\circ}]$ of $G$ should be close to
the corresponding continuous boundary arcs $[A,B]_{\partial{\Omega^{\ast}}}$,
$[B,C]_{\partial{\Omega^{\ast}}}$, $[C,D]_{\partial{\Omega^{\ast}}}$,
$[D,A]_{\partial{\Omega^{\ast}}}$ of $\Omega$. To be precise, since the
Carathéodory metric is the right notion of distance for a general simply
connected domain, a natural requirement is that the discrete boundary arcs are
close to the corresponding continuous boundary arcs in Carathéodory metric.
However, to define the Carathéodory metric, we need to introduce a reference
point, which a priori isn’t part of our setup. To avoid this, we will instead
use a closely related quantity, whose definition doesn’t require the
introduction of a reference point.

If $\gamma$ is a crosscut of $\Omega$, for any
$z\in{\Omega\setminus{\gamma}}$, let $\mathcal{N}_{\gamma}^{z}$ denote the
component of $\Omega\setminus{\gamma}$ containing $z$. Let $S,S^{\prime}$ be
disjoint compact subsets of $\partial{\Omega^{\ast}}$. Then for any
$z\in{\Omega^{\ast}}$, we define the crosscut distance
$d_{cc}^{\Omega}(z,S;S^{\prime})$ from $z$ to $S$, away from $S^{\prime}$ in
$\Omega$, by:

$d_{cc}^{\Omega}(z,S;S^{\prime}):=\inf\\{\text{length}(\gamma):\text{$\gamma$
is a crosscut of $\Omega$ that separates $z$ from $S^{\prime}$ such that
$\overline{\mathcal{N}_{\gamma}^{z}}\cap{S}\neq{\emptyset}$}\\}$

where $\overline{\mathcal{N}_{\gamma}^{z}}$ is the closure of
$\mathcal{N}_{\gamma}^{z}$ with respect to the Carathéodory metric. Similarly,
for a compact subset $C$ of $\Omega^{\ast}$, its crosscut distance to $S$,
away from $S^{\prime}$, is given by:

$d_{cc}^{\Omega}(C,S;S^{\prime}):=\sup_{z\in{C}}d^{\Omega}_{cc}(z,S;S^{\prime})$

The supremum on the right hand side is actually a maximum. To see this,
observe that $d^{\Omega}_{cc}(z,S;S^{\prime})$ is locally Lipschitz as a
function of $z$. Namely,

$|d_{cc}^{\Omega}(z,S;S^{\prime})-d^{\Omega}_{cc}(w,S;S^{\prime})|\leq{2d_{\Omega}(z,w)}$

where:

$d_{\Omega}(z,w):=\inf\\{\text{length}(\gamma):\text{$\gamma$ is a smooth
curve from $z$ to $w$ in $\Omega$}\\}$

In particular, if the line segment from $z$ to $w$ is contained in $\Omega$,
we have that:

$|d_{cc}^{\Omega}(z,S;S^{\prime})-d_{cc}^{\Omega}(w,S;S^{\prime})|\leq{2|z-w|}$

Given a conformal rectangle $(\Omega,A,B,C,D)$, we say that the orthodiagonal
rectangle $(G,A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet})$ is a
($\delta$, $\varepsilon$)- good interior approximation of $(\Omega,A,B,C,D)$
if $\widehat{G}\subseteq{\Omega}$, $|e|<\varepsilon$ for all edges $e\in{E}$,
and:

$\displaystyle
d_{cc}^{\Omega}([A^{\bullet},B^{\bullet}],[A,B]_{\partial{\Omega^{\ast}}};[C,D]_{\partial{\Omega^{\ast}}})<\delta,$
$\displaystyle
d_{cc}^{\Omega}([C^{\bullet},D^{\bullet}],[C,D]_{\partial{\Omega^{\ast}}};[A,B]_{\partial{\Omega^{\ast}}})<\delta$
$\displaystyle
d_{cc}^{\Omega}([B^{\circ},C^{\circ}],[B,C]_{\partial{\Omega^{\ast}}};[D,A]_{\partial{\Omega^{\ast}}})<\delta,$
$\displaystyle
d_{cc}^{\Omega}([D^{\circ},A^{\circ}],[D,A]_{\partial{\Omega^{\ast}}};[B,C]_{\partial{\Omega^{\ast}}})<\delta$

This notion of what it means for an orthodiagonal map with four distinguished
boundary points to approximate a conformal rectangle is more involved than the
approximation scheme used in [3] where it is only required that
$[A^{\bullet},B^{\bullet}]$, $[B^{\circ},C^{\circ}]$,
$[C^{\bullet},D^{\bullet}]$, and $[D^{\circ},A^{\circ}]$ are close to the
corresponding continuous boundary arcs $[A,B]_{\partial{\Omega^{\ast}}}$,
$[B,C]_{\partial{\Omega^{\ast}}}$, $[C,D]_{\partial{\Omega^{\ast}}}$,
$[D,A]_{\partial{\Omega^{\ast}}}$ in Hausdorff distance. Additionally, in [3],
the approximation doesn’t need to be be an interior approximation: it is not
required that $\widehat{G}\subseteq{\Omega}$.

The upside of our approach is that it is more general. Firstly, observe that
if we restrict our attention to interior approximations, if
$[A^{\bullet},B^{\bullet}]$ and $[A,B]_{\partial{\Omega^{\ast}}}$ are within
$\delta$ of one another in Hausdorff distance, then:

$d_{cc}^{\Omega}([A^{\bullet},B^{\bullet}],[A,B]_{\partial{\Omega^{\ast}}};[C,D]_{\partial{\Omega^{\ast}}})<2\delta$
(2.8)

The same is true for the other three boundary arcs. More importantly, using
this schema, we can approximate any conformal rectangle $(\Omega,A,B,C,D)$. If
we replace crosscut distance with Hausdorff distance, this is no longer true.
To unpack this statement, we’ll first need to make a brief digression. Recall
that a bounded, simply connected domain is Jordan if its boundary is a simple
closed curve. If $(\Omega,A,B,C,D)$ is a conformal rectangle, and
$\Omega^{\prime}$ is a Jordan subdomain of $\Omega$ with distinguished
boundary points $A^{\prime},B^{\prime},C^{\prime},D^{\prime}$ in
counterclockwise order, we say that
$(\Omega^{\prime},A^{\prime},B^{\prime},C^{\prime},D^{\prime})$ is a
$\delta$\- good interior approximation of $\Omega$ if:

$\displaystyle
d_{cc}^{\Omega}([A^{\prime},B^{\prime}]_{\partial{\Omega^{\prime}}},[A,B]_{\partial_{\Omega^{\ast}}};[C,D]_{\partial{\Omega^{\ast}}})$
$\displaystyle<\delta$ $\displaystyle
d_{cc}^{\Omega}([B^{\prime},C^{\prime}]_{\partial{\Omega^{\prime}}},[B,C]_{\partial_{\Omega^{\ast}}};[D,A]_{\partial{\Omega^{\ast}}})$
$\displaystyle<\delta$ $\displaystyle
d_{cc}^{\Omega}([C^{\prime},D^{\prime}]_{\partial{\Omega^{\prime}}},[C,D]_{\partial_{\Omega^{\ast}}};[A,B]_{\partial{\Omega^{\ast}}})$
$\displaystyle<\delta$ $\displaystyle
d_{cc}^{\Omega}([D^{\prime},A^{\prime}]_{\partial{\Omega^{\prime}}},[D,A]_{\partial_{\Omega^{\ast}}};[B,C]_{\partial{\Omega^{\ast}}})$
$\displaystyle<\delta$

Our reason for restricting to Jordan domains here has to do with the fact that
for a bounded, simply connected domain $\Omega$, $\partial{\Omega}$ and
$\partial{\Omega^{\ast}}$ are homeomorphic if and only if $\Omega$ is Jordan.
In other words, for a Jordan domain, the space of prime ends coincides with
its boundary in the usual sense (with respect to the Euclidean metric).
Otherwise, it is not clear what it means for a prime end of
$(\Omega^{\prime})^{\ast}$ to be close to a boundary arc of $\Omega$ in
crosscut distance, since a prime end of $(\Omega^{\prime})^{\ast}$ is not a
point in $\Omega$. With this in mind, we have the following result:

######  Proposition 2.14.

For any conformal rectangle $(\Omega,A,B,C,D)$ and any $\delta>0$ we can find
a Jordan subrectangle
$(\Omega^{\prime},A^{\prime},B^{\prime},C^{\prime},D^{\prime})$ with analytic
boundary arcs so that
$(\Omega^{\prime},A^{\prime},B^{\prime},C^{\prime},D^{\prime})$ is a
$\delta$\- good interior approximation of $(\Omega,A,B,C,D)$.

For a proof of Proposition 2.14, see Appendix A.1. Observe that if the
orthodiagonal rectangle $(G,A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet})$
is a $(\delta,\varepsilon)$\- good interior approximation of the conformal
rectangle $(\Omega,A,B,C,D)$, then the corresponding conformal rectangle
$(\widehat{G},A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet})$, is a
$(\delta+2\varepsilon)$-good interior approximation of $(\Omega,A,B,C,D)$.
Thus, at least heuristically, Proposition 2.14 tells us that given a conformal
rectangle $(\Omega,A,B,C,D)$ and $\delta,\varepsilon>0$, we can find a
$(\delta,\varepsilon)$\- good orthodiagonal approximation of
$(\Omega,A,B,C,D)$, provided the mesh of our orthodiagonal map is sufficiently
fine.

In contrast, consider the comb domain $\Omega$ defined as follows:

$\Omega:=(0,2)\times{(0,1)}\setminus{\Big{(}\big{(}\bigcup_{n=0}^{\infty}\\{2^{-2n}\\}\times{[2^{-2n-1},1)}\big{)}\bigcup{\big{(}\bigcup_{n=0}^{\infty}\\{2^{-2n-1}\\}\times{(0,1-2^{-2n-1}]}\big{)}}\Big{)}}$

The points $(1,\frac{1}{2})$, $(2,0)$, $(2,1)$ on the boundary of $\Omega$ as
well as the line segment $\\{0\\}\times{[0,1]}$ all correspond to a single
prime end. Take $A=\\{0\\}\times{[0,1]}$, $B=(1,\frac{1}{2})$, $C=(2,1)$,
$D=(2,0)$. Suppose we want to approximate the topological rectangle
$(\Omega,A,B,C,D)$ by a suborthodiagonal map of $\varepsilon\mathbb{Z}^{2}$
for some very small $\varepsilon>0$. Since the line segment
$\\{0\\}\times{[0,1]}$ corresponds to a single prime end $A$ of $\Omega$,
regardless of $\varepsilon$ and our choice of suborthodiagonal map $G$ of
$\varepsilon\mathbb{Z}^{2}$, it is not possible to choose
$A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet}\in{\partial{V}^{\bullet}}$ so
that the corresponding discrete boundary arcs are all within $\delta$ of their
continuum analogues, for $\delta$ sufficiently small.

Figure 3: The comb domain $\Omega$ with its four distinguished prime ends and
an orthodiagonal approximation $G$ of $\Omega$. Notice that regardless of how
fine the mesh of $G$ is, there is no choice of
$A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet}\in{\partial{V^{\bullet}}}$ so
that each of the discrete boundary arcs of
$(G,A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet})$ is close to its
continuous analogue.

In Section 2.3 we defined the tiling map associated with a discrete rectangle.
Without a fixed embedding, the faces of a discrete rectangle are purely
combinatorial objects. Having fixed an embedding, the faces of an
orthodiagonal rectangle are honest- to- goodness subsets of the plane. Hence,
we can think of the tiling map $\phi$ associated with an orthodiagonal
rectangle $G$, as a function $\phi:\widehat{G}\rightarrow{\mathbb{C}}$. We do
this by choosing for each interior face $f$ a homeomorphism that maps the
quadrilateral $f$ to the corresponding rectangle $\phi(f)$. Unfortunately,
this means that the tiling map $\phi:\widehat{G}\rightarrow{\mathbb{C}}$
depends on our choice of homeomorphism. There is also some ambiguity as to the
definition of $\phi$ on the edges of $G$, since each edge is shared by two
distinct faces. That said, by the regularity estimates in Section 3.2 we’ll
see that this isn’t a concern, since our choice of homeomorphism doesn’t
impact the convergence we are looking for.

Figure 4: The tiling map $\phi$ associated with an orthodiagonal rectangle.

Having established all the requisite terminology, we can now state precisely
the main theorem we intend to prove:

######  Theorem 2.15.

Suppose $(\Omega,A,B,C,D)$ is a conformal rectangle and
$\big{(}(G_{n},A^{\bullet}_{n},B^{\bullet}_{n},C^{\bullet}_{n},D^{\bullet}_{n})\big{)}_{n=1}^{\infty}$
is a sequence of orthodiagonal rectangles so that for each $n\in{\mathbb{N}}$,
$(G_{n},A^{\bullet}_{n},B^{\bullet}_{n},C^{\bullet}_{n},D^{\bullet}_{n})$ is a
$(\delta_{n},\varepsilon_{n})$\- good interior approximation of $\Omega$,
where:

$(\delta_{n},\varepsilon_{n})\rightarrow{(0,0)}\hskip 5.0pt\text{as
$n\rightarrow{\infty}$}$

Let $\phi_{n}:\widehat{G}_{n}\rightarrow{[0,L_{n}]\times{[0,1]}}$ be the
corresponding tiling maps, where $L_{n}$ is the discrete extremal length
between $[A_{n}^{\bullet},B_{n}^{\bullet}]$ and
$[C_{n}^{\bullet},D_{n}^{\bullet}]$ in $G_{n}^{\bullet}$. Then:

$\phi_{n}\rightarrow{\phi}\hskip 5.0pt\text{uniformly on compacts as
$n\rightarrow{\infty}$}$

where $\phi$ is the conformal map from $\Omega$ to the rectangle
$(0,L)\times{(0,1)}$ so that the prime ends $A,B,C,D$ are mapped to the four
corners of the rectangle and in particular, $\phi(A)=i$. Here, $L$ is the
extremal length between the arcs $[A,B]_{\partial{\Omega^{\ast}}}$ and
$[C,D]_{\partial{\Omega^{\ast}}}$ in $\Omega$. In particular, it follows that:

$L_{n}\rightarrow{L}\hskip 5.0pt\text{as $n\rightarrow{\infty}$}$

##  3 Precompactness of the Tiling Maps

To prove Theorem 2.15 following the framework outlined in Section 1, we need
to show that tiling maps corresponding to finer and finer orthodiagonal
approximations of a conformal rectangle $(\Omega,A,B,C,D)$ are equicontinuous
and uniformly bounded on compacts in $\Omega$. In this section, we address
this by proving estimates for the norm and modulus of continuity of our tiling
maps.

For both the norm and modulus of continuity, when doing this, we begin by
proving the corresponding result in the continuous setting. The proof in the
continuous setting motivates the proof in the discrete setting. Furthermore,
we will need the continuous analogues of our tiling map estimates to prove
Lemma A.1 in Appendix A.1 and thereby all the results thereafter that rely on
it.

###  3.1 Modulus of Continuity for the Limiting Conformal Map

Suppose $(\Omega,A,B,C,D)$ is a conformal rectangle. Using the notation of
Section 2.4, if $z,w\in{\Omega}$,

$\displaystyle d_{cc}^{\Omega}(z,w):=\min\\{$ $\displaystyle
d^{\Omega}_{cc}(\\{z,w\\},[A,B]_{\partial{\Omega^{\ast}}};[C,D]_{\partial{\Omega^{\ast}}}),d_{cc}^{\Omega}(\\{z,w\\},[B,C]_{\partial{\Omega^{\ast}}};[D,A]_{\partial{\Omega^{\ast}}}),$
$\displaystyle
d_{cc}^{\Omega}(\\{z,w\\},[C,D]_{\partial{\Omega^{\ast}}};[A,B]_{\partial{\Omega^{\ast}}}),d_{cc}^{\Omega}(\\{z,w\\},[D,A]_{\partial{\Omega^{\ast}}};[B,C]_{\partial{\Omega^{\ast}}})\\}$

In other words, $d_{cc}(z,w)$ is the length of the shortest crosscut of
$\Omega$ that joins $z$ and $w$ to one boundary arc of $(\Omega,A,B,C,D)$ and
separates them from the opposite boundary arc. Recall that for
$z,w\in{\Omega}$,

$d_{\Omega}(z,w)=\inf\\{\text{length}(\gamma):\text{$\gamma$ is a smooth curve
from $z$ to $w$ in $\Omega$}\\}$

That is, $d_{\Omega}$ is the ambient metric on $\Omega$. Let
$\phi:\Omega\rightarrow{\mathcal{R}_{L}}$ be the conformal map from $\Omega$
to the rectangle $\mathcal{R}_{L}$ such that the four prime ends $A,B,C,D$ of
$\Omega$, listed in counterclockwise order, are mapped to the four corners of
$\mathcal{R}_{L}$ and in particular, $\phi(A)=i$. The following theorem gives
us a modulus of continuity for the real and imaginary parts of $\phi$ and
therefore $\phi$ itself:

######  Theorem 3.1.

Suppose $(\Omega,A,B,C,D)$ is a conformal rectangle and
$\phi:\Omega\rightarrow{\mathcal{R}_{L}}$ be the conformal map from $\Omega$
to the rectangle $\mathcal{R}_{L}$ so that the four prime ends $A,B,C,D$ of
$\Omega$ are mapped to the four corners of $\mathcal{R}_{L}$ and in
particular, $\phi(A)=i$. Here, $L$ is the extremal length between the boundary
arcs $[A,B]_{\partial{\Omega^{\ast}}}$ and $[C,D]_{\partial{\Omega^{\ast}}}$
in $\Omega$. Define:

$\displaystyle d$ $\displaystyle=\inf\\{\text{diameter}(\gamma):\text{$\gamma$
is a crosscut of $\Omega$ joining $[A,B]_{\partial{\Omega^{\ast}}}$ and
$[C,D]_{\partial{\Omega^{\ast}}}$ in $\Omega$}\\}$ $\displaystyle d^{\prime}$
$\displaystyle=\inf\\{\text{diameter}(\gamma):\text{$\gamma$ is a crosscut of
$\Omega$ joining $[B,C]_{\partial{\Omega^{\ast}}}$ and
$[D,A]_{\partial{\Omega^{\ast}}}$ in $\Omega$}\\}$

Let $h$ and $\widetilde{h}$ be the real and imaginary parts of $\phi$,
respectively. Then for any $x,y\in{\Omega}$ so that
$(d_{\Omega}(x,y)\wedge{d_{cc}^{\Omega}(x,y)})<\frac{d\wedge{d^{\prime}}}{2}$,
we have that:

$\displaystyle|h(y)-h(x)|$
$\displaystyle\leq{\frac{2\pi}{\log\Big{(}\frac{d^{\prime}}{2(d_{\Omega}(x,y)\wedge{d_{cc}^{\Omega}(x,y)})}\Big{)}}},$
$\displaystyle|\widetilde{h}(y)-\widetilde{h}(x)|$
$\displaystyle\leq{\frac{2\pi{L}}{\log\Big{(}\frac{d}{2(d_{\Omega}(x,y)\wedge{d_{cc}^{\Omega}(x,y)})}\Big{)}}}$

###### Proof.

As per the theorem statement, let $h$ be the real part of $\phi$. Fix
$x,y\in{\Omega}$. If $h(x)=h(y)$, the desired result holds. Otherwise, suppose
WLOG that $h(x)<h(y)$. We now consider two cases:

Case 1: $d_{\Omega}(x,y)\leq{d_{cc}^{\Omega}(x,y)}$.

Consider the region:

$\Omega_{x,y}=\\{z\in{\Omega}:h(x)<h(z)<h(y)\\}$

This is simply connected. Furthermore, since $\phi$ maps $\Omega$ to the
rectangle $(0,L)\times{(0,1)}$, $\phi$ maps $\Omega_{x,y}$ to the rectangle
$(h(x),h(y))\times{(0,1)}$. Thus, if we think of $\Omega_{x,y}$ as a conformal
rectangle with the distinguished boundary arcs:

$\displaystyle N$
$\displaystyle=\Omega_{z,w}^{\ast}\cap{[D,A]_{\partial{\Omega^{\ast}}}},$
$\displaystyle E$ $\displaystyle=\\{z\in{\Omega}:h(z)=h(y)\\}$ $\displaystyle
S$ $\displaystyle=\Omega_{z,w}^{\ast}\cap{[D,A]_{\partial{\Omega^{\ast}}}},$
$\displaystyle W$ $\displaystyle=\\{z\in{\Omega}:h(z)=h(x)\\}$

Here, $N$, $E$, $S$ and $W$ stand for “North," “East," “South," and “West."
This is to emphasize that our picture is as follows:

Figure 5: On the left: The subrectangle $\Omega_{x,y}$ of $\Omega$ with its
distinguished boundary arcs $N,E,S$ and $W$. On the right: Notice that any
path from $N$ to $S$ in $\Omega_{x,y}$ must cross the annulus $\mathcal{A}$.

Then:

$\lambda(W\leftrightarrow{E};\Omega)=h(y)-h(x)$

Having reinterpreted the quantity we’re interested in as an extremal length,
we can bound it from above by bounding the dual extremal length from below.

Fix $\varepsilon$ so that
$0<\varepsilon<\frac{d^{\prime}}{2}-d_{\Omega}(x,y)$. By the definition of
$d_{\Omega}(x,y)$, we can find a smooth curve $\gamma$ in $\Omega$ from $x$ to
$y$ so that $\text{length}(\gamma)<d_{\Omega}(x,y)+\varepsilon$. Since $x$ and
$y$ lie on opposite boundary arcs of $\Omega_{x,y}$, while $\gamma$ may not be
a crosscut of $\Omega_{x,y}$, there must exist a subarc $\gamma^{\prime}$ of
$\gamma$ with endpoints $x^{\prime}\in{W}$, $y^{\prime}\in{E}$ so that
$\gamma^{\prime}$ is a crosscut of $\Omega_{x,y}$. Consider the annulus:

$\mathcal{A}=\\{u\in{\mathbb{C}}:d_{\Omega}(x,y)+\varepsilon<|u-x^{\prime}|<\frac{d^{\prime}}{2}\\}$

where:

$d^{\prime}=\inf\\{\text{diameter}(\gamma):\text{$\gamma$ is a curve joining
$[B,C]_{\partial{\Omega^{\ast}}}$ and $[D,A]_{\partial{\Omega^{\ast}}}$ in
$\Omega$}\\}$

Observe that:

  1. 1.

Since
$\text{length}(\gamma^{\prime})\leq{\text{length}(\gamma)}<d_{\Omega}(x,y)+\varepsilon$,
the diameter of $\gamma^{\prime}$ is at most $d_{\Omega}(x,y)+\varepsilon$.
Hence, $\gamma^{\prime}\subseteq{B(x^{\prime},d_{\Omega}(x,y)+\varepsilon)}$.
Since $\gamma^{\prime}$ separates the boundary arcs $N$ and $S$ in
$\Omega_{x,y}$, any path from $N$ to $S$ in $\Omega_{x,y}$ must intersect
$\gamma^{\prime}$ and therefore $B(x^{\prime},d_{\Omega}(x,y)+\varepsilon)$.

  2. 2.

On the other hand, since any curve from $N$ to $S$ in $\Omega_{x,y}$ is a
curve from $[B,C]_{\partial{\Omega^{\ast}}}$ to
$[D,A]_{\partial{\Omega^{\ast}}}$ in $\Omega$, such a curve must have diameter
$\geq{d^{\prime}}$. Hence, any curve from $N$ to $S$ in $\Omega_{x,y}$ must at
some point lie outside the ball $B(x^{\prime},\frac{d^{\prime}}{2})$.

Putting all this together, we see that any curve from $N$ to $S$ in
$\Omega_{x,y}$ must cross the annulus $\mathcal{A}$. Consider the metric:

$\rho(z)=\frac{1}{|z-x^{\prime}|}1_{\mathcal{A}\cap{\Omega_{x,y}}}(z)$

on $\Omega_{x,y}$. Observe that if $\eta$ is a $C^{1}$ curve that crosses the
annulus $\mathcal{A}$ at least once, then:

$l_{\rho}(\eta)=\int_{\eta}\frac{|dz|}{|z-x^{\prime}|}\geq{\int_{d_{\Omega}(x,y)+\varepsilon}^{d^{\prime}/2}}\frac{dr}{r}=\log\Big{(}\frac{d^{\prime}}{2(d_{\Omega}(x,y)+\varepsilon)}\Big{)}$

Furthermore:

$A(\rho)=\int_{\mathcal{A}\cap{\Omega_{x,y}}}\frac{1}{|z-x^{\prime}|^{2}}dz_{1}dz_{2}\leq{\int_{\mathcal{A}}\frac{1}{|z-x^{\prime}|^{2}}dz_{1}dz_{2}}=\int_{0}^{2\pi}\int_{d_{\Omega}(x,y)+\varepsilon}^{d^{\prime}/2}\frac{1}{r}drd\theta=2\pi\log\Big{(}\frac{d^{\prime}}{2(d_{\Omega}(x,y)+\varepsilon)}\Big{)}$

Hence, plugging $\rho$ into the variational problem for
$\lambda(N\leftrightarrow{S};\Omega_{x,y})$, we have that:

$\displaystyle\lambda(N\leftrightarrow{S};\Omega_{x,y})\geq{\frac{\inf\limits_{\eta}\big{(}l_{\rho}(\eta)\big{)}^{2}}{A(\rho)}}\geq{\frac{\Big{(}\log\Big{(}\frac{d^{\prime}}{2(d_{\Omega}(x,y)+\varepsilon)}\Big{)}\Big{)}^{2}}{2\pi\log\Big{(}\frac{d^{\prime}}{2(d_{\Omega}(x,y)+\varepsilon)}\Big{)}}}=\frac{1}{2\pi}\log\Big{(}\frac{d^{\prime}}{2(d_{\Omega}(x,y)+\varepsilon)}\Big{)}$

By duality for continuous extremal extremal length:

$\lambda(W\leftrightarrow{E};\Omega_{x,y})\cdot\lambda(N\leftrightarrow{S};\Omega_{x,y})=1$

Hence:

$\lambda(N\leftrightarrow{S};\Omega_{x,y})\leq{\frac{2\pi}{\log\Big{(}\frac{d^{\prime}}{2(d_{\Omega}(x,y)+\varepsilon)}\Big{)}}}$

Since $\varepsilon>0$ was arbitrary, letting $\varepsilon$ tend to $0$ in the
above inequality, the desired result follows.

Case 2: $d_{cc}^{\Omega}(x,y)\leq{d_{\Omega}(x,y)}$.

Fix $\varepsilon$ so that
$0<\varepsilon<\frac{d^{\prime}}{2}-d_{cc}^{\Omega}(x,y)$. By the definition
of $d_{cc}(x,y)$ we can find a crosscut $\gamma$ of $\Omega$ that joins $x$
and $y$ to one of the four distinguished boundary arcs of $(\Omega,A,B,C,D)$
and separates it from the opposite boundary arc, so that
$\text{length}(\gamma)<d_{cc}^{\Omega}(x,y)+\varepsilon$. We now split our
problem into two further cases, depending on whether the relevant boundary
arcs of $(\Omega,A,B,C,D)$ are Dirichlet arcs, where $h$ is constant, or
Neumann arcs, along which $h$ is monotone.

Case 2.1: $\gamma$ joins $x$ and $y$ to one of the Dirichlet arcs and
separates it from the opposite Dirichlet arc.

WLOG, suppose that $\gamma$ joins $x$ and $y$ to
$[A,B]_{\partial{\Omega^{\ast}}}$ and separates $x$ and $y$ from
$[C,D]_{\partial{\Omega^{\ast}}}$. Let $N_{x,y}^{\gamma}$ denote the connected
component of $\Omega^{\ast}\setminus{\gamma}$ containing $x$ and $y$. By the
maximum principle for harmonic functions:

$\sup_{z\in{N_{x,y}^{\gamma}}}h(z)=\max_{z\in{\gamma}}h(z)$

Let $v$ be a point of $\gamma$ so that:

$h(v)=\max_{z\in{\gamma}}h(z)$

Since $0<h(x)<h(y)<h(v)$, it follows that:

$h(y)-h(x)<h(v)$

Consider the region:

$\Omega_{v}=\\{z\in{\Omega}:0<h(z)<h(v)\\}$

$\phi$ maps $\Omega_{v}$ to the rectangle $(0,h(v))\times{(0,1)}$. Thinking of
$\Omega_{v}$ as a conformal rectangle with the distinguished boundary arcs:

$\displaystyle N$
$\displaystyle=\Omega_{v}^{\ast}\cap{[D,A]_{\partial{\Omega^{\ast}}}}$
$\displaystyle E$
$\displaystyle=\Omega_{v}^{\ast}\cap{[A,B]_{\partial{\Omega^{\ast}}}}$
$\displaystyle S$
$\displaystyle=\Omega_{v}^{\ast}\cap{[B,C]_{\partial{\Omega^{\ast}}}}$
$\displaystyle W$ $\displaystyle=\\{z\in{\Omega}:h(z)=h(v)\\}$

it follows that:

$\lambda(W\leftrightarrow{E};\Omega_{v})=h(v)$

Similar to case 1, having reinterpreted $h(v)$ as an extremal length, we can
bound it from above by bounding the dual extremal length from below.

Figure 6: On the left: The subrectangle $\Omega_{v}$ of $\Omega$ with its
distinguished boundary arcs $N,E,S$ and $W$. On the right: Observe that any
path from $N$ to $S$ in $\Omega_{v}$ must cross the annulus $\mathcal{A}$.

Since $v\in{W}$ and $v$ lies along a crosscut $\gamma$ of $\Omega$ that starts
and ends along $[A,B]_{\partial{\Omega^{\ast}}}$, there must exist a subarc
$\gamma^{\prime}$ of $\gamma$ of length at most
$\frac{1}{2}(d_{cc}^{\Omega}(x,y)+\varepsilon)$ travelling from $E$ to $W$ and
thereby separating $N$ and $S$ in $\Omega_{v}$. Let $v^{\prime}$ be the
endpoint of $\gamma^{\prime}$ that lies in $W$. Consider the annulus:

$\mathcal{A}=\\{u\in{\mathbb{C}}:\frac{d_{cc}^{\Omega}(x,y)+\varepsilon}{2}<|u-v^{\prime}|<\frac{d^{\prime}}{2}\\}$

Just as in case 1:

  * •

Since
$\text{length}(\gamma^{\prime})\leq{\frac{d_{cc}^{\Omega}(x,y)+\varepsilon}{2}}$,
the diameter of $\gamma^{\prime}$ is at most
$\frac{d_{cc}^{\Omega}(x,y)+\varepsilon}{2}$. Hence,
$\gamma\subseteq{B(v^{\prime},\frac{d_{cc}^{\Omega}(x,y)+\varepsilon}{2})}$.
Since $\gamma^{\prime}$ separates the boundary arcs $N$ and $S$ in
$\Omega_{v}$, any path from $N$ to $S$ in $\Omega_{v}$ must intersect
$\gamma^{\prime}$ and therefore
$B(v^{\prime},\frac{d_{cc}^{\Omega}(x,y)+\varepsilon}{2})$

  * •

Since any curve from $N$ to $S$ in $\Omega_{v}$ is a curve from
$[B,C]_{\partial{\Omega^{\ast}}}$ to $[D,A]_{\partial{\Omega^{\ast}}}$ in
$\Omega_{v}$, such a curve must have diameter $\geq{d^{\prime}}$. In
particular, any such curve must at some point lie outside the ball
$B(v^{\prime},\frac{d^{\prime}}{2})$.

Putting all this together, we conclude that any curve from $N$ to $S$ in
$\Omega_{v}$ must cross the annulus $\mathcal{A}$ at least once. Hence, by the
same argument as in case 1, plugging the metric:

$\rho(z)=\frac{1}{|z-v^{\prime}|}1_{\mathcal{A}\cap{\Omega_{v}}}(z)$

into the variational problem for $\lambda(N\leftrightarrow{S};\Omega_{v})$, we
have that:

$\lambda(N\leftrightarrow{S};\Omega_{v})\geq{\frac{1}{2\pi}\log\Big{(}\frac{d^{\prime}}{d_{cc}(x,y)+\varepsilon}\Big{)}}$

By duality for continuous extremal length:

$\lambda(W\leftrightarrow{E};\Omega_{v})\cdot\lambda(N\leftrightarrow{S};\Omega_{v})=1$

Hence:

$h(y)-h(x)<h(v)\leq{\frac{2\pi}{\log\Big{(}\frac{d^{\prime}}{d_{cc}(x,y)+\varepsilon}\Big{)}}}$

Since $\varepsilon>0$ was arbitrary, letting $\varepsilon$ tend to $0$ in the
above inequality, the desired result follows.

Case 2.2: $\gamma$ joins $x$ and $y$ to one of the Neumann arcs and separates
it from the opposite Neumann arc.

WLOG, suppose that $\gamma$ joins $x$ and $y$ to
$[B,C]_{\partial{\Omega^{\ast}}}$ and separates $x$ and $y$ from
$[D,A]_{\partial{\Omega^{\ast}}}$. Let $N_{x,y}^{\gamma}$ denote the connected
component of $\Omega^{\ast}\setminus{\gamma}$ containing $x$ and $y$. By the
maximum principle for harmonic functions:

$\displaystyle\sup_{z\in{N_{x,y}^{\gamma}}}h(z)$
$\displaystyle=\max_{z\in{\gamma}}h(z),$
$\displaystyle\inf_{z\in{N_{x,y}^{\gamma}}}h(z)$
$\displaystyle=\min_{z\in{\gamma}}h(z)$

Let $u$ and $v$ be points of $\gamma$ so that:

$\displaystyle h(u)$ $\displaystyle=\min_{z\in{\gamma}}h(z),$ $\displaystyle
h(v)$ $\displaystyle=\max_{z\in{\gamma}}h(z)$

Since $h(u)<h(x)<h(y)<h(v)$, it follows that:

$h(y)-h(x)<h(v)-h(u)$

Consider the region:

$\Omega_{u,v}=\\{z\in{\mathbb{C}}:h(u)<h(z)<h(v)\\}$

$\phi$ maps $\Omega_{u,v}$ to the rectangle $(h(u),h(v))\times{(0,1)}$.
Thinking of $\Omega_{u,v}$ as a conformal rectangle with the distinguished
boundary arcs:

$\displaystyle N$
$\displaystyle=\Omega_{u,v}^{\ast}\cap{[D,A]_{\partial{\Omega^{\ast}}}}$
$\displaystyle E$ $\displaystyle=\\{z\in{\Omega}:h(z)=h(v)\\}$ $\displaystyle
S$ $\displaystyle=\Omega_{u,v}^{\ast}\cap{[B,C]_{\partial{\Omega^{\ast}}}}$
$\displaystyle W$ $\displaystyle=\\{z\in{\Omega}:h(z)=h(u)\\}$

we have that:

$\lambda(W\leftrightarrow{E};\Omega_{u,v})=h(v)-h(u)$

Figure 7: The subrectangle $\Omega_{u,v}$ of $\Omega$ with its distinguished
boundary arcs $N,E,S$ and $W$.

Thus, to get an upper bound for $h(v)-h(u)$ and therefore $h(y)-h(x)$, it
suffices to bound the dual extremal length
$\lambda(N\leftrightarrow{S};\Omega_{u,v})$ from below. Observe that:

  * •

Since $u$ and $v$ both lie along a crosscut of $\Omega$ of length at most
$d_{cc}^{\Omega}(x,y)+\varepsilon$ and $u\in{W}$, $w\in{E}$, we can join $W$
and $E$ by a crosscut of $\Omega_{u,v}$ of length at most
$d_{cc}^{\Omega}(x,y)+\varepsilon$.

  * •

Since any curve from $N$ to $S$ in $\Omega_{u,v}$ is a curve from
$[A,D]_{\partial{\Omega^{\ast}}}$ to $[B,C]_{\partial{\Omega^{\ast}}}$ in
$\Omega$, such a curve will have diameter $\geq{d^{\prime}}$.

Putting all this together, by the same argument as in case 1, verbatim, it
follows that:

$\lambda(N\leftrightarrow{S};\Omega_{u,v})\geq{\frac{1}{2\pi}\log\Big{(}\frac{d^{\prime}}{2(d_{cc}^{\Omega}(x,y)+\varepsilon)}\Big{)}}$

Hence:

$h(y)-h(x)<h(v)-h(u)\leq{\lambda(W\leftrightarrow{E};\Omega_{u,v})}\leq{\frac{2\pi}{\log\Big{(}\frac{d^{\prime}}{2(d_{cc}^{\Omega}(x,y)+\varepsilon)}\Big{)}}}$

Letting $\varepsilon$ tend to $0$ in the above inequality, the desired result
follows. The analogous estimate for $\widetilde{h}$ follows by the same
argument. ∎

###  3.2 Modulus of Continuity for Tiling Maps

Suppose $(G,A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet})$ is an
orthodiagonal rectangle and let $\widetilde{h}$ be the unique solution to the
following boundary value problem on $G^{\circ}$:

$\displaystyle\widetilde{h}(x)$ $\displaystyle=0\hskip 5.0pt\text{for all
$x\in{[D^{\circ},A^{\circ}]}$}$ $\displaystyle\widetilde{h}(x)$
$\displaystyle=1\hskip 5.0pt\text{for all $x\in{[B^{\circ},C^{\circ}]}$}$
$\displaystyle\Delta^{\circ}\widetilde{h}(x)$ $\displaystyle=0\hskip
5.0pt\text{for all
$x\in{V^{\circ}\setminus{([D^{\circ},A^{\circ}]\cup{[B^{\circ},C^{\circ}]})}}$}$

Let $h$ be the solution to the following boundary value problem on
$G^{\bullet}$:

$\displaystyle h(x)$ $\displaystyle=0\hskip 5.0pt\text{for all
$x\in{[A^{\bullet},B^{\bullet}]}$}$ $\displaystyle h(x)$
$\displaystyle=L\hskip 5.0pt\text{for all $x\in{[C^{\bullet},D^{\bullet}]}$}$
$\displaystyle\Delta^{\bullet}h(x)$ $\displaystyle=0\hskip 5.0pt\text{for all
$x\in{V^{\bullet}\setminus{([A^{\bullet},B^{\bullet}]\cup{[C^{\bullet},D^{\bullet}]})}}$}$

where $L$ is the effective resistance between $[A^{\bullet},B^{\bullet}]$ and
$[C^{\bullet},D^{\bullet}]$ in $G^{\bullet}$. $h$ and $\widetilde{h}$ are the
conjugate discrete harmonic functions that correspond to the real and
imaginary parts of the tiling map
$\phi:\widehat{G}\rightarrow{(0,L)\times{(0,1)}}$. In this section we will
prove regularity estimates for $h$ and $\widetilde{h}$ analogous to the
regularity estimates we proved for the real and imaginary parts of the
corresponding conformal map in Section 3.1. We do this by adapting our
argument in Section 3.1 to the discrete, orthodiagonal setting. To do this, we
first need to establish the following lemma which gives us an estimate for the
gradients of $h$ and $\widetilde{h}$ across an edge:

######  Lemma 3.2.

Suppose $(G,A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet})$ is an
orthodiagonal rectangle so that the edges of $G$ all have length at most
$\varepsilon$. Let $\widetilde{h}$ be the unique solution to the following
boundary value problem on $G^{\circ}$:

$\displaystyle\widetilde{h}(x)$ $\displaystyle=0\hskip 5.0pt\text{for all
$x\in{[D^{\circ},A^{\circ}]}$}$ $\displaystyle\widetilde{h}(x)$
$\displaystyle=1\hskip 5.0pt\text{for all $x\in{[B^{\circ},C^{\circ}]}$}$
$\displaystyle\Delta^{\circ}\widetilde{h}(x)$ $\displaystyle=0\hskip
5.0pt\text{for all
$x\in{V^{\circ}\setminus{([D^{\circ},A^{\circ}]\cup{[B^{\circ},C^{\circ}]})}}$}$

Let $h$ be the harmonic conjugate of $\widetilde{h}$ which solves the
following boundary value problem on $G^{\bullet}$:

$\displaystyle h(x)$ $\displaystyle=0\hskip 5.0pt\text{for all
$x\in{[A^{\bullet},B^{\bullet}]}$}$ $\displaystyle h(x)$
$\displaystyle=L\hskip 5.0pt\text{for all $x\in{[C^{\bullet},D^{\bullet}]}$}$
$\displaystyle\Delta^{\bullet}h(x)$ $\displaystyle=0\hskip 5.0pt\text{for all
$x\in{V^{\bullet}\setminus{([A^{\bullet},B^{\bullet}]\cup{[C^{\bullet},D^{\bullet}]})}}$}$

where $L$ is the effective resistance between $[A^{\bullet},B^{\bullet}]$ and
$[C^{\bullet},D^{\bullet}]$ in $G^{\bullet}$. Define:

$\displaystyle d$ $\displaystyle=\inf\\{\text{diameter}(\gamma):\text{$\gamma$
is a curve in $\widehat{G}$ from
$[A^{\bullet},B^{\bullet}]_{\partial{\widehat{G}^{\ast}}}$ to
$[C^{\bullet},D^{\bullet}]_{\partial{\widehat{G}^{\ast}}}$}\\}$ $\displaystyle
d^{\prime}$ $\displaystyle=\inf\\{\text{diameter}(\gamma):\text{$\gamma$ is a
curve in $\widehat{G}$ from
$[B^{\bullet},C^{\bullet}]_{\partial{\widehat{G}^{\ast}}}$ to
$[D^{\bullet},A^{\bullet}]_{\partial{\widehat{G}^{\ast}}}$}\\}$

If $\varepsilon\leq{\frac{d\wedge{d^{\prime}}}{16}}$, there exists an absolute
constant $K>0$ so that if $x,y\in{V^{\bullet}}$ are neighboring edges in
$G^{\bullet}$ and $u,v\in{V^{\circ}}$ are neighboring edges in $G^{\circ}$,
then:

$\displaystyle|h(y)-h(x)|$
$\displaystyle\leq{\frac{K}{\log\big{(}\frac{d^{\prime}}{\varepsilon}\big{)}}}$
$\displaystyle|\widetilde{h}(u)-\widetilde{h}(v)|$
$\displaystyle\leq{\frac{K\hskip
1.0ptL}{\log\big{(}\frac{d}{\varepsilon}\big{)}}}$

###### Proof.

Suppose $(G,A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet})$ is a
orthodiagonal rectangle with edges of length at most $\varepsilon$ and
$h:V^{\bullet}\rightarrow{\mathbb{R}}$ be the solution to the Dirichlet-
Neumann problem on this orthodiagonal map that is $0$ on
$[A^{\bullet},B^{\bullet}]$, $L$ on $[C^{\bullet},D^{\bullet}]$ and harmonic
elsewhere, where $L$ is the discrete extremal length from
$[A^{\bullet},B^{\bullet}]$ to $[C^{\bullet},D^{\bullet}]$ in $G^{\bullet}$.
Define:

$\chi:=\max_{\\{x,y\\}\in{E^{\bullet}}}|h(y)-h(x)|$

If $\chi=0$, we’re done. Otherwise, select neighboring vertices
$x,y\in{V^{\bullet}}$ so that $\big{(}h(y)-h(x)\big{)}=\chi$. Consider the
sets $S_{x}$ and $S_{y}$ defined as follows:

$\displaystyle S_{x}$ $\displaystyle:=\\{z\in{V^{\bullet}}:h(z)\leq{h(x)}\\}$
$\displaystyle S_{y}$ $\displaystyle:=\\{z\in{V^{\bullet}}:h(z)\geq{h(y)}\\}$

By the maximum principle for harmonic functions, $S_{x}$, $S_{y}$ and
$V^{\bullet}\setminus{(S_{x}\sqcup{S_{y}})}$ are all connected subsets of
$G^{\bullet}$. Furthermore, $[A^{\bullet},B^{\bullet}]\subseteq{S_{x}}$,
$[C^{\bullet},D^{\bullet}]\subseteq{S_{y}}$. Let
$H=(V_{H}^{\bullet}\sqcup{V_{H}^{\circ}},E_{H})$ be the suborthodiagonal map
of $G$ formed by gluing together all the faces of $G$ that are incident to at
least one vertex of $V^{\bullet}\setminus{(S_{x}\sqcup{S_{y}})}$. By the
maximum principle, $H$ is simply connected with a unique, distinguished
exterior face. Moreover, let:

$\displaystyle O^{\bullet}$
$\displaystyle:=S_{x}\cap{\partial{V_{H}^{\bullet}}},$ $\displaystyle
W^{\bullet}$ $\displaystyle:=S_{y}\cap{\partial{V_{H}^{\bullet}}}$

Then $O^{\bullet}$ and $W^{\bullet}$ are primal boundary arcs of $H$ with
corresponding dual arcs:

$\displaystyle N^{\circ}$
$\displaystyle:=[B^{\circ},C^{\circ}]\cap{\partial{V_{H}^{\circ}}},$
$\displaystyle S^{\circ}$
$\displaystyle:=[D^{\circ},A^{\circ}]\cap{\partial{V_{H}^{\circ}}}$

Figure 8: The suborthodiagonal map $H$ of $G$ with its four distinguished
boundary arcs. Notice that any path from $N^{\circ}$ to $S^{\circ}$ in
$H^{\circ}$ has to use the dual edge $\\{u,v\\}\in{E^{\circ}}$ corresponding
to $\\{x,y\\}\in{E^{\bullet}}$. (The blue dotted lines are the edges of
$H^{\circ}$.)

Similar to the proof of Theorem 3.1, $N$, $O$, $S$ and $W$ stand for "North,"
"Orient," "South," and "West." We’d have used $E$ for "East," however in the
discrete setting, $E$ is already being used to denote the edges of $G$.
Proposition 2.4 tells us that for any function
$g:V^{\bullet}_{H}\rightarrow{\mathbb{R}}$ with
$\text{gap}_{O^{\bullet},W^{\bullet}}(g)\geq{0}$ and any flow $\theta$ from
$O^{\bullet}$ to $W^{\bullet}$ in $H^{\bullet}$:

$\text{strength}(\theta)\cdot\text{gap}_{O^{\bullet},W^{\bullet}}(g)\leq{\mathcal{E}^{\bullet}(\theta;H)^{1/2}\mathcal{E}^{\bullet}(g;H)^{1/2}}$

Plugging $g=h$ into the inequality above, we have that for any choice of flow
$\theta$ from $O^{\bullet}$ to $W^{\bullet}$ in $H^{\bullet}$:

$\chi=|h(y)-h(x)|\leq{\frac{\mathcal{E}^{\bullet}(\theta;H)^{1/2}\mathcal{E}^{\bullet}(h;H)^{1/2}}{\text{strength}(\theta)}}$

By Thomson’s principle, taking the infimum over all flows $\theta$ from
$O^{\bullet}$ to $W^{\bullet}$ in $H^{\bullet}$ in the expression on the RHS,
we have that:

$\chi\leq{\mathcal{E}^{\bullet}(h;H)^{1/2}\cdot{\lambda(O^{\bullet}\leftrightarrow{W^{\bullet}};H^{\bullet})^{1/2}}}$

In Section 2.3 we saw that $\mathcal{E}^{\bullet}(h)$ is the total area of
rectangles in the tiling associated with the orthodiagonal rectangle
$(G,A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet})$. Hence, by the
definition of $H$, the restriction $\mathcal{E}^{\bullet}(h;H)$ is the total
area of rectangles in our tiling that intersect $(h(x),h(y))\times{(0,1)}$.
Since $|h(y)-h(x)|=\chi$ and any rectangle in our tiling has width at most
$\chi$, it follows that:

$\mathcal{E}^{\bullet}(h;H)\leq{3\chi}$

Hence:

$\chi\leq{3\hskip
1.0pt\lambda(O^{\bullet}\leftrightarrow{W^{\bullet}};H^{\bullet})}$ (3.1)

By duality:

$\lambda(N^{\circ}\leftrightarrow{S^{\circ}};H^{\circ})\cdot\lambda(O^{\bullet}\leftrightarrow{W^{\bullet}};H^{\bullet})=1$

Thus, to bound $\lambda(O^{\bullet}\leftrightarrow{W^{\bullet}};H^{\bullet})$
and therefore $\chi$ from above, it suffices to bound the dual extremal length
$\lambda(N^{\circ}\leftrightarrow{S^{\circ}};H^{\circ})$ from below. We will
do this by picking by picking a good metric to plug into the variational
problem for $\lambda(N^{\circ}\leftrightarrow{S^{\circ}};H^{\circ})$.

Consider the metric $\rho:E_{H}^{\circ}\rightarrow{[0,\infty)}$ defined as
follows:

$\rho(e^{\circ})=\int_{e^{\circ}}\frac{|dz|}{|z-\frac{x+y}{2}|}$

for $e^{\circ}\in{E_{H}^{\circ}}$ contained in the annulus
$\widetilde{\mathcal{A}}=\\{z\in{\mathbb{C}}:4\varepsilon\leq{\left|z-\frac{x+y}{2}\right|}\leq{\frac{d^{\prime}}{2}}\\}$,
where we think of $e^{\circ}$ as a line segment in the plane. If $e^{\circ}$
is not contained in $\widetilde{\mathcal{A}}$, then $\rho(e^{\circ})=0$. Let
$\\{u,v\\}\in{E^{\circ}_{H}}$ be the dual edge corresponding to the primal
edge $\\{x,y\\}$. Observe that:

  1. 1.

By Lemma 2.7, minimal $O^{\bullet}$\- $W^{\bullet}$ cuts in $H^{\bullet}$
correspond to simple paths from $N^{\circ}$ to $S^{\circ}$ in $H^{\circ}$ and
vice versa. Since $\\{x,y\\}$ joins $O^{\bullet}$ and $W^{\bullet}$, this edge
is part of any $O^{\bullet}$\- $W^{\bullet}$ cut. Hence, any path from
$N^{\circ}$ to $S^{\circ}$ in $H^{\circ}$ must use the edge $\\{u,v\\}$.

  2. 2.

Any path from $N^{\circ}$ to $S^{\circ}$ in $H^{\circ}$ is a path from
$[D^{\circ},A^{\circ}]$ to $[B^{\circ},C^{\circ}]$ in $G^{\circ}$. Using the
notation in the statement of Lemma 3.2, it follows that the diameter of any
such path is at least $d^{\prime}$.

By 1, since $\\{u,v\\}\subseteq{B(\frac{x+y}{2},4\varepsilon)}$, any path in
$H^{\circ}$ from $N^{\circ}$ to $S^{\circ}$ must at some point lie in the
bounded component of $\mathbb{C}\setminus{\widetilde{\mathcal{A}}}$. By 2, any
path in $H^{\circ}$ from $N^{\circ}$ to $S^{\circ}$ must at some point lie in
the unbounded component of $\mathbb{C}\setminus{\widetilde{\mathcal{A}}}$.
Putting all this together, we conclude that any path from $N^{\circ}$ to
$S^{\circ}$ in $H^{\circ}$ crosses the annulus $\widetilde{\mathcal{A}}$ at
least once. If $\gamma$ is a piecewise $C^{1}$ curve that crosses the annulus
$\\{z\in{\mathbb{C}}:r<|z-z_{0}|<R\\}$ at least once, where
$z_{0}\in{\mathbb{C}}$, $R>r>0$, then:

$\int_{\gamma}\frac{|dz|}{|z-z_{0}|}\geq{\log\Big{(}\frac{R}{r}\Big{)}}$

If $\gamma$ is a path from $N^{\circ}$ to $S^{\circ}$ in $H^{\circ}$, we know
that $\gamma$ must cross the annulus $\widetilde{A}$ at least once. However,
when doing this, it is possible that $\gamma$ uses edges of $E^{\circ}_{H}$
that are not entirely contained in $\widetilde{\mathcal{A}}$ and so have zero
mass with respect to $\rho$. Since edges of $G$ all have length at most
$\varepsilon$, edges of $G^{\circ}$ all have length at most $2\varepsilon$.
Thus, we can be sure that all of the edges $e^{\circ}\in{E^{\circ}_{H}}$ that
we use when crossing the slightly smaller annulus
$\\{z\in{\mathbb{C}}:6\varepsilon\leq{\left|z-\frac{x+y}{2}\right|}\leq{\frac{d^{\prime}}{2}-2\varepsilon}\\}$
do indeed have positive mass. Hence:

$\ell_{\rho}(\gamma)=\int_{\gamma}\frac{|dz|}{|z-\frac{x+y}{2}|}\geq{\log{\Big{(}\frac{d^{\prime}-4\varepsilon}{12\varepsilon}\Big{)}}}$

Let $F_{in}(H)$ denote the set of inner faces of $H$. Given an inner face
$Q\in{F_{in}(H)}$, let
$e_{Q}^{\bullet}\in{E^{\bullet}_{H}},e_{Q}^{\circ}\in{E^{\circ}_{H}}$ denote
the primal and dual diagonals of $Q$. Next, we estimate the area of the metric
$\rho$:

$\displaystyle A(\rho)$
$\displaystyle=\sum_{\begin{subarray}{c}e^{\circ}\in{E_{H}^{\circ}}\\\
\rho(e^{\circ})>0\end{subarray}}\frac{|e^{\bullet}|}{|e^{\circ}|}\Big{(}\int_{e^{\circ}}\frac{|dz|}{|z-\frac{x+y}{2}|}\Big{)}^{2}\leq{\sum_{\begin{subarray}{c}e^{\circ}\in{E_{H}^{\circ}}\\\
\rho(e^{\circ})>0\end{subarray}}\frac{|e^{\bullet}|}{|e^{\circ}|}\cdot{\frac{|e^{\circ}|^{2}}{\min\limits_{z\in{e^{\circ}}}|z-\frac{x+y}{2}|^{2}}}}\stackrel{{\scriptstyle(i)}}{{\leq}}{2\sum_{\begin{subarray}{c}Q\in{F_{in}(H)}\\\
\rho(e_{Q}^{\circ})>0\end{subarray}}\frac{\text{Area}(Q)}{\min\limits_{z\in{Q}}|z-\frac{x+y}{2}|^{2}}}$
$\displaystyle\stackrel{{\scriptstyle(ii)}}{{\leq}}{2\sum_{\begin{subarray}{c}Q\in{F_{in}(H)}\\\
\rho(e_{Q}^{\circ})>0\end{subarray}}\int_{Q}\frac{1}{(|z-\frac{x+y}{2}|-2\varepsilon)^{2}}dz_{1}dz_{2}}\stackrel{{\scriptstyle(iii)}}{{\leq}}{2\int_{3\varepsilon<|z-\frac{x+y}{2}|<\frac{d^{\prime}}{2}+\varepsilon}\frac{1}{(|z-\frac{x+y}{2}|-2\varepsilon)^{2}}dz_{1}dz_{2}}$
$\displaystyle=4\pi\int_{3\varepsilon}^{\frac{d^{\prime}}{2}+\varepsilon}\frac{s}{(s-2\varepsilon)^{2}}ds=4\pi\int_{\varepsilon}^{\frac{d^{\prime}}{2}-\varepsilon}\frac{1}{u}du+8\pi\varepsilon\int_{\varepsilon}^{\frac{d^{\prime}}{2}-\varepsilon}\frac{1}{u^{2}}du$
$\displaystyle=4\pi\log\Big{(}\frac{d^{\prime}-2\varepsilon}{\varepsilon}\Big{)}+8\pi\varepsilon\left(\frac{1}{\varepsilon}-\frac{2}{(d^{\prime}-2\varepsilon)}\right)$
$\displaystyle\leq{4\pi\log\Big{(}\frac{d^{\prime}-2\varepsilon}{\varepsilon}\Big{)}+8\pi}$

In the calculations above:

  * •

(i) follows from the trivial observation that
$\min\limits_{z\in{Q}}|z-\frac{x+y}{2}|\leq{\min\limits_{z\in{e^{\circ}_{Q}}}|z-\frac{x+y}{2}|}$,
as well as the fact that for any inner face $Q$ of $H$, since $Q$ is a
quadrilateral with orthogonal diagonals,
$\text{Area}(Q)=\frac{1}{2}|e_{Q}^{\bullet}||e_{Q}^{\circ}|$.

  * •

(ii) follows from the fact that for any inner face $Q$ of $H$, since the edges
of $Q$ all have length at most $\varepsilon$, $Q$ has diameter at most
$2\varepsilon$. Hence,
$|z-\frac{x+y}{2}|-2\varepsilon\leq{\min\limits_{z\in{Q}}|z-\frac{x+y}{2}|}$
for any $z\in{Q}$.

  * •

(iii) follows from the fact that if $\rho(e^{\circ}_{Q})>0$ for some inner
face $Q$ of $H$, this means that the dual edge $e^{\circ}_{Q}$ is contained in
the annulus $\widetilde{\mathcal{A}}$. Since the corresponding primal edge
$e^{\bullet}_{Q}$ has length at most $\varepsilon$, any point of $Q$ lies in
an $\varepsilon$\- neighborhood of $e^{\circ}_{Q}$. In particular, any inner
face $Q$ of $H$ such that $\rho(e^{\circ}_{Q})>0$ is contained in the slightly
larger annulus
$\\{z\in{\mathbb{C}}:3\varepsilon<|z-\frac{x+y}{2}|<\frac{d^{\prime}}{2}+\varepsilon\\}$.

Since $d^{\prime}\geq{16\varepsilon}$, it follows that:

$A(\rho)\leq{4\pi\log\Big{(}\frac{d^{\prime}-2\varepsilon}{\varepsilon}\Big{)}+8\pi}\asymp{\log\Big{(}\frac{d^{\prime}}{\varepsilon}\Big{)}}$

If $\gamma$ is a path from $N^{\circ}$ to $S^{\circ}$ in $H^{\circ}$, using
the fact that $d^{\prime}\geq{16\varepsilon}$:

$\ell_{\rho}(\gamma)\geq{\log{\Big{(}\frac{d^{\prime}-4\varepsilon}{16\varepsilon}\Big{)}}}\asymp{\log\Big{(}\frac{d^{\prime}}{\varepsilon}\Big{)}}$

Putting all this together, we have that:

$\lambda(N^{\circ}\leftrightarrow{S^{\circ}};H^{\circ})\geq{\inf\limits_{\gamma}\frac{\ell^{2}_{\rho}(\gamma)}{A(\rho)}}\gtrsim{\log\Big{(}\frac{d^{\prime}}{\varepsilon}\Big{)}}$

where our infimum is over all paths $\gamma$ in $H^{\circ}$ from $N^{\circ}$
to $S^{\circ}$. By duality:

$\lambda(O^{\bullet}\leftrightarrow{W^{\bullet}};H^{\bullet})=\frac{1}{\lambda(N^{\circ}\leftrightarrow{S^{\circ}};H^{\circ})}\lesssim{\frac{1}{\log\big{(}\frac{d^{\prime}}{\varepsilon}\big{)}}}$

Plugging this into Equation 3.1, the desired result follows. The corresponding
estimate for $\widetilde{h}$ follows by the same argument. ∎

Having established our estimate for the gradient of $h$ and $\widetilde{h}$
across an edge, we are now ready to state and prove the discrete analogue of
Theorem 3.1:

######  Theorem 3.3.

Suppose $(G,A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet})$ is an
orthodiagonal rectangle with edges of length at most $\varepsilon$ and let $h$
and $\widetilde{h}$ be the real and imaginary parts of the associated tiling
map. That is, $\widetilde{h}$ is the unique solution to the following boundary
value problem on $G^{\circ}$:

$\displaystyle\widetilde{h}(x)$ $\displaystyle=0\hskip 5.0pt\text{for all
$x\in{[D^{\circ},A^{\circ}]}$}$ $\displaystyle\widetilde{h}(x)$
$\displaystyle=1\hskip 5.0pt\text{for all $x\in{[B^{\circ},C^{\circ}]}$}$
$\displaystyle\Delta^{\circ}\widetilde{h}(x)$ $\displaystyle=0\hskip
5.0pt\text{for all
$x\in{V^{\circ}\setminus{([D^{\circ},A^{\circ}]\cup{[B^{\circ},C^{\circ}]})}}$}$

and $h$ be the harmonic conjugate of $\widetilde{h}$ which solves the
following boundary value problem on $G^{\bullet}$:

$\displaystyle h(x)$ $\displaystyle=0\hskip 5.0pt\text{for all
$x\in{[A^{\bullet},B^{\bullet}]}$}$ $\displaystyle h(x)$
$\displaystyle=L\hskip 5.0pt\text{for all $x\in{[C^{\bullet},D^{\bullet}]}$}$
$\displaystyle\Delta^{\bullet}h(x)$ $\displaystyle=0\hskip 5.0pt\text{for all
$x\in{V^{\bullet}\setminus{([A^{\bullet},B^{\bullet}]\cup{[C^{\bullet},D^{\bullet}]})}}$}$

where $L$ is the effective resistance between $[A^{\bullet},B^{\bullet}]$ and
$[C^{\bullet},D^{\bullet}]$ in $G^{\bullet}$. Define:

$\displaystyle d$ $\displaystyle=\inf\\{\text{diameter}(\gamma):\text{$\gamma$
is a curve in $\widehat{G}$ joining
$[A^{\bullet},B^{\bullet}]_{\partial{\widehat{G}}}$ and
$[C^{\bullet},D^{\bullet}]_{\partial{\widehat{G}}}$}\\}$ $\displaystyle
d^{\prime}$ $\displaystyle=\inf\\{\text{diameter}(\gamma):\text{$\gamma$ is a
curve in $\widehat{G}$ joining
$[B^{\bullet},C^{\bullet}]_{\partial{\widehat{G}}}$ and
$[D^{\bullet},A^{\bullet}]_{\partial{\widehat{G}}}$}\\}$

Then there exists an absolute constant $K>0$ so that:

$\displaystyle|h(y)-h(x)|$
$\displaystyle\leq{\frac{K}{\log{\big{(}\frac{d^{\prime}}{(d_{\widehat{G}}(x,y)\wedge{d_{cc}^{\widehat{G}}(x,y)})\vee{\varepsilon}}\big{)}}}},$
$\displaystyle|\widetilde{h}(u)-\widetilde{h}(v)|$
$\displaystyle\leq{\frac{KL}{\log{\big{(}\frac{d}{(d_{\widehat{G}}(u,v)\wedge{d_{cc}^{\widehat{G}}(u,v)})\vee{\varepsilon}}\big{)}}}}$

for any $x,y\in{V^{\bullet}}$, $u,v\in{V^{\circ}}$ such that . In particular,
in the bulk (when
$|x-y|\leq{\text{dist}(x,\partial{\widehat{G}}),\text{dist}(y,\partial{\widehat{G}})}$
and
$|u-v|\leq{\text{dist}(u,\partial{\widehat{G}}),\text{dist}(v,\partial{\widehat{G}})}$),
we have that:

$\displaystyle|h(y)-h(x)|$
$\displaystyle\leq{\frac{K}{\log{\big{(}\frac{d^{\prime}}{|y-x|\vee{\varepsilon}}\big{)}}}},$
$\displaystyle|\widetilde{h}(u)-\widetilde{h}(v)|$
$\displaystyle\leq{\frac{KL}{\log{\big{(}\frac{d}{|u-v|\vee{\varepsilon}}\big{)}}}}$

A slightly weaker estimate (with $\sqrt{log(\cdot)}$ in place of
$\log(\cdot)$) is proven in [3] in the case where our orthodiagonal rectangle
$(G,A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet})$ is a good approximation
of a conformal rectangle $(\Omega,A,B,C,D)$, where $\Omega$ is a Jordan domain
and
$[A,B]_{\partial{\Omega}},[B,C]_{\partial{\Omega}},[C,D]_{\partial{\Omega}},[D,A]_{\partial{\Omega}}$
are analytic arcs that don’t meet at cusps. In the context of [3], “good"
means that the edges of $G$ have length at most $\varepsilon$ and discrete
boundary arcs are $\delta$\- close to the corresponding continuous boundary
arcs in Hausdorff distance for some small $\varepsilon,\delta>0$. For more
details, see Theorem 3 of [3].

We should also point out that while the estimates for the modulus of
continuity of $h$ and $\widetilde{h}$ in Theorem 3.3 are novel for points near
the boundary of $G$, even when our orthodiagonal map is just a chunk of the
square grid, in the bulk we can do significantly better. Namely, in [9],
Chelkak, Laslier and Russkikh show that bounded discrete harmonic and discrete
holomorphic functions on $t$\- embeddings whose corresponding origami map is
$\kappa$\- Lipschitz on large scales for some $\kappa\in{(0,1)}$, are
$\beta$\- Hölder in the bulk for some absolute constant $\beta\in{(0,1)}$ (see
Proposition 6.13 of [9]). The condition on the origami map here is known as
“$Lip(\kappa,\delta)$" where $\delta>0$ is the scale on which our origami map
is $\kappa$\- Lipschitz. $t$\- embeddings are a more general class of graphs
embedded in the plane that accommodate a notion of discrete complex analysis.
That is, every orthodiagonal map is a $t$\- embedding (see Section 8.1 of
[9]). While it is not explicitly stated in their paper, it is not hard to show
that if $G$ is an orthodiagonal map with edges of length at most
$\varepsilon$, then for any $\kappa\in{(0,1)}$ there exists an absolute
constant $c=c(\kappa)>0$ such that, as a $t$ -embedding, $G$ satisfies the
condition $Lip(\kappa,c\varepsilon)$. While the estimates in the bulk provided
by Theorem 3.3 are sufficient for proving our main result, Theorem 2.15, as an
immediate consequence of Proposition 6.13 of [9] we have that:

######  Proposition 3.4.

Suppose $(G,A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet})$ is an
orthodiagonal rectangle with edges of length at most $\varepsilon$ and $h$ and
$\widetilde{h}$ are the real and imaginary parts of the corresponding tiling
map. If $x\in{V}$, let $d_{x}$ denote the distance from $x$ to
$\partial{\widehat{G}}$. Then there exists absolute constants
$C,C^{\prime}>0$, $\beta\in{(0,1)}$ such that:

$\displaystyle|h(y)-h(x)|$
$\displaystyle\leq{CL\left(\frac{|x-y|}{d_{x}\wedge{d_{y}}}\right)^{\beta}}$
$\displaystyle|\widetilde{h}(u)-\widetilde{h}(v)|$
$\displaystyle\leq{C\left(\frac{|u-v|}{d_{u}\wedge{d_{v}}}\right)^{\beta}}$

for $x,y\in{V^{\bullet}}$, $u,v\in{V^{\circ}}$ such that
$|x-y|,|u-v|\geq{C^{\prime}\varepsilon}$.

All that having been said, we now turn to the proof of Theorem 3.3:

###### Proof.

Suppose $(G,A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet})$ is a
orthodiagonal rectangle with edges of length at most $\varepsilon$ and $h$ is
the real part of the corresponding tiling map. Fix $x,y\in{V^{\bullet}}$.
Since we are taking the maximum of
$d_{\widehat{G}}(x,y)\wedge{d_{cc}^{\widehat{G}}(x,y)}$ and $\varepsilon$ in
our estimate for $|h(y)-h(x)|$ in Theorem 3.3, we can assume WLOG that
$d_{\widehat{G}}(x,y)\wedge{d_{cc}^{\widehat{G}}(x,y)}\geq{\varepsilon}$,
since the relevant estimate in the case where
$d_{\widehat{G}}(x,y)\wedge{d_{cc}^{\widehat{G}}(x,y)}<\varepsilon$ follows
from the case where
$d_{\widehat{G}}(x,y)\wedge{d_{cc}^{\widehat{G}}(x,y)}\geq{\varepsilon}$, by
the maximum principle. If $h(y)=h(x)$, we’re done. Otherwise, suppose WLOG
that $h(x)<h(y)$. We now consider two cases:

Case 1: $d_{\widehat{G}}(x,y)\leq{d_{cc}^{\widehat{G}}(x,y)}$

Similar to the proofs of Theorem 3.1 and Lemma 3.2, we begin by reinterpreting
the quantity we want to estimate, $\big{(}h(y)-h(x)\big{)}$, as an extremal
length. Consider the sets $S_{x}$ and $S_{y}$ defined as follows:

$\displaystyle S_{x}$ $\displaystyle:=\\{z\in{V^{\bullet}}:h(z)\leq{h(x)}\\}$
$\displaystyle S_{y}$ $\displaystyle:=\\{z\in{V^{\bullet}}:h(z)\geq{h(y)}\\}$

By the maximum principle for harmonic functions, $S_{x}$, $S_{y}$ and
$V^{\bullet}\setminus{(S_{x}\sqcup{S_{y}})}$ are all connected subsets of
$G^{\bullet}$. Furthermore, $[A^{\bullet},B^{\bullet}]\subseteq{S_{x}}$,
$[C^{\bullet},D^{\bullet}]\subseteq{S_{y}}$. Let
$H=(V_{H}^{\bullet}\sqcup{V_{H}^{\circ}},E_{H})$ be the suborthodiagonal map
of $G$ formed by gluing together all the faces of $G$ that are incident to at
least one vertex of $V^{\bullet}\setminus{(S_{x}\sqcup{S_{y}})}$. By the
maximum principle, $H$ is simply connected with a unique, distinguished
exterior face. Furthermore, let:

$\displaystyle O^{\bullet}$
$\displaystyle:=S_{x}\cap{\partial{V_{H}^{\bullet}}},$ $\displaystyle
W^{\bullet}$ $\displaystyle:=S_{y}\cap{\partial{V_{H}^{\bullet}}}$

Then $O^{\bullet}$ and $W^{\bullet}$ are primal boundary arcs of $H$ with
corresponding dual arcs:

$\displaystyle N^{\circ}$
$\displaystyle:=[D^{\circ},A^{\circ}]\cap{\partial{V_{H}^{\circ}}},$
$\displaystyle S^{\circ}$
$\displaystyle:=[B^{\circ},C^{\circ}]\cap{\partial{V_{H}^{\circ}}}$

Proposition 2.4 tells us that for any function
$g:V^{\bullet}_{H}\rightarrow{\mathbb{R}}$ with
$\text{gap}_{O^{\bullet},W^{\bullet}}(g)\geq{0}$ and any flow $\theta$ from
$O^{\bullet}$ to $W^{\bullet}$ in $H^{\bullet}$:

$\text{strength}(\theta)\cdot\text{gap}_{O^{\bullet},W^{\bullet}}(g)\leq{\mathcal{E}^{\bullet}(\theta;H)^{1/2}\mathcal{E}^{\bullet}(g;H)^{1/2}}$

Plugging $g=h$ into the inequality above, we have that for any choice of flow
$\theta$ from $O^{\bullet}$ to $W^{\bullet}$ in $H^{\bullet}$:

$|h(y)-h(x)|\leq{\frac{\mathcal{E}^{\bullet}(\theta;H)^{1/2}\mathcal{E}^{\bullet}(h;H)^{1/2}}{\text{strength}(\theta)}}$

By Thomson’s principle, taking the infimum over all flows $\theta$ from
$O^{\bullet}$ to $W^{\bullet}$ in $H^{\bullet}$ in the expression on the RHS,
we have that:

$|h(y)-h(x)|\leq{\mathcal{E}^{\bullet}(h;H)^{1/2}\cdot{\lambda(O^{\bullet}\leftrightarrow{W^{\bullet}};H^{\bullet})^{1/2}}}$
(3.2)

In Section 2.3 we saw that $\mathcal{E}^{\bullet}(h)$ is the total area of
rectangles in the tiling associated with the orthodiagonal rectangle
$(G,A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet})$. Hence, by the
definition of $H$, the restriction $\mathcal{E}^{\bullet}(h;H)$ is the total
area of rectangles in our tiling that intersect $(h(x),h(y))\times{(0,1)}$.
Since the edges of $G$ have length at most $\varepsilon$, Lemma 3.2 tells us
that the width of any rectangle in our tiling is at most
$K\big{(}\log\big{(}\frac{d^{\prime}}{\varepsilon}\big{)}\big{)}^{-1}$, where
$K>0$ is an absolute constant. Hence:

$\mathcal{E}^{\bullet}(h;H)\leq{|h(y)-h(x)|+\frac{2K}{\log\big{(}\frac{d^{\prime}}{\varepsilon}\big{)}}}$
(3.3)

Using the shorthand
$\lambda^{\bullet}=\lambda(O^{\bullet}\leftrightarrow{W^{\bullet}};H^{\bullet})$,
plugging our estimate in Equation 3.3 into Equation 3.2 and rearranging, we
have that:

$|h(y)-h(x)|\leq{\frac{1}{2}\Big{(}\lambda^{\bullet}+\sqrt{(\lambda^{\bullet})^{2}+\frac{8K\lambda^{\bullet}}{\log\big{(}\frac{d^{\prime}}{\varepsilon}\big{)}}}\Big{)}}\leq{\lambda^{\bullet}+\frac{2K}{\log\big{(}\frac{d^{\prime}}{\varepsilon}\big{)}}}=\lambda^{\bullet}(O^{\bullet}\leftrightarrow{W^{\bullet}};H^{\bullet})+\frac{2K}{\log\big{(}\frac{d^{\prime}}{\varepsilon}\big{)}}$
(3.4)

By duality:

$\lambda(N^{\circ}\leftrightarrow{S^{\circ}};H^{\circ})\cdot\lambda(O^{\bullet}\leftrightarrow{W^{\bullet}};H^{\bullet})=1$

Thus, to bound $\lambda(O^{\bullet}\leftrightarrow{W^{\bullet}};H^{\bullet})$
and therefore $|h(y)-h(x)|$ from above, it suffices to bound the dual extremal
length $\lambda(N^{\circ}\leftrightarrow{S^{\circ}};H^{\circ})$ from below. We
will do this by picking by picking a good metric to plug into the variational
problem for $\lambda(N^{\circ}\leftrightarrow{S^{\circ}};H^{\circ})$.

Figure 9: On the left: The subrectangle $H$ of $G$ with its distinguished
boundary arcs $N^{\circ},O^{\bullet},S^{\circ}$, and $W^{\bullet}$. On the
right: Notice that any path from $N^{\circ}$ to $S^{\circ}$ in $H^{\circ}$
must cross the annulus $\widetilde{\mathcal{A}}$. For simplicity, our
orthodiagonal map $G$ here is just a chunk of the square grid but, in general,
this need not be the case.

By the definition of $d_{\widehat{G}}(x,y)$ we can find a smooth curve
$\gamma$ in $\widehat{G}$ from $x$ to $y$ so that
$\text{length}(\gamma)<d_{\Omega}(x,y)+\varepsilon$. By the maximum principle
for discrete harmonic functions, $\widehat{G}\setminus{\widehat{H}}$ consists
of two connected components $\Omega_{1}$ and $\Omega_{2}$ so that WLOG,
$[A^{\bullet},B^{\bullet}]\subseteq{\partial{\Omega_{1}}}$ and
$[C^{\bullet},D^{\bullet}]\subseteq{\partial{\Omega_{1}}}$. Thinking of
$\widehat{H}$ as a conformal rectangle with the distinguished boundary arcs:

$\displaystyle\widehat{W}$
$\displaystyle:=\partial{\Omega_{1}}\cap{\partial{\widehat{H}}}$
$\displaystyle\widehat{O}$
$\displaystyle:=\partial{\Omega_{2}}\cap{\partial{\widehat{H}}}$
$\displaystyle\widehat{N}$
$\displaystyle:=[D^{\bullet},A^{\bullet}]_{\partial{\widehat{G}}}\cap{\partial{\widehat{H}}}$
$\displaystyle\widehat{S}$
$\displaystyle:=[B^{\bullet},C^{\bullet}]_{\partial{\widehat{G}}}\cap{\partial{\widehat{H}}}$

our curve $\gamma$ from $x$ to $y$ in $\widehat{G}$ must have a subarc
$\gamma^{\prime}$ with endpoints
$x^{\prime}\in{\widehat{W}},y^{\prime}\in{\widehat{O}}$, so that
$\gamma^{\prime}$ is a crosscut of $\widehat{H}$. By the definition of
$\widehat{H}$, every point of $\widehat{W}$ lies within $\varepsilon$ of a
point of $W^{\bullet}$ and every point of $\widehat{O}$ lies within
$\varepsilon$ of a point in $O^{\bullet}$. Thus, perturbing $\gamma^{\prime}$
slightly, we can produce a crosscut $\gamma^{\prime\prime}$ of $\widehat{H}$
with endpoints
$x^{\prime\prime}\in{\widehat{W}},y^{\prime\prime}\in{\widehat{O}}$ so that
$\text{length}(\gamma^{\prime\prime})\leq{d_{\widehat{G}}(x,y)+3\varepsilon}$.
Since $\gamma^{\prime\prime}$ is a crosscut of $\widehat{H}$ joining
$O^{\bullet}$ and $W^{\bullet}$, $\gamma^{\prime\prime}$ separates $N^{\circ}$
and $S^{\circ}$ in $\widehat{H}$. Hence, any path in $\widehat{H}$ and
therefore $H^{\circ}$ from $N^{\circ}$ to $S^{\circ}$ must intersect
$\gamma^{\prime\prime}$. Consider the annulus:

$\widetilde{\mathcal{A}}:=\\{u\in{\mathbb{C}}:d_{\widehat{G}}(x,y)+3\varepsilon<|u-x^{\prime\prime}|<\frac{d^{\prime}}{2}\\}$

Observe that:

  1. 1.

Since
$\text{length}(\gamma^{\prime\prime})\leq{d_{\widehat{G}}(x,y)+3\varepsilon}$,
it follows that
$\gamma^{\prime\prime}\subseteq{B(x^{\prime\prime},d_{\widehat{G}}(x,y)+3\varepsilon)}$.
Since any path from $N^{\circ}$ to $S^{\circ}$ in $\widehat{H}$ must intersect
$\gamma^{\prime\prime}$, such a path must also intersect the ball
$B(x^{\prime\prime},d_{\widehat{G}}(x,y)+3\varepsilon)$.

  2. 2.

On the other hand, since any path from $N^{\circ}$ to $S^{\circ}$ in
$\widehat{H}$ is a also path from
$[D^{\bullet},A^{\bullet}]_{\partial{\widehat{G}}}$ to
$[B^{\bullet},C^{\bullet}]_{\partial{\widehat{G}}}$ in $\widehat{G}$, any such
curve has diameter $\geq{d^{\prime}}$. Hence, any curve from $N^{\circ}$ to
$S^{\circ}$ in $\widehat{H}$ must at some point exit the ball
$B(x^{\prime\prime},\frac{d^{\prime}}{2})$.

Putting all this together, we have that any path from $N^{\circ}$ to
$S^{\circ}$ in $H^{\circ}$ must cross the annulus $\widetilde{\mathcal{A}}$.
Having established this fact, consider the metric
$\rho:E^{\circ}_{H}\rightarrow{[0,\infty)}$ defined as follows:

$\rho(e^{\circ})=\int_{e^{\circ}}\frac{|dz|}{|z-x^{\prime\prime}|}$

for $e^{\circ}\in{E_{H}^{\circ}}$ contained in the annulus
$\widetilde{\mathcal{A}}$, where we think of $e^{\circ}$ as a line segment in
the plane. If $e^{\circ}$ is not contained in $\widetilde{\mathcal{A}}$, then
$\rho(e^{\circ})=0$. Suppose $\gamma$ is a path in $H^{\circ}$ from
$N^{\circ}$ to $S^{\circ}$. We saw earlier that $\gamma$ must cross the
annulus $\widetilde{\mathcal{A}}$ at least once. However, when it does this,
it is possible that $\gamma$ uses edges $e^{\circ}\in{E^{\circ}_{H}}$ that are
not entirely contained in $\widetilde{\mathcal{A}}$ and so have zero mass with
respect to $\rho$. Since edges of $G$ have length at most $\varepsilon$, edges
of $G^{\circ}$ have length at most $2\varepsilon$. Thus, we can be sure that
all of the edges we use when crossing the slightly smaller annulus
$\\{u\in{\mathbb{C}}:d_{\widehat{G}}(x,y)+5\varepsilon<|u-x^{\prime\prime}|<\frac{d^{\prime}}{2}-2\varepsilon\\}$
do indeed have positive mass. Hence:

$l_{\rho}(\gamma)=\int_{\gamma}\frac{|dz|}{|z-x^{\prime\prime}|}\geq{\log\Big{(}\frac{d^{\prime}-4\varepsilon}{2(d_{\widehat{G}}(x,y)+5\varepsilon)}\Big{)}}$

By the same argument as in the proof of Lemma 3.2:

$A(\rho)\leq{4\pi\log\Big{(}\frac{d^{\prime}-2\varepsilon}{2\hskip
1.0ptd_{\widehat{G}}(x,y)}\Big{)}}+\frac{8\pi\varepsilon}{d_{\widehat{G}}(x,y)}$

By our assumption that
$\varepsilon\leq{d_{\widehat{G}}(x,y)}<\frac{d^{\prime}}{4}$ and
$\varepsilon<\frac{d^{\prime}}{12}$, we have that:

$A(\rho)\leq{4\pi\log\Big{(}\frac{d^{\prime}-2\varepsilon}{2\hskip
1.0ptd_{\widehat{G}}(x,y)}\Big{)}}+\frac{8\pi\varepsilon}{d_{\widehat{G}}(x,y)}\lesssim{\log\Big{(}\frac{d^{\prime}}{d_{\widehat{G}}(x,y)}\Big{)}}$

Similarly, if $\gamma$ is a path from $N^{\circ}$ to $S^{\circ}$ in
$H^{\circ}$, we have that:

$l_{\rho}(\gamma)=\int_{\gamma}\frac{|dz|}{|z-x^{\prime\prime}|}\geq{\log\Big{(}\frac{d^{\prime}-4\varepsilon}{2(d_{\widehat{G}}(x,y)+5\varepsilon)}\Big{)}}\gtrsim{\log\Big{(}\frac{d^{\prime}}{d_{\widehat{G}}(x,y)}\Big{)}}$

Putting all this together, we have that:

$\lambda(N^{\circ}\leftrightarrow{S^{\circ}};H^{\circ})\geq{\inf\limits_{\gamma}\frac{\ell^{2}_{\rho}(\gamma)}{A(\rho)}}\gtrsim{\log\Big{(}\frac{d^{\prime}}{d_{\widehat{G}}(x,y)}\Big{)}}$

where our infimum is over all paths $\gamma$ in $H^{\circ}$ from $N^{\circ}$
to $S^{\circ}$. By duality:

$\lambda(O^{\bullet}\leftrightarrow{W^{\bullet}};H^{\bullet})=\frac{1}{\lambda(N^{\circ}\leftrightarrow{S^{\circ}};H^{\circ})}\lesssim{\frac{1}{\log\Big{(}\frac{d^{\prime}}{d_{\widehat{G}}(x,y)}\Big{)}}}$

Plugging this into Equation 3.4, the desired result follows.

Case 2: $d_{cc}^{\widehat{G}}(x,y)\leq{d_{\widehat{G}}(x,y)}$ By the
definition of $d_{cc}^{\widehat{G}}(x,y)$, we can find a crosscut $\gamma$ of
$\widehat{G}$ that joins $x$ and $y$ to one of the four distinguished boundary
arcs of $(\widehat{G},A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet})$ and
separates it from the opposite boundary arc, such that
$\text{length}(\gamma)<d_{cc}^{\widehat{G}}(x,y)+\varepsilon$. We now split
our problem into two further cases, depending on whether the relevant boundary
arcs of $(\widehat{G},A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet})$ are
Dirichlet arcs where $h$ is constant, or Neumann arcs along which $h$ is
monotone.

Case 2.1: $\gamma$ joins $x$ and $y$ to one of the Dirichlet arcs and
separates them from the opposite Dirichlet arc.

WLOG, suppose $\gamma$ joins $x$ and $y$ to
$[A^{\bullet},B^{\bullet}]_{\partial{\widehat{G}}}$ and separates them from
$[C^{\bullet},D^{\bullet}]_{\partial{\widehat{G}}}$. Let
$\widehat{N}_{x,y}^{\gamma}$ be the connected component of
$\widehat{G}\setminus{\gamma}$ containing $x$ and $y$. Define:

$\displaystyle(N_{x,y}^{\gamma})^{\bullet}$
$\displaystyle:=V^{\bullet}\cap{\widehat{N}_{x,y}^{\gamma}},$
$\displaystyle\gamma^{\bullet}$
$\displaystyle:=\\{z\in{V^{\bullet}}:\text{dist}(z,\gamma)\leq{\varepsilon}\\}$

By the maximum principle for discrete harmonic functions,

$\max_{z\in{(N_{x,y}^{\gamma})^{\bullet}}}h(z)\leq{\max_{z\in{\gamma^{\bullet}}}h(z)}$

Let $v$ be a vertex of $\gamma^{\bullet}$ so that
$h(v)=\max\limits_{z\in{\gamma^{\bullet}}}h(z)$ and for some neighboring
vertex $u$ of $v$ in $V^{\bullet}$, $h(u)<h(v)$. Since
$x,y\in{(N_{x,y}^{\gamma})^{\bullet}}$, we have that:

$(h(y)-h(x))<h(v)$

Similar to our argument in case 1, let $H$ be the suborthodiagonal map of $G$
formed by gluing together all of the quadrilaterals of $G$ that are tangent to
a vertex $z$ of $V^{\bullet}$ such that $0\leq{h(z)}<{h(v)}$. By the maximum
principle, $H$ is simply connected with a unique, distinguished exterior face.
Furthermore, let:

$\displaystyle O^{\bullet}$ $\displaystyle:=[A^{\bullet},B^{\bullet}],$
$\displaystyle W^{\bullet}$
$\displaystyle:=\\{z\in{V^{\bullet}}:h(z)\geq{h(v)}\\}\cap{\partial{V_{H}^{\bullet}}}$

Then $O^{\bullet}$ and $W^{\bullet}$ are primal boundary arcs of $H$ with
corresponding dual arcs:

$\displaystyle N^{\circ}$
$\displaystyle:=[B^{\circ},C^{\circ}]\cap{\partial{V_{H}^{\circ}}},$
$\displaystyle S^{\circ}$
$\displaystyle:=[D^{\circ},A^{\circ}]\cap{\partial{V_{H}^{\circ}}}$

With these pairs of distinguished primal and dual arcs, $H$ is an
orthodiagonal rectangle. By the same argument as in case 1, almost verbatim,
we have that:

$|h(y)-h(x)|\leq{h(v)}\leq{\lambda^{\bullet}(O^{\bullet}\leftrightarrow{W^{\bullet}};H^{\bullet})+\frac{2K}{\log\big{(}\frac{d^{\prime}}{\varepsilon}\big{)}}}$
(3.5)

By duality:

$\lambda(N^{\circ}\leftrightarrow{S^{\circ}};H^{\circ})\cdot\lambda(O^{\bullet}\leftrightarrow{W^{\bullet}};H^{\bullet})=1$

Thus, to bound $\lambda(O^{\bullet}\leftrightarrow{W^{\bullet}};H^{\bullet})$
and therefore $|h(y)-h(x)|$ from above, it suffices to bound the dual extremal
length $\lambda(N^{\circ}\leftrightarrow{S^{\circ}};H^{\circ})$ from below. We
will do this by picking a good metric to plug into the variational formula for
$\lambda(N^{\circ}\leftrightarrow{S^{\circ}};H^{\circ})$.

Figure 10: On the left: The subrectangle $H$ of $G$ with its distinguished
boundary arcs $N^{\circ},O^{\bullet},S^{\circ}$ and
$W^{\bullet}=[A^{\bullet},B^{\bullet}]$. On the right: Notice that any path
from $N^{\circ}$ to $S^{\circ}$ in $H^{\circ}$ must cross the annulus
$\widetilde{\mathcal{A}}$. For simplicity, our orthodiagonal map $G$ here is
just a chunk of the square grid but, in general, this need not be the case.

Observe that $\gamma$ is a crosscut of $\widehat{G}$ that starts and ends at a
point of $[A^{\bullet},B^{\bullet}]_{\partial{\widehat{G}}}$ and, by the
definition of $\gamma^{\bullet}$, $v$ lies within $\varepsilon$ of $\gamma$.
Since $v$ has a neighboring vertex $u$ with the property that $h(u)<h(v)$, we
also have that $v\in{W^{\bullet}}$. Since the edges of $G$ have length at most
$\varepsilon$, putting all this together, it follows that there exists a
crosscut $\gamma^{\prime}$ of $\widehat{H}$ of length at most
$\frac{1}{2}(d_{cc}^{\widehat{G}}(x,y)+5\varepsilon)$ joining $W^{\bullet}$
and $O^{\bullet}$ in $\widehat{H}$, thereby separating $N^{\circ}$ and
$S^{\circ}$ in $\widehat{H}$. Pick a point $x^{\prime}$ in $\gamma^{\prime}$
and consider the annulus:

$\widetilde{A}=\\{u\in{\mathbb{C}}:\frac{1}{2}(d_{cc}^{\widehat{G}}(x,y)+5\varepsilon)<{|u-x^{\prime}|<\frac{d^{\prime}}{2}}\\}$

Since any path in $\widehat{G}$ from
$[B^{\circ},C^{\circ}]_{\partial{\widehat{G}}}$ to
$[A^{\circ},D^{\circ}]_{\partial{\widehat{G}}}$ has diameter greater than or
equal to $d^{\prime}$, the same is true of any path in $\widehat{H}$ from
$N^{\circ}$ to $S^{\circ}$. On the other hand, since $\gamma^{\prime}$
separates $N^{\circ}$ and $S^{\circ}$ in $\widehat{H}$, any such path must
intersect $\gamma^{\prime}$ and therefore
$B(x^{\prime},\frac{1}{2}(d_{cc}^{\widehat{G}}(x,y)+5\varepsilon))$. Putting
all this together, we conclude that any path from $N^{\circ}$ to $S^{\circ}$
in $\widehat{H}$ must cross the annulus $\widetilde{A}$. Hence, if we define
the metric $\rho:E_{H}^{\circ}\rightarrow{(0,\infty)}$ by the formula:

$\rho(e^{\circ})=\begin{cases}\int_{e^{\circ}}\frac{|dz|}{|z-x^{\prime}|}\hskip
5.0pt\text{for edges $e^{\circ}\in{E^{\circ}_{H}}$ contained in the annulus
$\widetilde{A}$}\\\ 0\hskip 5.0pt\text{otherwise}\end{cases}$

by the same argument as in case 1, almost verbatim, plugging $\rho$ into the
variational formula for
$\lambda(N^{\circ}\leftrightarrow{S^{\circ}};H^{\circ})$, we have that:

$\lambda(N^{\circ}\leftrightarrow{S^{\circ}};H^{\circ})\gtrsim{\log\Big{(}\frac{d^{\prime}}{d_{cc}^{\widehat{G}}(x,y)}\Big{)}}$

By duality:

$\lambda(O^{\bullet}\leftrightarrow{W^{\bullet}};H^{\bullet})\lesssim{\frac{1}{\log\Big{(}\frac{d^{\prime}}{d_{cc}^{\widehat{G}}(x,y)}\Big{)}}}$

Plugging this into Equation 3.5, the desired result follows.

Case 2.2: $\gamma$ joins $x$ and $y$ to one of the Neumann arcs and separates
them from the opposite Neumann arc.

WLOG, suppose $\gamma$ joins $x$ and $y$ to
$[B^{\bullet},C^{\bullet}]_{\partial{\widehat{G}}}$ and separates them from
$[D^{\bullet},A^{\bullet}]_{\partial{\widehat{G}}}$. Similar to case 2.1, let
$\widehat{N}_{x,y}^{\gamma}$ be the connected component of
$\widehat{G}\setminus{\gamma}$ containing $x$ and $y$ and define:

$\displaystyle(N^{\gamma}_{x,y})^{\bullet}$
$\displaystyle:=V^{\bullet}\cap{\widehat{N}^{\gamma}_{x,y}},$
$\displaystyle\gamma^{\bullet}$
$\displaystyle:=\\{z\in{V^{\bullet}}:\text{dist}(z,\gamma)\leq{\varepsilon}\\}$

By the maximum principle for harmonic functions, since $h^{\bullet}$ is
harmonic on the part of the boundary of $(N^{\gamma}_{x,y})^{\bullet}$ that
intersects $[B^{\bullet},C^{\bullet}]$, we have that:

$\displaystyle\min_{z\in{\gamma^{\bullet}}}h(z)$
$\displaystyle\leq{\min_{z\in{(\widehat{N}^{\gamma}_{x,y})^{\bullet}}}h(z)},$
$\displaystyle\max_{z\in{(\widehat{N}^{\gamma}_{x,y})^{\bullet}}}h(z)$
$\displaystyle\leq{\max_{z\in{\gamma^{\bullet}}}h(z)}$

Let $v_{1}$ and $v_{2}$ be vertices of $\gamma^{\bullet}$ so that
$h(v_{1})=\min_{z\in{\gamma^{\bullet}}}h(z)$,
$h(v_{2})=\max_{z\in{\gamma^{\bullet}}}h(z)$ and $v_{1}$ and $v_{2}$ have
neighboring vertices $u_{1}$ and $u_{2}$ so that $h(v_{1})<h(u_{1})$ and
$h(v_{2})>h(u_{2})$. Since $x,y\in{(N_{\gamma}^{x,y})^{\bullet}}$, we have
that:

$\big{(}h(y)-h(x)\big{)}\leq{\big{(}h(v_{2})-h(v_{1})\big{)}}$

Similar to our argument in case 1, let $H$ be the suborthodiagonal map of $G$
formed by gluing together all of the quadrilaterals of $G$ that are tangent to
a vertex $z$ of $V^{\bullet}$ such that $h(v_{1})\leq{h(z)}<{h(v_{2})}$. By
the maximum principle, $H$ is simply connected with a unique, distinguished
exterior face. Furthermore, let:

$\displaystyle O^{\bullet}$
$\displaystyle:=\\{z\in{V^{\bullet}}:h(z)\geq{h(v_{1})}\\}\cap{\partial{V_{H}^{\bullet}}},$
$\displaystyle W^{\bullet}$
$\displaystyle:=\\{z\in{V^{\bullet}}:h(z)\leq{h(v_{2})}\\}\cap{\partial{V_{H}^{\bullet}}}$

Then $O^{\bullet}$ and $W^{\bullet}$ are primal boundary arcs of $H$ with
corresponding dual arcs:

$\displaystyle N^{\circ}$
$\displaystyle:=[D^{\circ},A^{\circ}]\cap{\partial{V_{H}^{\circ}}},$
$\displaystyle S^{\circ}$
$\displaystyle:=[B^{\circ},C^{\circ}]\cap{\partial{V_{H}^{\circ}}}$

With these pairs of distinguished primal and dual arcs, $H$ is an
orthodiagonal rectangle. By the same argument as in case 1, almost verbatim,
we have that:

$|h(y)-h(x)|\leq{h(v_{2})-h(v_{1})}\leq{\lambda^{\bullet}(O^{\bullet}\leftrightarrow{W^{\bullet}};H^{\bullet})+\frac{2K}{\log\big{(}\frac{d^{\prime}}{\varepsilon}\big{)}}}$
(3.6)

where $K>0$ is an absolute constant. By duality:

$\lambda(N^{\circ}\leftrightarrow{S^{\circ}};H^{\circ})\cdot\lambda(O^{\bullet}\leftrightarrow{W^{\bullet}};H^{\bullet})=1$

Thus, to bound $\lambda(O^{\bullet}\leftrightarrow{W^{\bullet}};H^{\bullet})$
and therefore $|h(y)-h(x)|$ from above, it suffices to bound the dual extremal
length $\lambda(N^{\circ}\leftrightarrow{S^{\circ}};H^{\circ})$ from below.

Figure 11: The subrectangle $H$ of $G$ with its distinguished boundary arcs
$N^{\circ},O^{\bullet},S^{\circ}$ and $W^{\bullet}$. For simplicity, our
orthodiagonal map $G$ here is just a chunk of the square grid but, in general,
this need not be the case.

Since $v_{1}\in{W^{\bullet}}$, $v_{2}\in{O^{\bullet}}$ and each of these
points lies within $\varepsilon$ of $\gamma$, there exists a crosscut
$\gamma^{\prime}$ of $\widehat{H}$ of length at most
$d_{cc}^{\widehat{G}}(x,y)+3\varepsilon$ joining $O^{\bullet}$ and
$W^{\bullet}$ in $\widehat{H}$ and thereby separating $N^{\circ}$ and
$S^{\circ}$ in $\widehat{H}$. Hence, by the same argument we used in cases 1
and 2.1, we have that:

$\lambda(N^{\circ}\leftrightarrow{S^{\circ}};H^{\circ})\gtrsim{\log\Big{(}\frac{d^{\prime}}{d_{cc}^{\widehat{G}}(x,y)}\Big{)}}$

By duality:

$\lambda(O^{\bullet}\leftrightarrow{W^{\bullet}};H^{\bullet})\lesssim{\frac{1}{\log\Big{(}\frac{d^{\prime}}{d_{cc}^{\widehat{G}}(x,y)}\Big{)}}}$

Plugging this into Equation 3.6, the desired result follows.

Having shown that $h$ has the prescribed modulus of continuity, observe that
the analogous estimate for $\tilde{h}$ follows by the same argument. This
completes our proof. ∎

###  3.3 Two- Sided Estimates for Extremal Length on Orthodiagonal Maps

In this section, we prove two- sided estimates for the discrete extremal
length between opposite boundary arcs for an orthodiagonal rectangle that is a
$(\delta,\varepsilon)$\- good approximation of some conformal rectangle
$(\Omega,A,B,C,D)$. This gives us uniform boundedness of the tiling maps
$(\phi_{n})_{n=1}^{\infty}$ in Theorem 2.15.

######  Proposition 3.5.

Suppose $(\Omega,A,B,C,D)$ is a conformal rectangle and
$(G,A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet})$ is a
$(\delta,\varepsilon)$\- good interior approximation of $(\Omega,A,B,C,D)$.
Suppose $\delta<\frac{\ell\wedge{\ell^{\prime}}}{2}$, where:

$\displaystyle\ell=\inf\\{\text{length}(\gamma):\text{$\gamma$ is a curve in
$\Omega$ joining $[A,B]_{\partial{\Omega^{\ast}}}$ and
$[C,D]_{\partial{\Omega^{\ast}}}$}\\},$
$\displaystyle\ell^{\prime}=\inf\\{\text{length}(\gamma):\text{$\gamma$ is a
curve in $\Omega$ joining $[B,C]_{\partial{\Omega^{\ast}}}$ and
$[D,A]_{\partial{\Omega^{\ast}}}$}\\}$

Then:

$\frac{(\ell-2\delta)^{2}}{2\cdot{\text{Area}(\Omega)}}\leq{\lambda([A^{\bullet},B^{\bullet}]\leftrightarrow{[C^{\bullet},D^{\bullet}]};G^{\bullet})}\leq{\frac{2\cdot\text{Area}(\Omega)}{(\ell^{\prime}-2\delta)^{2}}}$

###### Proof.

Let:

$\displaystyle\lambda^{\bullet}$
$\displaystyle=\lambda([A^{\bullet},B^{\bullet}]\leftrightarrow{[C^{\bullet},D^{\bullet}]};G^{\bullet}),$
$\displaystyle\lambda^{\circ}$
$\displaystyle=\lambda([B^{\circ},C^{\circ}]\leftrightarrow{[D^{\circ},A^{\circ}]};G^{\circ})$

Plugging the metric $\rho(e^{\bullet})=|e^{\bullet}|$ into the variational
problem for $\lambda^{\bullet}$, we have that:

$\displaystyle\lambda^{\bullet}$
$\displaystyle\geq{\frac{\Big{(}\inf\\{\sum\limits_{e^{\bullet}\in{\gamma}}|e^{\bullet}|:\text{$\gamma$
is a path from $[A^{\bullet},B^{\bullet}]$ to $[C^{\bullet},D^{\bullet}]$ in
$G^{\bullet}$}\\}\Big{)}^{2}}{\sum\limits_{e^{\bullet}\in{E^{\bullet}}}\frac{|e^{\circ}|}{|e^{\bullet}|}|e^{\bullet}|^{2}}}$
$\displaystyle=\frac{\Big{(}\inf\\{\text{length}(\gamma):\text{$\gamma$ is a
path from $[A^{\bullet},B^{\bullet}]$ to $[C^{\bullet},D^{\bullet}]$ in
$G^{\bullet}$}\\}\Big{)}^{2}}{2\cdot\text{Area}(\widehat{G})}$
$\displaystyle\geq{\frac{(\ell-2\delta)^{2}}{2\cdot{\text{Area}(\Omega)}}}$

The equality on the second line follows from the fact that if $Q$ is an inner
face of $G$ with primal diagonal $e^{\bullet}$ and dual diagonal $e^{\circ}$,
then $\text{Area}(Q)=\frac{1}{2}|e^{\bullet}|{|e^{\circ}|}$. Thus:

$\sum_{e^{\bullet}\in{E^{\bullet}}}|e^{\bullet}||e^{\circ}|=2\sum_{Q\in{F_{in}}}\text{Area}(Q)=2\cdot{\text{Area}(\widehat{G})}$

In the inequality on the third line, two things are going on. On the one hand,
since $G$ is an interior approximation of $\Omega$,
$\text{Area}(\widehat{G})\leq{\text{Area}(\Omega)}$. On the other hand,
suppose $\gamma$ is a path from $[A^{\bullet},B^{\bullet}]$ to
$[C^{\bullet},D^{\bullet}]$ in $G^{\bullet}$. Since
$(G,A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet})$ is a
$(\delta,\varepsilon)$\- good approximation of $(\Omega,A,B,C,D)$ (the
“$\delta$" is really the relevant part here), we can modify any such path to
get a path of length at most $\text{length}(\gamma)+2\delta$ from
$[A,B]_{\partial{\Omega^{\ast}}}$ to $[C,D]_{\partial{\Omega^{\ast}}}$ in
$\Omega$. By the definition of $\ell$:

$\text{length}(\gamma)+2\delta\geq{\ell}$

Since this is true of any curve $\gamma$ from $[A^{\bullet},B^{\bullet}]$ to
$[C^{\bullet},D^{\bullet}]$ in $G^{\bullet}$, the desired result follows.

Similarly, plugging the metric $\rho(e^{\circ})=|e^{\circ}|$ into the
variational problem for $\lambda^{\circ}$ we have that:

$\displaystyle\lambda^{\circ}$
$\displaystyle\geq{\frac{\Big{(}\inf\\{\sum\limits_{e^{\circ}\in{\gamma}}|e^{\circ}|:\text{$\gamma$
is a path from $[B^{\circ},C^{\circ}]$ to $[D^{\circ},A^{\circ}]$ in
$G^{\circ}$}\\}\Big{)}^{2}}{\sum\limits_{e^{\bullet}\in{E^{\circ}}}\frac{|e^{\bullet}|}{|e^{\circ}|}|e^{\circ}|^{2}}}$
$\displaystyle=\frac{\Big{(}\inf\\{\text{length}(\gamma):\text{$\gamma$ is a
path from $[B^{\circ},C^{\circ}]$ to $[D^{\circ},A^{\circ}]$ in
$G^{\circ}$}\\}\Big{)}^{2}}{2\cdot\text{Area}(\widehat{G})}$
$\displaystyle\geq{\frac{(\ell^{\prime}-2\delta)^{2}}{2\cdot{\text{Area}(\Omega)}}}$

By Ford- Fulkerson duality for discrete rectangles (Corollary 2.8):

$\lambda^{\bullet}=\frac{1}{\lambda^{\circ}}\leq{\frac{2\cdot\text{Area}(\Omega)}{(\ell^{\prime}-2\delta)^{2}}}$

∎

The two- sided estimate in Proposition 3.5 is very coarse, but it is
sufficient for our purposes. It is however worth noting that, as a consequence
of our estimates for the modulus of continuity of rectangle tiling maps as
well as the corresponding limiting conformal map, we can do significantly
better if we are dealing with continuous approximations of a continuous domain
or discrete approximations of a discrete domain.

######  Proposition 3.6.

If $(\Omega^{\prime},A^{\prime},B^{\prime},C^{\prime},D^{\prime})$ is a
$\delta$\- good approximation of the conformal rectangle $(\Omega,A,B,C,D)$,
$L$ is the extremal length between $[A,B]_{\partial{\Omega^{\ast}}}$ and
$[C,D]_{\partial{\Omega^{\ast}}}$ in $\Omega$, $L^{\prime}$ is the extremal
length between $[A^{\prime},B^{\prime}]_{\partial{\Omega^{\prime}}}$ and
$[C^{\prime},D^{\prime}]_{\partial{\Omega^{\prime}}}$ in $\Omega^{\prime}$,
and $\delta>0$ satisfies $\delta\leq{\frac{d^{\prime}}{2}e^{-\frac{4\pi}{L}}}$
and $\delta\leq{\frac{d}{2}e^{-16\pi{L}}}$, where:

$\displaystyle d$ $\displaystyle=\inf\\{\text{diameter}(\gamma):\text{$\gamma$