free-boundary tilings of $R$ that cover $R$ but do not extend to tilings of
all of $\mathbb{Z}^{3}$. (These might exist, for example, if $R$ is a concave
region.)

We define a metric on flows (Section 5.2), denoted $d_{W}$, which is an
adaptation of generalized Wasserstein distance from signed measures to flows.
Intuitively we want to consider two flows $f,g$ “close” if we don’t have to
change the flow of $f$ too much—either by moving flow over, or by adding or
deleting it—to transform it into $g$. This is what $d_{W}(f,g)$ will measure.
In terms of this metric, the main question of this section is: if $f_{n}\in
TF_{n}(R)$ for all $n\in\mathbb{N}$, what are the possible limits of the form
$\lim_{n\to\infty}f_{n}$?

We show (Theorem 5.3.1) that any fine-mesh limit of free-boundary tiling flows
on $R$ is an asymptotic flow on $R$, defined by:

######  Definition 5.0.3.

Let $R$ be a compact region which is the closure of a connected domain and has
piecewise smooth boundary. We say that $f$ is an asymptotic flow on $R$ if it
satisfies the following properties:

  * •

$f$ is a Borel-measureable vector field with support contained in $R$;

  * •

$f$ is valued in $\mathcal{O}$ (since $f$ is measurable, this means that $f$
is valued in $\mathcal{O}$ Lebesgue-a.e.);

  * •

$f$ is divergence-free in the interior of $R$, i.e. $\text{div}\,f=0$ as a
distribution (so for any smooth function $\phi$ compactly supported in the
interior of $R$,
$\int_{R}\phi\,\text{div}\,f:=\int_{R}\nabla\phi\,\cdot\,f=0$.)

We denote the set of asymptotic flows on $R$ by $AF(R)$.

In Theorem 5.3.4 we will show that $(AF(R),d_{W})$ is a compact metric space.
In Sections 5.4 and 5.5, we define a boundary value operator $T$ (trace
operator) which takes a flow to its boundary value on $\partial R$. In fact we
do something more general, and define the trace of a flow for any compact,
piecewise smooth surface contained in $R$. After defining $T$ for asymptotic
flows, we define the space of asymptotic flows with boundary value $b$,
denoted $AF(R,b)$, and show that it is compact with respect to $d_{W}$
(Corollary 5.4.9).

The boundary value operator is defined in different but analogous ways for
asymptotic flows (Section 5.4) and tiling flows (Section 5.5). The main
essential result about $T$ is that these definitions are compatible and that
$T$ is uniformly continuous (Theorem 5.5.7).

We remark that the main important property of Wasserstein distance in our
analysis is that it metrizes weak convergence, and that it thereby formalizes
the intuitive notions that the scaling limits of tiling flows are asymptotic
flows, and that boundary values depend continuously on the flow. While the
Wasserstein metric and other transportation metrics have a number of
additional special properties, we do not use this theory here. All the
properties of the Wasserstein metric that we use are described in Section 5.1.

###  5.1 Background on (generalized) Wasserstein distance

The original Wasserstein distance or earth-movers distance is a metric on
probability measures on a fixed metric space. Suppose that $(X,d)$ is a
compact, separable metric space. The $L^{1}$ Wasserstein distance $W_{1}$ is a
metric on $\mathcal{P}(X)$, the space of probability measures on $X$ and is
given by

$\displaystyle W_{1}(\mu,\nu):=\inf_{\gamma\in\Gamma(\mu,\nu)}\int_{X\times
X}d(x,y)d\gamma(x,y)$

where $\Gamma(\mu,\nu)$ is the collection of all couplings of $\mu$ and $\nu$.
Intuitively $W_{1}$ measures the cost—i.e. how much mass and how far it has to
be moved—required to transform $\mu$ into $\nu$ by redistributing the mass of
$\mu$. This metric was developed in the theory of optimal transport and has
been applied in many different contexts including probability, Riemannian
geometry, and image processing. For more see [Vil09].

We will define and use versions of Wasserstein distance for flows and their
boundary values. In the next section, we define a mapping between flows and
measures, where flow (with direction) corresponds to mass (with sign). The
measures corresponding to flows do not necessarily have the same mass and can
be signed. Given this, our Wasserstein distance on flows will be based on a
version of generalized Wasserstein distance.

Let $\mathcal{M}(\mathbb{R}^{d})$ denote the space of Borel regular measures
on $\mathbb{R}^{d}$ with finite total mass. In [PR14] and [PR16], they define
a generalized Wasserstein distance on $\mathcal{M}(\mathbb{R}^{d})$ by
introducing an $L^{1}$ cost for adding and deleting mass. It is denoted
$W_{1}^{1,1}$ and defined as

$\displaystyle
W_{1}^{1,1}(\mu,\nu)=\inf_{\tilde{\mu},\tilde{\nu}}|\mu-\tilde{\mu}|+|\nu-\tilde{\nu}|+W_{1}(\tilde{\mu},\tilde{\nu})$

where the infimum is taken over $\mathcal{M}(\mathbb{R}^{d})$.

In [AMS11] the $L^{1}$ Wasserstein distance was generalized to signed
probability measures. This metric is denoted $\mathbb{W}_{1}$. If $\mu,\nu$
are signed measures with Jordan decompositions $\mu=\mu_{+}-\mu_{-}$ and
$\nu=\nu_{+}-\nu_{-}$, then

$\displaystyle\mathbb{W}_{1}(\mu,\nu)=W_{1}(\mu_{+}+\nu_{-},\nu_{+}+\mu_{-}).$

In fact, note that this definition does not depend on the decomposition of the
measures $\mu,\nu$.

In [PRT19], they combine these to give a definition of Wasserstein distance
for signed measures of different masses. Let $\mathcal{M}^{s}(\mathbb{R}^{d})$
denote the space of signed Radon measures on $\mathbb{R}^{d}$, i.e. measures
$\mu$ that can be written $\mu_{+}-\mu_{-}$ for
$\mu\pm\in\mathcal{M}(\mathbb{R}^{d})$. Denoted $\mathbb{W}_{1}^{1,1}$, the
generalized Wasserstein distance on $\mathcal{M}^{s}(\mathbb{R}^{d})$ is
defined

$\displaystyle\mathbb{W}_{1}^{1,1}(\mu,\nu)=W_{1}^{1,1}(\mu_{+}+\nu_{-},\nu_{+}+\mu_{-}).$

This is the definition of Wasserstein distance that we will use in this paper.
We note a few important facts about $\mathbb{W}_{1}^{1,1}$ that we will use.

######  Lemma 5.1.1 (See [PRT19, Lemma 18]).

If $\mu,\nu,\rho\in\mathcal{M}^{s}(\mathbb{R}^{d})$, then

$\displaystyle\mathbb{W}_{1}^{1,1}(\mu,\nu)=\mathbb{W}_{1}^{1,1}(\mu+\rho,\nu+\rho).$

######  Proposition 5.1.2 (See [PRT19, Proposition 23]).

Let

$\displaystyle\mathcal{C}_{b}^{0,Lip}=\\{f:\mathbb{R}^{d}\to\mathbb{R}\leavevmode\nobreak\
:\leavevmode\nobreak\ f\text{ continuous, bounded, Lipschitz}\\}.$

Then

$\displaystyle\mathbb{W}_{1}^{1,1}(\mu,\nu)=\sup\bigg{\\{}\int_{\mathbb{R}^{d}}\varphi\,\mathrm{d}(\mu-\nu)\leavevmode\nobreak\
:\leavevmode\nobreak\
\varphi\in\mathcal{C}_{b}^{0,Lip},\|\varphi\|_{\infty}\leq
1,\|\varphi\|_{Lip}\leq 1\bigg{\\}}.$

From this it clearly follows that

######  Corollary 5.1.3.

If $\lim_{n\to\infty}\mathbb{W}_{1}^{1,1}(\mu_{n},\mu)=0$, then $\mu_{n}$
converges weakly to $\mu$.

The non-signed generalized Wasserstein distance $W^{1,1}_{1}$ metrizes weak
convergence for tight sequences of measures in $\mathcal{M}(\mathbb{R}^{d})$
[PR14, Theorem 13], as does the original $L^{1}$ Wasserstein distance for
probability measures [Vil09, Theorem 6.9]. With signed measures, slightly
stranger behavior can occur in general, see e.g. [PRT19, Remark 26]. However
we show that the Wasserstein distance for flows defined below does metrize
weak convergence, see Remark 5.2.5.

For $R\subset\mathbb{R}^{d}$, we define $\mathcal{M}(R)$ to be the Radon
measures supported in $R$, and $\mathcal{M}^{s}(R)$ to be signed Radon
measures supported in $R$. We let $\mathcal{M}_{\text{ac}}(R)$ (resp.
$\mathcal{M}^{s}_{\text{ac}}(R)$) denote the Radon measures (resp. signed
Radon measures) supported in $R$ and absolutely continuous with respect to
Lebesgue measure on $\mathbb{R}^{d}$. We say that
$\mathcal{M}_{\text{ac}}^{s}(R,a,b)$ denotes absolutely continuous signed
measures with densities valued between $a$ and $b$. By Lemma 5.1.1, for any
$a<0$ and $b>0$,

$(\mathcal{M}_{\text{ac}}^{s}(R,a,b),\mathbb{W}_{1}^{1,1})\cong(\mathcal{M}_{\text{ac}}(R,0,b-a),W_{1}^{1,1})$
(14)

as metric spaces. This identification has some useful consequences. From
Equation (14) and [PR14, Proposition 15] it follows that

######  Proposition 5.1.4.

If $R$ is compact, then
$(\mathcal{M}_{\text{ac}}^{s}(R,a,b),\mathbb{W}_{1}^{1,1})$ is a compact
metric space for $a,b\in\mathbb{R}$.

###  5.2 Wasserstein distance for flows

Let $R$ be a compact region which is the closure of a connected domain and has
piecewise smooth boundary. If $f_{\tau}\in TF_{n}(R)$, then $f_{\tau}$ is
supported in $B_{2/n}(R)=\\{x:d(x,R)\leq 2/n\\}$. We will define a
correspondence between

  1. 1.

vector fields $f$ on $R\subset\mathbb{R}^{3}$ or $f_{\tau}\in TF_{n}(R)$, and

  2. 2.

triples of signed measures $(\mu_{1},\mu_{2},\mu_{3})$ supported in
$B_{2/n}(R)$.

The idea is that the flow of the vector field in coordinate direction $i$
corresponds to mass of the $i^{th}$ measure, with sign coming from the
direction of the flow. We define Wasserstein distance on vector fields through
this correspondence:

######  Definition 5.2.1.

The Wasserstein distance on flows, denoted $d_{W}$, is the sum of the
generalized Wasserstein distances between the component measures. For any two
vector fields $f,g$ with corresponding triples of measures
$(\mu_{1},\mu_{2},\mu_{3})$ and $(\nu_{1},\nu_{2},\nu_{3})$, we define

$\displaystyle
d_{W}(f,g):=\mathbb{W}_{1}^{1,1}(\mu_{1},\nu_{1})+\mathbb{W}_{1}^{1,1}(\mu_{2},\nu_{2})+\mathbb{W}_{1}^{1,1}(\mu_{3},\nu_{3}).$

To complete the definition of the metric, we need to define the
correspondences between vector fields and triples of measures. There will be
two definitions, one for a measurable vector field on $R$ and one for a
discrete vector field $f_{\tau}\in TF_{n}(R)$. Let $x=(x_{1},x_{2},x_{3})$
denote a point in $\mathbb{R}^{3}$.

######  Definition 5.2.2.

(Measures corresponding to a measurable vector field.) Let $f$ be a measurable
vector field supported in $R$. The components of
$f(x)=(f_{1}(x),f_{2}(x),f_{3}(x))$ are measurable functions, and we define
the corresponding triple of measures $(\mu_{1},\mu_{2},\mu_{3})$ by

$\displaystyle\mathrm{d}\mu_{i}(x)=f_{i}(x)\,\mathrm{d}x_{1}\mathrm{d}x_{2}\mathrm{d}x_{3}\qquad\qquad
i=1,2,3$

where $\mathrm{d}x_{1}\mathrm{d}x_{2}\mathrm{d}x_{3}$ denotes Lebesgue measure
on $\mathbb{R}^{3}$.

######  Definition 5.2.3.

(Measures corresponding to a free-boundary tiling flow on $R$.) Suppose that
$f=f_{\tau}\in TF_{n}(R)$ for some $n$. Let $\eta_{1},\eta_{2},\eta_{3}$ be
the positively-oriented unit basis vectors. Orient all the edges $e$ of
$\frac{1}{n}\mathbb{Z}^{3}$ to be parallel to $\eta_{i}$, which we denote by
$e\parallel\eta_{i}$. (Recall that changing the orientation of $e$ changes the
sign of $f(e)$.) The triple of measures $(\mu_{1},\mu_{2},\mu_{3})$
corresponding to $f$ is given by

$\displaystyle\mathrm{d}\mu_{i}(x)=\sum_{{e}\parallel\eta_{i}}\frac{2}{n^{2}}f({e})\mathbbm{1}_{e}(x)\mathrm{d}x_{i}\qquad\qquad
i=1,2,3$

where $\mathbbm{1}_{e}$ denotes the indicator of the edge
${e}\in\frac{1}{n}\mathbb{Z}^{3}$, and $\mathrm{d}x_{i}$ is 1-dimensional
Lebesgue measure in the direction of $\eta_{i}$. Note that $\mu_{i}$ is
supported in $B_{2/n}(R)$.

######  Remark 5.2.4.

The scaling factor $\frac{2}{n^{2}}$ ensures that each edge $e$ such that
$e\parallel\eta_{i}$ contributes $\frac{2f(e)}{n^{3}}$ total mass to
$\mu_{i}$. The normalization is justified by looking at the extreme examples
corresponding to the brickwork tilings (i.e. tilings where all tiles are the
same type). Each cube in the $\frac{1}{n}\mathbb{Z}^{3}$ mesh can be viewed as
corresponding to its lower left edge. In the brickwork pattern, exactly half
of these cubes will have a dimer in the lower left edge.

######  Remark 5.2.5.

Now that Wasserstein distance on flows is defined, we can explain why it
metrizes weak convergence of the component measures. We do this by explaining
how we could “shift” everything to have positive mass and use Equation (14).
For asymptotic flows, we can just add a copy of the 3-dimensional Lebesgue
measure $\mathrm{d}x_{1}\mathrm{d}x_{2}\mathrm{d}x_{3}$. For tiling flows, we
note that we could have defined the corresponding measures to be positive by
translating the mean-current octahedron $\mathcal{O}$ by
$\eta_{1}+\eta_{2}+\eta_{3}$. After the translation, a scale $n$ tiling flow
measure would take values in
$\\{1/(3n^{2}),5/(3n^{2}),7/(3n^{2}),11/(3n^{2})\\}$ instead of
$\\{-5/(3n^{2}),-1/(3n^{2}),1/(3n^{2}),5/(3n^{2})\\}$.

In terms of adding measures, translating $\mathcal{O}$ corresponds to adding a
copy of 1-dimensional Lebesgue $\frac{2}{n^{2}}\mathrm{d}x_{i}$ along each
edge $e\parallel\eta_{i}$ in $\frac{1}{n}\mathbb{Z}^{3}$ to the scale $n$
tiling flow measure $\mathrm{d}\mu_{i}$. In the scaling limit as $n\to\infty$,
this sum of 1-dimensional Lebesgue measures converges in
$\mathbb{W}_{1}^{1,1}$ to $\mathrm{d}x_{1}\mathrm{d}x_{2}\mathrm{d}x_{3}$. By
Equation (14), this implies the “translated” tiling flows measures converge
(i.e. ones defined on the translated $\mathcal{O}$) to the “translated”
asymptotic flow measures (i.e. ones shifted by adding
$\mathrm{d}x_{1}\mathrm{d}x_{2}\mathrm{d}x_{3})$ in $W_{1}^{1,1}$ if and only
if the tiling flow measures converge to the asymptotic flow measures in
$\mathbb{W}_{1}^{1,1}$. Since $W_{1}^{1,1}$ metrizes weak convergence [PR14,
Proposition 15], $d_{W}$ metrizes weak convergence of the component measures
corresponding to tiling and asymptotic flows.

######  Proposition 5.2.6.

The measures corresponding to tiling flows are divergence-free on the interior
of $R$ in the sense of distributions, i.e. if $f$ is a tiling flow with
corresponding measures $(\mu_{1},\mu_{2},\mu_{3})$, then for any $\phi$ smooth
and supported in a compact set $C$ contained in the interior of $R$,

$\displaystyle\int_{R}\frac{\partial\phi}{\partial
x_{1}}\,\mathrm{d}\mu_{1}+\int_{R}\frac{\partial\phi}{\partial
x_{2}}\,\mathrm{d}\mu_{2}+\int_{R}\frac{\partial\phi}{\partial
x_{3}}\,\mathrm{d}\mu_{3}=0.$

###### Proof.

For $i=1,2,3$, let $e_{1}^{i},...,e_{k_{i}}^{i}$ be the edges from
$\frac{1}{n}\mathbb{Z}^{3}$ such that $e_{j}^{i}\parallel\eta_{i}$, is
oriented parallel to $\eta_{i}$, and which intersect $R$. Let
$(a_{j}^{i},b_{j}^{i})$ be the endpoints of $e_{j}^{i}$ such that
$b_{j}^{i}-a_{j}^{i}=\eta_{i}$. By the fundamental theorem of calculus,

$\displaystyle\sum_{i=1}^{3}\int_{R}\frac{\partial\phi}{\partial
x_{i}}\,\mathrm{d}\mu_{i}=\frac{2}{n^{2}}\sum_{i=1}^{3}\sum_{j=1}^{k_{i}}(\phi(b_{j}^{i})-\phi(a_{j}^{i}))f(e_{j}^{i})$

If $v=a_{j}^{i}$ or $b_{j}^{i}$ is not contained in the interior of $R$, then
$\phi(v)=0$. Therefore we can rewrite the above as a sum over vertices
$v\in\frac{1}{n}\mathbb{Z}^{3}$ contained in the interior of $R$:

$\displaystyle\sum_{i=1}^{3}\int_{R}\frac{\partial\phi}{\partial
x_{i}}\,\mathrm{d}\mu_{i}=\frac{2}{n^{2}}\sum_{v}\phi(v)F(v),$

where $F(v)$ is a sum (with appropriate signs) of the six $f(e)$ terms for $e$
incident to $v$. We show $F(v)=0$.

Let $e_{i}^{-},e_{i}^{+}$ denote the edges incident to $v$ and oriented
parallel to $\eta_{i}$. Let $e_{i}^{+}$ be the one for which the orientation
parallel to $\eta_{i}$ coincides with the orientation even to odd. Then

$\displaystyle F(v)=\sum_{i=1}^{3}f(e_{i}^{+})-f(e_{i}^{-}).$

But this is equal to $\sum_{\tilde{e}\ni v}f(\tilde{e})$, where the edges
$\tilde{e}$ incident to $v$ are all oriented even to odd. Therefore $F(v)=0$
since $f$ is divergence-free as a discrete vector field, see Equation (5). ∎

Next we prove a lemma about generalized Wasserstein distance for signed
measures, in the case that both signed measures correspond to either tiling or
asymptotic flows. This is an elementary result that we will use repeatedly.

######  Lemma 5.2.7.

Suppose that $\mu$ and $\nu$ are measures supported on a common compact set
$K$ corresponding to components of tiling or asymptotic flows. Suppose there
is a partition of $K$ into sets $\mathcal{B}=\\{B_{1},...,B_{M}\\}$ of
diameter at most $\epsilon$ such that $\bigg{|}\mu(B)-\nu(B)\bigg{|}<\delta$
for all $B\in\mathcal{B}$. If one of the measures corresponds to a scale $n$
tiling flow, then we require that $\frac{1}{n}\leq\epsilon$. Then

$\displaystyle\mathbb{W}^{1,1}_{1}(\mu,\nu)\leq M(10\epsilon^{4}+\delta).$

###### Proof.

Let $\mu=\mu_{+}-\mu_{-}$ and $\nu=\nu_{+}-\nu_{-}$ be decompositions into
positive measures and recall that

$\displaystyle\mathbb{W}_{1}^{1,1}(\mu,\nu)=W_{1}^{1,1}(\mu_{+}+\nu_{-},\mu_{-}+\nu_{+}).$

Let $\tilde{\mu}=\mu_{+}+\nu_{-}$ and $\tilde{\nu}=\mu_{-}+\nu_{+}$. To get an
upper bound for the distance, it suffices to give a method for redistributing
and deleting mass to transform $\tilde{\mu}$ into $\tilde{\nu}$.

We proceed as follows: transform $\tilde{\mu}\mid_{B_{1}}$ into
$\tilde{\nu}\mid_{B_{1}}$, the cost of this is at most
$W^{1,1}_{1}(\tilde{\mu}\mid_{B_{1}},\tilde{\nu}\mid_{B_{1}})$. Denote the new
version of $\tilde{\mu}$ by $\tilde{\mu}^{\prime}$. $\tilde{\mu}^{\prime}$
will agree with $\tilde{\mu}$ on $R\setminus B_{1}$ and with $\tilde{\nu}$ on
$B_{1}$. Next transform $\tilde{\mu}^{\prime}$ into $\tilde{\nu}$ on $B_{2}$.
This will cost at most
$W^{1,1}_{1}(\tilde{\mu}^{\prime}\mid_{B_{2}},\tilde{\nu}\mid_{B_{2}})\leq
W^{1,1}_{1}(\tilde{\mu}\mid_{B_{2}},\tilde{\nu}\mid_{B_{2}})$ with equality if
$B_{2}$ is disjoint from $B_{1}$. Iterating this we get that

$\displaystyle\mathbb{W}^{1,1}_{1}({\mu},{\nu})\leq\sum_{j=1}^{k}W^{1,1}_{1}(\tilde{\mu}\mid_{B_{j}},\tilde{\nu}\mid_{B_{j}})$

Now we just have to compute the distance for a single $B_{j}$. First spend
$\delta>0$ to delete the difference in mass on $B_{j}$. The total mass of
$\mu,\nu$ on any $B\in\mathcal{B}$ is bounded by $10\epsilon^{3}$, and the
furthest it would need to move is $\epsilon$. Therefore

$\displaystyle
W^{1,1}_{1}(\tilde{\mu}\mid_{B_{j}},\tilde{\nu}\mid_{B_{j}})\leq
10\epsilon^{4}+\delta.$

Summing over $j$ gives the result. ∎

An inverse version of the bound in Lemma 5.2.7 also holds, but with a constant
depending on the small region $B$.

######  Lemma 5.2.8.

Suppose $B\subset R$ is a connected region with piecewise smooth boundary. If
$\mu,\nu$ are component measures of tiling or asymptotic flows and
$\mathbb{W}_{1}^{1,1}(\mu,\nu)<\delta$, then there is a constant $C(B)$
depending only on $B$ such that

$\displaystyle\mathbb{W}_{1}^{1,1}(\mu\mid_{B},\nu\mid_{B})<\delta+(C(B)+1)\delta^{1/2}.$

######  Remark 5.2.9.

The constant $C(B)$ is not hard to understand and control. The term
$C(B)\delta^{1/2}$ is bounded by $2$ times the volume of the $\delta^{1/2}$
annulus with inner boundary $\partial B$.

###### Proof.

The redistribution, addition, and deletion of mass $\mu\to\nu$ gives a
redistribution $\mu\mid_{B}\to\nu\mid_{B}$, except any mass moved into or out
of $B$ now becomes an $L^{1}$ cost rather than a cost proportional to distance
moved. Let $f(r)$ be the amount of flow moved distance $r$ into or out of $B$
by the $\mu\to\nu$ redistribution. Then

$\displaystyle\mathbb{W}_{1}^{1,1}(\mu\mid_{B},\nu\mid_{B})\leq\delta+\int_{0}^{\infty}f(r)dr.$

On the other hand,

$\displaystyle\int_{0}^{\infty}rf(r)dr<\delta.$

We split the integral we want to bound into two pieces:

$\displaystyle\int_{0}^{\infty}f(r)dr=\int_{0}^{\delta^{1/2}}f(r)dr+\int_{\delta^{1/2}}^{\infty}f(r)dr.$

Since $\mu,\nu$ are measures corresponding to components of tiling or
asymptotic flows, we have

$\int_{0}^{\delta^{1/2}}f(r)dr\leq C(B)\delta^{1/2}$

(this quantity is proportional to the volume of the $\delta^{1/2}$ annulus
around $B$) and

$\displaystyle\delta^{1/2}\int_{\delta^{1/2}}^{\infty}f(r)dr\leq\int_{\delta^{1/2}}^{\infty}rf(r)dr<\delta.$

Combining these gives the desired bound. ∎

######  Lemma 5.2.10.

Let $\nu_{n}$ be a sequence of signed measures supported in $R$ which
converges in $\mathbb{W}_{1}^{1,1}$ to another measure $\nu$. Further suppose
the $\nu_{n}$ are absolutely continuous with respect to 3-dimensional Lebesgue
measure, and their densities $g_{n}(x)$ take values in $[-m,M]$. Then $\nu$ is
also absolutely continuous with respect to 3-dimensional Lebesgue measure.

###### Proof.

By Corollary 5.1.3, if $\mathbb{W}_{1}^{1,1}(\nu_{n},\nu)\to 0$ as
$n\to\infty$, then $\nu_{n}$ converges to $\nu$ in the weak topology.

Define an operator $Q:C^{\infty}(R)\to\mathbb{R}$ by integrating against
$\nu$:

$\displaystyle Q(h):=\int h\,\mathrm{d}\nu.$

Using Cauchy-Schwarz and the fact that $d\nu_{n}=g_{n}(x)\,\mathrm{d}x$ we
have that

$\displaystyle Q(h)=\lim_{n\to\infty}\int
h\,g_{n}\,\mathrm{d}x\leq\limsup_{n\to\infty}\|h\|_{L^{2}(R)}\|g_{n}\|_{L^{2}(R)}\leq\text{Vol}(R)^{1/2}(M+m)\|h\|_{L^{2}}$

Therefore $Q$ extends to an operator on $L^{2}(R)$. By the Riesz
representation theorem this means there exists an $L^{2}$ function $g$ such
that $Q(h)=\langle h,g\rangle=\int h\,g\,\mathrm{d}x$. Therefore
$\mathrm{d}\nu(x)=g(x)\mathrm{d}x$, and $\nu$ is absolutely continuous with
respect to Lebesgue measure on $\mathbb{R}^{3}$. ∎

######  Proposition 5.2.11.

Suppose that $f,g\in AF(R)$ are continuous and satisfy $|f(x)-g(x)|<\epsilon$
for all $x\in R$. Then $d_{W}(f,g)<\epsilon\,\text{vol}(R)$.

###### Proof.

The $d_{W}$-distance from $f$ to $g$ is bounded by adding and subtracting mass
for each of the component functions. Since pointwise they differ by at most
$\epsilon$, $d_{W}(f,g)\leq\epsilon\,\text{vol}(R)$. ∎

###  5.3 Main theorems

Here we prove two of the theorems mentioned at the beginning of the section.
First we show that fine-mesh limits of tiling flows are asymptotic flows.

######  Theorem 5.3.1.

Let $R\subset\mathbb{R}^{3}$ be a compact region which is the closure of a
connected domain and has piecewise-smooth boundary. Let $f_{n}\in
TF_{m_{n}}(R)$ be a free-boundary tiling flow on $R$ at scale $m_{n}$ with
$m_{n}$ going to infinity with $n$. Any $d_{W}$-subsequential limit of tiling
flows $f_{*}=\lim_{k\to\infty}f_{{n_{k}}}$ is in $AF(R)$.

Let $\mu_{k}=(\mu_{k}^{1},\mu_{k}^{2},\mu_{k}^{3})$ be the measures
corresponding to $f_{{n_{k}}}$ and let
$\mu_{*}=(\mu_{*}^{1},\mu_{*}^{2},\mu_{*}^{3})$ be the measures corresponding
to $f_{*}$. The main idea of the proof is to smoothen the measures
$(\mu_{k}^{1},\mu_{k}^{2},\mu_{k}^{3})$ corresponding to the tiling flow in an
especially nice way, then apply Lemma 5.2.10 to say that their limits
$(\mu_{*}^{1},\mu_{*}^{2},\mu_{*}^{3})$ are absolutely continuous with respect
to 3-dimensional Lebesgue measure. This shows that $f_{*}$ is a measurable
flow on $R$. Using our well-chosen smoothings, we will show that $f_{*}$ has
the other properties that an asymptotic flow must have. To make the argument
easier to digest, we break down the construction of the smoothing into two
lemmas.

######  Lemma 5.3.2.

Let $S_{N}=[0,N-1]^{3}$ (note that there are $N$ vertices from
$\mathbb{Z}^{3}$ on each edge of $S_{N}$). Let $\tau$ be a tiling of
$\mathbb{Z}^{3}$ with tiling flow $f_{\tau}$, and let
$(\mu_{1},\mu_{2},\mu_{3})$ be the corresponding measures. Then

$\displaystyle\frac{1}{\text{Vol}(S_{N})}\bigg{(}\int_{S_{N}}\mathrm{d}\mu_{1},\int_{S_{N}}\mathrm{d}\mu_{2},\int_{S_{N}}\mathrm{d}\mu_{3}\bigg{)}$

is valued in $(1+O(1/N))\mathcal{O}$.

###### Proof.

Let $E(S_{N})$ denote the edges intersecting in $S_{N}$. All edges
intersecting $S_{N}$ in more than one point are contained in it and have
length $1$.

The measure $\mu_{1}$ is supported on the edges parallel to $\eta_{1}$, and
thus

$\int_{S_{N}}\mathrm{d}\mu_{1}=\sum_{E(S_{N})\ni e\parallel\eta_{1}}2f(e),$

for edges $e$ oriented parallel to $\eta_{1}$. The results for
$\mu_{2},\mu_{3}$ are analogous.

For each $i=1,2,3$, there are $N^{2}(N-1)$ edges from $\mathbb{Z}^{3}$
contained in $S_{N}$ parallel to $\eta_{i}$. This number is always even, so
half of these edges from ($N^{2}(N-1)/2$) have even-to-odd orientation
parallel to $\eta_{i}$ and half have the opposite orientation. Let
$\alpha_{+}$ be the number of even-to-odd oriented tiles in $\tau\cap
E(S_{N})$ and $\alpha_{-}$ be the number of odd-to-even oriented tiles. Then

$\displaystyle\int_{S_{N}}\mathrm{d}\mu_{1}=\sum_{E(S_{N})\ni
e\parallel\eta_{i}}2f(e)=2(\alpha_{+}-\alpha_{-}).$

Let $s_{i}$ denote the fraction of tiles in the $+\eta_{i}$ direction in
$S_{N}$ minus the fraction of tiles in the $-\eta_{i}$ direction in $S_{N}$.
Note that irrespective of the tiling we have that number of tiles in $S_{N}$
is $N^{2}(N-1)/2+O(N^{2})$. Thus it follows that

$\displaystyle
s_{1}=\frac{2(\alpha_{+}-\alpha_{-})}{N^{2}(N-1)+O(N^{2})}=\frac{1}{N^{2}(N-1)+O(N^{2})}\int_{S_{N}}\mathrm{d}\mu_{1}.$

A similar equation holds for $s_{2},s_{3}$. We have that
$(s_{1},s_{2},s_{3})\in\mathcal{O}$. Thus

$\displaystyle\frac{1}{\text{Vol}(S_{N})}\bigg{(}\int_{S_{N}}\mathrm{d}\mu_{1},\int_{S_{N}}\mathrm{d}\mu_{2},\int_{S_{N}}\mathrm{d}\mu_{3}\bigg{)}$
$\displaystyle=\frac{N^{2}(N-1)+O(N^{2})}{(N-1)^{3}}(s_{1},s_{2},s_{3})$
$\displaystyle=(1+\frac{O(N^{2})}{(N-1)^{3}})(s_{1},s_{2},s_{3}),$

for some constant $C$ which is in $(1+O(1/N))\mathcal{O}$. ∎

If we smoothen the measures corresponding to the tiling flow over a partition
consisting of boxes of the form in Lemma 5.3.2, we can construct smoothings
that satisfy a very nice list of properties.

######  Lemma 5.3.3.

Let $(\mu_{1},\mu_{2},\mu_{3})$ be measures corresponding to a tiling flow
$f\in TF_{n}(R)$. For any $\epsilon>0$, there exists
$\nu=(\nu_{1},\nu_{2},\nu_{3})$ satisfying the following properties:

  1. 1.

$\nu_{i}$ is supported in $R$ for all $i=1,2,3$;

  2. 2.

$\nu_{i}$ has a density $g_{i}(x)$ with respect to Lebesgue measure on
$\mathbb{R}^{3}$ for $i=1,2,3$;

  3. 3.

$g=(g_{1},g_{2},g_{3})$ is valued in $(1+O(\epsilon))\mathcal{O}$ as a
distribution;

  4. 4.

$d_{W}(\mu,\nu)<C(R)(\epsilon n)^{-1}$ where $C(R)$ is a constant depending
only on $R$.

###### Proof.

Choose $N$ such $N=\lfloor\frac{1}{\epsilon}\rfloor$ and a partition of cubes
$\mathcal{B}=\\{B_{1},...,B_{M}\\}$ that cover $R$, where each $B_{i}$ is an
$N\times N\times N$ cube in $\frac{1}{n}\mathbb{Z}^{3}$ (we define the flow
$f$ to be $0$ outside $R$). This can be done so that $M\sim n^{3}/N^{3}$. For
$i=1,2,3$ and all $B\in\mathcal{B}$, define

$\displaystyle
C_{B}^{i}:=\frac{1}{\text{Vol}(B)}\int_{B}\mathrm{d}\mu_{i}=\frac{(N-1)^{3}}{n^{3}}\int_{B}\mathrm{d}\mu_{i}.$

Define the densities of $\nu_{i}$ by

$\displaystyle g_{i}(x)=C_{B}^{i}\qquad\forall\,x\in B\cap R.$

This satisfies properties 1 and 2. By Lemma 5.3.2, $(g_{1},g_{2},g_{3})$ is
valued in $(1+O(1/N))\mathcal{O}=(1+O(\epsilon))\mathcal{O}$ as a distribution
which completes property 3. Finally by Lemma 5.2.7 applied to the partition
$\mathcal{B}$, we have that

$\displaystyle d_{W}(\mu,\nu)\leq M(N/n)^{4}\leq C(R)(n/N)^{3}(N/n)^{4}\leq
C(R)\frac{1}{\epsilon n}$

where $C(R)$ is a constant depending only on $R$. ∎

We now return to the proof of the theorem.

###### Proof of Theorem 5.3.1.

Fix $\epsilon>0$. Recall that $\mu_{k}=(\mu_{k}^{1},\mu_{k}^{2},\mu_{k}^{3})$
is the triple of measures corresponding to $f_{{n_{k}}}$. Choose
$\epsilon_{k}=n_{k}^{-1/2}$ and let
$\nu_{k}=(\nu_{k}^{1},\nu_{k}^{2},\nu_{k}^{3})$ be the measures constructed in
Lemma 5.3.3 for $\epsilon_{k}>0$. For $k$ large enough so that
$d_{W}(\mu_{*},\mu_{k})<\epsilon$, by the triangle inequality

$\displaystyle d_{W}(\mu_{*},\nu_{k})\leq
d_{W}(\mu_{*},\mu_{k})+d_{W}(\mu_{k},\nu_{k})\leq\epsilon+C(R)n_{k}^{-1/2}.$

Therefore the triple of measures $\nu_{k}$ also converges to $\mu_{*}$ in
$d_{W}$. By Lemma 5.2.10, there are functions $f_{*}^{i}$ such that
$\mu_{*}^{i}=f_{*}^{i}(x)\mathrm{d}x$ for $i=1,2,3$, so $f_{*}$ is a
measurable vector field. It remains for us to check the additional properties
to show that $f_{*}$ is an asymptotic flow.

Since the $\nu_{k}$ are supported in $R$ for all $k$, so is $f_{*}$. By Lemma
5.3.3, the densities of $\nu_{k}$ are valued in
$(1+O(\epsilon_{k}))\mathcal{O}$, so the densities of $\mu_{*}$ are valued in
$\mathcal{O}$ (any open neighborhood is a continuity set, so we get that the
averages of $(f_{*}^{1},f_{*}^{2},f_{*}^{3})$ are valued in
$(1+O(\epsilon_{k}))\mathcal{O}$ for all $\epsilon_{k}$. This plus the
Lebesgue differentiation theorem imply that $f_{*}$ is valued in
$\mathcal{O}$). On the other hand, convergence in $d_{W}$ implies weak
convergence of the component measures (Corollary 5.1.3). Since $\mu_{k}$ is
divergence-free in the sense of distributions on the interior of $R$
(Proposition 5.2.6) for all $k$, $\mu_{*}$ is also divergence-free in the
sense of distributions on the interior of $R$. Thus $f_{*}\in AF(R)$. ∎

######  Theorem 5.3.4.

The metric space $(AF(R),d_{W})$ is compact.

###### Proof.

By Proposition 5.1.4, the space of triples of measures absolutely continuous
with respect to Lebesgue measure, supported in $R$, and valued in
$\mathcal{O}$ is compact. Since $(AF(R),d_{W})$ is a subspace of this, it
suffices to show that it is closed. Suppose that
$\mu_{n}=(\mu_{n}^{1},\mu_{n}^{2},\mu_{n}^{3})$ is a sequence in $AF(R)$ with
$d_{W}$-limit $\mu_{*}=(\mu_{*}^{1},\mu_{*}^{2},\mu_{*}^{3})$. By Lemma
5.2.10, $\mu_{*}^{i}$ has a density $g_{*}^{i}(x)$ for each $i=1,2,3$. Since
convergence in $\mathbb{W}_{1}^{1,1}$ implies weak convergence (Corollary
5.1.3), $g_{*}=(g_{*}^{1},g_{*}^{2},g_{*}^{3})$ is divergence-free. To show
that $g_{*}$ is valued in $\mathcal{O}$, note that any open ball $B\subset R$
is a continuity set, so since
$\frac{1}{\text{Vol}(B)}(\int_{B}\mathrm{d}\mu_{n}^{1},\int_{B}\mathrm{d}\mu_{n}^{2},\int_{B}\mathrm{d}\mu_{n}^{3})\in\mathcal{O}$,
the average of $g_{*}$ over $B$ is also valued in $\mathcal{O}$. Thus $g_{*}$
is valued in $\mathcal{O}$ by the Lebesgue differentiation theorem. Thus
$(AF(R),d_{W})$ is compact. ∎

Now we will prove that any asymptotic flow can be approximated by a smooth
flow which is divergence-free on a slightly smaller region. This is a standard
construction which we provide for completeness as it will be used in the next
subsection. Essentially all we need to do is to convolve the asymptotic flow
with an appropriate smooth bump function. For this, given a region $R$ and
$\epsilon>0$ we define

$R_{\epsilon}:=\\{x\in R\leavevmode\nobreak\ :\leavevmode\nobreak\
d(x,\partial R)\geq\epsilon\\}.$ (15)

We will denote the smooth asymptotic flows on a region $R$ by
$AF^{\infty}(R)\subset AF(R)$. Given $f\in AF(R),g\in AF(R_{\epsilon})$, we
say that $d_{W}(f,g)$ is the distance between $f$ and $g$, where $g$ is
extended to be $0$ on $R\setminus R_{\epsilon}$.

######  Proposition 5.3.5.

Fix $f\in AF(R)$. For all $\epsilon>0$ small enough, there is a smooth
asymptotic flow $g\in AF^{\infty}(R_{\epsilon})$ such that

$d_{W}(f,g)<K\sqrt{\epsilon}$

where $K$ is a constant depending only on $R$.

###### Proof.

Consider a bump function $\phi\in C^{\infty}(B_{\epsilon}(0))$, that is, it is
a non-negative smooth function such that $\phi|_{\partial B_{\epsilon}(0)}=0$
and $\int_{B_{\epsilon}(0)}\phi(x)\,\mathrm{d}x=1$.

Let $g=f\ast\phi|_{R_{\epsilon}}$ and suppose that $\psi$ is a smooth test
function with compact support in the interior of $R_{\epsilon}$. To check that
$g$ is divergence-free in the interior of $R_{\epsilon}$ we look at the
integral

$\displaystyle\int_{R_{\epsilon}}(\nabla\psi\cdot g)(x)\,\mathrm{d}x$
$\displaystyle=$
$\displaystyle\int_{R_{\epsilon}}\int_{B_{\epsilon}(0)}(\nabla\psi\cdot
f)(x-y)\phi(y)\,\mathrm{d}y\,\mathrm{d}x$ $\displaystyle=$
$\displaystyle\int_{B_{\epsilon}(0)}\int_{R_{\epsilon}}(\nabla\psi\cdot
f)(x-y)\phi(y)\,\mathrm{d}x\,\mathrm{d}y=0.$

Here the last equality uses that $f$ is divergence-free in the interior of
$R$. We have that $g(x)=f\ast\phi(x)$ is an average of elements in
$\mathcal{O}$. Since $\mathcal{O}$ is convex it follows that $g$ takes values
in $\mathcal{O}$. Finally we estimate $d_{W}(f,g)$. The amount of mass which
we might have to delete or add from $R\setminus R_{\epsilon}$ is bounded by
$6(\text{Vol}(R\setminus R_{\epsilon}))\leq c\epsilon^{2}$ where $c$ depends
only on $\partial R$. Now let $(\mu_{1},\mu_{2},\mu_{3})$ and
$(\nu_{1},\nu_{2},\nu_{3})$ be the measures corresponding to $f$ and $g$
respectively. Let $\delta>0$. We have that if $B$ is a box of side length
$\delta$ contained in $R_{\epsilon}$ then

$\nu_{i}(B)=\int_{B}\int_{B_{\epsilon}(0)}f_{i}(x-y)\phi(y)\,\mathrm{d}y\,\mathrm{d}x=\int_{B_{\epsilon}(0)}\mu_{i}(B-y)\phi(y)\,\mathrm{d}y.$

It follows that $|\nu_{i}(B)-\mu_{i}(B)|$ is less than the volume of the
annular region around $B$ of radius $\epsilon$. Thus we have that

$|\nu_{i}(B)-\mu_{i}(B)|<C\epsilon\delta^{2}$

where $C$ is independent of $\epsilon$ and $\delta$. Partition $R_{\epsilon}$
into boxes $B_{1},B_{2},\ldots B_{M}$; where $M\sim\delta^{-3}$. By Lemma
5.2.7 we have that

$\mathbb{W}^{1,1}_{1}(\mu_{i},\nu_{i})<M(10\delta^{4}+C\epsilon\delta^{2})+c\epsilon^{2}\sim\delta^{-3}(10\delta^{4}+C\epsilon\delta^{2})+c\epsilon^{2}$

Since $\delta$ is a free parameter, we can take $\delta\sim\sqrt{\epsilon}$ to
complete the proof. ∎

###  5.4 Boundary values of asymptotic flows

In this section we define the boundary values of asymptotic flows on $\partial
R$. In fact we do something slightly more general, and define the restriction
of an asymptotic flow (or tiling flow in the next subsection) on a whole class
of surfaces, namely

$\displaystyle\mathbb{S}(R)=\\{\text{compact piecewise smooth surfaces
contained in }R\\}$

Note that $R$ is closed, so $\partial R\in\mathbb{S}(R)$. This general set up
will make things easier to prove. We also use the trace operator for other
surfaces in Section 8.6. Recall that $AF^{\infty}(R)\subset AF(R)$ is the
smooth asymptotic flows.

######  Definition 5.4.1.

We define the trace operator on smooth asymptotic flows

$\displaystyle T:AF^{\infty}(R)\times\mathbb{S}(R)\to\mathcal{M}^{s}(R).$

by

$\displaystyle T(f,S)(x)=\langle
f(x),\xi_{S}(x)\rangle\mathrm{d}\sigma_{S}(x),\qquad x\in S,$

where $\mathrm{d}\sigma_{S}$ denotes surface area measure on $S$ and
$\xi_{S}(x)$ denotes the $L^{2}$ unit normal vector to $S$ at $x$.

We show that $T(\cdot,S)$ extends to a uniformly continuous map
$(AF(R),d_{W})\to(\mathcal{M}^{s}(R),\mathbb{W}_{1}^{1,1})$ for all
$S\in\mathbb{S}(R)$. We do this in three main steps:

  * •

Show that $T(\cdot,S)$ is uniformly continuous on $AF^{\infty}(R)$ when $S$ is
a small patch.

  * •

Extend this result to $AF(R)$ by approximation and compatibility results.
Since we don’t know if $AF^{\infty}(R)$ is dense in $AF(R)$, this requires
slightly more care.

  * •

Extend the uniform continuity result to general $S\in\mathbb{S}(R)$ by putting
together the patches.

######  Proposition 5.4.2.

Suppose $S\in\mathbb{S}(R)$ and is such that there exists a nonzero vector $v$
and a parameter $\theta>0$ for which $S+tv$ is contained in $R$ and disjoint
for all $0\leq t\leq\theta$. Then for all $\epsilon>0$ there exists $\delta>0$
such that for all $f,g\in AF^{\infty}(R)$ with $d_{W}(f,g)<\delta$, we have
that

$\mathbb{W}^{1,1}_{1}(T(f,S),T(g,S))<\epsilon.$

###### Proof.

Fix two parameters $\gamma_{1},\gamma_{2}>0$ which we will specify at the end
of the proof. Partition $S$ into patches $\alpha_{1},\ldots\alpha_{M}$ such
that

  * •

$\alpha_{i}$ is a smooth surface with piecewise smooth boundary for all
$i=1,...,M$.

  * •

$\alpha_{i}$ has diameter at most $\gamma_{1}$ for all $i$, and $M\leq
C\gamma_{1}^{-2}$ for some constant $C$ depending on $S$.

  * •

Let $\alpha_{i}(t):=\alpha_{i}+tv$ for $0\leq t\leq\theta$. For all
$i=1,...,M$, $\alpha_{i}(t)\cap\alpha_{i}(s)=\emptyset$ for $s\neq t$.

Let $\mu_{f}=T(f,S)$, $\mu_{g}=T(g,S)$ and define $\Delta>0$ by

$\Delta:=\sup_{1\leq i\leq
M}\bigg{|}\mu_{f}(\alpha_{i})-\mu_{g}(\alpha_{i})\bigg{|}.$ (16)

By the two-dimensional version of Lemma 5.2.7,

$\mathbb{W}^{1,1}_{1}(T(f,S),T(g,S))\leq M(10\gamma_{1}^{3}+\Delta)\leq
10C\gamma_{1}+C\gamma_{1}^{-2}\Delta.$ (17)

Note that the power of $\gamma_{1}$ is $3$ instead of $4$ because $S$ is two-
dimensional. It remains to find a bound for $\Delta$ in terms of $d_{W}(f,g)$.

If $h\in AF^{\infty}(R)$, then $h$ is divergence-free and hence its flux
through any closed surface is zero. This implies that there exists a threshold
$\theta_{\gamma_{2}}\in(0,1)$ such that for all $0\leq
t\leq\theta_{\gamma_{2}}$ and $h\in AF^{\infty}(R)$,

$\displaystyle\sup_{1\leq i\leq M}\bigg{|}\int_{\alpha_{i}}\langle
h(x),\xi(x)\rangle\,\mathrm{d}\sigma_{\alpha_{i}}(x)-\int_{\alpha_{i}(t)}\langle
h(x),\xi(x)\rangle\,\mathrm{d}\sigma_{\alpha_{i}(t)}(x)\bigg{|}\leq\gamma_{2}.$
(18)

Here $\xi(x)$ is the normal vector on the surfaces $\alpha_{i}(t)$ with
appropriate orientation and we are applying the divergence theorem to the
boundary of the region $U_{i}=\cup_{s=0}^{t}\alpha_{i}(t)$. Since $h$ takes
values in the compact set $\mathcal{O}$, the threshold $\theta_{\gamma_{2}}$
can be taken independent of $h$. Applying this to $f,g\in AF^{\infty}(R)$ it
follows that

$\displaystyle\sup_{1\leq i\leq M}\bigg{|}\int_{\alpha_{i}}\langle
f(x)-g(x),\xi(x)\rangle\,\mathrm{d}\sigma_{\alpha_{i}}(x)-\int_{\alpha_{i}(t)}\langle
f(x)-g(x),\xi(x)\rangle\,\mathrm{d}\sigma_{\alpha_{i}(t)}(x)\bigg{|}\leq
2\gamma_{2}$

for all $0\leq t\leq\theta_{\gamma_{2}}$. Observe that the first term in the
inequality is $\mu_{f}(\alpha_{i})-\mu_{g}(\alpha_{i})$. Integrating over
$t\in(0,\theta_{\gamma_{2}})$,

$\displaystyle\bigg{|}\bigg{|}\int_{0}^{\theta_{\gamma_{2}}}\bigg{(}\int_{\alpha_{i}(t)}\langle
f(x)-g(x),\xi(x)\rangle\,\mathrm{d}\sigma_{\alpha_{i}(t)}(x)\bigg{)}\,\mathrm{d}t\bigg{|}-\theta_{\gamma_{2}}(\mu_{f}(\alpha_{i})-\mu_{g}(\alpha_{i}))\bigg{|}\leq
2\theta_{\gamma_{2}}\gamma_{2}.$

Taking the supremum over $i=1,...,M$ we get that

$\displaystyle\Delta\leq 2\gamma_{2}+\frac{1}{\theta_{\gamma_{2}}}\sup_{1\leq
i\leq
M}\bigg{|}\int_{0}^{\theta_{\gamma_{2}}}\bigg{(}\int_{\alpha_{i}(t)}\langle
f(x)-g(x),\xi(x)\rangle\,\mathrm{d}\sigma_{\alpha_{i}(t)}(x)\bigg{)}\,\mathrm{d}t\bigg{|}.$

Suppose the supremum on the right hand side of the equation is achieved by the
index $i$, and let $\alpha(t):=\alpha_{i}(t)$ to simplify notation. Then
plugging this into (17) gives

$\mathbb{W}^{1,1}_{1}(T(f,S),T(g,S))$ (19) $\leq
10C\gamma_{1}+2C\gamma_{1}^{-2}\gamma_{2}+C\gamma_{1}^{-2}\theta_{\gamma_{2}}^{-1}\bigg{|}\int_{0}^{\theta_{\gamma_{2}}}\bigg{(}\int_{\alpha(t)}\langle
f(x)-g(x),\xi(x)\rangle\,\mathrm{d}\sigma_{\alpha(t)}(x)\bigg{)}\,\mathrm{d}t\bigg{|}.$

Since $\alpha_{i}$ is smooth, by an appropriate change of variables we can
rewrite the integral above as an integral over
$U=\cup_{t=0}^{\theta_{\gamma_{2}}}\alpha(t)$.

$\displaystyle\int_{0}^{\theta_{\gamma_{2}}}\bigg{(}\int_{\alpha(t)}\langle
f(x)-g(x),\xi(x)\rangle\,\mathrm{d}\sigma_{\alpha(t)}(x)\bigg{)}\,\mathrm{d}t$
$\displaystyle=\int_{U}\langle
f(x)-g(x),\xi(x)\rangle\varphi(x)\,\mathrm{d}x_{1}\mathrm{d}x_{2}\mathrm{d}x_{3}$
$\displaystyle=\sum_{k=1}^{3}\int_{U}(f_{k}(x)-g_{k}(x))\xi_{k}(x)\varphi(x)\,\mathrm{d}x_{1}\,\mathrm{d}x_{2}\,\mathrm{d}x_{3}.$

Here $\varphi(x)$ is the factor coming from the Jacobian in the change of
variables. Since $\alpha(t)$ is smooth for all $t\in[0,\theta_{\gamma_{2}}]$,
$\varphi(x)$ and $\xi(x)$ are both smooth functions on $U$, so
$\psi_{k}(x):=\xi_{k}(x)\varphi(x)$ is a smooth and therefore Lipschitz
function on $U$. Let $\lambda$ denote the maximum of Lipchitz constants of
$\psi_{k}$, $k=1,2,3$. Then by the dual definition of the Wasserstein metric
(Proposition 5.1.2),

$\displaystyle\int_{0}^{\theta_{\gamma_{2}}}\bigg{(}\int_{\alpha(t)}\langle
f(x)-g(x),\xi(x)\rangle\,\mathrm{d}\sigma_{\alpha(t)}(x)\bigg{)}\,\mathrm{d}t\leq\lambda\,d_{W}(f\mid_{U},g\mid_{U}).$

By Lemma 5.2.8, there is a constant $C(U)$ such that if $d_{W}(f,g)<\delta$
then $d_{W}(f\mid_{U},g\mid_{U})<\delta+C(U)\delta^{1/2}$. Therefore
substituting this in to Equation (19) gives that if $d_{W}(f,g)<\delta$ then

$\displaystyle\mathbb{W}^{1,1}_{1}(T(f,S),T(g,S))\leq
10C\gamma_{1}+2C\gamma_{1}^{-2}\gamma_{2}+C\gamma_{1}^{-2}\theta_{\gamma_{2}}^{-1}\lambda(\delta+C(U)\,\delta^{1/2}).$

Taking $\gamma_{2}=\gamma_{1}^{3}$ completes the proof. ∎

We will now prove that perturbing $S$ by a small amount does not change the
trace very much.

######  Proposition 5.4.3.

Let $S\in\mathbb{S}(R)$ be a surface satisfying the conditions of Proposition
5.4.2 for some vector $v$ and threshold $\theta>0$. For all $\epsilon>0$ there
exists $\delta>0$ such that for all $t<\delta$ and $f\in AF^{\infty}(R)$, we
have that

$\mathbb{W}^{1,1}_{1}(T(f,S),T(f,S+tv))<\epsilon.$

###### Proof.

This proof is much simpler than that of Proposition 5.4.2. As in that proof,
we take parameters $\gamma_{1},\gamma_{2}>0$ to be fixed later and choose a
partition of $\alpha_{1}\ldots,\alpha_{M}$ of $S$ such that $\alpha_{i}$ has
diameter less than $\gamma_{1}$ for all $i$, and $M\leq C\gamma_{1}^{-2}$ for
some $C>0$ independent of $\gamma_{1}$. In addition, by choosing a larger
constant $C$ if necessary we can assume that

$\sigma_{S}(\alpha_{i})\leq C\gamma_{1}^{2}.$

As in (18), there exists a threshold $\theta_{\gamma_{2}}\in(0,1)$ such that
for all $0\leq t\leq\theta_{\gamma_{2}}$ and $f\in AF^{\infty}(R)$,

$\displaystyle\sup_{1\leq i\leq
M}\bigg{|}T(f,S)(\alpha_{i})-T(f,S+tv)(\alpha_{i}(t))\bigg{|}\leq\gamma_{2}.$
(20)

To get an upper bound for the distance between the traces on each patch, we
give a method for redistributing, adding, and deleting mass to transform
$T(f,S)\mid_{\alpha_{i}}=T(f,\alpha_{i})$ into $T(f,\alpha_{i}(t))$. Note that
these are both are signed measures absolutely continuous with respect to
$\sigma_{S}$ and $\sigma_{S+tv}$ respectively, and both have densities bounded
between $-1$ and $1$. We can transform one to the other by (1) adding
$\gamma_{2}$ flow, (2) moving flow distance at most $t+\gamma_{1}$. There is
at most $4\sigma_{S}(\alpha_{i})$ total flow from both measures. Hence

$\mathbb{W}^{1,1}_{1}(T(f,S)|_{\alpha_{i}},T(f,S+tv)|_{\alpha_{i}(t)})\leq\gamma_{2}+4(t+\gamma_{1})(\sigma_{S}(\alpha_{i}))\leq\gamma_{2}+4C(t+\gamma_{1})\gamma_{1}^{2}.$

As in Lemma 5.2.7, by triangle inequality

$\displaystyle\mathbb{W}^{1,1}_{1}(T(f,S),T(f,S+tv))$ $\displaystyle\leq$
$\displaystyle M\left(\gamma_{2}+4C(t+\gamma_{1})\gamma_{1}^{2}\right)\leq
C\gamma_{1}^{-2}\gamma_{2}+4C^{2}(t+\gamma_{1}).$

As before, by setting $\gamma_{2}=\gamma_{1}^{3}$, the result follows. ∎

Using Proposition 5.4.2 and Proposition 5.4.3, when $S\in\mathbb{S}(R)$
satisfies the conditions of Proposition 5.4.2 for a vector $v$ and threshold
$\theta>0$, we can extend the definition of $T(S,\cdot)$ to all of $AF(R)$ as
follows. Given $t>0$, there exists $K$ such that for all $k\geq K$,
$S(t)\subset R_{1/k}$. Here recall that

$\displaystyle R_{1/k}=\\{x\in R\leavevmode\nobreak\ :\leavevmode\nobreak\
d(x,\partial R)\geq 1/k\\}.$

By Proposition 5.3.5, for any $f\in AF(R)$ we can find a sequence $g_{k}\in
AF^{\infty}(R_{1/k})$ such that $d_{W}(f,g_{k})\to 0$ as $k\to\infty$. For
$t>0$, we define

$T(f,S(t)):=\lim_{k\to\infty}T(g_{k},S(t))$ (21)

where the limit is taken with respect to the metric $d_{W}$ on flows supported
in $R$. Note that if $R^{\prime}\subset R$, then the projection map
$AF^{\infty}(R)\to AF^{\infty}(R^{\prime})$ given by $f\mapsto
f\mid_{R^{\prime}}$ is continuous. By Proposition 5.4.2, $T$ is uniformly
continuous on $AF^{\infty}(R_{1/k})$ for any $k>0$, so the limit in Equation
(21) is independent of the approximating sequence $(g_{k})_{k\geq 1}$ and
converges to $T(f,S+tv)$ if $f\in AF^{\infty}(R)$. Since $(AF(R),d_{W})$ is
compact (Theorem 5.3.4), $T(\cdot,S(t)):AF(R)\to\mathcal{M}^{s}(R)$ is a
uniformly continuous map for $t>0$.

Further, by a variant of Lemma 5.2.10 where 3-dimensional Lebesgue measure is
replaced by the surface area measure $\sigma_{S(t)}$, for any $f\in AF(R)$,
$T(f,S(t))$ is a signed measure absolutely continuous to $\sigma_{S(t)}$ with
density bounded between $-1$ and $1$. By Proposition 5.4.3, for any
$\epsilon>0$, there exists $\delta>0$ so that if $t,s>0$ and $|t-s|<\delta$
then

$\mathbb{W}_{1}^{1,1}(T(f,S(s)),T(f,S(t))<\epsilon.$ (22)

Thus we take another limit in the $d_{W}$ topology to define

$\displaystyle T(f,S):=\lim_{t\to 0}T(f,S(t)).$

Since $AF(R)$ is compact and the extension above is continuous, we get analogs
of Proposition 5.4.2 and Proposition 5.4.3 for $AF(R)$.

######  Proposition 5.4.4.

If $S\in\mathbb{S}(R)$ is a surface satisfying the conditions of Proposition
5.4.2 for a vector $v$ and threshold $\theta>0$, then

  1. 1.

Using the extension described above,

$\displaystyle
T(\cdot,S):(AF(R),d_{W})\to(\mathcal{M}^{s}(R),\mathbb{W}_{1}^{1,1})$

is uniformly continuous.

  2. 2.

Given $\epsilon>0$ there exists $\delta>0$ such that for all $t<\delta$ and
$f\in AF(R)$,

$\displaystyle\mathbb{W}_{1}^{1,1}(T(f,S),T(f,S(t))<\epsilon.$

  3. 3.

For any $f\in AF(R)$, $T(f,S)$ is a signed measure absolutely continuous with
respect to the surface area measure $\sigma_{S}$ with density bounded between
$-1$ and $1$.

We now prove a compatibility result for traces on overlapping surfaces. Using
this, we can extend the continuity theorems to the trace operator for any
$S\in\mathbb{S}(R)$ by cutting $S$ in patches which satisfy the conditions of
Proposition 5.4.2.

######  Lemma 5.4.5.

Suppose that $S,S^{\prime}\in\mathbb{S}(R)$ are surfaces satisfying the
conditions of Proposition 5.4.2 such that $S^{\prime}\subset S$. Then for any
$f\in AF(R)$, then

$T(f,S)\mid_{S^{\prime}}=T(f,S^{\prime}).$

###### Proof.

If $f\in AF^{\infty}(R)$, then the result follows immediately from the form
given in Definition 5.4.1. For general $f\in AF(R)$ this follows from
Proposition 5.4.4. ∎

Finally we put together the pieces to prove uniform continuity of $T(\cdot,S)$
on $AF(R)$ for any $S\in\mathbb{S}(R)$.

######  Proposition 5.4.6.

For all $S\in\mathbb{S}(R)$, $T(\cdot,S)$ extends to a uniformly continuous
map from $(AF(R),d_{W})$ to $(\mathcal{M}^{s}(R),\mathbb{W}_{1}^{1,1})$.
Further, for all $f\in AF(R)$, $T(f,S)$ is a signed measure on $S$ absolutely
continuous to the surface measure $\sigma_{S}$ with density bounded between
$-1$ and $1$.

######  Remark 5.4.7.

In particular this holds for $S=\partial R$.

###### Proof.

Fix $S\in\mathbb{S}(R)$. We can cover $S$ with finitely many open surfaces
$S_{1},...,S_{k}$ which all satisfy the conditions of Proposition 5.4.2. By
Lemma 5.4.5, if $S_{i}\cap S_{j}\neq\emptyset$, then for all $f\in AF(R)$,

$\displaystyle T(f,S_{i})\mid_{S_{i}\cap S_{j}}=T(f,S_{j})\mid_{S_{i}\cap
S_{j}}$

Therefore we can define the trace operator $T(S,\cdot)$ for any
$S\in\mathbb{S}(R)$ by

$\displaystyle T(f,S)\mid_{S_{i}}=T(f,S_{i}).$

By Proposition 5.4.4, $T(f,S)$ is uniformly continuous as a function of $f\in
AF(R)$ and has the desired form. ∎

With this machinery, we can define the space of asymptotic flows with a fixed
boundary value.

######  Definition 5.4.8.

We say that $b\in\mathcal{M}^{s}(R)$ is a boundary asmyptotic flow if $b\in
T(AF(R),\partial R)$. Further, we define $AF(R,b)$ to be the space of
asymptotic flows on $R$ with boundary value $b$, i.e. $f\in AF(R)$ such that
$T(f,\partial R)=b$.

As a corollary of Proposition 5.4.6 and Theorem 5.3.4 we get the following.

######  Corollary 5.4.9.

The metric space $(AF(R,b),d_{W})$ is compact.

###  5.5 Boundary values of tiling flows

Next we define the trace operator on tiling flows, and show that it is
compatible with the definition for asymptotic flows. Suppose $f\in TF_{n}(R)$
and that $S\in\mathbb{S}(R)$ is a surface which intersects the lattice
$\frac{1}{n}\mathbb{Z}^{3}$ transversely, i.e. $S$ does not contain any
vertices of $\frac{1}{n}\mathbb{Z}^{3}$ (if $S$ contains a vertex, we
translate the lattice slightly so that it does not). As usual let $e$ denote
an edge from $\frac{1}{n}\mathbb{Z}^{3}$ oriented from even to odd, and let
$\xi(x)$ denote the normal vector to $S$ at $x$.

######  Definition 5.5.1.

If $f\in TF_{n}(R)$ and $S\in\mathbb{S}(R)$ is a surface intersecting
$\frac{1}{n}\mathbb{Z}^{3}$ transversely, we define

$\displaystyle
T(f,S)=\sum_{e}\frac{2\,\text{sign}\langle\xi(x),e\rangle}{n^{2}}f(e)\delta(e\cap
S).$

Note that since $S$ is transverse to $\frac{1}{n}\mathbb{Z}^{3}$, if $e\cap S$
is nonempty it is a single point.

Using Definition 5.5.1 for tiling flows and Definition 5.4.1 for asymptotic
flows, the final goal of this section is to show that

$\displaystyle T(\cdot,S):(AF(R)\cup
TF(R),d_{W})\to(\mathcal{M}^{s}(R),\mathbb{W}_{1}^{1,1})$

is uniformly continuous for any $S\in\mathbb{S}(R)$ (see Theorem 5.5.7). The
sequence of results in this section building up to this mirrors the sequence
of results in the previous section. The discrete setting makes things slightly
more complicated. The main new step is that we start by proving a result for
the trace on planes, and extend to more general surfaces by approximating them
with planes. Throughout, we assume that any surface $S$ we consider intersects
$\frac{1}{n}\mathbb{Z}^{3}$ transversely. Any time it does not, the trace is
defined by perturbing the lattice slightly so that it does and then using
Definition 5.5.1.

######  Proposition 5.5.2.

Suppose $P\in\mathbb{S}(R)$ is a compact piece of a plane with normal vector
$\xi$, and there exists a threshold $\theta>0$ such that $P(t)=P+t\xi$ is
contained in $R$ for all $t\in[0,\theta]$. Let $f_{n}\in TF_{n}(R)$ be a
sequence of tiling flows such that $d_{W}(f_{n},f)\to 0$ as $n\to\infty$ for
some $f\in AF^{\infty}(R)$. Then

$\displaystyle\lim_{n\to\infty}\mathbb{W}_{1}^{1,1}(T(f_{n},P),T(f,P))=0.$

######  Remark 5.5.3.

The conditions here could be rephrased as saying that $P\in\mathbb{S}(R)$ is
contained in a plane and satisfies the conditions of Proposition 5.4.2 with
$v=\xi$.

###### Proof.

As in Proposition 5.4.2, fix two parameters $\gamma_{1},\gamma_{2}>0$ and
partition $P$ into patches $\alpha_{1},...,\alpha_{M}$ such that

  * •

$\alpha_{i}$ is a smooth surface with piecewise smooth boundary for all
$i=1,\ldots,M$;

  * •

$\alpha_{i}$ has diameter at most $\gamma_{1}$ for all $i=1,\ldots,M$, and
$M\leq C\gamma_{1}^{-2}$ for some constant $C$ depending on $P$;

  * •

Let $\alpha_{i}(t):=\alpha_{i}+t\xi$. For all $i=1,\ldots,M$,
$\alpha_{i}(t)\cap\alpha_{i}(s)=\emptyset$ for $s\neq t$.

We define

$\displaystyle\Delta_{n}=\sup_{1\leq i\leq
M}\bigg{\lvert}T(f_{n},P)(\alpha_{i})-T(f,P)(\alpha_{i})\bigg{\rvert}.$

By the two-dimensional version of Lemma 5.2.7,

$\displaystyle\mathbb{W}_{1}^{1,1}(T(f_{n},P),T(f,P))\leq
10C\gamma_{1}+C\gamma_{1}^{-2}\Delta_{n}.$

Let $U_{i}(s)=\cup_{t=0}^{s}\alpha_{i}(t)$ be the parallelopiped region
between $\alpha_{i}=\alpha_{i}(0)$ and $\alpha_{i}(s)$. Given $\gamma_{2}$, we
can find $\theta_{\gamma_{2}}$ small enough so that the number of edges from
$\frac{1}{n}\mathbb{Z}^{3}$ hitting $\partial
U_{i}(t)\setminus(\alpha_{i}(t)\cup\alpha_{i})$ for any $i$ is less than
$\gamma_{2}n^{2}+K^{\prime}n$, with constant $K^{\prime}$ depending on the
length of $\partial\alpha_{i}$. Since the magnitude of $f_{n}$ is of order
$1/n^{2}$, for $t<\theta_{\gamma_{2}}$ the flow of $f_{n}$ through $\partial
U_{i}(t)\setminus(\alpha_{i}(t)\cup\alpha_{i})$ is bounded by
$O(n^{-1})+\gamma_{2}$. Since any $f_{n}\in TF_{n}(R)$ is divergence-free as a
discrete vector field, there exists a constant $K>0$ so that for any $f_{n}$
and $t\in(0,\theta_{\gamma_{2}})$,

$\sup_{1\leq i\leq
M}\bigg{|}T(f_{n},\alpha_{i})(\alpha_{i})-T(f_{n},\alpha_{i}(t))(\alpha_{i}(t))\bigg{|}\leq\gamma_{2}+K/n.$
(23)

Possibly choosing a smaller $\theta_{\gamma_{2}}$, the same result holds for
$f$ without the $K/n$ in the upper bound. By the triangle inequality,

$\displaystyle\sup_{1\leq i\leq
M}\bigg{|}T(f_{n},\alpha_{i})(\alpha_{i})-T(f,\alpha_{i})(\alpha_{i})-T(f_{n},\alpha_{i}(t))(\alpha_{i}(t))+T(f,\alpha_{i}(t))(\alpha_{i}(t))\bigg{|}\leq
2\gamma_{2}+K/n.$

As in Proposition 5.4.2, integrating over $t\in(0,\theta_{\gamma_{2}})$ and
solving for $\Delta_{n}$ gives

$\displaystyle\Delta_{n}\leq
2\gamma_{2}+K/n+\frac{1}{\theta_{\gamma_{2}}}\sup_{1\leq i\leq
M}\bigg{|}\int_{0}^{\theta_{\gamma_{2}}}(T(f_{n},\alpha_{i}(t))(\alpha_{i}(t))-T(f,\alpha_{i}(t))(\alpha_{i}(t)))\,\mathrm{d}t\bigg{|}.$

Let $\alpha_{i}$ be the patch where the supermum is achieved, and let
$\alpha:=\alpha_{i}$ to simplify notation. Then

$\displaystyle\mathbb{W}_{1}^{1,1}(T(f_{n},P),T(f,P))\leq
10C\gamma_{1}+2C\gamma_{1}^{-2}\gamma_{2}+C\gamma_{1}^{-2}Kn^{-1}$
$\displaystyle+\frac{C\gamma_{1}^{-2}}{\theta_{\gamma_{2}}}\bigg{|}\int_{0}^{\theta_{\gamma_{2}}}T(f_{n},\alpha(t))(\alpha(t))-T(f,\alpha(t))(\alpha(t))\,\mathrm{d}t\bigg{|}.$

Finally we bound the integral in the last term. Let
$U=U_{i}(\theta_{\gamma_{2}})$ to simplify notation. Recall that
$\alpha(t)\subset P+tv$ is a piece of a plane, and has constant unit normal
vector $\xi$. By Definition 5.4.1, $T(f,\alpha(t))(x)=\langle
f(x),\xi\rangle$. Therefore letting $\sigma_{\alpha(t)}$ denote the surface
area measure on $\alpha(t)$, and applying change of variables,

$\displaystyle\int_{0}^{\theta}T(f,\alpha(t))(\alpha(t))\,\mathrm{d}t$
$\displaystyle=\int_{0}^{\theta}\int_{\alpha(t)}\langle
f(x),\xi\rangle\,\mathrm{d}\sigma_{\alpha(t))}(x)\mathrm{d}t=\int_{U}\langle
f(x),\xi\rangle\mathrm{d}x_{1}\,\mathrm{d}x_{2}\,\mathrm{d}x_{3}$
$\displaystyle=\sum_{j=1}^{3}\xi_{j}\mu_{j}(U),$

where $\xi=(\xi_{1},\xi_{2},\xi_{3})$, $f=(f_{1},f_{2},f_{3})$ and
$(\mu_{1},\mu_{2},\mu_{3})$ is the triple of measures corresponding to $f$. On
the other hand, for the tiling flow $f_{n}$,

$\displaystyle\int_{0}^{\theta}T(f_{n},\alpha(t))(\alpha(t))\,\mathrm{d}t=\sum_{j=1}^{3}\xi_{j}\,\mu_{j}^{n}(U),$

where $(\mu_{1}^{n},\mu_{2}^{n},\mu_{3}^{n})$ is the triple of measures
corresponding to $f_{n}$. Therefore

$\displaystyle\mathbb{W}_{1}^{1,1}(T(f_{n},P),T(f,P))$ $\displaystyle\leq$
$\displaystyle
10C\gamma_{1}+2C\gamma_{1}^{-2}\gamma_{2}+C\gamma_{1}^{-2}Kn^{-1}+C\gamma_{1}^{-2}\theta_{\gamma_{2}}^{-1}\sum_{j=1}^{3}|\xi_{j}||\mu_{j}^{n}(U)-\mu_{j}(U)|.$

By Lemma 5.2.8, $d_{W}(f_{n},f)\to 0$ implies that
$|\mu_{j}^{n}(U)-\mu_{j}(U)|\to 0$ as $n\to\infty$ for $j=1,2,3$. Taking
$n\to\infty$ gives

$\displaystyle\limsup_{n\to\infty}\mathbb{W}_{1}^{1,1}(T(f_{n},P),T(f,P))\leq
10C\gamma_{1}+2C\gamma_{1}^{-2}\gamma_{2}.$

Setting $\gamma_{2}=\gamma_{1}^{3}$ and taking $\gamma_{1}\to 0$ completes the
proof. ∎

Next we prove a version of Proposition 5.4.3 for tiling flows and small patch
surfaces as in Proposition 5.4.2.

######  Proposition 5.5.4.

Suppose that $S\in\mathbb{S}(R)$ satisfies the conditions of Proposition
5.4.2. For all $\epsilon>0$ there exists $\delta>0$ and $N>0$ such that for
all $t<\delta$, all $n\geq N$, and all $f\in TF_{n}(R)$,

$\displaystyle\mathbb{W}_{1}^{1,1}(T(f,S),T(f,S(t)))<\epsilon.$

###### Proof.

The proof is analogous to the proof of Proposition 5.4.3. Again we take
parameters $\gamma_{1},\gamma_{2}>0$ to be fixed later and a partition
$\alpha_{1},....,\alpha_{M}$ of $S$ into patches of diameter at most
$\gamma_{1}$ for all $i$, and such that $M\leq C\gamma_{1}^{-2}$ for some
constant $C$ independent of $\gamma_{1}$. Analogous to Equation (23), given
$\gamma_{2}$ we can find a threshold $\theta_{\gamma_{2}}>0$ such that for all
$0\leq t\leq\theta_{\gamma_{2}}$ and all $f\in TF_{n}(R)$,

$\displaystyle\sup_{1\leq i\leq
M}\bigg{\lvert}T(f,S)(\alpha_{i})-T(f,S(t))(\alpha_{i}(t))\bigg{\rvert}\leq\gamma_{2}+K/n.$
(24)

Using this, we get an upper bound for the distance by giving a method for
redistributing, adding, and deleting mass to transform
$T(f,S)\mid_{\alpha_{i}}$ into $T(f,S(t))\mid_{\alpha_{i}(t)}$. Both measures
are a sum of delta masses of weights $2/n^{2}(\pm 5/6)$ or $2/n^{2}(\pm 1/6)$.
The number of delta masses in $\alpha_{i}$ or $\alpha_{i}(t)$ is bounded above
by $\text{area}(\alpha_{i})n^{2}$. Since $\alpha_{i}$ has diameter bounded by
$\gamma_{1}$, there is a constant $A>0$ independent of $\alpha_{i}$ such that
$\text{area}(\alpha_{i})\leq A\gamma_{1}^{2}$. Hence the total mass in each
patch is bounded between $-2A\gamma_{1}^{2}$ and $2A\gamma_{1}^{2}$. Hence
adding $\gamma_{2}+K/n$ mass plus moving at most $8A\gamma_{1}^{2}$ mass
distance at most $t+\gamma_{1}$, we get the bound

$\displaystyle\mathbb{W}_{1}^{1,1}(T(f,S)\mid_{\alpha_{i}},T(f,S(t))\mid_{\alpha_{i}(t)})\leq\gamma_{2}+K/n+8A\gamma_{1}^{2}(t+\gamma_{1})$

Summing over $i$ we get that

$\displaystyle\mathbb{W}_{1}^{1,1}(T(f,S),T(f,S(t)))\ $ $\displaystyle\leq
M(\gamma_{2}+K/n+8\gamma_{1}^{2}(t+\gamma_{1}))$
$\displaystyle=C\gamma_{1}^{-2}\gamma_{2}+CK\gamma_{1}^{-2}n^{-1}+8A(t+\gamma_{1}).$

Take $\gamma_{2}=\gamma_{1}^{3}$ and $\gamma_{1},t$ small enough so that

$\displaystyle(C+8A)\gamma_{1}+8At<\epsilon/2.$ (25)

Then take $n$ large enough so that

$\displaystyle CK\gamma_{1}^{-2}n^{-1}<\epsilon/2,$

and the result follows with
$\delta=\min\\{\theta_{\gamma_{1}^{3}},\frac{1}{8A}(\epsilon/2-(C+8A)\gamma_{1})\\}$.
Here the first term in the minimum comes from (24) and the second comes from
(25). ∎

By approximation we can extend Proposition 5.5.2 to any surface $S\in
S(\mathbb{R})$. For technical reasons, we first show this for $S$ contained
strictly in the interior of $R$. Note that we also remove the condition that
the limiting flow is smooth.

######  Proposition 5.5.5.

Suppose that $S\in\mathbb{S}(R)$ is contained strictly in the interior of $R$.
Let $f_{n}\in TF_{n}(R)$ be a sequence of tiling flows such that
$d_{W}(f_{n},f)\to 0$ as $n\to\infty$ for some $f\in AF(R)$. Then

$\displaystyle\lim_{n\to\infty}\mathbb{W}_{1}^{1,1}(T(f_{n},S),T(f,S))=0.$

###### Proof.

As usual, let $\gamma_{1},\gamma_{2}>0$ be small parameters to be fixed later.
Since $S$ is contained strictly in the interior of $R$, we can cover $S$ by
very small patch surfaces $\alpha_{1},...,\alpha_{M}$ so that:

  * •

Each $\alpha_{i}$ is smooth with piecewise smooth boundary,

  * •

The diameter of $\alpha_{i}$ is at most $\gamma_{1}$.

  * •

There is a constant $C>0$ such that $M\leq C\gamma_{1}^{-2}$.

  * •

For all $i$, $\alpha_{i}$ satisfies the conditions of Proposition 5.4.2 for
some threshold $\theta>0$ with vector $v=\xi(q_{i})$, where $\xi(q_{i})$ is
the normal vector to the surface at $q_{i}$ for some $q_{i}\in\alpha_{i}$ with
the property that the distance between $q_{i}$ and any other $x\in\alpha_{i}$
is at most $\gamma_{1}$.

  * •

Let $P_{i}$ denote the tangent plane to $\alpha_{i}$ at $q_{i}$, and let
$\pi_{i}\subset P_{i}$ be the patch of the plane corresponding to projecting
$\alpha_{i}$ onto $P_{i}$. Potentially making $\theta>0$ or $\gamma_{1}$
smaller, we can assume that $\pi_{i}$ also satisfies the conditions of
Proposition 5.4.2 for $v=\xi(q_{i})$. This is where we are using the fact that
$S$ is contained in the interior of $R$.

As usual we denote $\alpha_{i}(t)=\alpha_{i}+t\xi(q_{i})$ and
$\pi_{i}(t)=\pi_{i}+t\xi(q_{i})$. Note by Definition 5.5.1 and Lemma 5.4.5
that

$\displaystyle
T(f_{n},S)\mid_{\alpha_{i}}=T(f_{n},\alpha_{i})\qquad\text{and}\qquad
T(f,S)\mid_{\alpha_{i}}=T(f,\alpha_{i})\qquad i=1,...,M.$

Define

$\displaystyle\Delta_{n}:=\sup_{1\leq i\leq
M}\bigg{\lvert}T(f_{n},S)(\alpha_{i})-T(f,S)(\alpha_{i})\bigg{\rvert}.$

By the two-dimensional version of Lemma 5.2.7,

$\mathbb{W}_{1}^{1,1}(T(f_{n},S),T(f,S))\leq
M(10\gamma_{1}^{3}+\Delta_{n})\leq 10C\gamma_{1}+C\gamma_{1}^{-2}\Delta_{n}.$
(26)

Since $f_{n}\in TF_{n}(R)$ is discrete divergence-free and $S$ is compact and
piecewise smooth, given $\gamma_{2}$ there exists $K>0$ depending on $S$ and a
threshold $\theta_{\gamma_{2}}$ such that for all $0\leq
t\leq\theta_{\gamma_{2}}$,

$\sup_{1\leq i\leq
M}\bigg{|}T(f_{n},\alpha_{i})(\alpha_{i})-T(f_{n},\alpha_{i}(t))(\alpha_{i}(t))\bigg{|}\leq\gamma_{2}+K/n.$
(27)

By Proposition 5.4.4, up to making $\theta_{\gamma_{2}}$ smaller, for all
$0\leq t\leq\theta_{\gamma_{2}}$,

$\sup_{1\leq i\leq
M}\bigg{|}T(f,\alpha_{i})(\alpha_{i})-T(f,\alpha_{i}(t))(\alpha_{i}(t))\bigg{|}\leq\gamma_{2}.$
(28)

Combining Equations (27) and (28), and as in Proposition 5.4.2 integrating
over $t\in(0,\theta_{\gamma_{2}})$ then solving for $\Delta_{n}$ gives

$\displaystyle\Delta_{n}\leq
2\gamma_{2}+K/n+\frac{1}{\theta_{\gamma_{2}}}\sup_{1\leq i\leq
M}\bigg{|}\int_{0}^{\theta_{\gamma_{2}}}T(f_{n},\alpha_{i}(t))(\alpha_{i}(t))-T(f,\alpha_{i}(t))(\alpha_{i}(t))\,\mathrm{d}t\bigg{|}.$

Let $i$ be the index where the supremum is achieved, and let
$\alpha(t):=\alpha_{i}(t)$ and $\pi(t):=\pi_{i}(t)$. We now bound

$\displaystyle T(f_{n},\alpha(t))(\alpha(t))-T(f,\alpha(t))(\alpha(t))$

using four terms. Let $(g_{m})\in AF^{\infty}(R_{\epsilon_{m}})$,
$\epsilon_{m}\to 0$ as $m\to\infty$, be a sequence of smooth asymptotic flows
such that $\lim_{m\to\infty}d_{W}(g_{m},f)=0$. Since $\alpha(t),\pi(t)$ are
contained strictly in the interior of $R$ for $0\leq t\leq\theta$, we can
assume they are all contained in $R_{\epsilon}$ for some $\epsilon>0$. In
particular, for $m$ large enough $\epsilon_{m}<\epsilon$ and hence $g_{m}$ is
defined on $\pi(t)$. We have for any $0\leq t\leq\theta_{\gamma_{2}}$,

$\displaystyle T(f_{n},\alpha(t))(\alpha(t))-T(f,\alpha(t))(\alpha(t))=$
$\displaystyle T(f_{n},\alpha(t))(\alpha(t))-T(f_{n},\pi(t))(\pi(t))$
$\displaystyle+$ $\displaystyle
T(f_{n},\pi(t))(\pi(t))-T(g_{m},\pi(t))(\pi(t))$ $\displaystyle+$
$\displaystyle T(g_{m},\pi(t))(\pi(t))-T(f,\pi(t))(\pi(t))$ $\displaystyle+$
$\displaystyle T(f,\pi(t))(\pi(t))-T(f,\alpha(t))(\alpha(t)).$

Consider the region $V(t)$ with boundary $\alpha(t)\cup\pi(t)$ plus sides to
enclose it. Since $\pi(t)$ is the tangent plane to $\alpha(t)$ at
$q(t)=q_{i}+tv$, the height of the sides needed to enclose this region is
bounded by $C_{2}\gamma_{1}^{2}$, where since $S$ is compact, $C_{2}>0$ is a
constant depending only on $S$ ($C_{2}$ is basically the maximum curvature at
a smooth point on $S$). On the other hand, the length of the boundary of
$\alpha(t)$ is bounded by $C_{3}\gamma_{1}$ for some constant $C_{3}>0$.
Therefore since $f_{n}$ is divergence-free, for some constant $K>0$,

$|T(f_{n},\alpha(t))(\alpha(t))-T(f_{n},\pi(t))(\pi(t))|\leq
C_{2}C_{3}\gamma_{1}^{3}+K/n.$ (29)

The analogous result holds for $f$ (without the $K/n$ term), controlling the
fourth term above. By Proposition 5.4.4,

$\displaystyle\lim_{m\to\infty}|T(g_{m},\pi(t))(\pi(t))-T(f,\pi(t))(\pi(t))|=0.$

As in Proposition 5.5.2, if $\mu_{j}^{n}$ denotes the component measures of
$f_{n}$, $\nu_{j}^{m}$ denotes the component measures of $g_{m}$, and
$\mu_{j}$ denotes the component measures of $f$, and letting
$U=\cup_{t=0}^{\theta_{\gamma_{2}}}\pi(t)$, we have

$\bigg{|}\int_{0}^{\theta_{\gamma_{2}}}T(f_{n},\pi(t))(\pi(t))-T(g_{m},\pi(t))(\pi(t))\,\mathrm{d}t\bigg{|}$
$\leq\sum_{j=1}^{3}|\xi_{j}(q)||\mu_{j}^{n}(U)-\nu_{j}^{m}(U)|\leq\sum_{j=1}^{3}|\xi_{j}(q)|(|\mu_{j}^{n}(U)-\mu_{j}(U)|+|\mu_{j}(U)-\nu_{j}^{m}(U)|).$

Taking the limit as $m\to 0$ gives

$\displaystyle\limsup_{m\to\infty}\bigg{|}\int_{0}^{\theta_{\gamma_{2}}}T(f_{n},\pi(t))(\pi(t))-T(g_{m},\pi(t))(\pi(t))\,\mathrm{d}t\bigg{|}\leq\sum_{j=1}^{3}|\xi_{j}(q)||\mu_{j}^{n}(U)-\mu_{j}(U)|.$

Therefore

$\displaystyle\bigg{|}\int_{0}^{\theta_{\gamma_{2}}}T(f_{n},\alpha(t))(\alpha(t))-T(f,\alpha(t))(\alpha(t))\,\mathrm{d}t\bigg{|}$
$\displaystyle\leq
2C_{2}C_{3}\theta_{\gamma_{2}}\gamma_{1}^{3}+\sum_{j=1}^{3}|\xi_{j}(q)||\mu_{j}^{n}(U)-\mu_{j}(U)|+K/n.$

Plugging back in to Equation (26), we get

$\displaystyle\mathbb{W}_{1}^{1,1}(T(f_{n},S),T(f,S))\leq$ $\displaystyle
10C\gamma_{1}+2C\gamma_{1}^{-2}\gamma_{2}+C\gamma_{1}^{-2}Kn^{-1}+2CC_{2}C_{3}\gamma_{1}+C\gamma_{1}^{-2}\theta_{\gamma_{2}}^{-1}Kn^{-1}+$
$\displaystyle
C\gamma_{1}^{-2}\theta_{\gamma_{2}}^{-1}\sum_{j=1}^{3}|\xi_{j}(q)||\mu_{j}^{n}(U)-\mu_{j}(U)|.$

Take $\gamma_{2}=\gamma_{1}^{3}$. Then taking $\gamma_{1}$ small makes terms
1, 2, and 4 small. Taking $n$ large makes terms 3 and 5 small and by Lemma
5.2.8 also makes term 6 small. ∎

Next we remove the condition that $S$ is contained in the interior of $R$.

######  Proposition 5.5.6.

For any $S\in\mathbb{S}(R)$ and any sequence of tiling flows $f_{n}\in
TF_{n}(R)$ such that $d_{W}(f_{n},f)\to 0$ as $n\to\infty$ for some $f\in
AF(R)$,

$\displaystyle\lim_{n\to\infty}\mathbb{W}_{1}^{1,1}(T(f_{n},S),T(f,S))=0.$

###### Proof.

We can cover $S$ with finitely many surfaces $S_{1},...,S_{M}$ which satisfy
the conditions of Proposition 5.4.2 for vectors $v_{1},...,v_{k}$ and a
threshold $\theta>0$. We can do this so that $d=\max_{1\leq j\leq
M}\text{diam}(S_{j})$ and there is a constant $C$ independent of $d$ such that
$M=Cd^{-2}$. Fix $\epsilon>0$. There exists $\delta>0$, $N>0$ such that for
all $j=1,...,M$, and all $0\leq t\leq\delta$, by Proposition 5.4.4,

$\displaystyle\mathbb{W}_{1}^{1,1}(T(f,S_{j}),T(f,S_{j}(t)))<\epsilon$

and by Proposition 5.5.4, for $n\geq N$ and $0\leq t\leq\delta$,

$\displaystyle\mathbb{W}_{1}^{1,1}(T(f_{n},S_{j}),T(f_{n},S_{j}(t)))<\epsilon.$

On the other hand, by Proposition 5.5.5, for all $j=1,..,M$ and all $t>0$, for
$n$ large enough

$\displaystyle\mathbb{W}_{1}^{1,1}(T(f_{n},S_{j}(t)),T(f,S_{j}(t)))<\epsilon.$

Hence by the triangle inequality, for all $j=1,...,M$

$\displaystyle\mathbb{W}_{1}^{1,1}(T(f,S_{j}),T(f_{n},S_{j}))<3\epsilon.$

By the two-dimensional version of Lemma 5.2.7, for $n$ large enough,

$\displaystyle\mathbb{W}_{1}^{1,1}(T(f,S),T(f_{n},S))\leq
M(10d^{3}+3\epsilon)\leq 10Cd+3d^{-2}\epsilon.$

Taking $d=\epsilon^{1/3}$ would complete the proof. ∎

Finally we can prove the main theorem about boundary values that we will refer
to later in paper.

######  Theorem 5.5.7.

For any $S\in\mathbb{S}(R)$, the trace operator

$\displaystyle T(\cdot,S):(AF(R)\cup
TF(R),d_{W})\to(\mathcal{M}^{s}(R),\mathbb{W}_{1}^{1,1})$

is uniformly continuous. In particular this holds for $S=\partial R$.

###### Proof.

By Theorem 5.3.4, $(AF(R),d_{W})$ is compact. Since $TF_{n}(R)$ is finite for
each $n$, Theorem 5.3.1 implies that the $d_{W}$ limit points of $TF(R)$ are
contained in $AF(R)$. Therefore $(AF(R)\cup TF(R),d_{W})$ is compact.

On the other hand, for any $S\in\mathbb{S}(R)$, Proposition 5.4.6 and
Proposition 5.5.6 combine to show that $T(\cdot,S)$ is a continuous map from
$(AF(R)\cup TF(R),d_{W})$ to $(\mathcal{M}^{s}(R),\mathbb{W}_{1}^{1,1})$.
Therefore by compactness $T(\cdot,S)$ is uniformly continuous. ∎

##  6 Patching

The main goal of this section is to prove a patching theorem (Theorem 6.3.5)
which will be an essential tool throughout this paper. We show that if the
flows associated to tilings $\tau_{1},\tau_{2}$ of $\mathbb{Z}^{3}$ are
nearly-constant (Definition 6.3.3 below) with value
$s\in\text{Int}(\mathcal{O})$ (which loosely speaking means the flows
associated with $\tau_{1},\tau_{2}$ both approximate the constant flow equal
to $s$), then we can remove a bounded piece from $\tau_{1}$, and patch it to
$\tau_{2}$ by tiling a thin (cubic) annulus.

Figure 21: Two dimensional schematic for patching.

Equivalently, we want to show that this annular region can be tiled by dimers
exactly so that it agrees with $\tau_{1}$ on one boundary and with $\tau_{2}$
on the other boundary. To do this, we need a condition to show that a region
is tileable.

A general condition for tileability, which works in any dimension, is given by
the classical Hall’s matching theorem ([Hal35], stated here as Theorem 6.1.2),
which says a region $R$ is tileable if and only if every $U\subset R$ has
certain properties; a $U$ that does not have such properties is a
“counterexample” to the condition that implies tilability, as we explain in
Section 6.1. In two dimensions, Hall’s matching theorem implies that a simple
condition on height function differences along the boundary of the region is
equivalent to tileabilty [Fou96]. In three dimensions, we show in Section 6.2
that it is sufficient to show that the region $R$ has no counterexample set
$U$ whose boundary is a certain type of minimal surface, built out of squares
from the $\mathbb{Z}^{3}$ lattice (Corollary 6.2.6). We call surfaces built
out of lattice squares discrete surfaces.

In Section 6.3 we give the statement of the patching theorem, and explain the
main ideas of the proof accompanied by a series of two dimensional figures.

The main new difficulty in higher dimensions is that the counterexample sets
$U$ can have more complicated geometry. In two dimensions, the boundary of the
counterexample region is a union of curves. In three dimensions it is a union
of surfaces. However the fact that these surfaces can be assumed to be in some
sense minimal gives us some control their geometry. In Section 6.4, we prove
some straightforward adaptations of the isoperimetric inequality for discrete
surfaces. In Section 6.5, we apply these to get useful bounds on the area
growth for minimal discrete surface (Proposition 6.5.1), and show that they
“spread out” (Lemma 6.5.3).

In Section 6.6, we prove an ergodic theorem for the flow of a tiling through a
coordinate plane (Theorem 6.6.3), and note that tilings sampled from ergodic
measures satisfy the conditions of the patching theorem with probability going
to $1$ as $n\to\infty$ (Corollary 6.6.5). We show that ergodic measures of any
mean current $s\in\mathcal{O}$ exist (Lemma 6.6.1), and prove some bounds for
their expected flow through discrete surfaces (Lemma 6.6.7). One of the ideas
in the proof of the patching theorem is to use a tiling sampled from an
ergodic measure as a “measuring stick” that we compare with the tilings we
want to patch.

Equipped with the lemmas from the previous sections, in Section 6.7 we give
the proof of the patching theorem (Theorem 6.3.5). Finally in Section 6.8 we
give some immediate corollaries of patching for ergodic Gibbs measures (EGMs).

We use the same tools and ideas developed in this section again in Section 8.6
to prove a generalized patching theorem (Theorem 8.6.2) where the flow the
tilings approximate is not required to be constant, and where the annular
region is allowed to have a more general shape. The main results of this paper
are two versions of a large deviation principle: one with soft boundary
conditions (Theorem 8.1.6) and one with hard boundary conditions (Theorem
8.2.4). The regular patching theorem proved here (Theorem 6.3.5) is sufficient
to prove the LDP with soft boundary conditions, but the generalized version
(Theorem 8.6.2) is needed in the final steps to prove the version with hard
boundary conditions.

###  6.1 Hall’s matching theorem and non-tileability

When can a finite region $R\subset\mathbb{Z}^{3}$ be exactly tiled by dimers,
i.e. without any tiles crossing the boundary, and with all cubes covered? This
is equivalent to asking: when does a finite subgraph $G\subset\mathbb{Z}^{3}$
have a perfect matching? A straightforward observation is that for any
bipartite graph $G$ with bipartition $(A,B)$, a necessary condition for $G$ to
have a perfect matching is that it is balanced, i.e. that $|A|=|B|$. The
balanced condition is not sufficient though, see Figure 22.

Figure 22: A region that is balanced (i.e. the number of black squares is
equal to the number of white squares) but has no dimer tilings.

Nonetheless, it turns out there is a very general necessary and sufficient
condition which characterizes whether or not a finite bipartite graph $G$ has
a perfect matching. In this section we explain these results from the graph
point of view where dimers are edges and dimer tilings are perfect matchings.
There are two perspectives, both of which can be useful:

  * •

the min-cut, max-flow principle

  * •

Hall’s matching theorem

We first describe the classical min-cut, max-flow principle. Let $G=(A,B)$ be
the bipartition of the graph ($A$ is “even” and $B$ is “odd”). If $G$ has a
perfect matching $\tau$, then there is a flow $v_{\tau}$ (the “pretiling
flow”) which sends a unit of current from each even vertex $a\in A$ to the odd
vertex $b\in B$ it is paired to. Note that $v_{\tau}$ has a source of $+1$ at
each $a\in A$ and a sink of $+1$ (or source of $-1$) at each $b\in B$. The
existence of a perfect matching is equivalent to the existence of a flow
$v_{\tau}$ with the desired source/sink values and a flow of $0$ or $1$ on
each even-to-odd edge.

A cut is a collection of edges in $G$ which, if deleted, separates $G$ into
two pieces $G_{1}$ and $G_{2}$. Let $F_{1}$ be the net total flow sourced in
$G_{1}$ (i.e. the number of even vertices in $G_{1}$ minus the number of odd
vertices), let $F_{2}$ be the net total flow sourced in $G_{2}$, and let $c$
be the number of cut edges. If $G$ has a perfect matching, we must have
$F_{1}+F_{2}=0$.

The value $F_{1}$ measures the amount of flow that would have to travel across
the cut if $G$ has a perfect matching, so we must have $F_{1}\leq c$. In other
words, if $G$ has a perfect matching, then it must be the case that for every
cut, the excess flow on either side must be less than the size of the cut. It
turns out that this is a sufficient condition too, so if $G$ does not have a
perfect matching then there is a cut of $c$ edges partitioning $G$ into
$G_{1}$ and $G_{2}$ such that the excess flow $F_{1}$ that needs to cross the
cut is more than $c$. In summary:

######  Theorem 6.1.1 (Min-cut, max-flow principle [FF56]).

A finite bipartite graph $G$ has a perfect matching (a.k.a. dimer tiling) if
and only if there is no cut consisting of $c$ edges partitioning $G$ into two
sets $G_{1}$ and $G_{2}$ such that $F_{1}>c$.

In Hall’s matching theorem, we shift our perspective from the cut to the sets
in the partition. Instead of looking at sets that are a mixture of even and
odd vertices, we consider a set $C$ of only even (resp. only odd) vertices,
plus their neighbors

$N(C)=\\{b\in B:a\in C,(a,b)\in E\\}.$

The set $C\cup N(C)$ is analogous to either $G_{1}$ or $G_{2}$ (without loss
of generality $G_{1}$) plus some of the endpoints of the edges in the cut. The
excess flow is now the number of even vertices (i.e. $|C|$) minus the number
of odd vertices (i.e. $|N(C)|$). Hall’s matching theorem is an analog of
Theorem 6.1.1 formulated in these terms:

######  Theorem 6.1.2 (Hall’s matching theorem [Hal35]).

Suppose that $G=(V,E)$ is a finite bipartite graph with bipartition $G=A\cup
B$. The graph $G$ admits a perfect matching consisting of $|A|$ edges if and
only if for all $C\subset A$,

$\displaystyle N(C)=\\{b\in B:a\in C,(a,b)\in E\\}$

satisfies $|N(C)|\geq|C|$.

An analogous result holds with $A$ and $B$ switched. If $G$ is balanced (i.e.
$|A|=|B|$), then the existence of a perfect matching with $|A|$ edges is
equivalent to the existence of a perfect matching of the whole graph $G$. If
$G$ is not balanced (i.e. $|A|\neq|B|$) then $G$ does not have a perfect
matching of the whole graph.

Note that if $C\subset A$ satisfies $|N(C)|<|C|$, then the set $U:=C\cup N(C)$
has more even than odd vertices, despite having only odd vertices on its
boundary within $G$. Therefore when $G$ is balanced, Theorem 6.1.2 is
equivalent to the following:

######  Theorem 6.1.3.

Suppose that $G=(V,E)$ is a finite bipartite graph with bipartition $G=A\cup
B$ with $|A|=|B|$. Then $G$ fails to have a perfect matching if and only if
there exists a connected set $U\subset V$ such that $|U\cap A|>|U\cap B|$ but
all boundary elements of $U$ (i.e., elements of $U$ that are adjacent to some
point in $V\setminus U$) belong to $B$.

We call the $U$ from Theorem 6.1.3 a counterexample to (the condition
equivalent to) tileability or just a counterexample. In our context, $A$ and
$B$ will always be sets of even and odd vertices in $\mathbb{Z}^{3}$, so for
us a counterexample to tileability for $R\subset\mathbb{Z}^{3}$ is any set $U$
that has more even than odd vertices, despite having only odd vertices on its
interior boundary, which we define to be the set of $x\in U$ that are incident
to some $y\in R\setminus U$.

To show that a graph $R$ has a dimer tiling (a.k.a. a perfect matching), we
check that it is balanced, and if it is, we have to show that there are no
counterexamples. We call the excess flow of a counterexample its imbalance,
given by

$\text{imbalance}(U)=\text{even}(U)-\text{odd}(U).$ (30)

Note that if $U\subset R\subset\mathbb{Z}^{3}$ has only odd vertices on its
interior boundary (within $R$) then $\text{imbalance}(U)>0$ if and only if $U$
is a counterexample for $R$.

###  6.2 Discrete surfaces and minimal counterexamples

As mentioned earlier, it is often intuitively useful to think about perfect
matchings as tilings by $2\times 1\times 1$ blocks. In this picture, a
counterexample set $U$ is a collection of unit cubes, each centered at a point
in $\mathbb{Z}^{3}$, and the edges out of it, i.e. its boundary $\partial U$,
is a collection of squares in the translated lattice
$(\frac{1}{2},\frac{1}{2},\frac{1}{2})+\mathbb{Z}^{3}$. In other words the
boundary region is a surface built out of squares from the lattice.

######  Definition 6.2.1.

A discrete surface in $\mathbb{Z}^{3}$ is a collection of squares from the
$(\frac{1}{2},\frac{1}{2},\frac{1}{2})+\mathbb{Z}^{3}$ lattice.

A discrete surface in $\mathbb{Z}^{3}$ is orientable if there is a well-
defined outward pointing normal vector to the surface. An orientable discrete
surface $S$ with a choice of outward pointing normal vector is called
oriented. For a square $s\subset S$, we call the side of $s$ that the outward
normal vector points toward the outside. If the outward pointing normal vector
to a square in an oriented surface is from even to odd, we color the outside
of the square white. Otherwise we color it black.

######  Definition 6.2.2.

An oriented discrete surface $S$ in $\mathbb{Z}^{3}$ is monochromatic if all
the outsides of all the squares in $S$ are black (resp. all are white).

We can rewrite Equation (30) for the imbalance of a counterexample $U$ in
terms of the black and white surface area of $\partial U$. Let $(A,B)$ be the
bipartition of $\mathbb{Z}^{3}$ into even and odd vertices respectively.

######  Proposition 6.2.3.

Suppose that $R$ is balanced but not tileable, and that $U\subset
R\subset\mathbb{Z}^{3}$ is a counterexample to tileability. Then

$0<\text{imbalance}(U)=\text{even}(U)-\text{odd}(U)=\frac{1}{6}\bigg{(}\text{white}(\partial
U)-\text{black}(\partial U)\bigg{)}.$ (31)

###### Proof.

Define a flow $r$ on $\mathbb{Z}^{3}$ such that $r({e})=\frac{1}{6}$ for every
dual edge (a.k.a. face) ${e}$ oriented from even to odd. Then

$\displaystyle\text{div}\,r(v)=\begin{cases}-1\quad&v\text{ is a odd cube}\\\
+1\quad&v\text{ is a even cube}\end{cases}$

By the divergence theorem, with all edges $e\in\partial U$ oriented out of
$U$,

$\displaystyle\text{imbalance}(U)$
$\displaystyle=\text{even}(U)-\text{odd}(U)=\sum_{v\in
U}\text{div}\,r(v)=\sum_{e\in\partial U}r({e})$
$\displaystyle=\frac{1}{6}\bigg{(}\text{white}(\partial
U)-\text{black}(\partial U)\bigg{)}.$

∎

By Proposition 6.2.3, if $U$ is a counterexample then it must have more white
surface area than black surface area. By the definitions in Section 6.1, $U$
must have only odd cubes along its interior boundary, i.e. cubes $x\in U$
which are adjacent to $y\in R\setminus U$. However $U$ also has an exterior
boundary consisting of cubes which are adjacent to
$y\in\mathbb{Z}^{3}\setminus R$. Exterior boundary cubes can be even or odd.

Correspondingly, the boundary $\partial U$ can be split into two pieces: the
exterior boundary surface $T=\partial R\cap\partial U$ and the interior
boundary surface $S=\partial U\setminus T$. The interior boundary surface $S$
must be built out of only black squares, while $T$ can be built out of a
mixture of white and black squares. Given this, only squares in $T$ contribute
positively to the imbalance of $U$. Intuitively to increase the imbalance of
$U$, one should minimize the area of $S$.

A surface $P$ embedded in $\mathbb{R}^{3}$ is said to locally minimize area if
given any point $p\in P$, there is a neighborhood $V\subset\mathbb{R}^{3}$
containing $p$ such that $P\cap V$ has the minimal area of any surface with
boundary $\partial(P\cap V)$. Surfaces that locally minimize area are called
minimal surfaces. We will be interested in certain discrete analogs of minimal
surfaces.

######  Definition 6.2.4.

A minimal discrete (monochromatic) surface with boundary $X$ is a surface $S$
that minimizes area subject to the constraint that it is discrete,
(monochromatic), and has $\partial S=X$. In particular, there is no way to
“tighten the surface locally” by changing a few faces in a way that maintains
the overall boundary conditions and reduces the overall area.

######  Proposition 6.2.5.

Let $R\subset\mathbb{Z}^{3}$ be a finite balanced region which is not
tileable, and suppose that $U$ is a counterexample to tileability in $R$. Let
$T=\partial U\cap\partial R$ and let $S=\partial U\setminus T$ be the interior
boundary surface. Then there exists another counterexample $U^{\prime}$ in $R$
such that $\partial U^{\prime}\cap\partial R=T$, and $S^{\prime}:=\partial
U^{\prime}\setminus T$ is a minimal monochromatic black discrete surface.

###### Proof.

The new set $U^{\prime}$ is defined so that $\partial U^{\prime}=T\cup
S^{\prime}$, where $S^{\prime}$ is a minimal monochromatic discrete surface.
Since $S$ is all black, by Proposition 6.2.3,

$6\cdot\text{imbalance}(U)=\text{white}(\partial U)-\text{black}(\partial
U)=\text{white}(T)-\text{black}(T)-\text{area}(S)$ (32)

replacing $S$ by $S^{\prime}$ only makes the imbalance larger, so $U^{\prime}$
is still a counterexample. ∎

We call counterexamples where the internal boundary surface is a minimal
surface minimal counterexamples. We immediately get the following corollary.

######  Corollary 6.2.6.

A finite balanced region $R\subset\mathbb{Z}^{3}$ is tileable if and only if
it has no minimal counterexamples.

###  6.3 Statement of patching theorem and outline of proof

We now give the statement of the patching theorem mentioned at the beginning
of the section. We also provide illustrations of the analogous constructions
in 2D (because they are easier to draw) in order to explain the 3D concepts.
Let $B_{n}=[-n,n]^{3}$, and for any $\delta>0$ define

$\displaystyle A_{n}=B_{n}\setminus B_{(1-\delta)n}.$

Given two tilings $\tau_{1},\tau_{2}\in\Omega$, we look at the region between
$\tau_{1}\mid_{\mathbb{Z}^{3}\setminus B_{n}}$ and
$\tau_{2}\mid_{B_{(1-\delta)n}}$. This will be the annulus $A_{n}$, with some
cubes removed along its boundary (see Figure 23). We call this $A_{n}$ with
boundary conditions $\tau_{1}$ and $\tau_{2}$.

Figure 23: Here is an example of a region of the form “$A_{n}$ with boundary
conditions $\tau_{1}$ and $\tau_{2}$” in two dimensions, which we call $A$.
The region shown is a subset of the “square annulus” bounded between the
boundary of a $12\times 12$ box and the boundary of a $32\times 32$ box. It is
obtained from the square annulus by removing some of the squares along the
outer boundary and some of the squares along the inner boundary. Given any
dimer tiling $\tau$ of a region containing the $32\times 32$ box, the union of
the dimers of $\tau$ that are strictly contained in the square annulus would
also be a region of this form.

The main question is: given a tiling $\tau_{1}$ restricted to
$\mathbb{Z}^{3}\setminus B_{n}$ and a tiling $\tau_{2}$ restricted to
$B_{(1-\delta)n}$, under what conditions can we patch them together, i.e. find
a tiling of $A_{n}$ with inner boundary condition $\tau_{2}$ and outer
boundary condition $\tau_{1}$? We are interested in showing that this is
possible when $n$ is large enough, when $\tau_{1},\tau_{2}$ satisfy a
consistency condition that they are nearly constant for the same
$s\in\text{Int}(\mathcal{O})$.

To specify the nearly constant condition, we give a few definitions.

######  Definition 6.3.1.

An $\epsilon$ patch $\alpha$ on $\partial B_{n}$ is an $\epsilon
n\times\epsilon n$ square contained in a face of $\partial B_{n}$.

We can then measure the flux of a discrete flow through a patch.

######  Definition 6.3.2.

Let $S$ be an oriented discrete surface with outward normal vector $\xi$. We
define the flux of a discrete vector field $v$ through $S$ by

$\displaystyle\text{flux}(v,S)=\sum_{e\in E(\mathbb{Z}^{3}),e\cap
S\neq\emptyset}\text{sign}\langle\xi(e\cap S),e\rangle v(e).$

Here $E(\mathbb{Z}^{3})$ denotes the edges of $\mathbb{Z}^{3}$ oriented from
even to odd.

We now use the definition of nearly-constant in terms of the flux of the pre-
tiling flow $v_{\tau}$ through patches.

######  Definition 6.3.3.

Fix $s\in\mathcal{O}$, let $B_{n}=[-n,n]^{3}$. A tiling $\tau\in\Omega$ is
$\epsilon$-nearly-constant with value $s$ if there exists $M=M(\epsilon)$ such
that for all $n>M$ and all $\epsilon$ patches $\alpha$ on $\partial B_{n}$,

$\displaystyle\text{flux}(v_{\tau},\alpha)=\frac{1}{2}\langle\xi_{\alpha},s\rangle\text{area}(\alpha)+o(\text{area}(\alpha))=\frac{1}{2}\langle\xi_{\alpha},s\rangle\epsilon^{2}n^{2}+o(\epsilon^{2}n^{2}),$

where $\xi_{\alpha}$ is the outward pointing unit normal vector to $\alpha$
(where outward means away from $B_{n}$).

######  Remark 6.3.4.

Any patch $\alpha$ is contained in a flat coordinate plane, so its area is
simply $\epsilon^{2}n^{2}$. The $\frac{1}{2}$ comes from the fact that the
mean current is actually measuring the average flow per even vertex.

Tilings satisfying the $\epsilon$-nearly-constant condition with value $s$
mimic the behavior of tilings sampled from ergodic measures of mean current
$s$. (This is made precise in Corollary 6.6.5 after Theorem 6.6.3.)

With the conditions defined, we can now state the patching theorem.

######  Theorem 6.3.5 (patching theorem).

Fix $\delta>0$ and a mean current $s\in\text{Int}(\mathcal{O})$. Let
$B_{n}=[-n,n]^{3}$ be the cube of radius $n$, and let $A_{n}=B_{n}\setminus
B_{(1-\delta)n}$ be the cube annulus of width $\delta n$. For $\epsilon>0$
small enough, if $\tau_{1},\tau_{2}\in\Omega$ are $\epsilon$-nearly-constant
with value $s$, then for $n$ large enough $A_{n}$ can be tiled with outer
boundary condition $\tau_{1}$ and inner boundary condition $\tau_{2}$.

The main tool in the proof is Hall’s matching theorem (Theorem 6.1.2). In this
section we explain the main steps of the proof, guided by a series of two
dimensional figures, and comment on differences between the two and three
dimensional versions of this story. After this, in the remaining subsections
we prove a series of lemmas (needed to control the more complicated geometric
situations that can occur in three dimensions) before giving a the formal
proof of Theorem 6.3.5 in Section 6.7.

Steps of proof

Given two tilings $\tau_{1},\tau_{2}$ of $\mathbb{Z}^{2}$, we want to know
whether they can be patched together. In other words, we want to know whether
a region $A$, which is a square annulus with some squares removed along the
outer boundary if they are connected by $\tau_{1}$ to the outside of the
annulus and some squares removed along the inner boundary if they are
connected by $\tau_{2}$ to the outside of the annulus, is tileable. See Figure
23 for an example of a region $A$ of this form. If $A$ is not tileable, then
there exists a counterexample set $U$ as in Figure 24.

Figure 24: A potential counterexample set $U\subset A$.

The red set $U\subset A$ in Figure 24 has the property that every square on
its inner boundary (i.e., every square of $U$ that is incident to a square in
$A\setminus U$) is black.

By Hall’s matching theorem (Theorem 6.1.3), there exists a dimer tiling of $A$
if and only if every $U$ of this form is not a counterexample. In other words,
every set $U$ of this form has
$\text{imbalance}(U)=\text{white}(U)-\text{black}(U)\leq 0$. We remark that
the colors white and black used here stand for even and odd vertices and not
for the colors we give to surfaces in 3 dimensions in the previous section. We
do this because it becomes easier to illustrate the main ideas using the
figures.

Given this, our strategy to show that $A$ is tileable for $n$ large enough
proceeds by contradiction. We suppose that for all $n$ there is a set $U$ of
the form above that is a counterexample, but then show that for $n$ large
enough, it must have $\text{white}(U)\leq\text{black}(U)$ and hence not be a
counterexample. To do this, we cut up $U$ into various smaller pieces, and
bound the white minus black in each piece.

First we divide the annular $A$ into regions as depicted in Figure 25. We call
these the “middle region” (blue), the “thin region” (orange) and the “corner
region” (green). The middle region is a centered square annulus whose size
will have to be appropriately tuned. The thin region is the union of the
“columns” obtained as straight-line paths of squares that go from the middle
square annulus boundary to the boundary of $A$. The corner region is the part
leftover.

We then define $U^{\prime}$ to be $U$ intersected with the middle layer,
depicted in Figure 26.

Figure 25: The middle (blue), thin (orange), and corner (green) regions of
$A$.
Figure 26: We can define $U^{\prime}$ to be the intersection of $U$ with the
middle region from Figure 25.

Given a tiling $\tau$ of a region containing $A$, we can define $U_{\tau}$ to
be the region covered by tiles from $\tau$ intersecting $U^{\prime}$. The set
$U^{\prime\prime}$ shown in Figure 27 is the subset of $U_{\tau}$ that
consists of the union of $U^{\prime}$ together with all of the squares covered
by tiles from $\tau$ that are contained in $U$ but in the complement of the
middle region. Note that $U_{\tau}\setminus U^{\prime\prime}$ consists of only
white squares and that $U_{\tau}$ is by construction evenly balanced between
black and white squares. If we can show that $|U_{\tau}\setminus
U^{\prime\prime}|$ is large then we know that $U^{\prime\prime}$ has many more
black squares than white.

Indeed, we show that we can choose this test tiling $\tau$ so that
$|U_{\tau}\setminus U^{\prime\prime}|$ is large and hence $U^{\prime\prime}$
has many more black than white squares. Since $U^{\prime\prime}\subset U$, it
remains to show that there cannot be enough white squares in $U\setminus
U^{\prime\prime}$ for $U$ to be a counterexample.

Figure 27: The region $U^{\prime\prime}$ (blue) and the tiles from $\tau$ that
intersect $U^{\prime}$ (green).

In order to prove that $U$ itself has more black than white squares, we will
divide the rest of $U$ into multiple pieces to treat separately, depicted in
Figure 28. Here $U^{\prime\prime}$ is as given in Figure 27. The “shadow
region” is $U\setminus U^{\prime\prime}$ restriced to the “thin region” from
Figure 25. It consists of the union of the columns that can be extended all
the way from $\partial A$ to $\partial U^{\prime\prime}$. The corner region
here is the intersection of $U$ with the corner region from Figure 25. The
“leftover pieces” are the parts of $U$ that do not belong to one of the other
three regions. Roughly speaking, we aim to show that $U$ has more black than
white by showing that (i) $U^{\prime\prime}$ has a lot more black than white,
(ii) the shadow region can only have a little more white than black (because
of the nearly-constant condition), (iii) the corner region can only have a
little excess white (since it has small volume), and (iv) the leftover pieces
have at least as much black as white.

Figure 28: Depiction of all the regions that we divide $U$ into: (i)
$U^{\prime\prime}$, (ii) shadow region, (iii) corner region, and (iv) leftover
pieces.

Ultimately to achieve (i), (ii), (iii), and (iv), we need to understand more
about the possible shapes of counterexample sets $U$. The main difference
between two dimensions and three dimensions is that the geometry of $U$ can be
more complicated.

In two dimensions, since $A$ is balanced, if $U$ is a counterexample then so
is $U^{c}$ (with even/odd switched). As a consequence, without loss of
generality $U$ and $U^{c}$ are both connected, and (since $U$ and $U^{c}$ must
both cross the annulus) the interior boundary curve $S$ between them consists
of two minimal monochromatic paths $S_{1},S_{2}$ from the inner boundary to
the outer boundary. There are then four points $a_{1},b_{1}$ and $a_{2},b_{2}$
which are the endpoints of $S_{1},S_{2}$ on the inner and outer boundaries
respectively. From here, one can argue as in [Fou96] to say that $A$ is
tileable as long as the boundary height differences (given by the boundary
condition tilings $\tau_{1},\tau_{2}$) at $a_{1}$ and $b_{1}$ and at $a_{2}$
and $b_{2}$ are in some sense compatible.

In three dimensions, since $A$ is balanced, it again follows that both $U$ and
$U^{c}$ are counterexamples, and that they are without loss of generality
connected and have interior boundary surface $S$ between them which is a
minimal monochromatic discrete surface. However the geometry (and topology) of
2D surfaces and connected sets $S$ in 3D can be much more complicated than
that of 1D curves in 2D. For example, the set $U$ in 3D need not be simply
connected, and instead of $S\cap\partial A$ consisting of four points, it
consists of a collection of curves.

To control the more complex geometric possibilities for counterexamples in
three dimensions, we prove that the interior boundary surface $S$ has area of
order $n^{2}$ (Lemma 6.5.2), and that there is a choice for the inward blue
layer where $S$ restricted to the layer (which is some union of curves) has
good properties (Lemma 6.5.3). This is the content of Section 6.4 and Section
6.5.

The other key tool is the notion of a test tiling. We show that we can use a
tiling $\tau$ sampled from an ergodic measure of mean current
$s\in\text{Int}(\mathcal{O})$ to define $U^{\prime\prime}$ so that
$U^{\prime\prime}$ will have order $n^{2}$ more black cubes than white cubes
by showing that $\tau$ on expectation has an order $n^{2}$ number of tiles
crossing $S$ (Lemma 6.6.7). This quantity is somewhat analogous to the height
difference in two dimensions. Further, a tiling sampled from ergodic measure
of mean current $s\in\text{Int}(\mathcal{O})$ is nearly constant with high
probability (Corollary 6.6.5). This is the content of Section 6.6.

###  6.4 Discrete isoperimetric inequalities

The classical isoperimetric inequality says that the minimal area of a region
$D$ bounded by a smooth closed curve $\gamma$ in $\mathbb{R}^{2}$ of length
$l$ is $\frac{1}{4\pi}(l^{2})$. The equality case is achieved when $\gamma$ is
a circle and $D$ is a disk. A lesser known fact is that this bound also holds
for curves in $\mathbb{R}^{3}$ [Alm86]. In this section we prove a discrete
version of the isoperimetric inequality.

######  Proposition 6.4.1.

Given any simple closed curve $\gamma$ in $\mathbb{Z}^{3}$, there is a surface
$S$ with $\partial S=\gamma$ and

$\text{area}(S)\leq\frac{1}{8}(\text{length}(\gamma))^{2}.$

######  Remark 6.4.2.

The constant $\frac{1}{8}$ that we get from the proof is not optimal but our
argument is simple and the result is sufficient for our purposes. Also note
that the boundary curve $\gamma$ can be replaced by a multicurve
$\Gamma=\gamma_{1}\cup\dots\cup\gamma_{k}$ and the same result holds.

###### Proof.

We proceed by induction. Suppose that for all simple closed curves $\gamma$ in
$\mathbb{Z}^{3}$ with $\text{length}(\gamma)=2m\leq 2n$, there exists a
surface $S$ with $\partial S=\gamma$ and

$\text{area}(S)\leq\frac{(m-1)m}{2}.$

This is sufficient because it implies that

$\text{area}(S)\leq\frac{1}{8}(\text{length}(\gamma))^{2}.$

Now suppose that $\beta$ is a simple closed curve of length $2n+2$, and equip
$\beta$ with an orientation. Choose two parallel edges
$e_{1}=(a_{1},b_{1}),e_{2}=(a_{2},b_{2})$ in $\beta$ with opposite
orientations. Removing $e_{1},e_{2}$ and identifying $a_{i}$ with $b_{i}$ for
$i=1,2$ results in a new curve $\beta^{\prime}$ of length $2n$. The
identification means that $\beta^{\prime}$ is a union simple curves
$\gamma_{1},...,\gamma_{k}$ of length $\leq 2n$ and double edges (note that
each double edge contributes $2$ to the length of $\beta^{\prime}$). By the
inductive hypothesis, there exist surfaces $S_{i}$ with boundary $\gamma_{i}$
satisfying the bound for each $i=1,...,k$. To get a surface with boundary
$\beta$, we find a path in $\mathbb{Z}^{3}$ from $a_{1}$ to $a_{2}$ along
$\cup_{i=1}^{k}S_{i}\cup D$, where $D$ is the double edges in
$\beta^{\prime}$. Since $\text{length}(\beta^{\prime})\leq 2n$, we can find a
path of length $\leq n$. We add back the edges $e_{1},e_{2}$, splitting the
path into two parallel paths. We then add squares along the path from $e_{1}$
to $e_{2}$ to construct a surface $S$ with boundary $\beta$ satisfying

$\displaystyle\text{area}(S)\leq\text{area}(\cup_{i=0}^{k}S_{i})+n\leq\frac{n(n+1)}{2}.$

∎

This can easily be extended to the monochromatic case; the only thing that
changes is the constant.

######  Corollary 6.4.3.

Given any collection of simple closed curves $\gamma_{1},...,\gamma_{k}$ in
$\mathbb{Z}^{3}$, there is a monochromatic surface $S$ with
$\Gamma=\gamma_{1}\cup\dots\cup\gamma_{k}$ as its boundary and

$\displaystyle\text{area}(S)\leq\frac{5}{8}\text{length}(\Gamma)^{2}$

###### Proof.

Using Proposition 6.4.1, we can find a surface $T$ with boundary
$\gamma_{1}\cup\dots\cup\gamma_{k}$ satisfying
$\text{area}(T)\leq\frac{1}{8}\text{length}(\Gamma)^{2}$. Replacing every
white square in $T$ by at most 5 black squares, we get a new surface $S$ which
is monochromatic (it is all black) satisfying the same bound with the constant
$\frac{5}{8}$. ∎

###  6.5 Area growth of minimal monochromatic discrete surfaces

We now use the discrete isoperimetric inequalities from the previous section
to show that minimal monochromatic discrete surfaces have quadratic area
growth (Proposition 6.5.1). We then apply this to the cube annulus
$A_{n}=B_{n}\setminus B_{(1-\delta)n}$ to get two results (Lemma 6.5.2 and
Lemma 6.5.3) which will serve as lemmas for the patching theorem (Theorem
6.3.5).

######  Proposition 6.5.1.

Let $S$ be a minimal monochromatic discrete surface. Let
$p\in\mathbb{Z}^{3}+(\frac{1}{2},\frac{1}{2},\frac{1}{2})$ be a vertex on $S$
and let $B_{n}(p)=p+[-n,n]^{3}$ be such that $S$ is not contained in
$B_{n}(p)$. Then there is a universal constant $\kappa>0$ (i.e., independent
of $S$ and $n$) such that

$\displaystyle\text{area}(S)\geq\kappa n^{2}.$

###### Proof.

To do this, we show that $\text{area}(S\cap B_{n}(p))$ grows quadratically in
$n$. Let $m=\lfloor{\frac{n}{2}}\rfloor$. For $k\leq m$, define annular
regions

$A_{k}=B_{m+k}(p)\setminus B_{m-k}(p)$

Let $S_{k}=S\cap A_{k}$ be $S$ restricted to $A_{k}$ and let
$\Gamma_{k}=\partial S_{k}$. By Corollary 6.4.3,

$\displaystyle\text{area}(S_{k})\leq\frac{5}{8}\text{length}(\Gamma_{k})^{2}.$

Note that $S_{k}\cap\partial A_{k}$ might be larger than $\Gamma_{k}$, since
there might be squares from $S_{k}$ contained in $\partial A_{k}$.

Any face in $S_{k+1}\setminus S_{k}$ corresponds to at most $4$ edges along
$\Gamma_{k}$. Therefore

$\displaystyle\text{area}(S_{k+1})-\text{area}(S_{k})\geq\frac{1}{4}\text{length}(\Gamma_{k})\geq\frac{1}{4}\sqrt{\frac{5}{8}}\sqrt{\text{area}(S_{k})}.$

Therefore the function $f(k):=\text{area}(S_{k})$ satisfies the inequality
$f(k+1)-f(k)\apprge\sqrt{f(k)}$. Extending $f$ linearly to a continuous
function that is differentiable from the left, this becomes
$f^{\prime}(k)\apprge\sqrt{f(k)}$, which implies that $f$ grows at least
quadratically in $k$. Applying this to $S$ itself we get that
$\text{area}(S)\geq\kappa n^{2}$, where the universal constant $\kappa$ comes
from the isoperimetric inequalities. ∎

######  Lemma 6.5.2.

Let $B_{n}=[-n,n]^{3}$ and suppose that $A=B_{n}\setminus B_{(1-\delta)n}$ for
some $\delta\in(0,1)$. Suppose that $S$ is a minimal monochromatic discrete
surface in $A$ which connects the inner and outer boundaries of $A$. Then
there exist constants $c_{1},c_{2}$ independent of $S$ and $n$ such that

$\displaystyle c_{1}n^{2}\geq\text{area}(S)\geq c_{2}n^{2}$

where $c_{2}\sim\delta^{2}$.

###### Proof.

For the upper bound, notice that $\partial S\subset\partial A$. From $\partial
A$, we construct a monochromatic surface $T$ by capping every white square on
$\partial A$ with a odd cube. There is a surface $S^{\prime}\subset T$ with
the same boundary as $S$. Since $\text{area}(T)\leq 5\,\text{area}(\partial
A)$, there exists a constant $c_{1}$ such that $\text{area}(T)<c_{1}n^{2}$.
Since $S$ is a minimal monochromatic surface,

$\displaystyle c_{1}n^{2}\geq\text{area}(S^{\prime})\geq\text{area}(S).$

Now we prove the lower bound. Since $S$ connects the inner and outer
boundaries, there is a point $p\in S$ where we can apply Proposition 6.5.1 to
$B_{(\delta/3)n}(p)$. Hence

$\displaystyle\text{area}(S)\geq\kappa((\delta/3)n)^{2}=:c_{2}n^{2}$

and $c_{2}$ is of order $\delta^{2}$. ∎

The next application of the area growth results is loosely that minimal
surfaces “spread out.” If $X\subset\partial B_{n}$ for some $n$ is a surface,
we define the $\epsilon$ covering area of $X$ to be the total area in disjoint
$\epsilon n\times\epsilon n$ size squares needed to contain $X$. We denote
this by $\text{Cov}_{\epsilon}(X)$.

######  Lemma 6.5.3 (Indenting lemma).

Fix $\delta>0$. Let $A=B_{n}\setminus B_{(1-\delta)n}$, and suppose that $S$
is a minimal monochromatic discrete surface connecting the inner and outer
boundaries of $A$. Let $\Gamma_{l}=\partial(S\cap\partial B_{l})$. There exist
constants $c,c^{\prime}>0$ independent of $S$ and $n$ such that for any
$\epsilon>0$, there exists $l\in((1-\epsilon^{1/2})n,n)$ such that

$\displaystyle\text{Cov}_{\epsilon}(S\cap\partial B_{l})\leq
c\epsilon^{1/2}n^{2}$

and

$\displaystyle\text{length}(\Gamma_{l})\leq c^{\prime}\epsilon^{-1/2}n.$

The analogous statements also hold for
$l\in((1-\delta)n,(1-\delta+\epsilon^{1/2})n)$.

###### Proof.

Cut the region between $\partial B_{n}$ and $\partial B_{(1-\epsilon^{1/2})n}$
into $M=\lfloor\frac{1}{3\epsilon^{1/2}}\rfloor$ layers $L_{1},...,L_{M}$ of
width $\frac{\epsilon^{1/2}n}{M}\geq 3\epsilon n$. By Lemma 6.5.2 plus the
pigeonhole principle, there exists $j$ such that

$\text{area}(S\cap L_{j})\leq\frac{c_{1}n^{2}}{M}\leq
3c_{1}\epsilon^{1/2}n^{2}.$ (33)

We further subdivide $L_{j}$ into three layers
$L_{j}^{(1)},L_{j}^{(2)},L_{j}^{(3)}$, where $L_{j}^{(2)}$ is the middle one.
These each have width at least $\epsilon n$. The $l$ we find satisfying the
conditions will have $\partial B_{l}\subset L_{j}^{(2)}$.

Let $J$ be the number of $(\epsilon n)^{3}$ sized boxes needed to cover $S\cap
L_{j}^{(2)}$. Then there are at least $J/9$ disjoint cubes of size $(3\epsilon
n)^{3}$ in $L_{j}$, centered on an $(\epsilon n)^{3}$ cube in $L_{j}^{(2)}$,
such that the central cube is in $J$ (i.e., $S$ intersects the $(\epsilon
n)^{3}$ central cube of this $(3\epsilon n)^{3}$ cube). Given this, we can
apply Proposition 6.5.1 to a point in each of the $J/9$ boxes with radius
$\epsilon n$ to get the bound

$\text{area}(S\cap L_{j})\geq\kappa\frac{J}{9}(\epsilon n)^{2}.$ (34)

Combining Equations (33) and (34) and solving for $J$ gives

$\displaystyle J\leq\frac{27c_{1}}{\kappa}\epsilon^{-3/2}.$

Therefore for any $l$ such that $\partial B_{l}\subset L_{j}^{(2)}$,

$\displaystyle\text{Cov}_{\epsilon}(S\cap\partial B_{l})\leq J(\epsilon
n)^{2}\leq\frac{27c_{1}}{\kappa}\epsilon^{1/2}n^{2}.$

so the first part holds with $c=27c_{1}/\kappa$.

For the second part, we note that $L_{j}^{(2)}$ has width $\epsilon n$. Any
square in $S\cap L_{j}^{(2)}$ contributes length at most four to the curves
$\Gamma_{l}$ for $l$ such that $\partial B_{l}\subset L_{j}^{(2)}$, thus
$\sum_{l:\partial B_{l}\subset L_{j}^{(2)}}\text{length}(\Gamma_{l})\leq
4\text{area}(S\cap L_{j}^{(2)})$. Therefore by Equation (33) and the
pigeonhole principle again, we can find $l$ with $\partial B_{l}\subset
L_{j}^{(2)}$ such that

$\displaystyle\text{length}(\Gamma_{l})\leq\frac{12c_{1}\epsilon^{1/2}n^{2}}{\epsilon
n}=12c_{1}\epsilon^{-1/2}n,$

so the second part holds with $c^{\prime}=12c_{1}$. ∎

###  6.6 Tilings sampled from ergodic measures

Recall that $\mathcal{P}_{e}$ denotes the measures on $\Omega$ which are
ergodic with respect to the action of $\mathbb{Z}^{3}_{\text{even}}$, and that
$\\{\eta_{i}\\}_{i=1}^{3}$ denote the standard unit basic vectors. In this
section we prove a few results for tilings sampled from ergodic measure of
mean current $s\in\mathcal{O}$. In the proof of the patching theorem, we use a
“test tiling” sampled from an ergodic measure which we compare with the two
other tilings we want to patch. First we note that there exist ergodic
measures of mean current $s$ for all $s\in\mathcal{O}$.

######  Lemma 6.6.1.

For every $s\in\mathcal{O}$, there exists an ergodic measure on dimer tilings
of $\mathbb{Z}^{3}$ of mean current $s$.

######  Remark 6.6.2.

We use methods called chain swapping described in Section 7.4 to construct
ergodic measures of any edge, then face, then interior mean current from the
ones for $s\in\mathcal{V}$. The only results that we use about chain swapping
here (Propositions 7.4.2 and 7.4.4) are essentially computations, and do not
rely on any results presented in this section.

Note that by Theorem 4.2.7, there exist ergodic measures for all mean currents
$s\in\partial\mathcal{O}$, so we only need to use chain swapping to show
existence for $s\in\text{Int}(\mathcal{O})$. However we choose not to rely on
this here, since the chain swapping techniques allow us to show existence
easily just from existence of ergodic measures at the vertices of
$\partial\mathcal{O}$.

###### Proof of Lemma 6.6.1.

Let $\mathcal{V}\subset\partial\mathcal{O}$ denote its vertices. For each
$s\in\mathcal{V}$, the atomic measure which samples the corresponding
brickwork pattern is an ergodic measure of mean current $s$.

Given any $s\in\text{Int}(\mathcal{O})$ (resp. $s$ contained in a face of
$\partial\mathcal{O}$, resp. an edge of $\partial\mathcal{O}$), there exists
$p\in(0,1)$ such that

$\displaystyle s=(1-p)s_{1}+ps_{2}$

for $s_{1},s_{2}\in\partial\mathcal{O}$ (resp. contained in the edges of
$\partial\mathcal{O}$, resp. contained in
$\mathcal{V}\subset\partial\mathcal{O}$). Let $\mu_{1}$ and $\mu_{2}$ be
ergodic measures of mean current $s_{1},s_{2}$ respectively. Let $\mu$ be a
measure on $\Omega\times\Omega$ which is an ergodic coupling of $\mu_{1}$ and
$\mu_{2}$, let $\mu^{\prime}$ be obtained from $\mu$ by chain swapping with
swap probability $p$, and let $\mu_{1}^{\prime}$ and $\mu_{2}^{\prime}$ denote
its marginals. By Proposition 7.4.4, the mean current of $\mu_{1}^{\prime}$ is

$\displaystyle s(\mu_{1}^{\prime})=(1-p)s_{1}+ps_{2}=s.$

By Proposition 7.4.2, $\mu^{\prime}$ is an ergodic measure on
$\Omega\times\Omega$. Therefore $\mu_{1}^{\prime}$ is an ergodic measure of
mean current $s$. ∎

Recall that the pretiling flow $v_{\tau}$ is defined for $e$ oriented from
even to odd by

$\displaystyle v_{\tau}(e)=\begin{cases}+1&\quad e\in\tau\\\ 0&\quad
e\not\in\tau.\end{cases}$

Let $S$ be an oriented discrete surface with outward normal vector $\xi$.
Applying Definition 6.3.2 to $v_{\tau}$, the flux of $v_{\tau}$ through $S$ by

$\displaystyle\text{flux}(v_{\tau},S)=\sum_{e\in E(\mathbb{Z}^{3}),e\cap
S\neq\emptyset}\text{sign}\langle\xi(e\cap S),e\rangle v_{\tau}(e).$

As in the definition, $E(\mathbb{Z}^{3})$ denotes the edges of
$\mathbb{Z}^{3}$ oriented from even to odd. Flux of $v_{\tau}$ has a simple
combinatorial interpretation. It counts the number of tiles in $\tau$ which
cross $S$, with sign corresponding to the parity of the tile. If $S$ is
monochromatic black, then $\text{flux}(v_{\tau},S)$ is minus the number of
tiles in $\tau$ which cross $S$.

######  Theorem 6.6.3.

Let $P$ be a coordinate plane with normal vector $\eta_{i}$ for some
$i=1,2,3$, and let $P_{n}=P\cap B_{n}$ (recall $B_{n}=[-n,n]^{3}$). If
$\mu\in\mathcal{P}_{e}$ has mean current $s\in\mathcal{O}$, then

$\displaystyle\lim_{n\to\infty}\frac{1}{|P_{n}|}\text{flux}(v_{\tau},P_{n})=\frac{1}{2}\langle
s,\eta_{i}\rangle$

where this limit converges almost surely and in probability.

######  Remark 6.6.4.

The completely equivalent statement holds with $v_{\tau}$ replaced by
$f_{\tau}$. In fact since $P_{n}$ is contained in a coordinate plane
$\text{flux}(f_{\tau},P_{n})=\text{flux}(v_{\tau},P_{n})+o(1)$. The reason for
the $\frac{1}{2}$ factor is that the mean current is the average current per
even vertex. The number of even vertices in $P_{n}$ is $|P_{n}|/2$.

###### Proof.

Without loss of generality let $P$ denote the $(x,y)$ coordinate plane, so the
normal vector is $\eta_{3}=(0,0,1)$. Recall that discrete surfaces consist of
squares in $\mathbb{Z}^{3}+(\frac{1}{2},\frac{1}{2},\frac{1}{2})$, so
$P=\\{(x,y,1/2)\leavevmode\nobreak\ :\leavevmode\nobreak\
x,y\in\mathbb{R}\\}$. Let
$\mathbb{Z}^{2}_{\text{even}}=(\mathbb{Z}^{2}\times\\{0\\})\cap\mathbb{Z}^{3}_{\text{even}}$.
Since $\mu$ is invariant under the $\mathbb{Z}^{2}_{\text{even}}$ action, we
can apply the ergodic theorem for this subaction. Let

$T=P\cap([-1/2,3/2]\times[-1/2,1/2]\times[0,1])$

and $\text{even}(B_{n})=\mathbb{Z}^{3}_{\text{even}}\cap B_{n}$. The set $T$
is defined so that it contains the intersection points of two adjacent edges,
namely $(0,0,0)$ to $(0,0,1)$ and $(1,0,0)$ to $(1,0,1)$. Consider the
function $F:\Omega\to\mathbb{R}$ given by
$F(\tau)=\frac{1}{2}\text{flux}(v_{\tau},T)$. We have that

$\left|\frac{1}{|\text{even}(B_{n})|}\sum_{\eta\in
B_{n}}F(\tau+\eta)-\frac{1}{|P_{n}|}\text{flux}(v_{\tau},P_{n})\right|=o(n^{2})$

and by the ergodic theorem

$\lim_{n\to\infty}\frac{1}{|P_{n}|}\text{flux}(v_{\tau},P_{n})$

exists in probability and almost surely. Temporarily we call the limit
$\text{flux}^{\star}(v_{\tau},P)$. We know that this is
$\mathbb{Z}^{2}_{\text{even}}$-invariant. By integrating the flux across $T$
we get that

$\int_{\Omega}\text{flux}^{\star}(v_{\tau},P)\;\mathrm{d}\mu(\tau)=\int_{\Omega}F(\tau)\>\mathrm{d}\mu(\tau)=\frac{1}{2}\langle
s,\eta_{3}\rangle.$

If we can now prove that the average $\text{flux}^{\star}(v_{\tau},P)$ is in
fact not just $\mathbb{Z}^{2}_{\text{even}}$-invariant but also
$\mathbb{Z}^{3}_{\text{even}}$-invariant it will follow that it is constant
almost surely and equal to $\frac{1}{2}\langle s,\eta_{3}\rangle$. To show
this, we use the fact that $v_{\tau}$ is essentially “divergence free”. Indeed
the tiling flow $f_{\tau}=v_{\tau}-r$ where $r$ is a reference flow and the
flux of $r$ across $T$ is zero. It follows that

$|\text{flux}(f_{\tau},P_{n})-\text{flux}(v_{\tau},P_{n})|=o(n)$

and thereby we have that

$\lim_{n\to\infty}\frac{1}{|P_{n}|}\text{flux}(v_{\tau},P_{n})=\lim_{n\to\infty}\frac{1}{|P_{n}|}\text{flux}(f_{\tau},P_{n}).$

Since $f_{\tau}$ is divergence free it follows that for all
$\gamma\in\mathbb{Z}^{3}_{\text{even}}\setminus\mathbb{Z}^{2}_{\text{even}}$,

$|\text{flux}(f_{\tau},P_{n})-\text{flux}(f_{\tau+\gamma},P_{n})|$

is given by the flux through the sides of parallelopiped formed by $P_{n}$ and
$P_{n}+\gamma$. Thus for a fixed $\gamma\in\mathbb{Z}^{3}_{\text{even}}$

$|\text{flux}(f_{\tau},P_{n})-\text{flux}(f_{\tau+\gamma},P_{n})|=o(n)$

and hence the same holds for $v_{\tau}$ in place of $f_{\tau}$. Thus
$\text{flux}^{\star}(v_{\tau},P)$ is $\mathbb{Z}^{3}_{\text{even}}$-invariant,
which completes the proof. ∎

As a straightforward corollary of Theorem 6.6.3, we see that tilings sampled
from ergodic measures satisfy the $\epsilon$-nearly-constant condition with
high probability.

######  Corollary 6.6.5.

Fix $\epsilon>1$. If $\mu$ is an ergodic measure of mean current $s$, then a
tiling $\tau$ sampled from $\mu$ is $\epsilon$-nearly-constant on $B_{n}$ with
value $s$ with probability arbitrarily close to $1$ for $n$ large enough.

The final goal of this section is to get an estimate on the expected flux of a
pretiling flow across a monochromatic discrete surface (e.g. the boundary of a
counterexample). To do this, we use the following combinatorial result.

######  Lemma 6.6.6.

Suppose that $S$ is a monochromatic discrete surface. Let
$X_{1},X_{2},X_{3}\subset S$ be the sets of squares with normal vectors
$\eta_{1},\eta_{2},\eta_{3}$ respectively.

If $S$ is a closed surface, then $|X_{1}|=|X_{2}|=|X_{3}|$. If $S$ has
boundary $\partial S$, then for all pairs $i\neq j$,

$|X_{i}|=|X_{j}|+O(\text{length}(\partial S)).$

###### Proof.

If $S$ is a closed discrete surface, then every edge of $S$ is contained in
either two or four squares from $S$ (four can happen if the edge is an edge of
non-manifold points). For an edge contained in two squares from the surface,
we say those squares are neighbors. For an edge contained in four squares, we
arbitrarily split the four into pairs, and say that the paired squares are
neighbors. With this definition, every square $f\in S$ has exactly four
neighbors. Since $S$ is monochromatic, if $f\in X_{1}$, then two of its
neighbors are in $X_{2}$ and two of its neighbors are in $X_{3}$ (and
similarly for any permutation of $1,2,3$).

View the set of all squares as a graph, where each square corresponds to a
vertex, and two vertices are connected by an edge if the corresponding squares
are neighbors. The number of edges connecting $X_{1}$ to $X_{2}$ must be equal
to $2|X_{1}|$ (since every $f\in X_{1}$ has two neighbors in $X_{2}$), and
analogously must be equal to $2|X_{2}|$ (since every $f\in X_{2}$ has two
neighbors in $X_{1}$). Therefore $|X_{1}|=|X_{2}|$. An analogous argument
shows that $|X_{3}|=|X_{1}|$ and completes the proof in the closed surface
case.

If $S$ is not closed, then squares $f\in S$ which contain an edge along
$\partial S$ do not have exactly four neighbors. Therefore the result holds up
to an error of $\text{length}(\partial S)$. ∎

######  Lemma 6.6.7.

Let $S$ be a monochromatic black surface with boundary $\partial S$, and let
$\Theta$ be the collection of odd cubes adjacent to $S$. For any tiling
$\tau$,

$\displaystyle|\text{flux}(v_{\tau},S)|\leq|\Theta|.$

If $\mu$ is an ergodic measure of mean current $s\in\text{Int}(\mathcal{O})$,
then there is a constant $K_{\mu}\in(0,1)$ independent of $S$ such that

$\displaystyle\mathbb{E}_{\mu}[|\text{flux}(v_{\tau},S)|]\geq
K_{\mu}|\Theta|+O(\text{length}(\partial S)).$

######  Remark 6.6.8.

If $\mu$ has mean current $s\in\partial\mathcal{O}$, then $K_{\mu}=0$.

######  Remark 6.6.9.

Since $S$ is monochromatic black, $\text{flux}(v_{\tau},S)$ is minus the
number of tiles from $\tau$ crossing $S$. This is why we add the absolute
value.

###### Proof.

Any tile $e\in\tau$ crossing $S$ contains a cube from $\Theta$. From this it
follows immediately that $|\text{flux}(v_{\tau},S)|\leq|\Theta|$.

Let $p_{1},\dots,p_{6}$ be the probabilities under $\mu$ for the six types of
tiles, ordered so that $s_{1}=p_{1}-p_{2}$, $s_{2}=p_{3}-p_{4}$ and
$s_{3}=p_{5}-p_{6}$. Similarly let $N_{1},\dots,N_{6}$ be the six types of
squares on $S$, where the tile type parallel to the outward pointing normal
vector at a square $f\in N_{i}$ has probability $p_{i}$. The random variable
$|\text{flux}(v_{\tau},S)|$ can also be written as $\sum_{f\in
S}\mathbbm{1}_{f}(\tau)$, where $\mathbbm{1}_{f}(\tau)$ is the indicator
variable which is $1$ if there is a tile in $\tau$ crossing $f$ and $0$
otherwise. From this, we see that

$\displaystyle\mathbb{E}_{\mu}[|\text{flux}(v_{\tau},S)|]=\sum_{i=1}^{6}p_{i}N_{i}.$

We minimize the right hand side to get a positive lower bound (clearly $0$ is
a lower bound). Let $\text{area}(S)=N=\sum_{i=1}^{6}N_{i}$. By Lemma 6.6.6,
$N_{1}+N_{2}$, $N_{3}+N_{4}$ and $N_{5}+N_{6}$ are equal to $N/3$ up to an
error of $O(\text{length}(\partial S))$. Thus

$\displaystyle\sum_{i=1}^{6}p_{i}N_{i}$
$\displaystyle\geq\max\\{p_{1}N_{1}+p_{2}N_{2},p_{3}N_{3}+p_{4}N_{4},p_{5}N_{5}+p_{6}N_{6}\\}$
$\displaystyle\geq\max\\{\min\\{p_{1},p_{2}\\},\min\\{p_{3},p_{4}\\},\min\\{p_{5},p_{6}\\}\\}(N/3+O(\text{length}(\partial
S)).$

Since $s\in\text{Int}(\mathcal{O})$, at least four of $\\{p_{i}\\}_{i=1}^{6}$
are nonzero, including one from each pair. Noting that $N=\text{area}(S)$, and
that

$\displaystyle|\Theta|\leq\text{area}(S)\leq 6|\Theta|,$

this proves the result with a constant of the form $K_{\mu}=p_{i}/3$ for some
$i$ such that $p_{i}\neq 0$. ∎

###  6.7 Proof of the patching theorem

We now give the proof of the patching theorem (Theorem 6.3.5), as described in
Section 6.3. We refer throughout to the corresponding figures from the outline
there.

###### Proof of Theorem 6.3.5.

Since $\tau_{1},\tau_{2}\in\Omega$, $A_{n}$ is balanced for all $n$. Thus by
Corollary 6.2.6, if $A_{n}$ is not tileable, it has a minimal counterexample
$U$, i.e. a set with

$\displaystyle\text{imbalance}(U)=\text{even}(U)-\text{odd}(U)>0,$

despite $U$ having only odd cubes on its interior boundary. Let $S$ denote the
interior boundary surface of $U$. Since $A_{n}$ is tileable with just the
$\tau_{1}$ boundary condition (resp. with just the $\tau_{2}$ booundary
condition), $S$ must connect the inner and outer boundaries of $A_{n}$. Thus
by Lemma 6.5.3, we can find $l_{+}\in((1-\epsilon^{1/2})n,n)$ and
$l_{-}\in((1-\delta)n,(1-\delta+\epsilon^{1/2})n)$ such that for $l=l_{+}$ or
$l=l_{-}$,

$\text{Cov}_{\epsilon}(S\cap\partial B_{l})\leq c\epsilon^{1/2}n^{2}$ (35)
$\text{length}(\partial(S\cap\partial B_{l}))\leq c^{\prime}\epsilon^{-1/2}n$
(36)

where $c,c^{\prime}$ are constants. We define the “middle region”
$A_{\text{mid}}=(B_{l_{+}}\setminus B_{l_{-}})$, see Figure 25. Then we let

$\displaystyle U^{\prime}=U\cap A_{\text{mid}}.$

See Figure 26. Let $\mu$ be an ergodic measure of mean current $s$ (this
exists by Lemma 6.6.1). Let $\Theta$ be the collection of odd cubes adjacent
to $S^{\prime}=S\cap U^{\prime}$. Note that

$\displaystyle\text{area}(S^{\prime})\leq|\Theta|\leq
6\text{area}(S^{\prime})$

By Lemma 6.5.2, $|\Theta|\geq\text{area}(S^{\prime})\geq c_{2}^{\prime}n^{2}$,
where $c_{2}^{\prime}\sim(\delta-2\epsilon^{1/2})^{2}$. Thus by Lemma 6.6.7
and Equation (36), there is a constant $K_{\mu}\in(0,1)$ such that

$\displaystyle\mathbb{E}_{\mu}[|\text{flux}(v_{\tau},S^{\prime})|]\geq
K_{\mu}c_{2}^{\prime}n^{2}+O(\epsilon^{-1/2}n).$

By Theorem 6.6.3, for any coordinate plane $P$, a tiling $\tau$ sampled from
$\mu$ is $\epsilon$-nearly-constant on $P$ with value $s$ with probability
approaching $1$ as $n$ goes to $\infty$. Therefore we can sample a tiling
$\tau$ from $\mu$ which is $\epsilon$-nearly-constant with value $s$ on
$A_{\text{mid}}$ for $n$ large enough and satisfies

$|\text{flux}(v_{\tau},S^{\prime})|\geq
K_{\mu}c_{2}^{\prime}n^{2}+O(\epsilon^{-1/2}n).$ (37)

We fix this choice of $\tau$ for the rest of the proof. Define $U_{\tau}$ to
be the region covered by the tiles from $\tau$ which intersect $U^{\prime}$,
see Figure 27. Since $U_{\tau}$ is tileable,

$\displaystyle\text{imbalance}(U_{\tau})=\text{even}(U_{\tau})-\text{odd}(U_{\tau})=0.$

We next define a new region $U^{\prime\prime}$, which is $U_{\tau}$ minus the
even cubes adjacent to $S^{\prime}$ (see Figure 27 again). Note that the
region $U_{\tau}\setminus U^{\prime\prime}$ consists of only even cubes. By
Equation (37),

$\displaystyle|U_{\tau}\setminus U^{\prime\prime}|\geq
K_{\mu}c_{2}^{\prime}n^{2}+O(\epsilon^{-1/2}n).$

Therefore

$\displaystyle\text{imbalance}(U^{\prime\prime})\leq-
K_{\mu}c_{2}^{\prime}n^{2}+O(\epsilon^{-1/2}n).$

It remains to show that $\text{imbalance}(U)$ is very close to
$\text{imbalance}(U^{\prime\prime})$, and this is where we use the
$\epsilon$-nearly-constant condition. We split $A_{n}\setminus A_{\text{mid}}$
into two regions. First we define the “thin region” $A_{\text{thin}}$ to be
the union of columns parallel to one of the coordinate directions which
connect $\partial A_{n}$ and $\partial A_{\text{mid}}$. See Figures 25 and 28.
The complement of $A_{\text{mid}}\cup A_{\text{thin}}$ we call the “corner
region” $A_{\text{corner}}$, and consists of a neighborhood of the edges of
the outer boundary cube of $\partial A_{n}$, and the inner boundary cube of
$\partial A_{\text{mid}}$. We note that

$\displaystyle\text{area}(A_{\text{corner}}\cap\partial A_{n})\leq
24\epsilon^{1/2}n^{2}.$

The constant factor comes from the fact that the cube has 12 edges. Therefore

$\displaystyle\text{imbalance}(U\cap A_{\text{corner}})\leq
24\epsilon^{1/2}n^{2}.$

We define $U_{\text{shadow}}=(U\setminus U^{\prime\prime})\cap
A_{\text{thin}}$. We bound the imbalance of $U_{\text{shadow}}$ column-by-
column, where each column $C$ consists of a straight line path of single cubes
from $\partial A_{\text{mid}}$ to $\partial A_{n}$. Recall $U_{\text{shadow}}$
is defined with boundary condition $\tau$ on $\partial A_{\text{mid}}$ and
boundary condition $\tau_{1}$ or $\tau_{2}$ on $\partial A_{n}$. We also note
that for any column $C$, since $S$ is monochromatic black,
$\text{imbalance}(U\cap C)\leq+1$.

We cut $\partial A_{\text{mid}}$ into $(\epsilon n)\times(\epsilon n)$ patches
$\alpha$. For each $\alpha$, there is a corresponding patch $\beta$ on
$\partial A_{n}$ matched to $\alpha$ by columns. For any patch
$\alpha\subset\partial A_{\text{mid}}$, since $\tau$ is sampled from an
ergodic measure, it is $\epsilon$-nearly-constant with value $s$ on $\partial
A_{\text{mid}}$ (Corollary 6.6.5). Thus we have that

$\text{flux}(v_{\tau},\alpha)=\frac{1}{2}\langle
s,\xi_{\alpha}\rangle(\epsilon n)^{2}+o(\epsilon^{2}n^{2}).$ (38)

Let $v_{\ast}$ be equal to $v_{\tau_{1}}$ on the outer boundary of $\partial
A_{n}$ and $v_{\tau_{2}}$ on the inner boundary. For $\beta\subset\partial
A_{n}$ the patch connected by a column to $\alpha$, since $\tau_{1},\tau_{2}$
are $\epsilon$-nearly-constant with value $s$,

$\text{flux}(v_{\ast},\beta)=\frac{1}{2}\langle s,\xi_{\alpha}\rangle(\epsilon
n)^{2}+o(\epsilon^{2}n^{2}).$ (39)

(Note that $\xi_{\alpha}=\xi_{\beta}$.) For a patch $\alpha\subset\partial
A_{\text{mid}}$, let $C(\alpha)$ be the union of columns incident to $\alpha$.
The set $U_{\text{shadow}}$ is covered by these column sets, so it remains to
control imbalance of $U_{\text{shadow}}\cap C(\alpha)$ for each patch
$\alpha$.

By Equation (35), at most $c\epsilon^{1/2}n^{2}$ of the area of $\partial
A_{\text{mid}}$ is in patches $\alpha$ which intersect $U$ and $U^{c}$. The
total imbalance contribution from these is bounded by $c\epsilon^{1/2}n^{2}$.
If $\alpha\subset U^{c}$ but $C(\alpha)$ still intersects $U$, then all the
columns in $C(\alpha)$ have at least one end on $S$, and the imbalance in
$C(\alpha)\cap U_{\text{shadow}}$ is at most $0$.

Now we look at the cases where $\alpha\subset U$. If $C(\alpha)\subset U$,
then the total imbalance in $C(\alpha)\cap U$ is $o(\epsilon^{2}n^{2})$ by
Equations (38) and (39). If $\alpha\subset U$ but $C(\alpha)\not\subset U$,
then some of the columns starting from $\alpha$ hit $S$ before hitting
$\beta$. This means they end on an odd cube. Extending the column all the way
to $\beta$ would only make the imbalance larger, however then the imbalance in
$C(\alpha)\cap U$ is bounded above by $o(\epsilon^{2}n^{2})$ by Equations (38)
and (39). The number of columns is a bounded by a constant independent of $n$
times $\epsilon^{-2}$, hence in total the imbalance in $U_{\text{shadow}}$ is
bounded by $c\epsilon^{1/2}n^{2}+o(n^{2})$.

Putting everything together, we have that

$\displaystyle\text{imbalance}(U)$
$\displaystyle\leq\text{imbalance}(U^{\prime\prime})+\text{imbalance}(U_{\text{shadow}})+\text{imbalance}(U\cap
A_{\text{corner}})$ $\displaystyle\leq-
K_{\mu}c_{2}^{\prime}n^{2}+24\epsilon^{1/2}n^{2}+c\epsilon^{1/2}n^{2}+O(\epsilon^{-1/2}n)+o(n^{2}).$

Recall that $c_{2}^{\prime}\sim(\delta-2\epsilon^{1/2})^{2}$, so it gets
larger as $\epsilon$ gets smaller. Fixing $\epsilon$ small enough as a
function of the constants, for $n$ large enough $\text{imbalance}(U)\leq 0$
and hence $U$ is not a counterexample. Therefore by Proposition 6.2.6, for $n$
large enough, $A_{n}$ is tileable with boundary conditions
$\tau_{1},\tau_{2}$. ∎

###  6.8 Corollaries for ergodic Gibbs measures

######  Corollary 6.8.1.

If $\mu_{1},\mu_{2}$ are EGMs of the same mean current
$s\in\text{Int}(\mathcal{O})$, then $h(\mu_{1})=h(\mu_{2})$.

######  Remark 6.8.2.

The proof of the patching theorem (Theorem 6.3.5) uses that
${s}\not\in\partial\mathcal{O}$ since this is a condition of Lemma 6.6.7. This
corollary shows that $s\not\in\partial\mathcal{O}$ is a necessary condition
and not just an artifact of the proof, since if
$s\in\mathcal{\partial}\mathcal{O}$ then not all ergodic Gibbs measures of
mean current $s$ have the same specific entropy (see Proposition 4.2.10).

###### Proof.

Fix $\delta>0$, and let $B_{n}=[-n,n]^{3}$ and $A_{n}=B_{n}\setminus
B_{(1-\delta)n}$. By the patching theorem (Theorem 6.3.5) with outer boundary
condition sampled from $\mu_{1}$ and inner boundary condition sampled from
$\mu_{2}$, we get that

$\displaystyle h(\mu_{2})\leq(1+O(\delta))h(\mu_{1}).$

Switching them, we find that

$\displaystyle h(\mu_{1})\leq(1+O(\delta))h(\mu_{2}).$

Therefore $h(\mu_{1})=h(\mu_{2})$. ∎

Another useful result comes from applying patching to a sequence of
$\epsilon$-nearly-constant tilings and a sample from an EGM. This relates the
number of tilings of a region with fixed $\epsilon$-nearly-constant boundary
conditions to the specific entropy of an EGM. This serves as a lemma in the
proof of the lower bound in the large deviation principle (Theorem 8.1.10).
Recall that $\Omega$ denotes the set of dimer tilings of $\mathbb{Z}^{3}$.

######  Proposition 6.8.3.

Fix $\delta>0$, $s\in\text{Int}(\mathcal{O})$, $\epsilon>0$ small enough,
$B_{n}=[-n,n]^{3}$, and let $A_{n}=B_{n}\setminus B_{(1-\delta)n}$. Suppose
that $(\tau_{n})_{n\geq 1}\subset\Omega$ is such that $\tau_{n}$ is
$\epsilon$-nearly-constant on $\partial B_{n}$ with value $s$ for $n$ large
enough. Let $\pi_{n}$ be the uniform probability measure on tilings $\sigma$
of $B_{n}$ such that $\sigma\mid_{\partial B_{n}}=\tau_{n}$. Then for any EGM
$\mu$ of mean current $s$ and $n$ large enough,

$\displaystyle|B_{n}|^{-1}H(\pi_{n})\geq h(\mu)(1+O(\delta)).$

######  Remark 6.8.4.

Since $\pi_{n}$ is a uniform measure, $H(\pi_{n})=\log Z_{n}$, where $Z_{n}$
is the partition function of $\pi_{n}$.

###### Proof.

We apply the patching theorem (Theorem 6.3.5) to $A_{n}=B_{n}\setminus
B_{(1-\delta)n}$ with $\tau_{n}$ on the outer boundary and a sample from $\mu$
on the inner boundary. For $n$ large enough, patching is possible with
$\mu$-probability $(1-\epsilon)$ on an annulus of width $\delta$.

For $\Lambda\subset\mathbb{Z}^{3}$, let $\Omega(\Lambda)$ denote the free-
boundary dimer tilings of $\Lambda$. For $\sigma\in\Omega(\Lambda)$, recall
that $X(\sigma)$ is the set of extensions of $\sigma$, i.e.

$X(\sigma)=\\{\tilde{\sigma}\in\Omega:\tilde{\sigma}\mid_{\Lambda}=\sigma\\}.$

Then we compute

$\displaystyle H_{B_{(1-\delta)n}}(\mu)$
$\displaystyle=-\sum_{\sigma\in\Omega(B_{(1-\delta)n})}\mu(X(\sigma))\log\mu(X(\sigma))$
$\displaystyle=-\sum_{\begin{subarray}{c}\sigma\in\Omega(B_{(1-\delta)n})\\\
\sigma,\tau_{n}\text{
patchable}\end{subarray}}\mu(X(\sigma))\log\mu(X(\sigma))+O(\epsilon\log\epsilon)$

Let $U_{n}$ denote the uniform probability measure on

$\displaystyle\\{\sigma\in\Omega(B_{(1-\delta)n}):\sigma,\tau_{n}\text{
patchable}\\}.$

Since uniform measure maximizes entropy,

$\displaystyle H_{B_{(1-\delta)n}}(\mu)\leq
H(U_{n})+O(\epsilon\log\epsilon)\leq H(\pi_{n})+O(\epsilon\log\epsilon).$

Thus for $n$ large enough such that the patching theorem applies for
$\tau_{n}$ and a sample from $\mu$ on $A_{n}=B_{n}\setminus B_{(1-\delta)n}$,
with probability $(1-\epsilon)$ we have that

$\displaystyle|B_{n}|^{-1}H(\pi_{n})\geq(1+O(\delta))|B_{(1-\delta)n}|^{-1}H_{B_{(1-\delta)n}}(\mu)\geq(1+O(\delta))h(\mu).$

∎

##  7 Properties of entropy

In this section we prove results about the entropy functions ent and Ent
introduced in Section 2.3. Recall that $\Omega$ is the set of dimer tilings of