\input domhead.tex

\centerline{\vbigbf Spaces of domino tilings}
\smallskip
\centerline{\it Nicolau C. Saldanha, Carlos Tomei,}
\centerline{\it Mario A. Casarin Jr., Domingos Romualdo}

\bigskip
{

\narrower

{\nobf Abstract:}
We consider the set of all tilings by dominoes
($2 \times 1$ rectangles)
of a surface, possibly with boundary,
consisting of unit squares.
Convert this set into a graph by joining two tilings
by an edge if they differ by a {\sl flip}, i.e.,
a $90^\circ$ rotation of a pair of side-by-side dominoes.
We give a criterion to decide if two tilings are in the same
connected component, a simple formula for distances
and a method to construct geodesics in this graph.
For simply connected surfaces, the graph is connected.
By naturally adjoining to this graph higher dimensional cells,
we obtain a CW-complex whose
connected components are homotopically equivalent to
points or circles.
As a consequence, for any region different from
a torus or Klein bottle,
all geodesics with common endpoints are equivalent
in the following sense.
Build a graph whose vertices are these geodesics,
adjacent if they differ
only by the order of two flips on disjoint squares:
this graph is connected.

}

\goodbreak
\bigskip
{\nobigbf 0. Introduction}

\nobreak
\smallskip
\nobreak

In this paper we consider tilings of a region consisting of unit
squares by {\sl dominoes}, i.e., pairs of adjacent squares.
Tilings of a rectangle of integral sides
were counted by Kasteleyn ([5]).
More recently, Lieb and Loss ([6]) showed how to count tilings of
general regions by making use of determinants.
Conway and Lagarias ([1]) studied the problem
of tiling a subset of $\RR^2$ with a given set of tiles, by
group-theoretical techniques.
Thurston ([9]) adapted these techniques
to study domino tilings, producing a
necessary and sufficient condition
for a simply connected region of the plane to be tileable by dominoes.

We are interested in studying $T$,
the set of all possible tilings of a fixed region.
Given a tiling, we perform a {\sl flip} by lifting two dominoes
and placing them back in a different position:
clearly, the two dominoes must form a square of side 2.
Two tilings are {\sl adjacent} in $T$
if we move from one to the other by a flip.
Turn $T$ into a graph by joining adjacent tilings by edges
and define connected components of $T$ and distance between
tilings in the usual way.
We obtain a very operational criterion
(the equivalent Theorems 1.1, 1.2 and 3.1)
for two tilings to belong to the same connected component of $T$;
as a corollary, if the region is simply connected, $T$ is connected.
Our techniques provide us with a fair understanding
of the combinatorial, topological and metric structure of $T$:
thus, for example, each connected component of $T$ is a lattice and
we describe in Theorem 3.2 a simple formula for the distance
between tilings
and a characterization of shortest routes between points.
In a sense to be detailed in Section 3, all such routes are equivalent:
a topological version of this statement is that
$T$ induces naturally a CW-complex whose connected components
are contractible (Theorem 3.4).
More generally,
we consider quadriculated surfaces (defined in Section 4)
and obtain analogous results to Theorems 3.1, 3.2 (Theorem 4.1)
and 3.4 (Theorem 4.3).

{\nobf Acknowledgements:}
The first two authors receive support from SCT and CNPq, Brazil.
The other two were supported by a grant for undergraduates by CNPq.
We would like to thank one referee for a very careful reading of the
manuscript, often leading to a much clearer text.

\goodbreak
\bigskip
{\nobigbf 1. Connected components of \bigit T}

\nobreak
\smallskip
\nobreak

Let $A$ be a finite subset of the lattice $\ZZ^2$.
We say that two points of $A$ are {\sl adjacent}
if the distance between them is equal to 1.
In this case we say they are connected by an {\sl edge},
the line segment joining them.
The set $A$ thus receives a graph structure.
Closely related to $A$ is the set $\hat A \subseteq \RR^2$,
the interior of the union of
closed squares of side 1 (in the usual position) with centers in $A$:
we often identify $A$ and $\hat A$ and a point $p$ of $A$ with the unit square
whose center is $p$. The graph $A$ is called {\sl connected}
(or {\sl simply connected}) if $\hat A$ is.
Without real loss, we always assume $A$ to be connected.
A {\sl covering of $A$ by edges} is a set of edges
such that each point of $A$ is the extremity of precisely one edge.
Equivalently, we speak of {\sl tilings of $\hat A$ by dominoes},
each domino covering two unit squares connected by an edge.

A point of $\ZZ^2$ shall be called {\sl white} (resp., {\sl black})
if the sum of its coordinates is even (resp., odd);
$\hat A$ is therefore painted black and white like a chessboard.
Edges of $A$ connect points of different colours.
Clearly, if $A$ admits a tiling,
the number of white squares equals that of black squares.
In Figure 1.1, $A$ is not tileable even though
the colour condition is satisfied.

\input domfig11.tex

Our first aim is to state
a necessary and sufficient condition for two tilings
to be in the same connected component of $T$.
In order to do this, we define combinatorial invariants
for these components.
We start with an explicit and easily computable description
of such invariants,
which is then rephrased in the vocabulary of homology theory.

A {\sl cut} of $\hat A$ is a simple oriented polygonal line
in $\hat A$ consisting of a sequence of edges of squares and
joining two points in the boundary of $\hat A$.
The {\sl flow} of a tiling across a cut is defined
to be the number of dominoes crossing the cut,
where the domino is counted positively (resp., negatively)
if its white square is to the left (resp., right) of the cut.

Consider a cut of $\hat A$
disconnecting it into two sets $\hat A_\ell$ and $\hat A_r$
to the left and right of the cut, respectively.
The flow of any tiling of $\hat A$ across this cut
is clearly given by the number of white squares in $\hat A_\ell$
minus the number of black squares in $\hat A_\ell$:
this must be equal to the number of black squares in $\hat A_r$
minus the number of white squares in $\hat A_r$.
For a cut which does not disconnect $\hat A$,
on the other hand, the flow may admit different values
for different tilings, as in Figure 1.2.
It is easy to see, however, that for a fixed cut,
adjacent tilings in $T(A)$ have the same flow:
these are therefore invariants for the connected components of $T$. This
is the easy part of Theorem 1.1 below.

\input domfig12.tex

{\nobf Theorem 1.1:} (combinatorial version)
{\sl
Assume $\hat A$ has genus $n$.
Choose $n$ disjoint cuts in $\hat A$
which jointly do not disconnect $\hat A$.
Two tilings $t_1$ and $t_2$ are in the same connected component of $T$
if and only if
their flows across each of the $n$ chosen cuts are equal.}

In particular, if $A$ is simply connected, $T$ is connected.
We will give two other equivalent versions of this theorem, and will
prove the last one, after constructing the necessary tools.

Let us see how we can associate to two tilings
$t_1$ and $t_2$ an element of $H_1(\hat A;\ZZ)$,
which we shall denote by $[t_1 - t_2]$.
We first build two CW-complexes $\bA$ and ${\bA^\star}$
with $\bA \subseteq \hat A \subseteq {\bA^\star}$
and such that the inclusions are homotopy equivalences.
For $\bA$, the 0-cells are the points of $A$ 
(the centers of the squares of $\hat A$),
the 1-cells are the edges between points of $A$,
the 2-cells are the open unit squares with all vertices in $A$.
For ${\bA^\star}$, the 2-cells are the squares of $\hat A$,
the 1-cells are their sides and the 0-cells their vertices,
where the common side of two adjacent squares gives us only one 1-cell,
as do the common vertices of adjacent squares
but common vertices of non-adjacent squares are {\sl not} identified
unless the two squares are adjacent to a third one.
In Figure 1.3, we show $A$ (big dots),
$\bA$ (big dots and thin lines) and ${\bA^\star}$ (thick lines);
notice that point $p$ gives rise to two 0-cells in ${\bA^\star}$.
For future use, call $A^\star$ the set of 0-cells of ${\bA^\star}$.

\input domfig13.tex

We shall be interested in a few related
homology and cohomology moduli of the above spaces.
Since the two CW-complexes are homotopy equivalent to $\hat A$,
$H_1({\bA^\star};\ZZ) = H_1(\bA;\ZZ)$.
By Poincar\'e-Lefschetz duality (See Sections 26 and 28 of [4]),
on the other hand,
$H_1(\hat A;\ZZ) = H^1({\bA^\star},\partial {\bA^\star};\ZZ)$.
The equality is induced by
the natural identification between $C_1(\bA;\ZZ)$
and $C^1({\bA^\star}, \partial {\bA^\star};\ZZ)$,
where $C_k$ and $C^k$ are the usual spaces of
$k$-complexes or cocomplexes.

%
% Dar uma referencia melhor acima e talvez explicar mais.
%

Consider each edge (or domino) as a 1-cell in $\bA$
and always orient it from black to white;
domino tilings correspond therefore to elements of $C_1(\bA;\ZZ)$
with boundary always equal to the sum of all white vertices
minus the sum of all black vertices.
The difference between two domino tilings $t_1$ and $t_2$ is therefore
a closed element of $C_1(\bA;\ZZ)$: call the corresponding homology class
$[t_1 - t_2]$. In Figure 1.4, we show how, given two tilings $t_1$ and
$t_2$ ((a) and (b), resp.), we represent the class $[t_1 - t_2]$ in (c),
consisting of a sum of cycles in $H_1(\hat A,\ZZ)$.

\input domfig14.tex

This homology class turns out to have the following properties:

{\parskip=0pt
\item{(a)} {$[t_1 - t_1] = 0$,}
\item{(b)} {$[t_1 - t_2] = - [t_2 - t_1]$,}
\item{(c)} {$[t_1 - t_2] + [t_2 - t_3] = [t_1 - t_3]$,}
\item{(d)} {if $t_1$ and $t_2$ are adjacent, $[t_1 - t_2] = 0$.}

}

Properties (a), (b) and (c) are obvious.
If $t_1$ and $t_2$ are adjacent,
the cycle $[t_1 - t_2]$ is precisely the boundary of one of the
2-cells introduced in our construction, hence is exact, and (d) follows.

We defined $[t_1 - t_2]$ as an element of $H_1(\hat A;\ZZ)$.
From the duality stated above, we can also think of $[t_1 - t_2]$
as an element of $H^1({\bA^\star},\partial {\bA^\star};\ZZ)$;
let us see a direct way of interpreting this cohomology class.
In Figure 1.5.a we show the cocomplex corresponding to the tiling
in Figure 1.4.a: notice that the cocomplex is represented in ${\bA^\star}$
while the complex was represented in $\bA$.
The way to obtain the cocomplex from the tiling should be clear:
take the 1-cells (edges) on boundaries of dominoes to zero
and 1-cells crossing dominoes,
when oriented so that the white square is at the left, to 1.
The cocycle corresponding to the cycle in Figure 1.4.c
is shown in Figure 1.5.b;
notice that it is zero at the boundary and therefore corresponds to
a class in $H^1({\bA^\star},\partial {\bA^\star};\ZZ)$, as claimed.

Consider the cocomplex in $C^1({\bA^\star};\ZZ)$
taking any edge to 1 if oriented with white at the left:
the difference $\langle t \rangle$ between this cocomplex
and four times the cocomplex associated
with a tiling $t$ is a cocycle 
since it takes the boundary of any square to zero;
notice that its value on $\partial {\bA^\star}$
does not depend on $t$.
Since ${\bA^\star}$ is a closed disk with holes,
the map induced by inclusion from
$H^1({\bA^\star};\ZZ)$ to $H^1(\partial {\bA^\star};\ZZ)$ is injective.
The cohomology class in $H^1({\bA^\star};\ZZ)$
corresponding to $\langle t \rangle$ does not therefore
depend on the tiling $t$.
In Figure 1.5.c we represent $\langle t \rangle$
for the tiling in Figure 1.4.a:
in order to recover the tiling from the cocomplex,
place central edges of dominoes over the triple arrows;
in particular, different tilings correspond to different cocomplexes.
We shall see the uses of the cocycle $\langle t \rangle$
in the next section.

\input domfig15.tex

To a cut $\Gamma$ we associate an element $[\Gamma] \in
H^1(\hat A,\ZZ) = H^1(\bA,\ZZ)$ as follows.
Each element of $C_1$ (for $\bA$) is mapped to an integer:
a 1-cell which does not cross $\Gamma$ is taken to $0$; if the 1-cell
crosses $\Gamma$, it is taken to $\pm 1$, according to orientation
(if the 1-cell crosses $\Gamma$ from right to left it is taken to 1).
This map is a cocycle, i.e., the boundary of a 2-cell is taken to 0.
This is the usual
construction of a cohomology class from a curve such as a cut.
Since the cohomology of a disk with $n$ holes $D$ is known to be generated
by the classes of $n$ such curves not disconnecting $D$, the $n$ cuts
mentioned in Theorem 1.1 form a basis of $H^1(\hat A, \ZZ)$.
Notice that the flow of a tiling $t$ across a cut $\Gamma$ is
$[\Gamma](t)$;
the difference between flows for two tilings $t_1$ and $t_2$
is the usual pairing (between cohomology and homology)
$[\Gamma] \frown [t_1 - t_2]$.
In particular, if $[t_1 - t_2] = 0$,
the flows of $t_1$ and $t_2$ coincide on any cut.

{\nobf Theorem 1.2:} (homological version)
{\sl
The tilings
$t_1$ and $t_2$ are in the same connected component of $T$
if and only if $[t_1 - t_2] = 0$.}

Both versions, Theorems 1.1 and 1.2, are equivalent.
Indeed, from the remarks above, triviality of the class $[t_1 - t_2]$
is equivalent to the equality of the corresponding flows
across the cuts mentioned in Theorem 1.1.

\goodbreak
\bigskip
{\nobigbf 2. Height sections}

\nobreak
\smallskip
\nobreak

We begin this section by discussing
height functions, originally presented by Thurston ([9]).
Height sections,
which are appropriate extensions of the concept of height functions,
are the main tools in our proofs of Theorems 3.1 and 3.2.
The height function (or section) corresponding to $t$
is in fact obtained by integrating $\langle t \rangle$;
we nevertheless give an elementary and independent description
of these objects.

Consider a (parametrized) polygonal line consisting of edges of unit
squares with vertices in $(\ZZ + 1/2)^2$.
We assign numerical values to the parametrized vertices
by a sort of integration process:
in particular, it may happen that to a point on the line
correspond two different values.
Take an initial value (say 0) and assign it to
the origin of the polygonal line.
When walking along an edge with a white (resp., black) square to its left,
add (resp., subtract) 1 to the value at the starting point of the edge
in order to get the value at the endpoint.
Notice that if the line joins $P$ to $Q$
and the integration process starting with $a$ for $P$
leads to $b$ for $Q$ then integration from $Q$ to $P$
along the same line starting with $b$ yields the same value $a$ at $P$.

If the endpoint coincides with the starting point of the line,
how do the two values assigned to this point relate?
It is not hard to see that we add (resp., subtract) 4
each time we surround a white (resp., black) square counter-clockwise,
with reversed signs for opposite orientation.
By the obvious additivity properties with respect to paths of
integration, the value obtained when returning to the original point
is the following.
For each white (resp., black) square,
take $4$ (resp., $-4$) times the winding of the path 
around it and sum over all squares.

Thus, the value mod 4 at the endpoint does not depend on
the integration path, and is given (up to a global additive constant)
by the function $\phi : (\ZZ + 1/2)^2 \to \ZZ/{(4)}$ defined as
$\phi(x,y) = 0$ if $\floor{x} = x - 1/2$ and
$\floor{y} = y - 1/2$ are both even,
$\phi(x,y) = 1$ if $\floor{x}$ is odd and $\floor{y}$ is even,
$\phi(x,y) = 2$ if $\floor{x}$ and $\floor{y}$ are both odd,
$\phi(x,y) = 3$ if $\floor{x}$ is even $\floor{y}$ is odd.
However, integration along the boundary of a domino, or,
more generally, of a simply connected tileable region
assigns the same value to the starting point and endpoint.
Notice that the situation above is very similar to two other more familiar
constructions: the calculation of the area of
a planar region by Green's Theorem
and the computation of a complex integral by adding residues.

We now discuss height functions
and their relation to tilings in the case when
the closure of $\hat A$ is a closed disk.
Assume therefore that the closure of $\hat A$ is a (topological) closed disk.
Let $A^\star \subseteq (\ZZ + 1/2)^2$ be,
as above, the set of vertices of squares in $\hat A$.
Choose a {\sl basepoint} $p_0 = (x_0,y_0) \in A^\star$,
$p_0$ in the exterior boundary of $\hat A$,
and a {\sl base value} $v_0 \in \ZZ$ so that $v_0 \bmod 4 = \phi(x_0,y_0)$.
Given a tiling $t$,
we define a function $\theta$ from $A^\star$ to $\ZZ$
at a typical point $p$ by integrating
along any path contained in boundaries of dominoes,
starting from the basepoint $p_0$ with initial value $v_0 = \theta(p_0)$
and reaching $p$ with value $\theta(p)$.
This function does not depend on choices of paths.
Indeed, as in the paragraph above, $\theta$ is locally well defined;
our hypothesis on the global topology of $\hat A$
guarantees that $\theta$ is also globally well defined.
Given any path contained in boundaries of dominoes
joining points $p_1$ and $p_2$,
integration along this path starting with $\theta(p_1)$ yields $\theta(p_2)$.
Also, different choice of basepoint or base value produce
the same height function up to an additive constant in $4\ZZ$.
We call $\theta$ the {\sl height function} of $t$;
in Figure 2.1 we show an example of a domino tiling
and the corresponding height function.

\input domfig21.tex

Two points in $A^\star$ are called {\sl adjacent}
if the distance between them is 1
and the segment joining them is in the closure of $\hat A$.
It is easy to see that a height function 
satisfies the following properties:

{ \parskip=0pt
\item{(a)}{$\theta(x,y) \bmod 4 = \phi(x,y)$,}
\item{(b)}{the values of $\theta$ at adjacent points
never differ by more than 3,}
\item{(c)}{the values of $\theta$ at points
which are adjacent along a segment contained in the boundary
of $\hat A$ always differ by exactly 1.}

}

Conversely, given a function $\tau$ satisfying conditions (a), (b)
and (c) as above, we obtain a tiling $t$ as follows:
join two adjacent points in $A^\star$
if the values of $\tau$ at such points differ by precisely 1,
thus obtaining the contours of the dominoes of $t$.
It remains to prove that the construction actually gives rise
to a tiling by dominoes.
Indeed, each square of $\hat A$ is surrounded by four points of $A^\star$
and from conditions (a) and (b)
exactly three of these sides are drawn in the above process:
the fourth one (which cannot lie on the boundary, by (c))
indicates which way the domino covering our square goes.
Furthermore, the height function $\theta$ corresponding to $t$
is equal to $\tau$, up to an additive constant in $4\ZZ$:
these two constructions are the inverse of each other.
We thus defined a bijection between the space of tilings $T$
and the class of functions satisfying the three above conditions,
i.e., height functions, modulo additive constants in $4\ZZ$.

Let us consider
how to extend these concepts to the general case.
First, there can be nasty points in $(\ZZ + 1/2)^2$
with all four edges arriving at it being part of the boundary of $\hat A$:
as we have already seen in the construction of ${\bA^\star}$,
such a point ought to be interpreted as {\sl two} points in $A^\star$
with adjacency relations defined in the obvious way that assures
the local good behaviour of $A$ and $A^\star$.
Of course, height functions are free to assume different values
at these two points.

A more serious problem comes from the consistency of $\theta$
along boundaries if $\hat A$ is not simply connected.
If inside one of the holes of $\hat A$ the number of white squares
is different from the number of black squares,
no height function can exist
because we get multivaluedness when following the boundary.
Still, it is easy to construct such regions
which admit tilings, as in Figure 1.2.

In cohomological terms, it is clear what is going on.
The height function $\theta$ was obtained
by integrating $\langle t \rangle$:
this was possible because this cocycle is {\sl exact},
i.e., corresponds to the cohomology class 0 in $H^1({\bA^\star};\ZZ)$.
In other words, $\langle t \rangle$ is the coboundary of $\theta$.
Now, the cohomology of a disk is trivial
but if $\hat A$ is not simply connected 
$H^1({\bA^\star};\ZZ)$ is non-trivial and it may well happen
(as in the example mentioned in the previous paragraph)
that the cohomology class of $\langle t \rangle$ is non-zero.

What we need is not height functions but height sections
of a certain fibre bundle with base space $A^\star$ and fibre $\ZZ$.
In this bundle, a fibre is not an additive group:
there is no natural 0 nor addition on each fibre.
We are allowed to add an integer to an element of a fibre
(thus getting another element of the same fibre)
or to subtract elements of the same fibre (thus getting an integer).
We are also allowed to compare elements of the same or neighbouring
fibres, but otherwise we are not allowed to compare
elements of different fibres.
The congruence class mod 4
of an element of a fibre {\sl is}, however, well defined.

We begin the construction by choosing a basepoint $p_0$ in $A^\star$.
Consider next the set $\sP$ of all paths in $A^\star$
going from $p_0$ to some other point of $A^\star$,
i.e.,  functions $\xi$ from sets of the form $\{0,1,\ldots,m\}$
to $A^\star$ such that $\xi(0)$ is the basepoint
and $\xi(i)$ and $\xi(i+1)$ are always neighbours in $A^\star$.
Our bundle shall be obtained from $\sP \times \ZZ$ by a quotient:
the projection from $\sP \times \ZZ$ to $A^\star$
just takes a pair (path, integer) to the endpoint $\xi(m)$ of the path.
Two pairs $(\xi_1,k_1)$ and $(\xi_2,k_2)$ are identified
if the following conditions hold.
First, $\xi_1$ and $\xi_2$ must have the same endpoint.
Second, consider $\xi$ the path obtained by following $\xi_1$
and then following $\xi_2$ backwards;
let $\ell$ be the sum of the windings of $\xi$ around
white squares {\sl not in $\hat A$} minus 
the sum of the windings of $\xi$ around
black squares, again not in $\hat A$:
identify the two pairs if $k_1 - k_2 =  4\ell$.
This defines the desired {\sl height bundle}, or $\cal H$.
The allowed operations on this bundle have the obvious definitions
in terms of representatives of the equivalence classes.

Another essentially equivalent interpretation for $\cal H$
is as a (not necessarily connected) covering space for $\hat A$,
or, equivalently, ${\bA^\star}$.
Indeed, take the fibres as defined over $A^\star$
and extend them to edges of ${\bA^\star}$
by the provided identification between neighbouring points.
Finally, define fibres over the squares of ${\bA^\star}$
in the essentially unique possible way:
it {\sl is} always possible to do it for each such square
because the four identifications around it are compatible.
The name `height section' should generate no confusion:
it is always to be understood as a section of $\cal H$
restricted to $A^\star$.

We construct the height bundle for the region shown in Figure 2.2(a),
in a manner which is slightly different from the one described above.
Start by drawing cuts as indicated, and consider the
sub-CW-complex $\hat\bB$ of ${\bA^\star}$, obtained by removing the
1- and 2-cells intersecting the cuts. Take now the cartesian product
$\hat\bB \times \ZZ$. This is necessarily isomorphic to
the restriction of $\cal H$ to $\hat\bB$, since $\hat\bB$ is
contractible. In order to construct $\cal H$, it suffices to extend
this bundle to the missing cells. This shall be done by choosing
appropriate additive shifts between consecutive fibres, indicated again
in Figure 2.2(a).
How are those shifts obtained? Consider, for example, the two paths
$\xi_1$ and $\xi_2$ in the picture; the equivalence relations defined in
the construction of $\cal H$ yield $(\xi_1,-4) = (\xi_2,0)$.
Of course, a different choice of cuts would give rise to
isomorphic bundles.
To construct an isomorphism, start by identifying (arbitrarily,
but respecting orientation) a pair of fibres with the same base point
and extend (in the only possible way)
the identification to the entire bundle.
It is only necessary to check that the above construction
yields a well defined map:
this follows from the definition of the bundles.
Thus, from now on, we speak of {\sl the} height bundle over $A^\star$.

\input domfig22.tex

We define the {\sl height section} corresponding to a tiling $t$
by integration just as we did for height functions:
the bundle $\cal H$ is constructed in such a way
that the definition of the section does not depend on choices of paths.
Indeed, for two arbitrary paths along boundaries of dominoes
with same starting point and endpoint,
integration yields two values at the endpoint
which are identified in the construction of the bundle.
As with height functions,
basepoint and base value contribute only with a constant in $4\ZZ$.
Figure 2.2(b) shows an example of a tiling and its height section;
Figure 2.2(c) shows how we would write the same height section
with a different set of cuts.
The large difference between numbers on neighbouring points at opposite
sides of cuts does {\sl not} correspond to a jump of the height section:
remember that such neighbouring fibres are attached with an additive
shift. Again, a different choice of cuts would not have changed the
height section itself, but only the notation employed.

A height section satisfies conditions (a), (b) and (c) with the
appropriate invariant interpretation: in conditions (b) and (c),
the difference between values of the section at neighbouring points
is to be computed using the identification of neighbouring fibres
intrinsic to the definition of the bundle, and {\sl not} as a difference
between the (cut-dependent) integers used in our examples.
We again have a natural bijection between $T$
and the class of sections of $\cal H$ satisfying the three conditions,
modulo constants in $4\ZZ$.

We now list some convenient properties of height sections.
The difference of two height sections is a function with domain
$A^\star$ and values in $4\ZZ$; this is well defined up to an additive
constant.
Consider a cut $\Gamma$ connecting two points $x_a$ and $x_b$
in different boundary components of $A^\star$.
Let $f_1$ and $f_2$ be the flows across $\Gamma$ of two tilings $t_1$ and $t_2$.
We claim that 
$$\theta_2(x_b) - \theta_1(x_b) - \theta_2(x_a) + \theta_1(x_a) =
-4(f_2 - f_1).$$
From the previous remark, the left hand side
is a well defined integer. Indeed, the function
$\theta_2(x) - \theta_1(x) - \theta_2(x_a) + \theta_1(x_a)$, from the
cut $\Gamma$ to the integers, can be computed, starting at $x = x_a$
(where it is clearly equal to 0) and ending at $x = x_b$ by the
integration processes, yielding the result.
Also, two tilings are adjacent iff their corresponding height sections
differ at a single (interior) point by $\pm 4$
once the additive constants have been chosen
so that they agree at a boundary point.
Finally, the maximum or minimum of two or more
height sections is again a height section, since properties (a), (b) and
(c) are preserved.

It is clear from property (c) of height sections (in particular,
functions), that if two height sections for the same region agree at one
point of the boundary, they agree on the entire connected component of
the boundary containing that point. We may thus assume without loss that
height sections always agree on the exterior boundary. Our main
interest, however, is on relating height sections for tilings $t_1$ and
$t_2$ with $[t_1 - t_2] = 0$. In this case, the corresponding height
sections agree on the {\sl entire} boundary. Indeed, consider two
sections $\theta_1$ and $\theta_2$ which agree on the exterior boundary.
The identity above, which relates the values of the sections at one
point of the boundary to the values at another point, immediately yields
the equality of the sections at all boundary components, since flows for
both tilings are equal, by the equivalence between Theorems 1.1 and 1.2.

There is then a lattice structure (and a partial order) on $T(A)$,
induced by the corresponding order on height sections: remember that
height sections are assumed to agree on the exterior boundary.

\goodbreak
\bigskip
{\nobigbf 3. Distances in \bigit T}

\nobreak
\smallskip
\nobreak

We state yet a different version of Theorems 1.1 and 1.2.

{\nobf Theorem 3.1:} (height section version)
{\sl
The tilings
$t_1$ and $t_2$ are in the same connected component of $T$
if and only if their corresponding height sections
$\theta_1$ and $\theta_2$ coincide on the whole boundary.}

As shown at the end of the previous section, $[t_1 - t_2] = 0$ iff the
corresponding height sections agree on the whole boundary:
the equivalence between Theorems 1.2 and 3.1 is now clear.
A corollary of this theorem is that each connected component of $T$
is a lattice, with a maximum and minimum height sections.
The non-trivial part of Theorem 3.1 reduces therefore to the following:
there always exists a path (in $T(A)$) joining two height sections
coinciding on the boundary of ${\bA^\star}$, in which consecutive
height sections differ at a single point.

{\nobf Theorem 3.2:}
{\sl
Suppose the tilings $t_1$ and $t_2$ are 
such that their corresponding height sections
$\theta_1$ and $\theta_2$ coincide on the whole boundary.
Then $t_1$ and $t_2$ are in the same component of $T(A)$ and
$$d(t_1,t_2) =
{1 \over 4} \sum_{p \in A^\star}{|\theta_1(p) - \theta_2(p)|}.$$
Also, the diameter of a connected component of $T(A)$
is the distance between the minimum and the maximum
of all height sections in that component.
}

As we shall see, Theorem 3.1 and Theorem 3.2 follow easily
from the lemma below.

{\nobf Lemma 3.3:}
{\sl
Let $\theta_1 < \theta_2$ be two height sections
coinciding on the boundary of ${\bA^\star}$.
Then there exists $\theta_3$, $\theta_3$ adjacent to $\theta_1$,
with $\theta_1 < \theta_3 \le \theta_2$.
}

{\nobf Proof:}

\nobreak

To satisfy adjacency,
the new section $\theta_3$ must be constructed as follows: choose a
point $p_0$ in $A^\star$
and define $\theta_3(p) = \theta_1(p)$ for $p \ne p_0$
and $\theta_3(p_0) = \theta_1(p_0) + 4$.
The section $\theta_3$ is a height section iff conditions (a),
(b) and (c) hold. Condition (a) is trivially satisfied.
Condition (c) is satisfied provided $p_0$ lies
in the interior of ${\bA^\star}$. Condition (b) holds if and only
if $p_0$ is a local minimum of $\theta_1$. Indeed, local minimality
and condition (a) guarantee that, if $p$ is a neighbour of $p_0$,
$\theta_1(p) - \theta_1(p_0) = 1$ or 3.
Thus, $\theta_3(p) - \theta_3(p_0) = - 3$ or $-1$.
Finally, to obtain $\theta_1 <
\theta_3 \le \theta_2$, we must choose $p_0$ with
$\theta_1(p_0) < \theta_2(p_0)$. We only have to prove then that
such a point $p_0$ exists.
Consider that (non-empty) part $B$ of the domain where
$\theta_2 - \theta_1$ is maximum; we will see that there must be such a
$p_0$ in $B$.

When height sections are just height functions,
select $p_0$ so that $\theta_1(p_0)$ is minimum in $B$.
We prove that $p_0$ is a local minimum in $A^\star$.
Let $p$ be a neighbour of $p_0$. If $p$ is in $B$, we have
$\theta_1(p) > \theta_1(p_0)$ by hypothesis.
If $p$ is not in $B$, let $x_i = \theta_i(p_0)$;
conditions (a), (b) and (c) allow two possible values for each of
$\theta_i(p)$: call these $y_i$ and $z_i$ with $y_i < x_i < z_i$.
Clearly, $z_2 - z_1 = x_2 - x_1 = y_2 - y_1$ and,
since $p \notin B$,
$\theta_2(p) - \theta_1(p) < \theta_2(p_0) - \theta_1(p_0)$,
whence $\theta_1(p) = z_1$ and $\theta_2(p) = y_2$,
proving our claim in this second case.

The difficulty in the proof for sections lies in the fact that it
doesn't make sense to look for global minima. From the previous
arguments, however, a local minimum in $B$ (which is necessarily a local
minimum in $A^\star$) is what we need. Suppose by contradiction
that no such local minima exist:
every point in $B$ has a neighbour where $\theta_1$ is smaller.
Since $B$ is finite, there exists a cycle
$p_0, p_1, \ldots, p_{N-1}, p_N = p_0$ of points of $B$
with $\theta_1(p_i) > \theta_1(p_{i+1})$.
Assume without loss that the cycle is simple
(i.e., has no self-intersections),
turns counterclockwise in the plane and encloses minimum area.
It is clear that this minimum area is greater than 1.

We claim that $\theta_1(p_i) - \theta_1(p_{i+1}) = 1$.
Indeed, if this is not the case, the difference equals 3.
The edge joining $p_i$ and $p_{i+1}$ is the central edge of a domino
which is common to both tilings $t_1$ and $t_2$. The two points
to the left of the oriented segment $p_i p_{i+1}$
also belong to $B$, and we may therefore insert these
two points between $p_i$ and $p_{i+1}$
thus obtaining a new cycle with smaller enclosed area.
If the new cycle is not simple, take a simple subcycle of it.

Also, the segments $p_i p_{i+1}$ and $p_{i+1} p_{i+2}$ form a right angle,
since otherwise we would have a difference of 3 on one of the two edges.
Finally, we cannot have $p_i$, $p_{i+1}$, $p_{i+2}$ and $p_{i+3}$
vertices of the same square traversed counterclockwise:
otherwise, omit $p_{i+1}$ and $p_{i+2}$ to get a cycle
with smaller area.
It follows that the polygonal line joining midpoints
between consecutive points of the cycle never turns left
and this contradicts the fact that the cycle turns counterclockwise.
{\hfill \vrule height3pt width6pt depth3pt \smallskip} %qed

\medbreak

{\nobf Proof of Theorems 3.1 and 3.2:}

\nobreak

This lemma (plus induction) tells us that we can move from
a smaller to a larger height section by flips;
in particular, we can go from any height section to the maximum,
thus proving the connectivity of classes of height sections
with given boundary values (Theorem 3.1).
The inequality $d(t_1,t_2) \ge
{1 \over 4} \sum_{p \in A^\star}{|\theta_1(p) - \theta_2(p)|}$
is an obvious consequence of the fact at the end of the previous section
relating height sections of adjacent tilings.
As to the non-trivial half of the distance formula,
a shortest path is to move from one section to the maximum of the two
and then to the other;
we could equally well have first moved to the minimum
and the distance would be the same.
Our claim about the diameter follows from the distance formula;
this, of course, finishes the proof of Theorem 3.2.
{\hfill \vrule height3pt width6pt depth3pt \smallskip} %qed

It is clear from the proof above that we know which flips to perform in
order to get closer to a tiling $t_2$ starting from a tiling $t_1$:
simply compute both height sections and look for local minima of $t_1$
below $t_2$, or local maxima of $t_1$ above $t_2$. In this sense, there
is a local characterization of the shortest paths in the graph $T(A)$.

Some of these paths should clearly be considered equivalent.
For instance, let $t_1$ and $t_2$ be the tilings (a) and (d)
in Figure 3.1: the two paths (abd) and (acd) are such an example.
We render this notion precise by turning $T$ into a CW-complex.
The 0-cells are just the elements of $T$ and
the 1-cells connect adjacent tilings so that the notion
of a connected component of $T$ remains unaltered.
The 2-cells are glued along squares whose edges
are two independent flips (i.e., ocurring on disjoint squares);
in Figure 3.1, there is a 2-cell whose boundary is composed
of the four 1-cells connecting the tilings in 
(a) to (b), (b) to (d), (d) to (c) and (c) to (a).
Similarly, 3-cells correspond to three independent flips,
and $k$-cells to $k$ independent flips.
The above mentioned equivalence of paths is of course homotopy
and it turns out that all shortest routes between tilings
are homotopic, as follows from the theorem below.

\input domfig31.tex

{\nobf Theorem 3.4:}
{\sl Each connected component of $T$ is contractible.}

\medbreak

{\nobf Proof:}

\nobreak

For an arbitrary tiling $t_0$ (i.e., a 0-cell)
we contruct a homotopy from the identity to a constant function
taking the entire connected component to $t_0$.
Start by those points which are furthest from $t_0$:
from each such point $t$ there are a few (say $k$) possible flips.
All these flips must necessarily approach the base tiling
and must be independent; $t$ is therefore the vertex of a $k$-cell
corresponding to these $k$ flips and it is easy to deform this cell,
a $k$-dimensional cube, onto the walls of the cell that do not touch $t$
without moving these walls.
Repeat the process for all tilings different from $t_0$,
taking distances in decreasing order.
{\hfill \vrule height3pt width6pt depth3pt \smallskip} %qed

\goodbreak
\bigskip
{\nobigbf 4. Quadriculated surfaces}

\nobreak
\smallskip
\nobreak

In this section we generalize the constructions and results
of the previous sections to the situation where $\bA^\star$ 
is not a subset of the plane but a {\sl quadriculated surface}.
The idea of a quadriculated surface is very natural
but its definition is somewhat technical:
start with a finite collection of squares of unit side
and glue certain pairs of sides
(taking orientation of the sides into account)
in such a way that the following two conditions hold.
First, two sides of the same square are never identified.
Two vertices of different squares are identified if they are
the corresponding extremes of identified sides.
Given an edge of a square and an incident vertex
we can either replace the edge by the other edge
on which the vertex lies or, if the edge is identified
with an edge of some other square, pass to that edge
and to the corresponding vertex.
Performing these two operations in alternation,
we see that a vertex in the surface
(i.e., after identifications)
corresponds to a sequence of vertices of squares;
it is clear that such a sequence is either finite
(if we reach the boundary, i.e., a non-identified side) or periodic.
Our second condition is that periodic sequences must have length 4;
intuitively, this says that the angles at vertices of squares
are $\pi/2$ so that it is impossible to surround a point
with less than or more than 4 squares.
Of course, the surface may be non-orientable
or not consistently colourable in black and white.
As in the planar case, we consider only connected surfaces.

We can easily construct a quadriculated torus 
and a quadriculated Klein bottle by identifying opposite sides
of a (quadriculated) rectangle in the usual way.
More generally, any quadriculated torus can be constructed by
taking the quotient of $\RR^2$ by a sublattice of $\ZZ^2$;
the construction of the general quadriculated Klein bottle is similar.
It is easy to see that these are the only
quadriculated surfaces with no boundary.
Quadriculated cylinders and quadriculated M\"obius bands
are even easier to construct:
start with any simply connected region in the plane
and glue along congruent boundaries.

As for Euclidean manifolds, it is easy to define
a developing map ([10]) from the universal cover of
a quadriculated surface to the plane.
Similarly, define the holonomy of a quadriculated surface:
it is a homomorphism from the fundamental group of the surface
to the group of isometries of $\ZZ^2$.
If the surface is a topological disk, it has
trivial holonomy and may be thought of as some kind of
Riemann surface over $\ZZ^2$.

The notion of a domino tiling of a quadriculated surface
is clear, as is the notion of a flip.
As in the previous simpler situation,
we want to characterize the connected components of $T$.

We now describe the correct generalization of $[t_1 - t_2]$,
$\langle t \rangle$ and height sections.
The cell complexes $\bA$ and ${\bA^\star}$ are easily defined,
as are $A$ and $A^\star$.
Consider the original homological construction of $[t_1 - t_2]$:
we draw an edge for each domino of either tiling,
orienting those in $t_1$ from black to white
and those in $t_2$ from white to black.
The obvious difficulty in generalizing this construction is:
there are no white or black squares now,
and it may even be impossible to assign colour globally
in a coherent way.
This makes it clear that $H_1(\bA^\star;\ZZ)$ 
is not the right place to try to define $[t_1 - t_2]$:
we must instead use homology with local coefficients.
Homology and cohomology with local coefficients
are briefly described for the situation of interest
in the Appendix to this Section.
More precisely, let ${\cal Z}_1$ be a $\ZZ$-bundle
over ${\bA^\star}$ constructed  as follows:
put in $\ZZ$ fibres over each square
and glue fibres on neighbouring squares by
identifying $k$ on one fibre with $-k$ on the other.
The gluing instructions provide us with fibres
over edges and create no obstruction towards defining the fibre
over a vertex because of our second condition on quadriculated surfaces;
notice that on each square there is a privileged generator for the fibre,
originally labeled 1, which we call {\sl positive}.
A more global characterization of ${\cal Z}_1$ is that its fibre twists
along a given closed curve in ${\bA^\star}$
iff this curve passes through an odd number of squares.
If we try to colour squares alternatedly black and white
we find that this is similar to constructing a section of ${\cal Z}_1$:
in particular, $\bA^\star$ is bicolourable iff ${\cal Z}_1$ is trivial.
It is now clear that our definition of $[t_1 - t_2]$
makes sense as an element of $H_1(\bA;{\cal Z}_1)$:
edges of any tiling $t$ (i.e., edges connecting the centres
of the two squares composing a domino) are oriented so that
their boundaries come out as two points with positive coefficients.

The cohomological construction of $[t_1 - t_2]$ or $\langle t \rangle$
is of course similar but it has to be performed
with different coefficients,
as is to be expected from duality anyway.
Let therefore ${\cal Z}_2$ be a $\ZZ$-bundle
constructed over ${\bA^\star}$ as follows:
first put in fibres over each square as before,
but now each generator of the fibre corresponds
to a possible orientation for the square.
Glue fibres on neighbouring squares so that orientations don't match
(thus constructing the fibres over edges);
again, our second condition on quadriculated surfaces
states that the fibre is well defined on vertices.
Equivalently, ${\cal Z}_2$ twists along a given closed curve
iff the curve inverts either colour or orientation, but not both.
The (very general) version of Poincar\'e duality for sheaves
(as in [8]) guarantees that
$H_1(\bA^\star;{\cal Z}_1) =
H^1({\bA^\star}, \partial {\bA^\star};{\cal Z}_2)$;
we provide a sketch of a direct proof of this isomorphism
in the Appendix.
Also, over any edge of a square, there is a natural correspondence
between orientations for the edge and generators
of the fibre of ${\cal Z}_2$ over the edge:
choose an adjacent square, orient the square,
and take the corresponding generator of the fibre of ${\cal Z}_2$
to correspond to the counterclockwise orientation for the edge.
It is now easy to define
$\langle t \rangle \in C^1({\bA^\star};{\cal Z}_2)$:
for edges not crossing dominoes, take the corresponding generator;
for edges crossing dominoes, take $-3$ times the same generator.
Again, this gives us an element of $H^1({\bA^\star};{\cal Z}_2)$
whose restriction to the boundary does not depend on the tiling $t$ but
it is important to notice that since the map induced by the inclusion
from $H^1({\bA^\star};{\cal Z}_2)$ to $H^1(\partial{\bA^\star};{\cal Z}_2)$
is usually not injective,
this does not mean that the cohomology class of $\langle t \rangle$
does not depend on $t$:
in Figure 4.1, the two tilings of the torus 
produce $\langle t \rangle$'s
which are not cohomologous in $H^1({\bA^\star};{\cal Z}_2) = \ZZ^2$
(notice that ${\cal Z}_2$ is trivial).
On the other hand, the hypothesis $[t_1 - t_2] = 0$
(in $H_1(\bA^\star;{\cal Z}_1) =
H^1({\bA^\star}, \partial {\bA^\star};{\cal Z}_2)$)
guarantees that $\langle t_1 \rangle$ and $\langle t_2 \rangle$
are cohomologous (in $H^1({\bA^\star};{\cal Z}_2)$).

\input domfig41.tex

As an additional example, consider the cylinders in Figures 4.2(a)
and (b) and the M\"obius bands in Figures 4.2(c) and (d).
In (a), ${\cal Z}_1$ and ${\cal Z}_2$ are both non-trivial and
$H_1(\bA;{\cal Z}_1)$ (which by Poincar\'e duality equals
$H^1({\bA^\star}, \partial {\bA^\star};{\cal Z}_2)$)
is trivial (as discussed in the Appendix);
all tilings are therefore homologous
and the reader can easily check that $T$ is connected.
In (b), ${\cal Z}_1$ and ${\cal Z}_2$ are both trivial and
$H_1(\bA;{\cal Z}_1) = \ZZ$;
there are 4 connected components in $T$
classified by $[t_1 - t_2]$.
In (c), ${\cal Z}_1$ is trivial, ${\cal Z}_2$ is not
and $H_1(\bA;{\cal Z}_1) = \ZZ$;
$T$ has 3 components in (c), again classified by $[t_1 - t_2]$.
Finally, in (d), ${\cal Z}_1$ is non-trivial, ${\cal Z}_2$ is trivial,
$H_1(\bA;{\cal Z}_1) = 0$ and $T$ is connected.

\input domfig42.tex

Our next step is to construct the height bundle $\cal H$
and the height section $\theta$ in it;
examples will be given in Figure 4.3.
It is convenient to construct both simultaneously
and, unlike the previous simpler situation of planar regions,
the structure of $\cal H$ depends to a certain extent on the tiling $t$.
The fibres of $\cal H$ are to be copies of $\ZZ$
with no distinguished zero and not even a privileged orientation
defining order on fibres;
we are allowed to add to an element of a fibre of $\cal H$
an element of the corresponding fibre of ${\cal Z}_2$
and we are allowed to compare for `equality'
elements of neighbouring fibres.
In order to build $\cal H$ and $\theta$,
take ${\cal Z}_2$ on the vertices of $\bA^\star$
and `forget' the zero section
and the exact way of identifying two neighbouring fibres:
we shall take the old zero section to be $\theta$ (by definition)
and glue neighbouring fibres with an additive shift.
This shift is described, of course, by $\langle t \rangle$,
so that $\langle t \rangle$ is the `derivative' of $\theta$
by construction. As before,
$\cal H$ can be thought of as a covering space over ${\bA^\star}$.

Since $\cal H$ is not the same for all $t$ we have to explain
how we can ever compare different height sections.
The first observation is that the structure of $\cal H$
depends on the cohomology class of $\langle t \rangle$
in $H^1({\bA^\star};{\cal Z}_2)$ only.
Indeed, if $\langle t_1 \rangle$ and $\langle t_2 \rangle$
are cohomologous, their difference is by definition a coboundary
and therefore a sum of coboundaries of `delta functions',
i.e., functions with support given by a single vertex.
Construct a discrete path from $t_1$ to $t_2$
by adding one such `delta function' at each step.
The intermediate cocomplexes in this path usually
do not correspond to tilings at all but
they still allow for the construction of $\cal H$ (and even $\theta$)
at intermediate steps.
The isomorphism of consecutive height bundles (but not sections)
is clear and our claim follows.
If, furthermore, $[t_1 - t_2] = 0$
(as an element of $H^1({\bA^\star}, \partial {\bA^\star};{\cal Z}_2)$),
the `delta functions' are all in the interior of $\bA^\star$
and, for the same procedure of taking intermediate bundles and sections,
consecutive sections coincide on the boundary.
Thus, in this case, $\theta_1$ and $\theta_2$
are sections coinciding on $\partial{\bA^\star}$
of the same bundle $\cal H$.
We are thus ready to compare $\theta_1$ and $\theta_2$
in the relevant case $[t_1 - t_2] = 0$
if $\partial{\bA^\star}$ is non-empty,
by the connectedness of $\bA^\star$.
If $\partial{\bA^\star}$ is empty, however,
we have to consider if the above construction of intermediate
bundles and sections introduces any ambiguity
in the identification of the two height bundles.
For planar regions,
height sections were well defined up to a constant.
If ${\cal Z}_2$ is trivial, i.e., if it admits
at least one non-zero section, the same thing happens.
Otherwise, height sections are well defined given $\cal H$:
the difference between two of them
(obtained by integration from the same tiling)
is a section of ${\cal Z}_2$, hence 0.
Thus, if $\partial{\bA^\star}$ is empty and ${\cal Z}_2$ is non-trivial,
there is no ambiguity in comparing height sections,
but if $\partial{\bA^\star}$ is empty and ${\cal Z}_2$ trivial
we are free to add constants (i.e., sections of ${\cal Z}_2$)
to any of the two sections.
In any case, $[t_1 - t_2] = 0$ if and only if
the height bundles for $t_1$ and $t_2$ are isomorphic
and the height sections $\theta_1$ and $\theta_2$
coincide on boundaries.

We should be able to characterize height section by properties
similar to (a), (b) and (c) above.
Properties (b) and (c) do not change:
just remember to interpret them as taking place inside $\cal H$
(and not some cut-dependent system of coordinates you may want to use).
Property (a), however, has to be rephrased a bit more carefully.
Inside each fibre of $\cal H$ there exists a class of elements
with the `right' congruence mod 4, i.e.,
those elements which differ from the height section
used for the construction of $\cal H$ by a multiple of 4.
We call the union of such subsets ${\cal H}_\phi$,
a subset of $\cal H$;
${\cal H}_\phi$ is not quite a fibre bundle however
since it is defined only over $A^\star$ and can not be naturally
extended to ${\bA^\star}$ since the height section itself
is only defined over $A^\star$.
Property (a) now says that a height section
must assume values in ${\cal H}_\phi$.
It should be clear that again
these properties characterize height sections.

In Figure 4.3 we show the height sections
for the four tilings in Figure 4.2.
Notice how simple it is to construct such height sections:
work as if the region were planar and at the end
the identifications will be automatically provided.
In these examples a value $x$ for the height section
on a point at the left cut corresponds to a value of
$1-x$, $x$, $-3-x$ and $x$
in Figures (a), (b), (c) and (d), respectively,
for the corresponding point at the right cut.

\input domfig43.tex

The M\"obius band in example (c) in Figure 4.3
illustrates an interesting point when compared to
the tilings in Figure 4.4.
Here, the cohomology group $H^1({\bA^\star};{\cal Z}_2)$
is isomorphic to ${\ZZ/(2)}$.
As discussed, the structure of $\cal H$
depends on the cohomology class of $\langle t \rangle$
in $H^1({\bA^\star};{\cal Z}_2)$ only.
However, two different tiling $t_1$ and $t_2$ for this region induce
cohomology classes $\langle t_1 \rangle$ and $\langle t_2 \rangle$
differing by a multiple of 4 in $H^1({\bA^\star};{\cal Z}_2)$
and being therefore equal.
The height section if Figure 4.4(a) does not appear
at first to be in the same bundle as Figure 4.3(c)
but an appropriate renaming of the fibres as in 4.4(b) shows that,
as predicted, the bundles are indeed isomorphic
(they have the same gluing instructions)
even though the subsets ${\cal H}_\phi$ are different in the two cases.
Similarly, the height section in 4.4(c) can be renamed as in 4.4(d)
to fit inside the bundle for 4.3(c) but the values of the section
at the boundary are different.

\input domfig44.tex

This shows that if $t_1$ and $t_2$
are in the same connected component then $[t_1 - t_2] = 0$.
It is disconcerting at this point to realize that
the converse is {\sl false}:
in the simple example shown in Figure 4.4,
$\bA^\star$ is a cylinder,
${\cal Z}_1$ and ${\cal Z}_2$ are both trivial
and the reader will have no trouble checking that
$[t_1 - t_2] = 0$ (or in computing height sections).
No flip, however, is possible.

\input domfig45.tex

If we try to follow the proof of Lemma 3.3
in this example, we see what the problem is.
The height sections differ by 4 along the entire central zig-zag
(which actually contains all points of $A^\star$ not on the boundary).
No point is, however, a local minimum or maximum
for any of the two height sections.

Let us consider this counter-example
from a slightly different point of view.
By a {\sl ladder} we mean a sequence of parallel dominoes
side by side such that:
two neighbouring dominoes always touch along one edge of the longer side,
each domino in the ladder has two neighbours in it and
these two neighbours touch the domino at different squares.
In Figure 4.4 the two tilings consist of two ladders each.
The important thing about ladders is that they are totally immune to flips.
So, if $t_1$ and $t_2$ are in the same connected component
then $[t_1 - t_2] = 0$ and $t_1$ and $t_2$ have precisely the same ladders.
It may surprise the reader that this rather ad-hoc condition
is actually necessary and sufficient.

{\nobf Theorem 4.1:}
{\sl
Two domino tilings $t_1$ and $t_2$ are in the same connected component of $T$
if and only if $[t_1 - t_2] = 0$
and $t_1$ and $t_2$ have precisely the same ladders.
Furthermore, if this is the case,
the distance between them is given by
$$d(t_1,t_2) =
{1 \over 4} \sum_{p \in A^\star}{|\theta_1(p) - \theta_2(p)|};$$
in the case where there is no boundary and ${\cal Z}_2$ is trivial,
the additive constants in the height sections
are to be chosen so that the right hand side is minimum.
}

The right hand side of the distance formula makes sense
(and is an integer):
$\theta_1(p)$ and $\theta_2(p)$ are in the same fibre of $\cal H$,
$\theta_1(p) - \theta_2(p)$ is an element of the corresponding
fibre of ${\cal Z}_2$, whose absolute value is in $\ZZ$.
As with Theorems 3.1 and 3.2, we isolate the inductive step in a lemma.

{\nobf Lemma 4.2:}
{\sl
Let ${\bA^\star}$ be a quadriculated surface
and let $t_1$ and $t_2$ be two different tilings of it
with $[t_1 - t_2] = 0$ and such that neither of them has ladders;
let $\theta_1$ and $\theta_2$ be the corresponding height sections.
Assume that $\theta_1$ and $\theta_2$ coincide on a non-empty set
(possibly the boundary).
Then there exists a tiling $t_3$ of the same region,
obtained from $t_1$ by a flip
and such that the corresponding height section $\theta_3$
always lies between $\theta_1$ and $\theta_2$.
}

{\nobf Proof:}

\nobreak

As in Lemma 3.3, let $B$ be that part of the domain where
$|\theta_1 - \theta_2|$ is maximum;
by hypothesis, $B$ is neither empty nor equal to $A^\star$.
We claim there is a point of $B$ where we can perform
a flip on $t_1$ in order to obtain $t_3$:
we call such a point (with a certain abuse of notation)
a {\sl local minimum} of $\theta_1$.
Again as in Lemma 3.3, therefore,
our aim is to prove the existence of such a local minimum.
Suppose by contradiction there is no such point:
we show the existence of a ladder.

Let $p$ and $p'$ be two neighbouring points in $B$.
We say that, when moving from $p$ to $p'$,
$\theta_1$ changes {\sl as if trying to get further} from $\theta_2$
if $|x_1 - \theta_2(p')| < |\theta_1(p') - \theta_2(p')|$, where
$x_1$ is the element of the fibre of $\cal H$
over $p'$ which belongs to ${\cal H}_\phi$,
is different from $\theta_1(p')$
and satisfies $|x_1 - \theta_1(p)| \le 3$.
A point $p$ in $B$ is a local minimum of $\theta_1$ if and only if
it has no neighbour $p'$ in $B$ such that, when moving from $p$ to $p'$,
$\theta_1$ changes as if trying to get further from $\theta_2$.
Thus, since $B$ is finite, there exists a cycle
$p_0, p_1, \ldots, p_{N-1}, p_N = p_0$ of points of $B$
such that, when going from $p_i$ to $p_{i+1}$,
$\theta_1$ changes as if trying to get further from $\theta_2$.
Call such cycles {\sl monotonic}.
We may interpret a cycle as a 1-complex;
we call two monotonic cycles {\sl adjacent} if their difference
is the boundary of a square in $\bA^\star$.
Two monotonic cycles are {\sl homotopic}
if they can be joined by a sequence of adjacent monotonic cycles;
thus, monotonic cycles break into homotopy classes.
If a cycle does not reverse orientations,
we can consistently speak of left and right;
since an orientation reversing cycle
yields an orientation preserving one by running along it twice
we assume from now on, without loss,
that we are dealing with orientation preserving cycles.
It makes sense therefore to speak of left and right of a cycle and,
given two adjacent cycles, we can naturally order them by
saying that one is to the left and the other one to the right.

\noindent{\bf Claim: }Inside each homotopy class
there are a leftmost and a rightmost monotonic cycles.

Supposing the opposite, it would always
be possible to push a cycle to the left (say),
obtaining a closed sequence
$c_0, c_1, \ldots, c_{M-1}, c_M = c_0$
of adjacent monotonic cycles
such that $c_{i+1}$ is to the left of $c_i$.
The contradiction arises from proving that
the existence of a closed sequence of cycles as above implies that
${\bA^\star}$ is a torus or a Klein bottle
and that the height sections $\theta_1$ and $\theta_2$ never coincide.
By going to the universal cover and using the developing map
as in [10],
each cycle $c_i$ becomes a periodic line $\tilde{c_i}$ in $\ZZ^2$,
the period being an orientation preserving isometry of $\RR^2$
preserving $\ZZ^2$, thus either a translation
or a rotation of period 2 or 4.
If the period of $\tilde{c_0}$ is not a translation,
the curve $\tilde{c_0}$
surrounds a certain signed area, which decreases in the process
of passing from $\tilde{c_i}$ to $\tilde{c_{i+1}}$,
contradicting the fact that the isometric curves
$\tilde{c_N}$ and $\tilde{c_0}$ enclose equal areas;
thus, the period is a translation.
Also, any isometry taking the infinite curve $\tilde{c_0}$
to $\tilde{c_N}$ is another translation,
since rotations would move remote points by distances
far greater than $M$;
also, the two translations are linearly independent
since passing from $\tilde{c_i}$ to $\tilde{c_{i+1}}$
moves curves to the left.
The cycle $c_0$ gives rise to a closed curve
in $\bA^\star$ by connecting neibouring points;
similarly, the points $c_i(0)$ are joined to produce a second
closed curve, based on the same point $c_0(0)$.
These curves can be interpreted as elements of
$\pi_1(\bA^\star,c_0(0))$ and the above translations are
their representations under holonomy.
By extending the discrete homotopy of cycles
to a map from the rectangle $[0,N] \times [0,M]$ to $\bA^\star$
we see that these two elements of $\pi_1(\bA^\star,c_0(0))$ commute,
thus generating a copy of $\ZZ^2$ inside $\pi_1(\bA^\star,c_0(0))$.
The only compact surfaces, however, for which the fundamental group
contains a copy of $\ZZ^2$ are a torus or a Klein bottle,
since any other surface is hyperbolic and there is no copy
of $\ZZ^2$ inside the isometries of the hyberbolic plane (see [10]).
Since this construction is performed in $B$, $B = A^\star$
and the two height sections never meet.
The proof of the claim is thus complete.

Consider these two extreme cycles:
they behave very similarly to the least area cycle
in the proof of Lemma 3.3.
In fact, repeating the same steps, we see that
the polygonal line joining midpoints
between consecutive points of the leftmost (resp., rightmost)
cycle never turns left (resp., right).
Now, since these two cycles are homotopic
these two polygonal lines turn by the same angle
and it follows that neither turns at all:
both cycles are zig-zag lines exactly like boundaries of ladders.
Furthermore, to the left of the leftmost cycle
or to the right of the rightmost cycle
$t_1$ and $t_2$ must each have a ladder
since we cannot have arrived at the boundary.
This contradicts the hypothesis and ends the proof of the Lemma.
{\hfill \vrule height3pt width6pt depth3pt \smallskip} %qed

\medbreak

{\nobf Proof of Theorem 4.1:}

\nobreak

All we have to prove is that if $[t_1 - t_2] = 0$
and $t_1$ and $t_2$ have the same ladders
then $t_1$ and $t_2$ are in the same component
and the distance between them 
is smaller than or equal to the expression
at the right hand side in the statement of the theorem.
Let therefore $t_1$ and $t_2$ be tilings as above.
Start by removing all ladders from ${\bA^\star}$:
we have to prove that the tilings
on each connected component of whatever remains
are in the same connected of $T$.
It is clear that on each such connected component
height bundles for $t_1$ and $t_2$ are isomorphic
and the height sections
coincide on whatever remains of the old boundary
and differ by a constant on boundaries of removed ladders.
We claim that we can never have a connected component
of the boundary consisting of the boundary of a ladder only:
indeed, if this happened, the only way to tile
the neighbourhood of this boundary component
would be with a new ladder.
It follows that $\theta_1$ and $\theta_2$
coincide on the entire boundary of each connected component
of whatever remains after removing ladders.
We can therefore assume without loss of generality
that $t_1$ and $t_2$ have no ladders.

When ${\bA^\star}$ has boundary,
Lemma 4.2 (plus induction) finishes with the proof.
If ${\bA^\star}$ has no boundary,
we must consider two cases.
If ${\cal Z}_2$ is non-trivial,
the two height sections must coincide at some point
by topological reasons:
if they did not, their difference would yield
a global choice of generators for ${\cal Z}_2$
(the difference is to be a positive multiple of the chosen generator)
hence a trivialization of ${\cal Z}_2$
(since the fibre is one-dimensional).
If ${\cal Z}_2$ is trivial,
add a constant to $\theta_2$ in order to make the
right hand side of the distance formula minimum:
it is clear that now $\theta_1$ and $\theta_2$ coincide at some point.
{\hfill \vrule height3pt width6pt depth3pt \smallskip} %qed

As in the planar case, we know which flips to perform in
order to get closer to a tiling $t_2$ starting from a tiling $t_1$,
assuming, of course, $[t_1 - t_2] = 0$.
Start by computing the (isomorphic) height bundles
and the height sections $\theta_1$ and $\theta_2$
which must coincide on the boundary.
In case there is no boundary and ${\cal Z}_2$ is trivial,
adjust the additive constant to make distance minimum
(this may allow for one or two answers).
Now flip at any local extremum of $\theta_1$
if that takes the section closer to $\theta_2$.
Again, there is a local characterization of shortest paths in $T$.
However, not all paths are homotopic anymore.

{\nobf Theorem 4.3:}
{\sl
If ${\bA^\star}$ has boundary or ${\cal Z}_2$ is non-trivial,
all connected components of $T$ are contractible.
If ${\bA^\star}$ is a torus or a Klein bottle
and ${\cal Z}_2$ is trivial, there are two kinds of
connected components of $T$:
some consist of one single isolated point which
corresponds to a tiling constructed entirely from ladders;
others are homotopy equivalent to $\SS^1$.
}

\medbreak

{\nobf Proof:}

\nobreak

When ${\bA^\star}$ has boundary or ${\cal Z}_2$ is non-trivial,
the proof is entirely analogous to the planar case.
From now on, assume the other situation.
Notice first that if a tiling contains a ladder,
it must consist of ladders only:
only a ladder fits beside a ladder.
We now prove that a tiling with height section $\theta_1$ 
which admits no flips must be of this type.
Assume first that ${\bA^\star}$ is a torus,
the quotient of $\RR^2$ (quadriculated by $\ZZ^2$)
by a 2-dimensional sublattice $L$ of $\ZZ^2$.
Raise the tiling to the universal cover
in order to obtain an $L$-periodic tiling of the plane.
Taking $\theta_2$ to be $\theta_1 + 4$,
as in Lemma 4.2, there must exist a monotonic cycle $c$ which,
raised to the universal cover $\RR^2$,
must connect the origin to some other point of $L$;
without loss, this point is of the form $(x,y)$ with $x \ge y \ge 0$.
By the triviality of ${\cal Z}_2$,
$x$ and $y$ must be of the same parity.
Raise $\theta_1$ to a height function $\tilde\theta_1$ in the plane:
we can assume without loss that $\tilde\theta_1(0,0) = 0$.
Also, the value of $\tilde\theta_1$ decreases along
the lifted monotonic cycle $\tilde c$.
The value of $\tilde\theta_1(x,y)$ must be precisely $-2x$:
a smaller value is impossible for any height function
by conditions (a), (b) and (c) and
a larger value does not allow for a monotonic decreasing path
from the origin to $(x,y)$.
If $x = y$ (in which case $x > 0$)
the monotonic path $\tilde c$ must be a zig-zag
going from the origin to $(x,y)$
which can not cross dominoes and must therefore be a side of a ladder.
Otherwise, the values of the height function at 0 and $(x,y)$
are enough to dictate the values on a parallellogram
with vertices at these points.
Since, as in the proof of Lemma 4.2,
there exist monotonic cycles through every point,
the whole height function is well determined
and the tiling must look like a garden variety brick wall,
constructed from ladders going both ways.
We take care of the Klein bottle by going to the orientable
double cover, which is a torus.

For the other cases we claim that the universal cover
of the corresponding connected component of $T$
consists of all height sections
{\sl without} identifying sections which differ by a constant.
It is clear that this is a covering map
and what the CW-complex structure for this space must be.
It is enough to prove that this space is connected and contractible
since the quotient group will obviously be $\ZZ$,
or, more precisely, $4\ZZ$.
Since after identifications this space is known to be connected,
it is enough to prove that we can move by flips
from a section $\theta_1$ to $\theta_1 + 4$.
From the previous paragraph, we can perform {\sl some}
flip on $\theta_1$, without loss an increasing one,
to obtain $\theta_2$;
but now $\theta_2$ intersects $\theta_1 + 4$
at the flipped point and by Lemma 4.2 and Theorem 4.1
we can move $\theta_2$ to $\theta_1 + 4$ by flips.
The proof that the space of sections is contractible
is similar to what we already saw in the previous cases,
the fact that there are infinitely many cells being
no source of trouble:
a point contained in a cell such that its furthest vertex
from the base section is at a distance $d$ 
starts moving at time $1/{2^d}$.

This argument also shows that the generator of the fundamental group
of a connected component of $T$ is the path from $\theta$ to $\theta+4$.
Actually, such a closed path is a deformation retract of the connected
component, but we give no details.
In Figure 4.6 we show the four steps of such a cycle
for the only non-trivial component of $T({\bA^\star})$,
where ${\bA^\star} = {\RR^2}/{(2\ZZ)^2}$
(move from (a) to (b) to (c) to (d) to (e)).
{\hfill \vrule height3pt width6pt depth3pt \smallskip} %qed

\input domfig46.tex

\goodbreak
\bigskip
{{\nobf Appendix:} Homology and cohomology with local coefficients}

\nobreak\smallskip\nobreak

This Appendix contains a brief review of the main facts
about homology and cohomology with local coefficients
which are necessary or convenient for us.
More specifically, we apply the general constructions to our examples.
Readers which know enough about the subject
to compute homology and cohomology in simple examples
and who are acquainted with Poincar\'e duality in this context
are encouraged to skip the Appendix altogether.
There are good expositions of the subject in [11] and [7];
for the more general theory of sheaves,
the reader may consult [8].

We begin with a description of $H_1(\bA,{\cal Z}_1)$;
as in the usual homology, this is obtained from a chain complex
of additive groups $C_2 \to C_1 \to C_0$
by taking the quotient $Z_1/B_1$,
where $Z_1$ is the kernel of the second boundary map
and $B_1$ is the image of the first.
Recall that the fibre of ${\cal Z}_1$
over a square of $\bA^\star$ has a {\sl positive}
and a {\sl negative} generator;
thus, the fibre of ${\cal Z}_1$ over vertices of $\bA$
also has a positive and a negative generator.
Thus, the generators of the fibre of ${\cal Z}_1$ over an edge of $\bA$
are positive on one of extrema and negative on the other.
Over a square of $\bA$,
the generators are alternatedly positive and negative
over the four vertices.
The additive groups $C_i$ are generated by (formal) products
of an oriented $i$-cell in $\bA$ by a generator of the fibre over it.
Thus, generators of $C_0$ are vertices with an orientation,
i.e., a sign, as in Figure A.1(a).
Similarly, generators of $C_1$ and $C_2$ are indicated
in Figures A.1(b) and (c);
the second equality in (b) is a notational convenience.
The action of the boundary maps $C_{i+1} \to C_i$ over generators
is indicated in Figures A.1(d) and (e);
notice that the composition of both is zero.

\input domfiga1.tex

We provide a similar description of
the relative cohomology group
$H^1(\bA^\star,\partial\bA^\star;{\cal Z}_2)$.
Again, our first task is to construct $C^2$, $C^1$, $C^0$
and the coboundary maps.
The fibre of ${\cal Z}_2$ over a square of $\bA^\star$ is 
isomorphic to $\ZZ$.
An orientation for the square and a sign 
(which again alternates between neighbouring squares)
determine a generator of the fibre:
changing one of these ingredients alters the generator.
Thus, the additive group $C^2$ is generated by
the map taking a given oriented square of $\bA^\star$
to the generator of ${\cal Z}_2$ over this same square
corresponding to the orientation of the square and the plus sign:
we denote such generators as in Figure A.2(a).
Generators of $C^1$ are maps taking a given non-boundary
oriented edge of $\bA^\star$ to the generator
of ${\cal Z}_2$ defined as follows:
choose any of the two adjacent squares to the edge,
orient it so that the
induced orientation on the boundary
equals the original orientation of the edge,
and take the generator of ${\cal Z}_2$ over it corresponding
to its orientation and the plus sign.
It is easy to check that this map does not depend on the choice
of the adjacent square; we denote the generators of $C^1$  as in
Figure A.2(b).
Generators of $C^0$ are maps taking an interior vertex of $\bA^\star$
to one of the generators of the fibre of ${\cal Z}_2$ over it;
the choice of the generator is indicated
by an orientation and signs for the neighbouring squares
as in Figure A.2(c).
Coboundary maps over generators are indicated in Figures A.2(d) and (e).
In order to obtain the cohomology group $H^1(\bA^\star;{\cal Z}_2)$,
drop the restrictions that edges or vertices must be interior.

\input domfiga2.tex

We recall the basic facts concerning Poincar\'e duality.
Ordinary Poincar\'e duality ([4]) works by identifying
$C_k(M)$ for a given triangulation with
$C^{n-k}(M)$ for the dual triangulation,
where $M$ is an $n$-dimensional oriented closed manifold
(the orientation is used in the identification procedure).
This identification commutes (up to signs)
with boundary and coboundary operations
and thus induces, by taking quotients,
the identifications between $H_k(M)$ and $H^{n-k}(M)$.
In Lefschetz duality ([4]),
we work with oriented $n$-manifolds with boundary
and identify $C_k(M)$ with $C^k(M,\partial M)$
again by looking at dual triangulations.
More generally, we can consider local coefficients $\cal Z$
and, still for an oriented manifold with boundary,
essentially the same construction yields an identification
between $C_k(M;{\cal Z})$ and $C^{n-k}(M,\partial M;{\cal Z})$.
One way to get rid of the orientability hypothesis
is to let the cohomology coefficients take care of the problem ([8]):
if ${\cal Z}_0$ is the $\ZZ$-bundle over $M$
with generators corresponding to (local) orientations,
the appropriate generalization of Poincar\'e's construction
provides the identification
$$C_k(M;{\cal Z}) =
C^{n-k}(M,\partial M;{\cal Z} \otimes {\cal Z}_0).$$
By taking quotients, we obtain the duality we need:
$$H_k(M;{\cal Z}) =
H^{n-k}(M,\partial M;{\cal Z} \otimes {\cal Z}_0).
\eqno{(\ast)}$$

Going back to our context, it is easy to see that
${\cal Z}_2 = {\cal Z}_1 \otimes {\cal Z}_0$,
where ${\cal Z}_0$ is constructed as above.
The identification
$$H_1(\bA;{\cal Z}_1) =
H^1(\bA^\star,\partial\bA^\star;{\cal Z}_2)$$
is a special case of $(\ast)$.
Notice, however, that our descriptions
of the chain and cochain complexes yield an
explicit construction of this bijection:
just match corresponding letters in Figures A.1 and A.2.

We compute the homology groups which appear in Figure 4.2.
Whenever the coefficient bundle is trivial,
we are dealing with the usual homology group
with coefficients in $\ZZ$ ([11]),
and this takes care of (b) and (c).
Otherwise, by invariance of homology under deformation retracts,
we are reduced to computing $H_1(S^1;{\cal Z})$,
where $\cal Z$ is the non-trivial $\ZZ$-bundle over $S^1$.
Consider the very simple CW-decomposition of the circle
with a single edge having both extrema attached to the same 0-cell.
The groups $C_0$ and $C_1$ are both cyclic and the boundary map
takes a generator of $C_1$ to twice a generator of $C_2$;
thus, $H_1(S^1;{\cal Z}) = 0$, as claimed,
and $H_0(S^1;{\cal Z}) = \ZZ/(2)$.
In the comments concerning Figure 4.4,
we state that $H^1(\bA^\star;{\cal Z}_2) = \ZZ/(2)$,
where $\bA^\star$ is a M\"obius band and ${\cal Z}_2$ is non-trivial:
again, by invariance under deformation retracts,
it suffices to compute $H^1(S^1;{\cal Z})$ with $\cal Z$ as above.
The groups $C^0$ and $C^1$ are both cyclic and $C^2$ is trivial;
the coboundary takes a generator of $C^0$
to twice a generator of $C^1$,
so that $H^1(S^1;{\cal Z}) = \ZZ/(2)$ and $H^0(S^1;{\cal Z}) = 0$.

\goodbreak
\bigskip
{\nobigbf 5. Final remarks}

\nobreak\smallskip\nobreak

{\nobf A. Calisson tilings}

\nobreak\smallskip\nobreak

A {\sl calisson} is the union of two equilateral triangles
with a common side.
Calisson tilings of simply connected regions in the plane
admit height functions ([9]) with a strong visual interpretation:
by looking at a calisson tiling, you can see it as a figure of
a pile of (3-dimensional) cubes,
the calissons being their faces ([2]).
In close analogy with what we did in this paper,
we can define height sections for calisson tilings of other regions.
In this context, we perform a {\sl flip} by lifting
three calissons forming a hexagon and placing them back
in the only possible different configuration.
Clearly, two tilings are adjacent by a flip
if their height sections differ at a single point.
Under the pile-of-cubes interpretation,
a flip corresponds to adding or removing a cube.
Thus, for simply connected regions,
the space of tilings is connected and the distance
between two tilings is given by the number of non-common cubes.
Height sections might be useful for a more careful study
of the space of calisson tilings of more complicated regions.

{\nobf B. The adjacency matrix of $A$}

\nobreak\smallskip\nobreak

The adjacency matrix of $A$ is of the form
$$\pmatrix{0&M \cr M^T&0 \cr},$$
provided white vertices are listed before black vertices.
The sign of $\det M$ is not natural:
it depends on the order in which the vertices are listed.
Tilings of $\bA^\star$ correspond to monomials
in the expansion of the determinant of $M$.
Indeed, such a monomial (up to sign)
corresponds to a set of 1's in $M$
with exactly one element in each row or column:
each 1 gives rise to an edge of $A$ and it is clear
that the associated set is a covering by edges.
Tilings are thus naturally divided into two classes
according to the sign of the corresponding monomial
and we shall say that two tilings have the same or opposite parities
if the corresponding monomials have the same or opposite signs
(there is, however, no natural definition of an `even' and an `odd' tiling).
It is easy to see that adjacent tilings always have opposite parities.

\input domfig51.tex

The absolute value of the determinant of $M$ is the difference
between the number of tilings of each parity:
in [3], it is shown that, when
$\bA^\star$ is a simply connected surface, this difference is 0 or 1.
On the other hand, if $A$ is not simply connected,
this difference can have any value
(see [3] or consider instead a $4 \times (2n - 1)$ rectangle
with $n-1$ vertical isolated dominoes removed from its interior).
Since $A$ being simply connected implies the connectivity of $T(A)$,
it might be thought that the correct generalization
to non-simply connected regions would be that,
on each connected component of $T(A)$, this difference still is 0 or 1.
In the examples shown in Figure 5.1, however, there are always three
connected components with differences of 1, $-2$ and 1
(in the natural order).
Indeed, in both cases it is easy to see that $\det(M) = 0$
by considering the element of the kernel indicated in the Figure;
the fact that there are three connected components follows from
Theorem 1.1, by merely constructing tilings with different flows,
and it is just as easy to see that
two of these components have a single element each,
always with the same parity: our claim follows.

\smallskip
{\nobf C. Higher dimensions}

\nobreak\smallskip\nobreak

The obvious generalization of Theorem 1.1 to higher dimensions is false
even if ${\bA^\star}$ is a topological closed ball contained in $\ZZ^n$
(although the definition of $[t_1 - t_2]$ still works,
and properties (a), (b), (c) and (d) as above still hold).
In dimension 3, let
$$A = \{
(0,0,0), (0,0,1), (0,0,2), (0,1,0), (0,1,1), (0,1,2),
(1,0,1), (1,0,2), (1,1,0), (1,1,1)\}.$$
The tiling
$$([0,1],1,0), ([0,1],0,2), (1,[0,1],1), (0,0,[0,1]), (0,1,[1,2])$$
has {\sl no} adjacent tilings (since there is no square to flip)
but is not the only one: consider
$$(0,[0,1],0), (0,[0,1],1), (0,[0,1],2), (1,0,[1,2]), (1,1,[0,1]).$$
As another example, now in dimension 4,
let $A = \{0,1\}^4$ be the cube of side 2. The tiling
$$\matrix{
([0,1],0,0,0), ([0,1],1,1,1), (0,0,1,[0,1]), (0,1,[0,1],0),\cr
(0,[0,1],0,1), (1,1,0,[0,1]), (1,0,[0,1],1), (1,[0,1],1,0) \cr}
$$
again has no neighbours but is not the only one.
By the way, we know of no satisfactory extension
of the idea of height sections to higher dimensions:
the definition of $\langle t \rangle$
as a $(n-1)$-cocycle still works but, even if this is exact,
its integral is not close to being unique.

\goodbreak
\bigskip
{\nobigbf References}

\nobreak
\smallskip
\nobreak

\item{[1]}{J. H. Conway and J. C. Lagarias,
{Tilings with polyominoes and combinatorial group theory,}
J. Comb. Theor. {\bf A53}, 183-208 (1990).}

\item{[2]}{G. David and C. Tomei,
{The problem of the calissons,}
Amer. Math. Monthly, {\bf 96}, 429-431 (1989).}

\item{[3]}{P. A. Deift and C. Tomei,
{On the determinant of the adjacency matrix for a planar
sublattice,} J. Comb. Theor. {\bf B35}, 278-289 (1983).}

\item{[4]}{M. J. Greenberger and J. R. Harper,
{Algebraic Topology, A First Course (Revised)},
Benjamin-Cummings, Menlo Park, California, 1981.}

\item{[5]}{P. W. Kasteleyn,
{The statistics of dimers on a lattice I. The number of dimer
arrangements on a quadratic lattice,}
Phisica {\bf 27}, 1209-1225 (1961).}

\item{[6]}{E. H. Lieb and M. Loss,
{Fluxes, Laplacians and Kasteleyn's theorem,}
Duke Math. Jour., {\bf 71}, 337-363 (1993).}

\item{[7]}{E. H. Spanier,
{Algebraic Topology,}
New York: McGraw-Hill, 1966.}

\item{[8]}{R. G. Swan,
{The Theory of Sheaves,}
The University of Chicago Press, Chicago, 1964.}

\item{[9]}{W. P. Thurston,
{Conway's tiling groups,}
Amer. Math. Monthly, {\bf 97}, 8, 757-773 (1990).}

\item{[10]}{W. P. Thurston,
{Geometry and Topology of Three-manifolds,}
Department of Mathematics, Princeton University, 1979.}

\item{[11]}{G. W. Whitehead,
{Elements of Homotopy Theory,}
GTM 61, Springer-Verlag, New York, 1978.}

\bigskip\bigskip

\goodbreak

\line{Nicolau C. Saldanha, IMPA and PUC-Rio \hfill}
\line{e-mail: nicolau@impa.br, http://www.impa.br/$\sim$nicolau/ \hfill}

\smallskip

\line{Carlos Tomei, IMPA and PUC-Rio \hfill}
\line{e-mail: tomei@impa.br \hfill}

\smallskip

\line{Mario A. Casarin Jr., NYU \hfill}
\line{e-mail: casarin@math1.cims.edu \hfill}

\smallskip

\line{Domingos Romualdo, Harvard University \hfill}
\line{e-mail: romualdo@husc.harvard.edu \hfill}

\medskip

\line{Instituto de Matematica Pura e Aplicada,
Estrada Dona Castorina, 110 \hfill}
\line{Jardim Bot\^anico, Rio de Janeiro, RJ 22460-320, BRAZIL\hfill}

\smallskip

\line{Departamento de Matem\'atica, PUC-Rio,
Rua Marqu\^es de S\~ao Vicente, 225 \hfill}
\line{G\'avea, Rio de Janeiro, RJ 22453-900, BRAZIL \hfill}

\smallskip

\line{Courant Institute of Mathematical Sciences,
251, Mercer St. \hfill}
\line{New York, NY 10012, USA \hfill}

\smallskip

\line{Departament of Economics, Harvard University,
One Oxford St., Science Center 325\hfill}
\line{Cambridge, MA 02138, USA \hfill}

\bye
