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Definitions and main results
In this section, we prove that under very general conditions, Brillouin zones
tile (as defined below) the space in which they are defined,
generalizing an old result of Bieberbach
[Bi]. With stronger assumptions, we prove that these tiles are
in fact well-behaved sets: they are equal to the closure of their interior.
Notation:
Throughout this paper, we shall assume X is a path connected, proper (see
below) metric space (with metric
).
We will make use the following notation:
Definition 2.1
A metric space X is proper if the distance function
is a proper map for every fixed . In particular, for every and r>0
, the closed ball Dr(
x)
is compact. Such a metric space is also
sometimes called a geometry (See [Ca]).
Note if X is proper, it is locally compact and complete. The converse
also holds if X is a geodesic metric space (see Thm. 1.10 of
[Gr]). The metric spaces considered here need not be geodesic.
This property is satisfied for any Riemannian metric.
Note that any mediatrix La,b separates X, that is: X-Labcontains at least two components (one containing the point a and the other
b).
Let L0a be a mediatrix. If it is minimally separating, then X-L has
exactly two components. Also note that if a separating set
contains a non-empty open set V, it cannot be minimally
separating. For if A and B are disjoint open sets containing
X-L, then so are
and
.
Now let x be any point in V. Then
it is easy to see that the disjoint open sets
and
cover X-(L-x). Thus L-x separates.
We define the following sets
Usually, there will be a discrete set of points
in
X which will be of interest. By discrete we mean that any compact subset
of X contains finitely many points of S. Note that if
,
then S is discrete.
Definition 2.5
We say a proper, path connected metric space
X is Brillouin if it
satisfies the following conditions:
- 1:
- X is metrically consistent.
- 2:
- For all a, b in X, the mediatrices Lab are minimally
separating sets.
- 3:
- For any three distinct points 0, a, and b in X, the
mediatrices L0a and L0b are topologically transversal.
The last two conditions in the above definition may be weakened to apply
only to those mediatrices Lab where a and b in S. In this
case, we will say that X is Brillouin over S, if it is not obvious
from the context.
Figure:
set
L(0,0),(a,a) contains
two quarter-planes.
Figure:
L(0,0),(4,6) (thin solid line) and
L(0,0),(2,4) (thick grey line) are not transverse.
Figure:
The mediatrices L0a for
with the Manhattan
metric and a in the lattice
.
|
Example 2.6
Equip
with the ``Manhattan metric'', that is,
d(
p,
q) = |
p1 -
q1| + |
p2 -
q2|. In this metric, a circle
Cr(
p)is a diamond of side length
centered at
p, so condition
1 is satisfied. However, condition 2 fails: if the coordinates of a point
a are equal, then
L0a consists of a line segment and two
quarter-planes (see Fig.
2.1). If the discrete set
Scontains no such points, we can still run into trouble with topological
transversality. For example, the mediatrices
L(0,0),(2,4) and
L(0,0),(4,6) both contain the ray
(Fig.
2.2). But, if we are careful, we can ensure that the
space is Brillouin over
S. To achieve this, we must have that for
all pairs (
a1,
a2) and (
b1,
b2) in
S,
.
For example,
take
S to be an irrational lattice such as
.
It is easy to check that for all
,
the properties of
definition
2.5 are true. (From this example, we see that to do
well in Manhattan, one should be carefully irrational.)
As mentioned in the introduction, for each ,
the
mediatrices Lx0 a give a partition of X. Informally, those elements of
the partition which are reached by crossing n-1 mediatrices
from x0 form the n-th Brillouin zone, Bn(x0). This definition is
impractical, in part because a path may cross several mediatrices
simultaneously, or the same mediatrix more than once. Instead, we will use
a definition given in terms of the number of elements of S which are
nearest to x.
This definition is equivalent to the informal one
when X is Brillouin over S (see Prop. 2.13 below).
We use the notation
to denote the cardinality of the set S.
Figure:
we illustrate the definition of the sets
bn(x0) and Bn(x0) for the lattice
in
.
In both
pictures, the circle
is drawn, and the
basepoint x0 lies in the center of the square at the lower left.
On the left side, the point x (marked by a small cross) lies in
b5, and
,
while x0 is
the only point of S on the circle. On the right, we have m=4 and
,
so x lies in all of the sets
.
|
Definition 2.7
Let
,
let n be a positive integer,
, and
let
r=
d(
x,
x0)
. Then define the sets bn(
x0)
and Bn(
x0)
as follows.
Here the point x0 is called the base point, and the set Bn(x0)is the
n-th Brillouin zone with base point x0. Note that in the second part,
if m=n-1 and
,
then
.
So
.
Note
also that the complement of bn(x0) in Bn(x0) consists of subsets
of mediatrices (see Def. 1.1).
Note also that bn(x0) is open and that
Bn(x0) is closed. Finally, observe that for fixed x0 the sets
bn(x0) are disjoint, but the sets Bn(x0) are not.
In what follows it will be proved that
cover the space X. Thus we can assign to
each point x its Brillouin index as the the largest n for
which
.
This definition was first given in
[Pe1].
The following lemma, which follows immediately from Def. 2.7,
explains a basic feature of the zones, namely that they are concentric in in
a weak sense. This property is also apparent from the figures.
Lemma 2.8
Any continuous path from
x0 to Bn(
x0)
intersects
Bn-1(
x0)
.
The Brillouin zones actually form a covering of X by
non-overlapping closed sets in various ways. This is proved
in parts. The next two results assert that the zones B cover
X, but the zones b do not. The first of these is an immediate
consequence of the definitions. The second is more surprising and leads
to corollary 3.4. The fact that the Bi(xn)
is the closure of bi(xn) (and thus that the interiors do
not overlap) is proved in proposition 2.12. We
will use the word ``tiling'' for a covering by non-overlapping
closed sets.
Lemma 2.9
For fixed n the Brillouin zones tile
X in the
following sense:
Figure:
example illustrates Lemma 2.9 and
Thm. 2.10. Let S be the discrete set
in the Euclidean plane. On the left is the
tiling given by Bi(0,0) and in the middle is the tiling by
Bi(2,0). In both cases, b2 is shaded. On the right is the
tiling given by B2(xi) as in Thm. 2.10. The sets
b2(0,0), b2(1,0), and b2(2,0) have been shaded. Note that this
S does not correspond to a group, nor does it satisfy the hypotheses
of Prop. 2.11, because there are no isometries
which permute S and do not fix the origin.
|
Theorem 2.10
Let
X be a proper, path connected metric space let
be a discrete set. Then, for fixed
, the sets
tile X in the following sense:
Proof:
First, we show that for any fixed n>0 and each ,
there is an
with
.
Re-index S so that if
and
i < j, then
.
This can be done; since S is a discrete subset and closed
balls Dc(xi) are compact,
the subsets of S with
are all
finite. Let
ri = d(x,xi). We will show that
.
Note that
.
Suppose first that
rn > rn-1, then
contains exactly n-1 points, and
.
Thus
.
Note that if
rn+1 > rn, then we
would have
.
If, on the other hand,
rn = rn-1,
then there is a k>0 so that
,
and so
.
But then
,
and hence
as desired.
For the second part, we show that
.
If not, then there is a point x in their intersection.
If ri = rj, then xi = xj, because by the definition of
bn(xk),
.
If not, then
ri < rj. In this case,
. Thus, since
,
Nrj(x) must
contain at least n points of S, a contradiction.
Next: Brillouin zones in spaces
Up: ON BRILLOUIN ZONES
Previous: Introduction
Translated from LaTeX by Scott Sutherland
1998-06-12