arXiv:math/9806154v1 [math.MG] 29 Jun 1998
ON BRILLOUIN ZONES
A. C. Rocha
S. Sutherland
J. J. P. Veerman
M. M. Peixoto
veerman@dmat.ufpe.br
peixoto@impa.br
acr@dmat.ufpe.br scott@math.sunysb.edu
Physics Department
UFPE
Recife, Brazil
IMPA
Rio de Janeiro, Brazil
Mathematics Dept.
Mathematics Dept.
SUNY
UFPE
Stony Brook, NY, USA
Recife, Brazil
Abstract
Brillouin zones were introduced by Brillouin [Br] in the thirties to describe quantum
mechanical properties of crystals, that is, in a lattice in Rn . They play an important role in
solid-state physics. It was shown by Bieberbach [Bi] that Brillouin zones tile the underlying
space and that each zone has the same area. We generalize the notion of Brillouin Zones
to apply to an arbitrary discrete set in a proper metric space, and show that analogs of
Bieberbach’s results hold in this context.
We then use these ideas to discuss focusing of geodesics in orbifolds of constant curvature.
In the particular case of the Riemann surfaces H2 /Γ(k) (k = 2, 3, or 5), we explicitly count
the number of geodesics of length t that connect the point i to itself.
1
Introduction
In solid-state physics, the notion of Brillouin zones is used to describe the behavior of an electron
in a perfect crystal. In a crystal, the atoms are often arranged in a lattice; for example, in NaCl,
the sodium and chlorine atoms are arranged along the points of the simple cubic lattice Z3 .
If we pick a specific atom and call it the origin, its first Brillouin zone consists of the points
in R3 which are closer to the origin than to any other element of the lattice. This same zone
can be constructed as follows: for each element a in the lattice, let L0a be the perpendicular
bisecting plane of the line between 0 and a (this plane is called a Bragg plane). The volume
about the origin enclosed by these intersecting planes is the first Brillouin zone, b1 (0). This
construction also allows us to define the higher Brillouin zones as well: a point x is in bn if the
line connecting it to the origin crosses exactly n − 1 planes L0a , counted with multiplicity.
This notion was introduced by Brillouin in the 1930s ([Br]), and plays an important
role in solid-state theory (see, for example, [AM] or [Jo2], also [Ti]). The construction which
gives rise to Brillouin zones is not limited to consideration of crystals, however. For example, in
computational geometry, the notion of the Voronoi cell corresponds exactly to the first Brillouin
zone described above (see [PS]). We shall also see below how, after suitable generalization, this
construction coincides with the Dirichlet domain of Riemannian geometry, and in many cases,
to the focal decomposition introduced in [Pe1] (see also [Pe3]).
Stony Brook IMS Preprint #1998/7
June 1998
2
Veerman, Peixoto, Rocha, and Sutherland
4
3
4
3
2
4
4
3
3
4
4
2
2
4
4
1
3
3
4
2
3
4
4
3
4
Figure 1.1: On the right are the Brillouin zones for the lattice Z2 in R2 . On the left is the outer boundary of
the third Brillouin zone for the lattice Z3 in R3 .
With some slight hypotheses (see section 2), we generalize the construction of Brillouin
zones to any discrete set S in a path-connected, proper metric space X. We generalize the
Bragg planes above as mediatrices, defined here.
Definition 1.1 For a and b distinct points in S, define the mediatrix (also called the equidistant
set or bisector) Lab of a and b as:
Lab = {x ∈ X d(x, a) = d(x, b)} .
Now choose a preferred point x0 in S, and consider the collection of mediatrices {Lx0 ,s }s∈S .
These partition X into Brillouin zones as above: roughly, the n-th Brillouin zone Bn (x0 ) consists of those points in X which are accessible from x0 by crossing exactly n − 1 mediatrices.
(There is some difficulty accounting for multiple crossings— see definition 2.7 for a precise
statement.)
One basic property of the zones Bn is that they tile the space X:
[
Bn (xi ) = X
and
Bn (x0 ) ∩ Bn (x1 ) is small .
xi ∈S
Here, with some extra hypotheses, “small” means of measure zero. Furthermore, again with
some extra hypothesis, each zone Bn has the same area. (This property was “obvious” to
Brillouin). Both results were proved by Bieberbach in [Bi] in the case of a lattice in R2 .
Indeed, he proves (as we do) that each zone forms a fundamental set for the group action of
the lattice. His arguments rely heavily on planar Euclidean geometry, although he remarks
that his considerations work equally well in Rd and can be extended to “groups of motions in
non-Euclidean spaces”. In [Jo1], Jones proves these results for lattices in Rd , as well as giving
asymptotics for both the distance from Bn to the basepoint, and for the number of connected
components of the interior of Bn . In section 2, we show that the tiling result holds for arbitrary
3
On Brillouin Zones
discrete sets in a metric space. If the discrete set is generated by a group of isometries, we show
that each Bn forms a fundamental set, and consequently all have the same area (see Prop. 2.11).
We now discuss the relationship of Brillouin zones and focal decomposition of Riemannian
manifolds.
If x1 (t) and x2 (t) are two solutions of a second order differential equation with x1 (0) =
x2 (0) and there is some T 6= 0 so that x1 (T ) = x2 (T ), then the trajectories x1 and x2 are
said to focus at time T . One can ask how the number of trajectories which focus varies with
the endpoint x(T )— this gives rise to the concept of a focal decomposition (originally called a
sigma decomposition). This concept was introduced in [Pe1] and has important applications in
physics, for example when computing the semiclassical quantization using the Feynman path
integral method (see [Pe3]). There is also a connection with the arithmetic of positive definite
quadratic forms (see [Pe2], [KP], and [Pe3]). Brillouin zones have a similar connection with
arithmetic, as can be seen in section 4.
More specifically, consider the two-point boundary problem
ẍ = f (t, x, ẋ),
x(t0 ) = x0 ,
x(t1 ) = x1 ,
x, t, ẋ, ẍ ∈ R.
Associated with this equation, there is a partition of R4 into sets Σk , where (x0 , x1 , t0 , t1 ) is in
Σk if there are exactly k solutions which connect (x0 , t0 ) to (x1 , t1 ). This partition is the focal
decomposition with respect to the boundary value problem. In [PT], several explicit examples
are worked out, in particular the fundamental example of the pendulum ẍ = − sin x. Also,
using results of Hironaka ([Hi]) and Hardt ([Ha]), the possibility of a general, analytic theory
was pointed out. In particular, under very general hypotheses, the focal decomposition yields
an analytic Whitney stratification.
Later, in [KP], the idea of focal decomposition was approached in the context of geodesics
of a Riemannian manifold M (in addition to a reformulation of the main theorem of [PT]). Here,
one takes a basepoint x0 in the manifold M : two geodesics γ1 and γ2 focus at some point y ∈ M
if γ1 (T ) = y = γ2 (T ). This gives rise to a decomposition of the tangent space of M at x into
regions where the same number of geodesics focus.
In order to study focusing of geodesics on an orbifold (M, g) with metric g via Brillouin
zones, we do the following. Choose a base-point p0 in M and construct the universal cover X,
lifting p0 to a point x0 in X. Let γ be a smooth curve in M with initial point p0 and endpoint
p. Lift γ to γ̃ in X with initial point x0 . Its endpoint will be some x ∈ π −1 (p). The metric g
on M is lifted to a metric g̃ on X by setting g̃ = π ∗ g. Under the above conditions, the group
G of deck transformations is discontinuous and so π −1 (p0 ) ⊂ X is a discrete set. One can ask
how many geodesics of length t there are which start at p0 end in p, or translated to (X, γ̃),
this becomes: How many mediatrices Lx0 ,s intersect at x, as s ranges over π −1 (p0 )?
Notice that if the universal cover of M coincides with the tangent space T Mx , the focal
decomposition of [KP] and that given by Brillouin zones will be the same. If the universal
cover and the tangent space are homeomorphic (as is the case for a manifold of constant negative curvature), the two decompositions are not identical, but there is a clear correspondence.
However, if the universal cover of the manifold is not homeomorphic to the tangent space at
the base point, the focal decomposition and that given by constructing Brillouin zones in the
4
Veerman, Peixoto, Rocha, and Sutherland
universal cover are completely different. For example, let M be Sn , and let x be any point in it.
The focal decomposition with respect to x gives a collection of nested n − 1-spheres centered at
x; on each of these infinitely many geodesics focus (each sphere is mapped by the exponential
to either x or its antipodal point). Between the spheres are bands in which no focusing occurs.
(See [Pe3]). However, using the construction outlined in the previous paragraph gives a very
different result. Since Sn is simply connected, it is its own universal cover. There is only one
point in our discrete set, and so the entire sphere Sn is in the first zone B1 .
The organization of this paper is as follows. In section 2, we set up the general machinery
we need, and prove the main theorems in the context of a discrete set S in a proper metric
space.
Section 3 explores this in the context of manifolds of constant curvature. The universal
cover is Rn , Sn , or Hn , and the group G of deck transformations is a discrete group of isometries
(see [doC]). The discrete set S is the orbit of a point not fixed by any element of G under this
discontinuous group. It is easy to see that the mediatrices in this case are totally geodesic
spaces. From the basic property explained above, one can deduce that the n-th Brillouin zone
is a fundamental domain for the group G in X.
In section 4, we calculate exactly the number of geodesics of length t that connect the
origin to itself in two cases: the flat torus R2 /Z2 and the Riemann surfaces H2 /Γ(p), for
p ∈ {2, 3, 5}. While these calculations could, of course, be done independent of our construction,
the Brillouin zones help visualize the process.
In the final section, we give a nontrivial example in the case of a non-Riemannian metric,
and mention a connection to the question of how many integer solutions there are to the equation
ak + bk = n, for fixed k.
Acknowledgments: It is a pleasure to acknowledge useful conversations with Federico Bonetto, Johann Dupont, Irwin Kra, Bernie Maskit, John Milnor, Chi-Han Sah, and Duncan
Sands. Part of this work was carried out while Peter Veerman was visiting the Center for
Physics and Biology at Rockefeller University and the Mathematics Department at SUNY
Stony Brook; the authors are grateful for the hospitality of these institutions.
2
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 d(·, ·)). We will make use the following notation:
• Write an open r-neighborhood of a point x0 as Nr (x0 ) = {x ∈ X d(x0 , x) < r}.
• Define the circumference as Cr (x0 ) = {x ∈ X d(x0 , x) = r}.
5
On Brillouin Zones
• Their union is the closed disk of radius r, denoted by Dr (x0 ) = {x ∈ X d(x0 , x) ≤ r}.
Definition 2.1 A metric space X is proper if the distance function d(x, ·) is a proper map for
every fixed x ∈ X. In particular, for every x ∈ X 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.
Definition 2.2 The space X is called metrically consistent if, for all x in X, R > r >
0 in R, and for each a ∈ CR (x), there is a z ∈ Cr (x) satisfying Nd(z,a) (z) ⊆ NR (x) and
Cd(z,a) (z) ∩ CR (x) = {a}.
This property is satisfied for any Riemannian metric.
Note that any mediatrix La,b separates X, that is: X − Lab contains at least two components (one containing the point a and the other b).
Definition 2.3 We say that the mediatrix Lab is minimally separating if for any subset L̃ ⊂ Lab
with L̃ 6= Lab , the set X − L̃ has one component.
If a mediatrix L it is minimally separating, then X − L has exactly two components. Note that
if a separating set L ⊆ X 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 à = A − (A ∩ V ) and
B̃ = B − (B ∩ V ). Now let x be any point in V . Then it is easy to see that the disjoint open sets
à ∪ V and B̃ cover X − (L − x). Thus L − x separates X, so L could not have been minimal.
We define the following sets
L−
0a = {x ∈ X d(0, x) − d(a, x) < 0} ,
L+
0a = {x ∈ X d(0, x) − d(a, x) > 0} .
Definition 2.4 Two minimally separating sets L0a and L0b are topologically transversal if they
+
are disjoint, or if for each x ∈ L0a ∩ L0b and every neighborhood V of x, the sets L+
0a ∩ L0b ∩ V ,
+
−
−
+
−
−
L0a ∩ L0b ∩ V , L0a ∩ L0b ∩ V , L0a ∩ L0b ∩ V are all nonempty.
Usually, there will be a discrete set of points S = {xi }i∈I 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 lim inf d(a, b) > 0, then S is discrete.
a,b∈S
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.
6
Veerman, Peixoto, Rocha, and Sutherland
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 2.1: The set L(0,0),(a,a) contains two
quarter-planes.
2
L√
0a for R with the Manhattan
Figure 2.2: L(0,0),(4,6) (thin solid line) and Figure 2.3: The mediatrices
L(0,0),(2,4) (thick grey line) are not transverse. metric and a in the lattice (m, n 2) .
Example 2.6 Equip R2 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 r 2 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 S contains
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 {(t, 1) t ≥ 4} (Fig. 2.2). But, if we
are careful, we can ensure that the space is Brillouin over S. To achieve this, if (0, 0) is the
basepoint, we must have that for all pairs (a1 , a2 ) and (b√
1 , b2 ) in S, a1 − a2 6= b1 − b2 . For
example, take S to be an irrational lattice such as (m, n 2) m, n ∈ Z . It is easy to check
that for all a, b ∈ S, 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 x0 ∈ S, 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
7
On Brillouin Zones
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 #(S) to denote the cardinality of the set S.
x0
x0
Figure 2.4: Here we illustrate the definition of the sets bn (x0 ) and Bn (x0 ) for the lattice Z2 in R2 . In both
pictures, the circle Cd(x,x0 ) (x) 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 # (Nr (x) ∩ S) = 4, while x0 is the only
point of S on the circle. On the right, we have m = 4 and ℓ = 8, so x lies in all of the sets B5 , B6 , . . . , B12 .
Definition 2.7 Let x ∈ X, let n be a positive integer, n ≤ #(S), and let r = d(x, x0 ). Then
define the sets bn (x0 ) and Bn (x0 ) as follows.
• x ∈ bn (x0 ) ⇐⇒ # (Nr (x) ∩ S) = n − 1
• x ∈ Bn (x0 ) ⇐⇒ # (Nr (x) ∩ S) = m
with m + 1 ≤ n ≤ m + ℓ.
and
and
Cr (x) ∩ S = {x0 }.
# (Cr (x) ∩ S) = ℓ,
where l, m ∈ Z+
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 ℓ = 1, then x ∈ bn (x0 ).
So bn (x0 ) ⊆ Bn (x0 ). 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 {Bn (x0 )}n>0 cover the space X. Thus we can assign to each point
x its Brillouin index as the the largest n for which x ∈ Bn (x0 ). 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.
8
Veerman, Peixoto, Rocha, and Sutherland
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:
[
Bi (xn ) = X
and
bi (xn ) ∩ bj (xn ) = ∅ if i 6= j.
i
Figure 2.5:
This example illustrates Lemma 2.9 and Thm. 2.10. Let S be the discrete set {(m, 0)} ∪
{(0, n)} , m, n ∈ Z 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 S = {xi }i∈I be a discrete
set. Then, for fixed n ≤ #(S), the sets {Bn0 (xi )}i∈I tile X in the following sense:
[
Bn (xi ) = X
and
bn (xi ) ∩ bn (xj ) = ∅ if i 6= j.
i
Proof: First, we show that for any fixed n > 0 and each x ∈ X, there is an xi ∈ S with
x ∈ Bn (xi ). Re-index S so that if S = {x1 , x2 , x3 , . . .} and i < j, then d(x, xi ) ≤ d(x, xj ). This
can be done; since S is a discrete subset and closed balls Dc (xi ) are compact, the subsets of S
with d(x, xi ) ≤ c are all finite. Let ri = d(x, xi ). We will show that x ∈ Bn (xn ).
Note that rn ≥ rn−1 . Suppose first that rn > rn−1 , then Nrn (x) ∩ S contains exactly
n − 1 points, and xn ∈ Crn (x) ∩ S. Thus x ∈ Bn (xn ). Note that if rn+1 > rn , then we would
have x ∈ bn (xn ) ⊂ Bn (xn ).
If, on the other hand, rn = rn−1 , then there is a k > 0 so that rn = rn−1 = . . . = rn−k ,
and so # (Nrn (x) ∩ S) = n − k − 1 ≤ n − 1. But then # (Crn (x) ∩ S) ≥ k + 1, and hence
x ∈ Bn (xn ) as desired.
9
On Brillouin Zones
For the second part, we show that bn (xi )∩bn (xj ) = ∅. If not, then there is a point x in their
intersection. If ri = rj , then xi = xj , because by the definition of bn (xk ), {xk } = Crk (x) ∩ S.
If not, then ri < rj . In this case, xi ∈ Dri (x) ⊂ Nrj (x) . Thus, since # (Nri (x) ∩ S) = n − 1,
Nrj (x) must contain at least n points of S, a contradiction.
The next result indicates how this notion of tiling is related to the notion of fundamental
domain.
Proposition 2.11 Let S be a discrete set in a metric space X as in theorem 2.10. Suppose
that for each xi in S there is an isometry gi of X such that gi (x0 ) = xi , gi permutes S and
the only gi which leaves x0 fixed is the identity. Then there is a set F (the fundamental set),
satisfying:
bn (x0 ) ⊆
[
i
gi (F ) = X
F
and
⊆ Bn (x0 )
with
gi (F ) ∩ gj (F ) = ∅ (i 6= j).
Proof: Suppose that x ∈ bn (x0 ). From definition 2.7 and the fact that the gi are isometries,
we see that this is equivalent to gi (x) ∈ bn (xi ). Thus gi (bn (x0 )) = bn (xi ). Now apply theorem
2.10. A similar reasoning proves the statement for Bn (x0 ).
Remark: The fundamental set F is not necessarily connected. Also, note that it follows from
this proposition that Bi (x0 ) is scissors congruent to Bj (x0 ) (see [Sah] for a discussion of scissors
congruence). In particular, this implies immediately that the Bi all have the same area. Note
that this result does not hold if S is not generated by a group of isometries. For example,
the “face-centered cubic” structures common to many crystals do not satisfy this, and the
corresponding Bn have different volumes, although they still tile the space. See also figure 2.5.
The preceding results are valid without the transversality conditions on the metric, as the
reader can check. We will now use the additional properties of the metric to derive a regularity
result.
Proposition 2.12 If X is Brillouin (over S), then Bn (xi ) is the closure of bn (xi ).
Proof: Let x ∈ Bn (x0 ). If x ∈ bn (x0 ), we are done. So suppose x 6∈ bn (x0 ), and then
by Def. 2.7 (with r = d(x, x0 )), Nr (x) contains m ≤ n − 1 points of S and Cr (x) contains
ℓ ≥ n − m with ℓ ≥ 2 points of S, including x0 . Thus x lies in the intersection of ℓ − 1 ≥ 1
ℓ−1
mediatrices {Lx0 ,xi }i=1
.
Suppose ℓ = 2, so x lies on a single mediatrix L1 . Then by condition 2, x is in the
+
−
+
′
′
′
closure of L−
1 and L1 . Thus if x ∈ L1 and close enough to x, then x ∈ bm+1 (x0 ). If x ∈ L1
and sufficiently close to x, then x′ ∈ bm+2 (x0 ). From our choices, we see that m = n − 1 or
m = n − 2. In either case, we are done.
The proof for ℓ > 2 is slightly more involved (we have to use transversality), but the idea
is the same. The point x lies in the intersection of at least two mediatrices. Essentially, we
10
Veerman, Peixoto, Rocha, and Sutherland
need show that the mediatrices disentangle in every neighborhood of x. We will prove that for
all j ∈ {1, . . . , ℓ}, x ∈ bm+j (x0 ). Note that n − ℓ ≤ m ≤ n − 1, and so showing this is sufficient.
Again, since mediatrices are minimally separating, we have that for all i ∈ {1, . . . , ℓ − 1}
+
there exist open sets L−
i and Li in any neighborhood of x for which
d(x, x0 ) < d(x, xi ) for x ∈ L−
i
and
d(x, x0 ) > d(x, xi ) for x ∈ L+
i .
Noting that xi ∈ Cr (x), then from condition 1 of Def. 2.7, we have that for any small ρT> 0,
there is a point z − ∈ Cρ (x) for which d(z − , x0 ) < d(z − , xi ) for all i ∈ {1, . . . , ℓ}. Thus i L−
i
is nonempty, and is contained in bm+1 (x0 ). By the same principle,
we
can
choose
a
point
T
z + ∈ Cr (x) for which d(z + , x0 ) > d(z + , xi ). As a consequence, i L+
i is nonempty, and is a
subset of bm+ℓ (x0 ).
Now, bm+1 (x0 ) and bm+ℓ (x0 ) are disjoint. If ℓ > 1, their complement in a neighborhood of
x must contain at least two mediatrices. By transversality (condition 3), these cannot coincide.
Thus V must be intersected by some bm+i (x0 ) for i ∈ {1, . . . ℓ}. By induction one finishes the
argument.
Remark: The references to mediatrices in the above proof may make it appear that we are
relying on the informal definition of Bn rather than Definition 2.7. In fact, we are only using
the fact that Bn \ bn consists of subsets of mediatrices, which was noted following Def. 2.7. In
fact, for Brillouin spaces, the informal definition of bn by mediatrices and Definition 2.7 are
equivalent, as the following proposition shows.
Proposition 2.13 Let X be Brillouin over S. Then there is a path γ from x to x0 which
crosses exactly n − 1 mediatrices with no multiple crossings if and only if x ∈ bn (x0 ). By “no
multiple crossings” we mean that if Li and Lj are distinct mediatrices, γ intersects Li in at
most a single point, and γ ∩ Li 6= γ ∩ Lj .
Proof: First, suppose there is such a path γ from x0 to x. As long as γ(t) crosses no mediatrices,
the number of elements of S contained in Dr (γ(t)) remains constant. If this number is i, then
γ(t) ∈ bi (x0 ). Each time γ(t) crosses a single mediatrix, this number increases by 1. Thus,
after crossing n − 1 mediatrices, γ(t) will be in bn (x0 ).
Now, let x ∈ bn (x0 ). We must construct a path from x to x0 which crosses exactly n − 1
mediatrices. Let b∗n denote the connected component of bn (x0 ) which contains x, and choose a
point y in Bn−1 (x0 ) ∩ ∂b∗n . Such a point always exists by Lemma 2.8 and Prop. 2.12. In fact,
because ∂bn consists of subsets of mediatrices, topological transversality of mediatrices allows
us to choose y so that it lies on a single mediatrix.
There certainly is a path connecting x to y which crosses no mediatrices; by similar
reasoning, we can choose x′ ∈ bn−1 (x0 ) and a path from x′ to y which crosses no mediatrices.
We have now constructed a path from x ∈ bn to x′ ∈ bn−1 which crosses exactly one mediatrix—
repeating the argument n − 2 times finishes the proof.
On Brillouin Zones
3
11
Brillouin zones in spaces of constant curvature
In this section X will be assumed to be one of Rn , Sn , or Hn , all equipped with the standard
metric, and let G be a discontinuous group of isometries of X. Denote the quotient X/G with
the induced metric by (M, g). Then the construction of lifting to the universal cover, as outlined
in the introduction, applies naturally to (M, g). In this section we describe focusing of geodesics
in (M, g) by Brillouin zones in X. The discrete set S is given by the orbit of a chosen point in
X (which we will call the origin) under the group of deck-transformations G. The fact that the
Brillouin zones are fundamental domains is now a direct corollary of proposition 2.11.
The regularity conditions of Def. 2.5 are easily verified in the present context. We do this
first.
Lemma 3.1 If X is either Rn , Sn , or Hn , then a mediatrix Lab in X is an (n−1)-dimensional,
totally geodesic subspace consisting of one component, and X − Lab has two components.
Proof: This is easy to see if we change coordinates by an isometry of X, putting a and b
in a convenient position, say as x and −x. The mediatrix Lx,−x is easily seen to satisfy the
conditions (in the case of Sn , it is the equator, and for the others, it is a hyperplane). The
conclusion follows.
Proposition 3.2 All such spaces X are Brillouin (see definition 2.5).
Proof: As remarked before, the first condition is satisfied for any Riemannian metric. The
second condition is also easy. It suffices to observe that the subspaces of Lemma 3.1 are
minimally separating.
To prove the topological transversality condition, note that lemma 3.1 implies that if L0a
and L0b coincide in some open set, then they are equal. One easily sees that then we must have
L0a = L0b = Lab = L. Applying the second part of the lemma to L0a , L0b , and Lab , we see that
they must separate X in at least 3 components, namely one for each point a, b, and 0. This
leads to a contradiction.
Remark: Note that in fact, the mediatrices are transversal in the usual sense. If L0a and L0b
coincide in an open set, then their tangent spaces also coincide at some point. Uniqueness of
solutions of second order differential equations then implies L0a = L0b .
Recall that a metric space X is called rigid if the only isometry which fixes each point
of a nonempty open subset of X is the identity. It is not hard to see that Sn , Hn , and Rn are
rigid spaces. See [Ra] for more details of rigid metric spaces and for the proof of the following
result. Recall that the stabilizer in G of a point x ∈ X consists of those elements of G that fix
x.
Proposition 3.3 Let G be a discontinuous group of isometries of a rigid metric space X. Then
there exists a point y of X whose stabilizer Gy is consists of the identity.
12
Veerman, Peixoto, Rocha, and Sutherland
We now return to Brillouin zones as defined in the last section. Recall that G is a group
of isometries of X that acts discontinuously on points in X. Let x0 be a point in X whose
stabilizer under the action of G is trivial. Let x ∈ X be a point in the orbit of x0 under G, that
is to say, x = γ(x0 ) for some γ ∈ G − {e}, where e is the identity of G. Let [x0 , x] be a geodesic
segment of minimal length whose endpoints are x0 and x. Then Bn (x0 ), the n-th Brillouin
zone relative to x0 , is the set of points in X such that the segment [x0 , x] intercepts exactly
n − 1 mediatrices Lx0 ,y , where y is in the orbit of x0 under the group G. Proposition 2.11
immediately implies the most important fact about Brillouin zones in this setting.
Figure 3.1:
Brillouin zones for P SL(2, Z) in the hyperbolic disk. We have transported the “usual” upper
half-plane representation using the map z 7→ iz+1
. On the left is shown the sets Bn ( 4i ), which give fundamental
z+i
domains as in Cor. 3.4. On the right, 0 is taken as a basepoint. Since the origin has a non-trivial stabilizer, the
corresponding Brillouin zones give a double cover of the fundamental domains.
Corollary 3.4 Let X be Rn , Sn , or Hn , and let G be a discontinuous group of isometries of
X. Let x0 ∈ X be such that its stabilizer Gx0 under G is trivial. Then for every positive integer
n, the n-th Brillouin zone Bn (x0 ) is a fundamental domain for the action of G on points in X.
Its boundary is the union of pieces of totally geodesic subspaces and equals the boundary of its
interior.
Remark: The first Brillouin zone B1 (0) is the usual Dirichlet fundamental domain for the action
of G. Furthermore, even when Gx0 is not trivial, Bn (x0 ) is a k-fold cover of a fundamental
domain.
As pointed out in the introduction, the number of geodesics that focus in a certain point
is counted in the lift. So if a given point x ∈ X is intersected by n mediatrices, it is reached by
n + 1 geodesics of length d(0, x) emanating from the reference point (the origin). In the next
section, we give more specific examples of this.
Finally, we state a conjecture.
13
On Brillouin Zones
Conjecture 3.5 Let (X, g̃) be the universal cover of a d-dimensional smooth Riemannian manifold (M, g) as described in the construction. For a generic metric g on M , no more than d
mediatrices intersect in any given point y of X.
This conjecture acquires perhaps even more interest (and certainly more structure), when
one restricts the collection of metrics on M to conformal ones ([Mas]). A result in this direction
for M = R2 /Z2 can be found in [Jo1].
4
Focusing in two Riemannian examples
In this section, we give two examples (one of them new as far as we know) of focusing. Suppose
that at t = 0 geodesics start emanating in all possible directions from the a point. At certain
times t1 , t2 , ...., we may see geodesics returning to that point. We derive expressions for the
number of geodesics returning at tn in two cases. First, as introductory example we will discuss
this for M = R2 /Z2 (a more complete discussion of this example can be found in [Pe3]).
Second, we will deal with a much more unusual example, namely M = H2 /Γ(k), where Γ(k) is
a subgroup of P SL(2, Z) called the principal congruence subgroup of level k (defined in more
detail below). We note that it seems to be considerably harder to count geodesics that focus in
points other than our basepoint.
Before continuing, consider the classical problem of counting Rg (n), the number of solutions in Z2 of
p2 + q 2 = n.
Let
n = 2α
k
Y
pβi i
i=1
ℓ
Y
qiγi
j=1
be the prime decomposition of the number n, where pi ≡ 1(mod 4) and qi ≡ 3(mod 4). The
following classical result of Gauss (see [NZM]) will be very useful.
Lemma 4.1 Rg (n) is zero whenever n is not an integer, or any of the γi is odd. Otherwise,
Rg (n) = 4
k
Y
(1 + βi ).
i=1
Example 4.2 Choose an origin in M = R2 /Z2 and lift it to the origin in R2 . Our discrete set
S is then Z2 . Let ρx (t) be the number of geodesics of length t that connect the origin to the
point x ∈ M .
Proposition 4.3 In the flat torus, the number of geodesics of length t that connect the the
origin to itself is given by
Rg (t2 ), if t2 ∈ N
ρ0 (t) =
0,
otherwise.
14
Veerman, Peixoto, Rocha, and Sutherland
Proof: Notice that by definition geodesics of length t leaving from the origin in R2 reach the
points contained in Ct (0). Only if t2 is an integer does this circle intersect points in Z2 .
Example 4.4 We now turn to the next example. Recall that P SL(2, Z) can be identified
with the group of two by two matrices with integer entries and determinant one, and with
multiplication by −1 as equivalence. For each k, the group Γ(k) is the subgroup of P SL(2, Z)
given by
Γ(k) =
a b
c d
∈ P SL(2, Z) a ≡ d ≡ 1 (mod k), b ≡ c ≡ 0 (mod k), .
This group has important applications in number theory. The action of Γ(k) on H2 is given by
the Möbius transformations
az + b
a b
g(z) =
where
∈ Γ(k).
c d
cz + d
We point out that for k = 2, 3, or 5, the surface H2 /Γ(k) is a sphere with 3, 4, or 12 punctures
(see [FK]).
We will find it more convenient to work in the hyperbolic disk D2 , which is the universal
cover of H2 /Γ(k). We shall choose a representation of Γ(k) in the disk so that i ∈ H2 corresponds
to the origin. This will allow us to determine the focusing of the geodesics which emanate from
i. Note that the surface H2 /Γ(k) has special symmetries with respect to i: for example, i is the
unique point fixed by the order 2 element of P SL(2, Z).
Figure 4.1: The orbit of i under Γ(2) transported to the hyperbolic disk, and the corresponding Brillouin
zones. Each zone Bn forms a fundamental domain for a 3-punctured sphere.
15
On Brillouin Zones
Lemma 4.5 The action of the fundamental group of the surface H2 /Γ(k) can be represented
as
r − is p + iq
r + p ≡ 1 (mod k), r − p ≡ 1 (mod k)
2
2
2
2
and p + q + 1 = r + s ,
p − iq r + is
s + q ≡ 0 (mod k), s − q ≡ 0 (mod k)
acting on D2 . We shall denote this particular representation as the group Γ⊙ (k).
Proof: Following the conventions in [Be], define
φ : D2 → H2 ,
φ(z) = i
z+1
.
−z + 1
Push back the transformation g ∈ Γ(k) from H2 to D2 by g → φ−1 gφ to obtain a representation
of g ∈ Γ(k) as a transformation acting on D2 . The matrix representation of this transformation
is given by:
a+d
b−c
a−d
b+c
2 +i 2
2 −i 2
,
Ag =
b+c
a+d
b−c
a−d
2 +i 2
2 −i 2
where det Ag = 1, since this matrix is conjugate to g, whose determinant is equal to 1. Let
p = (a − d)/2
r = (a + d)/2
q = −(b + c)/2
s = −(b − c)/2
and Ag now written as
Ag =
r − is p + iq
p − iq r + is
.
Here the numbers p, q, r, s are in Z and must satisfy the following congruence conditions:
r + p ≡ 1 (mod k), r − p ≡ 1 (mod k)
s + q ≡ 0 (mod k), s − q ≡ 0 (mod k)
Since the determinant of Ag is equal to 1, we must also have
p2 + q 2 + 1 = r 2 + s 2 .
We need another auxiliary result before we state the main result of this section.
Lemma 4.6 Let (p, q) and (r, s) be two points in Z2 such that the integers A = p2 + q 2 and
B = r 2 + s2 are relatively prime, and let ϕ be a rotation. Now ϕ(p, q) = (p′ , q ′ ) and ϕ(r, s) =
(r ′ , s′ ) are in Z2 if and only if ϕ is a rotation by a multiple of π/2.
16
Veerman, Peixoto, Rocha, and Sutherland
Proof: Let c be the cosine of the angle of rotation. We have
c=
p′ p + q ′ q
r ′ r + s′ s
=
.
A
B
Thus if p′ p + q ′ q and r ′ r + s′ s are not both equal to zero,
A
p′ p + q ′ q
= .
′
′
rr+ss
B
Because A and B are relatively prime and surely |p′ p + q ′ q| is less than or equal to A, and
similarly for B, we have that
p′ p + q ′ q = ±A and r ′ r + s′ s = ±B.
This implies the result.
Now we define a counter just as before. Choose a lift of M = D2 /Γ⊙ (k) so that 0 ∈ M
lifts to 0 ∈ D2 . Let γx (t) be the number of geodesics of length t that connect the origin to the
point x ∈ M .
512
512
256
256
128
128
64
64
32
32
16
16
8
8
4
4
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
0 1 2 3 4 5 6 7 8
Figure 4.2: The non-zero values of ρ0 (t) for t ≤ 25 (left) and γ0 (t) for t ≤ 8 (right), which count how many
geodesics of length t connect the origin to itself in the R2 /Z2 and D2 /Γ⊙ (2), respectively.
Theorem 4.7 In the surface H2 /Γ(k), the number of geodesics which connect the point i ∈ H2
to itself is given by
2
1
t − 1)Rg (cosh2 t)
for k = 2
4 Rg (cosh
2
1
cosh2 t−1
Rg (cosh t)
for k = 3
4 Rg
9
cosh2 t−1
1
Rg (cosh2 t)
for k = 5
4 Rg
25
Note that in all cases, the number is nonzero only if cosh2 t ∈ N.
17
On Brillouin Zones
Proof: We shall work in the disk, rather than in H2 . Let S be the orbit of 0 ∈ D2 under Γ⊙ (k).
Then the number of such geodesics is exactly the number of distinct points of S which lie on
on the circle Ct (0) of radius t and centered at the origin.
If x ∈ S, then by Lemma 4.5, it is of the form p+iq
r+is with p, q, r, s integers satisfying
2
2
2
2
p + q = r + s − 1. Let n be their common value, that is,
n = p2 + q 2 = r 2 + s2 − 1.
We will first count the number of 4-tuples (p, q, r, s) that solve this equation, momentarily
ignoring the congruence conditions.
Note that the point x has Euclidean distance to the origin given by
|x|2e =
p2 + q 2
n
.
=
r 2 + s2
n+1
The hyperbolic length of the geodesic connecting the origin to this point is
t = arctanh (|x|e ).
q
n
, or, equivalently, n = cosh2 t − 1.
This gives that γ0 (t) is only non-zero when t = arctanh n+1
To count the number of intersections of Ct (0) with S for these values of t, observe that
we can use Gauss’ result to count the number of pairs (p, q) such that p2 + q 2 = n. This number
is given by Rg (n). For each such pair (p, q), we have a number of choices to form
x=
p + iq
r + is
By the above, this number is equal to Rg (n + 1). Thus, γ0 (t) is at most Rg (n)Rg (n + 1).
However, we have over-counted: some of our choices for p, q, r, s represent the same point
x ∈ S, and some of them may not satisfy the congruence conditions, which we have so far
ignored. We will first account for the multiple representations, and then account for the congruence relations.
q
n
Let p, q, r, s ∈ Z be as above, giving a point x = p+iq
at
distance
t
=
arctanh
r+is
n+1
from the origin. If we multiply the numerator and denominator of x by eiθ , then x will remain
unchanged. Because of the requirement that p2 + q 2 = r 2 + s2 − 1 = n, this is the only invariant,
and by Lemma 4.6, θ must be a multiple of π2 for the numerator and denominator to remain
Gaussian integers. We see that in our counting, we have represented our point x in 4 different
ways:
−q + ip
−p − iq
q − ip
p + iq
=
=
=
x=
r + is
−s + ir
−r − is
s − ir
meaning we have over-counted by a factor of at least 4.
Now we account for the congruence conditions.
First, consider the case k = 2. Note that q + s ≡ 0 (mod 2) if and only if p + r ≡
1 (mod 2), because p2 + q 2 + 1 = r 2 + s2 , so we need only check this one condition. If the
representation p+iq
r+is fails to satisfy our parity condition, then q+s ≡ 1 ( mod 2) and consequently
18
Veerman, Peixoto, Rocha, and Sutherland
p + r ≡ 0 ( mod 2). This means that the representation −q+ip
−s+ir of this same point does satisfy the
2
1
parity conditions, giving exactly 4 Rg (cosh t − 1)Rg (cosh2 t) distinct points of S at distance t
from the origin.
If k = 3, then since k is odd, the congruence conditions on p, q, r, and s imply that
r ≡ 1(mod 3)
and
p ≡ q ≡ s ≡ 0(mod 3).
Note that the equation
p2 + q 2 = n
and
p ≡ q ≡ 0 (mod 3)
will be satisfied exactly Rg (n/32 ) times. (Recall that if n/9 is not an integer, Rg (n/9) = 0.)
For fixed n, let (p, q) be any one of the solutions. We need to decide how many solutions
the equation
r 2 + s2 = n + 1
with
r ≡ 1 (mod 3)
and
s ≡ 0 (mod 3)
admits. The solution of the first equation implies that 3 divides n. Thus r 2 + s2 ≡ 1 (mod 3).
Consequently, we have 4 choices mod 3 for the pair (r, s), namely (0, 1), (1, 0), (0, 2), and (2, 0).
Let (p, q, r, s) ∈ Z2 ×Z2 be any solution to n = p2 +q 2 = r 2 +s2 −1 with p ≡ q ≡ 0( mod 3).
For each choice of (p, q), we have exactly Rg (n+1) choices of (r, s). Now let R denote the product
of the rotations by π/2 on each of the components of Z2 × Z2 . Using Lemma 4.6, we see that all
such solutions can be obtained from just one by applying R repeatedly. It is easy to check that
each quadruple of solutions thus constructed runs exactly once through the above list. Since
precisely one out of the four associated solutions is compatible with the conditions, the total
number of solutions is exactly:
1 n
Rg (n + 1).
Rg
4
9
Using the relationship between the Euclidean distance and the Poincaré length as before gives
the result.
If k = 5, the proof for k = 3 can be literally transcribed to obtain the result.
Remark: Note that the above results do not hold if k is not one of the cases mentioned. The
primary difficulty is that for prime k ≥ 7, there are solutions which are not related by applying
the rotation R. However, the argument does give 14 Rg (cosh2 t − 1)/k2 Rg (cosh2 t) as an upper
bound for H2 /Γ(k) when k is an odd prime. Note that the surface H2 /Γ(k) is of genus 0 if and
only if k ≤ 5 (see [FK]).
5
Non-Riemannian examples
The present context is certainly not restricted to Riemannian metrics. As an indicator of this
we now discuss a different set of examples.
19
On Brillouin Zones
Let k be a positive number greater than one. Equip R2 with the distance function
1/k
k x − y k= |x1 − y1 |k + |x2 − y2 |k
and let the discrete set S be given by Z2 . For k not equal to 2, this is not a Riemannian metric,
yet all conclusions of section 2 hold. In particular, each Brillouin zone forms a fundamental
domain. Note that determining the zones by inspecting the picture requires close attention!
10
2
5
1
y
0
y
0
-1
-5
-2
-10
-2
-1
0
1
2
-10
-5
0
5
10
x
x
Brillouin zones for the lattice Z2 in R2 with the metric |x1 − y1 |4 + |x2 − y2 |4
Figure 2.3 and Example 2.6, which deal with the case k = 1, the “Manhattan metric”.
Figure 5.1:
1/4
. See also
Now the problem of determining Ct (0) ∩ S for any given t is unsolved for general k. In
fact, even for certain integer values of k greater than 2, it is not known whether Ct (0) ∩ S ever
contains at least two points that are not related by the symmetries of the problem. For k = 4,
the smallest t for which Ct (0) ∩ S has at least two (unrelated) solutions is given by
t4 = 1334 + 1344 = 1584 + 594 .
However, for k ≥ 5, it unknown whether this can happen at all (see [SW]).
There are some things that can be said, however. In the situation where k ∈ {3, 4, 5, . . .},
the mediatrices intersect the coordinate axes only in irrational points or in multiples of 1/2.
For if x = (p/q, 0) is a point of a mediatrix L(0,0),(a1 ,a2 ) , we have
|p|k = |p − qa1 |k + |a2 q|k
(p 6= 0, q 6= 0)
.
By Fermat’s Last Theorem, this has no solution unless either p = qa1 or a2 = 0. In the first
case, pq = ±a2 , which can only occur if the lattice point is of the form (a2 , ±a2 ). If a2 = 0, then
p
a1
q = 2 . In particular, there is no nontrivial focusing along the axes.
20
Veerman, Peixoto, Rocha, and Sutherland
To compute Figure 5.1, we took advantage of the smoothness of the metric. Not all
metrics are sufficiently smooth for this procedure to work. Even for Riemannian metrics, in
general the distance function is only Lipschitz, which will not be sufficiently smooth.
For each a = (a1 , a2 ) ∈ Z2 , define a Hamiltonian:
Ha (x) =k x − a k − k x k .
The mediatrix L0a corresponds to the level set Ha (x) = 0. Because Ha (x) is smooth, we have
uniqueness of solutions to Hamilton’s equations. In the current situation, where the dimension
is two, the level set consists of one orbit. Thus, one can produce the mediatrix by numerically
tracing the zero energy orbits of the above Hamiltonian.
As mentioned above, for a general Riemannian metric, the distance function is only
Lipschitz. This means we have no guarantee that the solutions of the above differential equation
are unique. Indeed, there are examples of multiply connected Riemannian manifolds with selfintersecting mediatrices, as will be shown in a forthcoming work.
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