3Triangulations and the Euler number
IB Geometry
3 Triangulations and the Euler number
We shall now study the idea of triangulations and the Euler number. We aren’t
going to do much with them in this course — we will not even prove that the
Euler number is a well-defined number. However, we need Euler numbers in order
to state the full Gauss-Bonnet theorem at the end, and the idea of triangulations
is useful in the IID Algebraic Topology course for defining simplicial homology.
More importantly, the discussion of these concepts is required by the schedules.
Hence we will get some exposure to these concepts in this chapter.
It is convenient to have an example other than the sphere when discussing
triangulations and Euler numbers. So we introduce the torus
Definition ((Euclidean) torus). The (Euclidean) torus is the set
R
2
/Z
2
of
equivalence classes of (x, y) ∈ R
2
under the equivalence relation
(x
1
, y
1
) ∼ (x
2
, y
2
) ⇔ x
1
− x
2
, y
1
− y
2
∈ Z.
It is easy to see this is indeed an equivalence relation. Thus a point in
T
represented by (x, y) is a coset (x, y) + Z
2
of the subgroup Z
2
≤ R
2
.
Of course, there are many ways to represent a point in the torus. However, for
any closed square
Q ⊆ R
2
with side length 1, we can obtain
T
is by identifying
the sides
Q
We can define a distance d for all P
1
, P
2
∈ T to be
d(P
1
, P
2
) = min{kv
1
− v
2
k : v
i
∈ R
2
, v
i
+ Z
2
= P
i
}.
It is not hard to show this definition makes (
T, d
) into a metric space. This
allows us to talk about things like open sets and continuous functions. We will
later show that this is not just a metric space, but a smooth surface, after we
have defined what that means.
We write
˚
Q
for the interior of
Q
. Then the natural map
f
:
˚
Q → T
given
by v
7→
v +
Z
2
is a bijection onto some open set
U ⊆ T
. In fact,
U
is just
T \ {two circles meeting in 1 point}
, where the two circles are the boundary of
the square Q.
Now given any point in the torus represented by
P
+
Z
2
, we can find a square
Q
such that
P ∈
˚
Q
. Then
f
:
˚
Q → T
restricted to an open disk about
P
is an
isometry (onto the image, a subset of
R
2
). Thus we say
d
is a locally Euclidean
metric.
One can also think of the torus
T
as the surface of a doughnut, “embedded”
in Euclidean space R
3
.
Given this, it is natural to define the distance between two points to be the
length of the shortest curve between them on the torus. However, this distance
function is not the same as what we have here. So it is misleading to think of
our locally Euclidean torus as a “doughnut”.
With one more example in our toolkit, we can start doing what we really
want to do. The idea of a triangulation is to cut a space
X
up into many smaller
triangles, since we like triangles. However, we would like to first define what a
triangle is.
Definition (Topological triangle). A topological triangle on
X
=
S
2
or
T
(or
any metric space
X
) is a subset
R ⊆ X
equipped with a homeomorphism
R →
∆,
where ∆ is a closed Euclidean triangle in R
2
.
Note that a spherical triangle is in fact a topological triangle, using the radial
projection to the plane R
2
from the center of the sphere.
Definition (Topological triangulation). A topological triangulation
τ
of a metric
space
X
is a finite collection of topological triangles of
X
which cover
X
such
that
(i)
For every pair of triangles in
τ
, either they are disjoint, or they meet in
exactly one edge, or meet at exactly one vertex.
(ii) Each edge belongs to exactly two triangles.
These notions are useful only if the space
X
is “two dimensional” — there is
no way we can triangulate, say
R
3
, or a line. We can generalize triangulation to
allow higher dimensional “triangles”, namely topological tetrahedrons, and in
general,
n
-simplices, and make an analogous definition of triangulation. However,
we will not bother ourselves with this.
Definition (Euler number). The Euler number of a triangulation
e
=
e
(
X, τ
) is
e = F − E + V,
where
F
is the number of triangles;
E
is the number of edges; and
V
is the
number of vertices.
Note that each edge actually belongs to two triangles, but we will only count
it once.
There is one important fact about triangulations from algebraic topology,
which we will state without proof.
Theorem. The Euler number e is independent of the choice of triangulation.
So the Euler number
e
=
e
(
X
) is a property of the space
X
itself, not a
particular triangulation.
Example. Consider the following triangulation of the sphere:
This has 8 faces, 12 edges and 6 vertices. So e = 2.
Example. Consider the following triangulation of the torus. Be careful not
to double count the edges and vertices at the sides, since the sides are glued
together.
This has 18 faces, 27 edges and 9 vertices. So e = 0.
In both cases, we did not cut up our space with funny, squiggly lines. Instead,
we used “straight” lines. These triangles are known as geodesic triangles.
Definition (Geodesic triangle). A geodesic triangle is a triangle whose sides
are geodesics, i.e. paths of shortest distance between two points.
In particular, we used spherical triangles in
S
2
and Euclidean triangles in
˚
Q
.
Triangulations made of geodesic triangles are rather nice. They are so nice that
we can actually prove something about them!
Proposition. For every geodesic triangulation of
S
2
(and respectively
T
) has
e = 2 (respectively, e = 0).
Of course, we know this is true for any triangulation, but it is difficult to
prove that without algebraic topology.
Proof.
For any triangulation
τ
, we denote the “faces” of ∆
1
, ··· ,
∆
F
, and write
τ
i
=
α
i
+
β
i
+
γ
i
for the sum of the interior angles of the triangles (with
i = 1, ··· , F ).
Then we have
X
τ
i
= 2πV,
since the total angle around each vertex is 2
π
. Also, each triangle has three
edges, and each edge is from two triangles. So 3
F
= 2
E
. We write this in a
more convenient form:
F = 2E − 2F.
How we continue depends on whether we are on the sphere or the torus.
–
For the sphere, Gauss-Bonnet for the sphere says the area of ∆
i
is
τ
i
− π
.
Since the area of the sphere is 4π, we know
4π =
X
area(∆
i
)
=
X
(τ
i
− π)
= 2πV −F π
= 2πV −(2E − 2F )π
= 2π(F − E + V ).
So F − E + V = 2.
– For the torus, we have τ
i
= π for every face in
˚
Q. So
2πV =
X
τ
i
= πF.
So
2V = F = 2E − 2F.
So we get
2(F − V + E) = 0,
as required.
Note that in the definition of triangulation, we decomposed
X
into topological
triangles. We can also use decompositions by topological polygons, but they are
slightly more complicated, since we have to worry about convexity. However,
apart from this, everything works out well. In particular, the previous proposition
also holds, and we have Euler’s formula for
S
2
:
V −E
+
F
= 2 for any polygonal
decomposition of
S
2
. This is not hard to prove, and is left as an exercise on the
example sheet.