Part IB — Geometry
Based on lectures by A. G. Kovalev
Notes taken by Dexter Chua
Lent 2016
These notes are not endorsed by the lecturers, and I have modified them (often
significantly) after lectures. They are nowhere near accurate representations of what
was actually lectured, and in particular, all errors are almost surely mine.
Parts of Analysis II will be found useful for this course.
Groups of rigid motions of Euclidean space. Rotation and reflection groups in two and
three dimensions. Lengths of curves. [2]
Spherical geometry: spherical lines, spherical triangles and the Gauss-Bonnet theorem.
Stereographic projection and M¨obius transformations. [3]
Triangulations of the sphere and the torus, Euler number. [1]
Riemannian metrics on open subsets of the plane. The hyperbolic plane. Poincar´e
models and their metrics. The isometry group. Hyperbolic triangles and the Gauss-
Bonnet theorem. The hyperboloid model. [4]
Embedded surfaces in
R
3
. The first fundamental form. Length and area. Examples. [1]
Length and energy. Geodesics for general Riemannian metrics as stationary points of
the energy. First variation of the energy and geodesics as solutions of the corresponding
Euler-Lagrange equations. Geodesic polar coordinates (informal proof of existence).
Surfaces of revolution. [2]
The second fundamental form and Gaussian curvature. For metrics of the form
du
2
+
G
(
u, v
)
dv
2
, expression of the curvature as
√
G
uu
/
√
G
. Abstract smooth surfaces
and isometries. Euler numbers and statement of Gauss-Bonnet theorem, examples and
applications. [3]
Contents
0 Introduction
1 Euclidean geometry
1.1 Isometries of the Euclidean plane
1.2 Curves in R
n
2 Spherical geometry
2.1 Triangles on a sphere
2.2 M¨obius geometry
3 Triangulations and the Euler number
4 Hyperbolic geometry
4.1 Review of derivatives and chain rule
4.2 Riemannian metrics
4.3 Two models for the hyperbolic plane
4.4 Geometry of the hyperbolic disk
4.5 Hyperbolic triangles
4.6 Hyperboloid model
5 Smooth embedded surfaces (in R
3
)
5.1 Smooth embedded surfaces
5.2 Geodesics
5.3 Surfaces of revolution
5.4 Gaussian curvature
6 Abstract smooth surfaces
0 Introduction
In the very beginning, Euclid came up with the axioms of geometry, one of
which is the parallel postulate. This says that given any point
P
and a line
`
not containing
P
, there is a line through
P
that does not intersect
`
. Unlike
the other axioms Euclid had, this was not seen as “obvious”. For many years,
geometers tried hard to prove this axiom from the others, but failed.
Eventually, people realized that this axiom cannot be proved from the others.
There exists some other “geometries” in which the other axioms hold, but the
parallel postulate fails. This was known as hyperbolic geometry. Together with
Euclidean geometry and spherical geometry (which is the geometry of the surface
of a sphere), these constitute the three classical geometries. We will study these
geometries in detail, and see that they actually obey many similar properties,
while being strikingly different in other respects.
That is not the end. There is no reason why we have to restrict ourselves
to these three types of geometry. In later parts of the course, we will massively
generalize the notions we began with and eventually define an abstract smooth
surface. This covers all three classical geometries, and many more!
1 Euclidean geometry
We are first going to look at Euclidean geometry. Roughly speaking, this is the
geometry of the familiar
R
n
under the usual inner product. There really isn’t
much to say, since we are already quite familiar with Euclidean geometry. We
will quickly look at isometries of
R
n
and curves in
R
n
. Afterwards, we will try
to develop analogous notions in other more complicated geometries.
1.1 Isometries of the Euclidean plane
The purpose of this section is to study maps on R
n
that preserve distances, i.e.
isometries of
R
n
. Before we begin, we define the notion of distance on
R
n
in the
usual way.
Definition ((Standard) inner product). The (standard) inner product on
R
n
is
defined by
(x, y) = x · y =
n
X
i=1
x
i
y
i
.
Definition (Euclidean Norm). The Euclidean norm of x ∈ R
n
is
kxk =
p
(x, x).
This defines a metric on R
n
by
d(x, y) = kx − yk.
Note that the inner product and the norm both depend on our choice of
origin, but the distance does not. In general, we don’t like having a choice of
origin — choosing the origin is just to provide a (very) convenient way labelling
points. The origin should not be a special point (in theory). In fancy language,
we say we view R
n
as an affine space instead of a vector space.
Definition (Isometry). A map f : R
n
→ R
n
is an isometry of R
n
if
d(f(x), f(y)) = d(x, y)
for all x, y ∈ R
n
.
Note that
f
is not required to be linear. This is since we are viewing
R
n
as an affine space, and linearity only makes sense if we have a specified point
as the origin. Nevertheless, we will still view the linear isometries as “special”
isometries, since they are more convenient to work with, despite not being special
fundamentally.
Our current objective is to classify all isometries of
R
n
. We start with the
linear isometries. Recall the following definition:
Definition (Orthogonal matrix). An
n × n
matrix
A
is orthogonal if
AA
T
=
A
T
A = I. The group of all orthogonal matrices is the orthogonal group O(n).
In general, for any matrix A and x, y ∈ R
n
, we get
(Ax, Ay) = (Ax)
T
(Ay) = x
T
A
T
Ay = (x, A
T
Ay).
So A is orthogonal if and only if (Ax, Ay) = (x, y) for all x, y ∈ R
n
.
Recall also that the inner product can be expressed in terms of the norm by
(x, y) =
1
2
(kx + yk
2
− kxk
2
− kyk
2
).
So if
A
preserves norm, then it preserves the inner product, and the converse
is obviously true. So
A
is orthogonal if and only if
kA
x
k
=
k
x
k
for all x
∈ R
n
.
Hence matrices are orthogonal if and only if they are isometries.
More generally, let
f(x) = Ax + b.
Then
d(f(x), f(y)) = kA(x − y)k.
So any
f
of this form is an isometry if and only if
A
is orthogonal. This is not
too surprising. What might not be expected is that all isometries are of this
form.
Theorem. Every isometry of f : R
n
→ R
n
is of the form
f(x) = Ax + b.
for A orthogonal and b ∈ R
n
.
Proof. Let f be an isometry. Let e
1
, ··· , e
n
be the standard basis of R
n
. Let
b = f(0), a
i
= f(e
i
) − b.
The idea is to construct our matrix
A
out of these a
i
. For
A
to be orthogonal,
{a
i
} must be an orthonormal basis.
Indeed, we can compute
ka
i
k = kf(e
i
) − f(0)k = d(f(e
i
), f(0)) = d(e
i
, 0) = ke
i
k = 1.
For i 6= j, we have
(a
i
, a
j
) = −(a
i
, −a
j
)
= −
1
2
(ka
i
− a
j
k
2
− ka
i
k
2
− ka
j
k
2
)
= −
1
2
(kf(e
i
) − f(e
j
)k
2
− 2)
= −
1
2
(ke
i
− e
j
k
2
− 2)
= 0
So a
i
and a
j
are orthogonal. In other words,
{
a
i
}
forms an orthonormal set. It
is an easy result that any orthogonal set must be linearly independent. Since we
have found n orthonormal vectors, they form an orthonormal basis.
Hence, the matrix
A
with columns given by the column vectors a
i
is an
orthogonal matrix. We define a new isometry
g(x) = Ax + b.
We want to show
f
=
g
. By construction, we know
g
(x) =
f
(x) is true for
x = 0, e
1
, ··· , e
n
.
We observe that g is invertible. In particular,
g
−1
(x) = A
−1
(x − b) = A
T
x − A
T
b.
Moreover, it is an isometry, since
A
T
is orthogonal (or we can appeal to the
more general fact that inverses of isometries are isometries).
We define
h = g
−1
◦ f.
Since it is a composition of isometries, it is also an isometry. Moreover, it fixes
x = 0, e
1
, ··· , e
n
.
It currently suffices to prove that h is the identity.
Let x ∈ R
n
, and expand it in the basis as
x =
n
X
i=1
x
i
e
i
.
Let
y = h(x) =
n
X
i=1
y
i
e
i
.
We can compute
d(x, e
i
)
2
= (x − e
i
, x −e
i
) = kxk
2
+ 1 − 2x
i
d(x, 0)
2
= kxk
2
.
Similarly, we have
d(y, e
i
)
2
= (y − e
i
, y − e
i
) = kyk
2
+ 1 − 2y
i
d(y, 0)
2
= kyk
2
.
Since
h
is an isometry and fixes 0
,
e
1
, ··· ,
e
n
, and by definition
h
(x) = y, we
must have
d(x, 0) = d(y, 0), d(x, e
i
) = d(y, e
i
).
The first equality gives
k
x
k
2
=
k
y
k
2
, and the others then imply
x
i
=
y
i
for all
i
.
In other words, x = y = h(x). So h is the identity.
We now collect all our isometries into a group, for convenience.
Definition (Isometry group). The isometry group
Isom
(
R
n
) is the group of all
isometries of R
n
, which is a group by composition.
Example (Reflections in an affine hyperplane). Let
H ⊆ R
n
be an affine
hyperplane given by
H = {x ∈ R
n
: u · x = c},
where
k
u
k
= 1 and
c ∈ R
. This is just a natural generalization of a 2-dimensional
plane in
R
3
. Note that unlike a vector subspace, it does not have to contain the
origin (since the origin is not a special point).
Reflection in H, written R
H
, is the map
R
H
: R
n
→ R
n
x 7→ x − 2(x · u − c)u
It is an exercise in the example sheet to show that this is indeed an isometry.
We now check this is indeed what we think a reflection should be. Note that
every point in
R
n
can be written as a +
t
u, where a
∈ H
. Then the reflection
should send this point to a − tu.
0
a
a − tu
H
a + tu
This is a routine check:
R
H
(a + tu) = (a + tu) −2tu = a − tu.
In particular, we know R
H
fixes exactly the points of H.
The converse is also true — any isometry
S ∈ Isom
(
R
n
) that fixes the points
in some affine hyperplane H is either the identity or R
H
.
To show this, we first want to translate the plane such that it becomes a
vector subspace. Then we can use our linear algebra magic. For any a
∈ R
n
, we
can define the translation by a as
T
a
(x) = x + a.
This is clearly an isometry.
We pick an arbitrary a
∈ H
, and let
R
=
T
−a
ST
a
∈ Isom
(
R
n
). Then
R
fixes
exactly
H
0
=
T
−a
H
. Since 0
∈ H
0
,
H
0
is a vector subspace. In particular, if
H = {x : x · u = c}, then by putting c = a · u, we find
H
0
= {x : x · u = 0}.
To understand
R
, we already know it fixes everything in
H
0
. So we want to see
what it does to u. Note that since
R
is an isometry and fixes the origin, it is in
fact an orthogonal map. Hence for any x ∈ H
0
, we get
(Ru, x) = (Ru, Rx) = (u, x) = 0.
So
R
u is also perpendicular to
H
0
. Hence
R
u =
λ
u for some
λ
. Since
R
is an
isometry, we have
kR
u
k
2
= 1. Hence
|λ|
2
= 1, and thus
λ
=
±
1. So either
λ
= 1, and
R
=
id
; or
λ
=
−
1, and
R
=
R
H
0
, as we already know for orthogonal
matrices.
It thus follow that S = id
R
n
, or S is the reflection in H.
Thus we find that each reflection R
H
is the (unique) isometry fixing H but
not id
R
n
.
It is an exercise in the example sheet to show that every isometry of
R
n
is a composition of at most
n
+ 1 reflections. If the isometry fixes 0, then
n
reflections will suffice.
Consider the subgroup of
Isom
(
R
n
) that fixes 0. By our general expression
for the general isometry, we know this is the set
{f
(x) =
A
x :
AA
T
=
I}
∼
=
O(
n
),
the orthogonal group.
For each
A ∈
O(
n
), we must have
det
(
A
)
2
= 1. So
det A
=
±
1. We use this
to define a further subgroup, the special orthogonal group.
Definition (Special orthogonal group). The special orthogonal group is the
group
SO(n) = {A ∈ O(n) : det A = 1}.
We can look at these explicitly for low dimensions.
Example. Consider
A =
a b
c d
∈ O(2)
Orthogonality then requires
a
2
+ c
2
= b
2
+ d
2
= 1, ab + cd = 0.
Now we pick 0 ≤ θ, ϕ ≤ 2π such that
a = cos θ b = −sin ϕ
c = sin θ d = cos ϕ.
Then
ab
+
cd
= 0 gives
tan θ
=
tan ϕ
(if
cos θ
and
cos ϕ
are zero, we formally say
these are both infinity). So either θ = ϕ or θ = ϕ ± π. Thus we have
A =
cos θ −sin θ
sin θ cos θ
or A =
cos θ sin θ
sin θ −cos θ
respectively. In the first case, this is a rotation through
θ
about the origin. This
is determinant 1, and hence A ∈ SO(2).
In the second case, this is a reflection in the line
`
at angle
θ
2
to the
x
-axis.
Then det A = −1 and A 6∈ SO(2).
So in two dimensions, the orthogonal matrices are either reflections or rota-
tions — those in SO(2) are rotations, and the others are reflections.
Before we can move on to three dimensions, we need to have the notion of
orientation. We might intuitively know what an orientation is, but it is rather
difficult to define orientation formally. The best we can do is to tell whether
two given bases of a vector space have “the same orientation”. Thus, it would
make sense to define an orientation as an equivalence class of bases of “the same
orientation”. We formally define it as follows:
Definition (Orientation). An orientation of a vector space is an equivalence
class of bases — let v
1
, ··· ,
v
n
and v
0
1
, ··· ,
v
0
n
be two bases and
A
be the change
of basis matrix. We say the two bases are equivalent iff
det A >
0. This is an
equivalence relation on the bases, and the equivalence classes are the orientations.
Definition (Orientation-preserving isometry). An isometry
f
(x) =
A
x + b
is orientation-preserving if
det A
= 1. Otherwise, if
det A
=
−
1, we say it is
orientation-reversing.
Example. We now want to look at O(3). First focus on the case where
A ∈ SO(3), i.e. det A = 1. Then we can compute
det(A − I) = det(A
T
− I) = det(A) det(A
T
− I) = det(I − A) = −det(A −I).
So
det
(
A − I
) = 0, i.e. +1 is an eigenvalue in
R
. So there is some v
1
∈ R
3
such
that Av
1
= v
1
.
We set W = hv
1
i
⊥
. Let w ∈ W . Then we can compute
(Aw, v
1
) = (Aw, Av
1
) = (w, v
1
) = 0.
So
A
w
∈ W
. In other words,
W
is fixed by
A
, and
A|
W
:
W → W
is well-defined.
Moreover, it is still orthogonal and has determinant 1. So it is a rotation of the
two-dimensional vector space W .
We choose
{
v
2
,
v
3
}
an orthonormal basis of
W
. Then under the bases
{v
1
, v
2
, v
3
}, A is represented by
A =
1 0 0
0 cos θ −sin θ
0 sin θ cos θ
.
This is the most general orientation-preserving isometry of
R
3
that fixes the
origin.
How about the orientation-reversing ones? Suppose
det A
=
−
1. Then
det(−A) = 1. So in some orthonormal basis, we can express A as
−A =
1 0 0
0 cos θ −sin θ
0 sin θ cos θ
.
So A takes the form
A =
−1 0 0
0 cos ϕ −sin ϕ
0 sin ϕ cos ϕ
,
where
ϕ
=
θ
+
π
. This is a rotated reflection, i.e. we first do a reflection, then
rotation. In the special case where ϕ = 0, this is a pure reflection.
That’s all we’re going to talk about isometries.
1.2 Curves in R
n
Next we will briefly look at curves in
R
n
. This will be very brief, since curves in
R
n
aren’t too interesting. The most interesting fact might be that the shortest
curve between two points is a straight line, but we aren’t even proving that,
because it is left for the example sheet.
Definition (Curve). A curve Γ in R
n
is a continuous map Γ : [a, b] → R
n
.
Here we can think of the curve as the trajectory of a particle moving through
time. Our main objective of this section is to define the length of a curve. We
might want to define the length as
Z
b
a
kΓ
0
(t)k dt,
as is familiar from, say, IA Vector Calculus. However, we can’t do this, since our
definition of a curve does not require Γ to be continuously differentiable. It is
merely required to be continuous. Hence we have to define the length in a more
roundabout way.
Similar to the definition of the Riemann integral, we consider a dissection
D = a = t
0
< t
1
< ··· < t
N
= b of [a, b], and set P
i
= Γ(t
i
). We define
S
D
=
X
i
k
−−−−→
P
i
P
i+1
k.
P
0
P
1
P
2
P
N−1
P
N
Notice that if we add more points to the dissection, then
S
D
will necessarily
increase, by the triangle inequality. So it makes sense to define the length as the
following supremum:
Definition (Length of curve). The length of a curve Γ : [a, b] → R
n
is
` = sup
D
S
D
,
if the supremum exists.
Alternatively, if we let
mesh(D) = max
i
(t
i
− t
i−1
),
then if ` exists, then we have
` = lim
mesh(D)→0
s
D
.
Note also that by definition, we can write
` = inf{
˜
` :
˜
` ≥ S
D
for all D}.
The definition by itself isn’t too helpful, since there is no nice and easy way to
check if the supremum exists. However, differentiability allows us to compute
this easily in the expected way.
Proposition. If Γ is continuously differentiable (i.e.
C
1
), then the length of Γ
is given by
length(Γ) =
Z
b
a
kΓ
0
(t)k dt.
The proof is just a careful check that the definition of the integral coincides
with the definition of length.
Proof.
To simplify notation, we assume
n
= 3. However, the proof works for all
possible dimensions. We write
Γ(t) = (f
1
(t), f
2
(t), f
3
(t)).
For every s 6= t ∈ [a, b], the mean value theorem tells us
f
i
(t) − f
i
(s)
t − s
= f
0
i
(ξ
i
)
for some ξ
i
∈ (s, t), for all i = 1, 2, 3.
Now note that
f
0
i
are continuous on a closed, bounded interval, and hence
uniformly continuous. For all
ε ∈
0, there is some
δ >
0 such that
|t − s| < δ
implies
|f
0
i
(ξ
i
) − f
0
(ξ)| <
ε
3
for all ξ ∈ (s, t). Thus, for any ξ ∈ (s, t), we have
Γ(t) − Γ(s)
t − s
− Γ
0
(ξ)
=
f
0
1
(ξ
1
)
f
0
2
(ξ
2
)
f
0
3
(ξ
3
)
−
f
0
1
(ξ)
f
0
2
(ξ)
f
0
3
(ξ)
≤
ε
3
+
ε
3
+
ε
3
= ε.
In other words,
kΓ(t) − Γ(s) − (t −s)Γ
0
(ξ)k ≤ ε(t − s).
We relabel t = t
i
, s = t
i−1
and ξ =
s+t
2
.
Using the triangle inequality, we have
(t
i
− t
i−1
)
Γ
0
t
i
+ t
i−1
2
− ε(t
i
− t
i−1
) < kΓ(t
i
) − Γ(t
i−1
)k
< (t
i
− t
i−1
)
Γ
0
t
i
+ t
i−1
2
+ ε(t
i
− t
i−1
).
Summing over all i, we obtain
X
i
(t
i
− t
i−1
)
Γ
0
t
i
+ t
i−1
2
− ε(b − a) < S
D
<
X
i
(t
i
− t
i−1
)
Γ
0
t
i
+ t
i−1
2
+ ε(b − a),
which is valid whenever mesh(D) < δ.
Since Γ
0
is continuous, and hence integrable, we know
X
i
(t
i
− t
i−1
)
Γ
0
t
i
+ t
i−1
2
→
Z
b
a
kΓ
0
(t)k dt
as mesh(D) → 0, and
length(Γ) = lim
mesh(D)→0
S
D
=
Z
b
a
kΓ
0
(t)k dt.
This is all we are going to say about Euclidean space.
2 Spherical geometry
The next thing we are going to study is the geometry on the surface of a sphere.
This is a rather sensible thing to study, since it so happens that we all live on
something (approximately) spherical. It turns out the geometry of the sphere is
very different from that of R
2
.
In this section, we will always think of
S
2
as a subset of
R
3
so that we can
reuse what we know about R
3
.
Notation. We write
S
=
S
2
⊆ R
3
for the unit sphere. We write
O
= 0 for the
origin, which is the center of the sphere (and not on the sphere).
When we live on the sphere, we can no longer use regular lines in
R
3
, since
these do not lie fully on the sphere. Instead, we have a concept of a spherical
line, also known as a great circle.
Definition (Great circle). A great circle (in
S
2
) is
S
2
∩
(
a plane through O
).
We also call these (spherical) lines.
We will also call these geodesics, which is a much more general term defined
on any surface, and happens to be these great circles in S
2
.
In
R
3
, we know that any three points that are not colinear determine a
unique plane through them. Hence given any two non-antipodal points
P, Q ∈ S
,
there exists a unique spherical line through P and Q.
Definition (Distance on a sphere). Given
P, Q ∈ S
, the distance
d
(
P, Q
) is the
shorter of the two (spherical) line segments (i.e. arcs)
P Q
along the respective
great circle. When
P
and
Q
are antipodal, there are infinitely many line segments
between them of the same length, and the distance is π.
Note that by the definition of the radian,
d
(
P, Q
) is the angle between
−−→
OP
and
−−→
OQ, which is also cos
−1
(P · Q) (where P = OP, Q = OQ).
2.1 Triangles on a sphere
One main object of study is spherical triangles – they are defined just like
Euclidean triangles, with
AB, AC, BC
line segments on
S
of length
< π
. The
restriction of length is just merely for convenience.
A
B C
c
b
a
α
β
γ
We will take advantage of the fact that the sphere sits in R
3
. We set
n
1
=
C × B
sin a
n
2
=
A × C
sin b
n
3
=
B × A
sin c
.
These are unit normals to the planes
OBC, OAC
and
OAB
respectively. They
are pointing out of the solid OABC.
The angles
α, β, γ
are the angles between the planes for the respective sides.
Then 0
< α, β, γ < π
. Note that the angle between n
2
and n
3
is
π
+
α
(not
α
itself — if α = 0, then the angle between the two normals is π). So
n
2
· n
3
= −cos α
n
3
· n
1
= −cos β
n
1
· n
2
= −cos γ.
We have the following theorem:
Theorem (Spherical cosine rule).
sin a sin b cos γ = cos c −cos a cos b.
Proof. We use the fact from IA Vectors and Matrices that
(C × B) · (A ×C) = (A ·C)(B ·C) −(C ·C)(B · A),
which follows easily from the double-epsilon identity
ε
ijk
ε
imn
= δ
jm
δ
kn
− δ
jn
δ
km
.
In our case, since C ·C = 1, the right hand side is
(A · C)(B · C) −(B ·A).
Thus we have
−cos γ = n
1
· n
2
=
C × B
sin a
·
A × C
sin b
=
(A · C)(B · C) −(B ·A)
sin a sin b
=
cos b cos a − cos c
sin a sin b
.
Corollary (Pythagoras theorem). If γ =
π
2
, then
cos c = cos a cos b.
Analogously, we have a spherical sine rule.
Theorem (Spherical sine rule).
sin a
sin α
=
sin b
sin β
=
sin c
sin γ
.
Proof. We use the fact that
(A × C) × (C ×B) = (C ·(B ×A))C,
which we again are not bothered to prove again. The left hand side is
−(n
1
× n
2
) sin a sin b
Since the angle between n
1
and n
2
is
π
+
γ
, we know n
1
×
n
2
= C
sin γ
. Thus
the left hand side is
−C sin a sin b sin γ.
Thus we know
C · (A × B) = sin a sin b sin γ.
However, since the scalar triple product is cyclic, we know
C · (A × B) = A ·(B ×C).
In other words, we have
sin a sin b sin γ = sin b sin c sin α.
Thus we have
sin γ
sin c
=
sin α
sin a
.
Similarly, we know this is equal to
sin β
sin b
.
Recall that for small a, b, c, we know
sin a = a + O(a
3
).
Similarly,
cos a = 1 −
a
2
2
+ O(a
4
).
As we take the limit
a, b, c →
0, the spherical sine and cosine rules become the
usual Euclidean versions. For example, the cosine rule becomes
ab cos γ = 1 −
c
2
2
−
1 −
a
2
2
1 −
b
2
2
+ O(k(a, b, c)k
3
).
Rearranging gives
c
2
= a
2
+ b
2
− 2ab cos γ + O(k(a, b, c, )k
3
).
The sine rule transforms similarly as well. This is what we would expect, since
making
a, b, c
small is equivalent to zooming into the surface of the sphere, and
it looks more and more like flat space.
Note that if
γ
=
π
, it then follows that
C
is in the line segment given by
AB
.
So c = a + b. Otherwise, we get
cos c > cos a cos b − sin a sin b = cos(a + b),
since cos γ < 1. Since cos is decreasing on [0, π], we know
c < a + b.
Corollary (Triangle inequality). For any P, Q, R ∈ S
2
, we have
d(P, Q) + d(Q, R) ≥ d(P, R),
with equality if and only if Q lies is in the line segment P R of shortest length.
Proof.
The only case left to check is if
d
(
P, R
) =
π
, since we do not allow our
triangles to have side length π. But in this case they are antipodal points, and
any Q lies in a line through P R, and equality holds.
Thus, we find that (S
2
, d) is a metric space.
On
R
n
, straight lines are curves that minimize distance. Since we are calling
spherical lines lines, we would expect them to minimize distance as well. This is
in fact true.
Proposition. Given a curve Γ on
S
2
⊆ R
3
from
P
to
Q
, we have
`
=
length
(Γ)
≥
d
(
P, Q
). Moreover, if
`
=
d
(
P, Q
), then the image of Γ is a spherical line segment
P Q.
Proof. Let Γ : [0, 1] → S and ` = length(Γ). Then for any dissection D of [0, 1],
say 0 = t
0
< ··· < t
N
= 1, write P
i
= Γ(t
i
). We define
˜
S
D
=
X
i
d(P
i−1
, P
i
) > S
D
=
X
i
|
−−−−→
P
i−1
P
i
|,
where the length in the right hand expression is the distance in Euclidean 3-space.
Now suppose
` < d
(
P, Q
). Then there is some
ε >
0 such that
`
(1 +
ε
)
<
d(P, Q).
Recall from basic trigonometric that if θ > 0, then sin θ < θ. Also,
sin θ
θ
→ 1 as θ → 0.
Thus we have
θ ≤ (1 + ε) sin θ.
for small θ. What we really want is the double of this:
2θ ≤ (1 + ε)2 sin θ.
This is useful since these lengths appear in the following diagram:
2 sin θ
2θ
2θ
This means for P, Q sufficiently close, we have d(P, Q) ≤ (1 + ε)|
−−→
P Q|.
From Analysis II, we know Γ is uniformly continuous on [0
,
1]. So we can
choose D such that
d(P
i−1
, P
i
) ≤ (1 + ε)|
−−−−→
P
i−1
P
i
|
for all i. So we know that for sufficiently fine D,
˜
S
D
≤ (1 + ε)S
D
< d(P, Q),
since
S
D
→ `
. However, by the triangle inequality
˜
S
D
≥ d
(
P, Q
). This is a
contradiction. Hence we must have ` ≥ d(P, Q).
Suppose now
`
=
d
(
P, Q
) for some Γ : [0
,
1]
→ S
,
`
=
length
(Γ). Then for
every t ∈ [0, 1], we have
d(P, Q) = ` = length Γ|
[0,t]
+ length Γ|
[t,1]
≥ d(P, Γ(t)) + d(Γ(t), Q)
≥ d(P, Q).
Hence we must have equality all along the way, i.e.
d(P, Q) = d(P, Γ(t)) + d(Γ(t), Q)
for all Γ(t).
However, this is possible only if Γ(
t
) lies on the shorter spherical line segment
P Q, as we have previously proved. So done.
Note that if Γ is a curve of minimal length from
P
to
Q
, then Γ is a
spherical line segment. Further, from the proof of this proposition, we know
length
Γ
|
[0,t]
=
d
(
P,
Γ(
t
)) for all
t
. So the parametrisation of Γ is monotonic.
Such a Γ is called a minimizing geodesic.
Finally, we get to an important theorem whose prove involves complicated
pictures. This is known as the Gauss-Bonnet theorem. The Gauss-Bonnet
theorem is in fact a much more general theorem. However, here we will specialize
in the special case of the sphere. Later, when doing hyperbolic geometry, we
will prove the hyperbolic version of the Gauss-Bonnet theorem. Near the end of
the course, when we have developed sufficient machinery, we would be able to
state the Gauss-Bonnet theorem in its full glory. However, we will not be able
to prove the general version.
Proposition (Gauss-Bonnet theorem for
S
2
). If ∆ is a spherical triangle with
angles α, β, γ, then
area(∆) = (α + β + γ) − π.
Proof.
We start with the concept of a double lune. A double lune with angle
0
< α < π
is two regions
S
cut out by two planes through a pair of antipodal
points, where α is the angle between the two planes.
A
A
0
It is not hard to show that the area of a double lune is 4
α
, since the area of the
sphere is 4π.
Now note that our triangle ∆ =
ABC
is the intersection of 3 single lunes,
with each of
A, B, C
as the pole (in fact we only need two, but it is more
convenient to talk about 3).
A
A
0
B
B
0
C
C
0
Therefore ∆ together with its antipodal partner ∆
0
is a subset of each of the 3
double lunes with areas 4
α,
4
β,
4
γ
. Also, the union of all the double lunes cover
the whole sphere, and overlap at exactly ∆ and ∆
0
. Thus
4(α + β + γ) = 4π + 2(area(∆) + area(∆
0
)) = 4π + 4 area(∆).
This is easily generalized to arbitrary convex
n
-gons on
S
2
(with
n ≥
3).
Suppose
M
is such a convex
n
-gon with interior angles
α
1
, ··· , α
n
. Then we
have
area(M) =
n
X
1
α
i
− (n − 2)π.
This follows directly from cutting the polygon up into the constituent triangles.
This is very unlike Euclidean space. On
R
2
, we always have
α
+
β
+
γ
=
π
.
Not only is this false on
S
2
, but by measuring the difference, we can tell the
area of the triangle. In fact, we can identify triangles up to congruence just by
knowing the three angles.
2.2 M¨obius geometry
It turns out it is convenient to identify the sphere
S
2
withe the extended complex
plane
C
∞
=
C ∪{∞}
. Then isometries of
S
2
will translate to nice class of maps
of C
∞
.
We first find a way to identify
S
2
with
C
∞
. We use coordinates
ζ ∈ C
∞
. We
define the stereographic projection π : S
2
→ C
∞
by
N
P
π(P )
π(P ) = (line P N) ∩ {z = 0},
which is well defined except where P = N, in which case we define π(N ) = ∞.
To give an explicit formula for this, consider the cross-section through the
plane ONP .
O
N
r
P
z
π(P)
R
If
P
has coordinates (
x, y
), then we see that
π
(
P
) will be a scalar multiple of
x
+
iy
. To find this factor, we notice that we have two similar triangles, and
hence obtain
r
R
=
1 − z
1
.
Then we obtain the formula
π(x, y, z) =
x + iy
1 − z
.
If we do the projection from the South pole instead, we get a related formula.
Lemma. If
π
0
:
S
2
→ C
∞
denotes the stereographic projection from the South
Pole instead, then
π
0
(P ) =
1
π(P)
.
Proof. Let P (x, y, z). Then
π(x, y, z) =
x + iy
1 − z
.
Then we have
π
0
(x, y, z) =
x + iy
1 + z
,
since we have just flipped the z axis around. So we have
π(P )π
0
(P ) =
x
2
+ y
2
1 − z
2
= 1,
noting that we have x
2
+ y
2
+ z
2
= 1 since we are on the unit sphere.
We can use this to infer that
π
0
◦ π
−1
:
C
∞
→ C
∞
takes
ζ 7→
1
/
¯
ζ
, which is
the inversion in the unit circle |ζ| = 1.
From IA Groups, we know M¨obius transformations act on
C
∞
and form a
group G by composition. For each matrix
A =
a b
c d
∈ GL(2, C),
we get a M¨obius map C
∞
→ C
∞
by
ζ 7→
aζ + b
cζ + d
.
Moreover, composition of M¨obius map is the same multiplication of matrices.
This is not exactly a bijective map between
G
and
GL
(2
, C
). For any
λ ∈ C
∗
=
C \ {
0
}
, we know
λA
defines the same map M¨obius map as
A
.
Conversely, if
A
1
and
A
2
gives the same M¨obius map, then there is some
λ
1
6
= 0
such that A
1
= λA
2
.
Hence, we have
G
∼
=
PGL(2, C) = GL(2, C)/C
∗
,
where
C
∗
∼
=
{λI : λ ∈ C
∗
}.
Instead of taking the whole
GL
(2
, C
) and quotienting out multiples of the
identities, we can instead start with
SL
(2
, C
). Again,
A
1
, A
2
∈ SL
(2
, C
) define
the same map if and only if
A
1
=
λA
2
for some
λ
. What are the possible values
of λ? By definition of the special linear group, we must have
1 = det(λA) = λ
2
det A = λ
2
.
So
λ
2
=
±
1. So each M¨obius map is represented by two matrices,
A
and
−A
,
and we get
G
∼
=
PSL(2, C) = SL(2, C)/{±1}.
Now let’s think about the sphere. On
S
2
, the rotations
SO
(3) act as isometries.
In fact, the full isometry group of
S
2
is O(3) (the proof is on the first example
sheet). What we would like to show that rotations of
S
2
correspond to M¨obius
transformations coming from the subgroup SU(2) ≤ GL(2, C).
Theorem. Via the stereographic projection, every rotation of
S
2
induces a
M¨obius map defined by a matrix in SU(2) ⊆ GL(2, C), where
SU(2) =
a −b
¯
b ¯a
: |a|
2
+ |b|
2
= 1
.
Proof.
(i)
Consider the
r
(
ˆ
z, θ
), the rotations about the
z
axis by
θ
. These corresponds
to the M¨obius map ζ 7→ e
iθ
ζ, which is given by the unitary matrix
e
iθ/2
0
0 e
−iθ/2
.
(ii) Consider the rotation r(
ˆ
y,
π
2
). This has the matrix
0 0 1
0 1 0
−1 0 0
x
y
z
=
z
y
−x
.
This corresponds to the map
ζ =
x + iy
1 − z
7→ ζ
0
=
z + iy
1 + x
We want to show this is a M¨obius map. To do so, we guess what the
M¨obius map should be, and check it works. We can manually compute
that −1 7→ ∞, 1 7→ 0, i 7→ i.
1
−1
i
0
∞
The only M¨obius map that does this is
ζ
0
=
ζ − 1
ζ + 1
.
We now check:
ζ − 1
ζ + 1
=
x + iy − 1 + z
x + iy + 1 − z
=
x − 1 + z + iy
x + 1 − (z −iy)
=
(z + iy)(x − 1 + z + iy)
(x + 1)(z + iy) − (z
2
+ y
2
)
=
(z + iy)(x − 1 + z + iy)
(x + 1)(z + iy) + (x
2
− 1)
=
(z + iy)(x − 1 + z + iy)
(x + 1)(z + iy + x −1)
=
z + iy
x + 1
.
So done. We finally have to write this in the form of an SU(2) matrix:
1
√
2
1 −1
1 1
.
(iii) We claim that SO(3) is generated by r
ˆ
y,
π
2
and r(
ˆ
z, θ) for 0 ≤ θ < 2π.
To show this, we observe that
r
(
ˆ
x, ϕ
) =
r
(
ˆ
y,
π
2
)
r
(
ˆ
z, ϕ
)
r
(
ˆ
y, −
π
2
). Note that
we read the composition from right to left. You can convince yourself this
is true by taking a physical sphere and try rotating. To prove it formally,
we can just multiply the matrices out.
Next, observe that for v
∈ S
2
⊆ R
3
, there are some angles
ϕ, ψ
such that
g
=
r
(
ˆ
z, ψ
)
r
(
ˆ
x, ϕ
) maps v to
ˆ
x
. We can do so by first picking
r
(
ˆ
x, ϕ
) to
rotate v into the (x, y)-plane. Then we rotate about the z-axis to send it
to
ˆ
x.
Then for any
θ
, we have
r
(v
, θ
) =
g
−1
r
(
ˆ
x, θ
)
g
, and our claim follows by
composition.
(iv)
Thus, via the stereographic projection, every rotation of
S
2
corresponds
to products of M¨obius transformations of
C
∞
with matrices in
SU
(2).
The key of the proof is step (iii). Apart from enabling us to perform the
proof, it exemplifies a useful technique in geometry — we know how to rotate
arbitrary things in the
z
axis. When we want to rotate things about the
x
axis
instead, we first rotate the sphere to move the
x
axis to where the
z
axis used
to be, do those rotations, and then rotate it back. In general, we can use some
isometries or rotations to move what we want to do to a convenient location.
Theorem. The group of rotations
SO
(3) acting on
S
2
corresponds precisely
with the subgroup
PSU
(2) =
SU
(2)
/ ±
1 of M¨obius transformations acting on
C
∞
.
What this is in some sense a converse of the previous theorem. We are saying
that for each M¨obius map from
SU
(2), we can find some rotation of
S
2
that
induces that M¨obius map, and there is exactly one.
Proof. Let g ∈ PSU(2) be a M¨obius transformation
g(z) =
az + b
¯
bz + ¯a
.
Suppose first that
g
(0) = 0. So
b
= 0. So
a¯a
= 1. Hence
a
=
e
iθ/2
. Then
g
corresponds to r(
ˆ
z, θ), as we have previously seen.
In general, let
g
(0) =
w ∈ C
∞
. Let
Q ∈ S
2
be such that
π
(
Q
) =
w
. Choose a
rotation
A ∈ SO
(3) such that
A
(
Q
) =
−
ˆ
z
. Since
A
is a rotation, let
α ∈ PSU
(2)
be the corresponding M¨obius transformation. By construction we have
α
(
w
) = 0.
Then the composition
α ◦ g
fixes zero. So it corresponds to some
B
=
r
(
z, θ
).
We then see that g corresponds to A
−1
B ∈ SO(3). So done.
Again, we solved an easy case, where
g
(0) = 0, and then performed some
rotations to transform any other case into this simple case.
We have now produced a 2-to-1 map
SU(2) → PSU(2)
∼
=
SO(3).
If we treat these groups as topological spaces, this map does something funny.
Suppose we start with a (non-closed) path from
I
to
−I
in
SU
(2). Applying
the map, we get a closed loop from I to I in SO(3).
Hence, in
SO
(3), loops are behave slightly weirdly. If we go around this loop
in
SO
(3), we didn’t really get back to the same place. Instead, we have actually
moved from
I
to
−I
in
SU
(2). It takes two full loops to actually get back to
I
.
In physics, this corresponds to the idea of spinors.
We can also understand this topologically as follows: since
SU
(2) is defined by
two complex points
a, b ∈ C
such that
|a|
2
+
|b|
2
, we can view it as as three-sphere
S
3
in SO(3).
A nice property of
S
3
is it is simply connected, as in any loop in
S
3
can be
shrunk to a point. In other words, given any loop in
S
3
, we can pull and stretch
it (continuously) until becomes the “loop” that just stays at a single point.
On the other hand,
SO
(3) is not simply connected. We have just constructed
a loop in
SO
(3) by mapping the path from
I
to
−I
in
SU
(2). We cannot deform
this loop until it just sits at a single point, since if we lift it back up to
SU
(2), it
still has to move from I to −I.
The neat thing is that in some sense,
S
3
∼
=
SU
(2) is just “two copies” of
SO
(3). By duplicating
SO
(3), we have produced
SU
(2), a simply connected
space. Thus we say SU(2) is a universal cover of SO(3).
We’ve just been waffling about spaces and loops, and throwing around terms
we haven’t defined properly. These vague notions will be made precise in the
IID Algebraic Topology course, and we will then (maybe) see that
SU
(2) is a
universal cover of SO(3).
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.
4 Hyperbolic geometry
At the beginning of the course, we studied Euclidean geometry, which was not
hard, because we already knew about it. Later on, we studied spherical geometry.
That also wasn’t too bad, because we can think of S
2
concretely as a subset of
R
3
.
We are next going to study hyperbolic geometry. Historically, hyperbolic
geometry was created when people tried to prove Euclid’s parallel postulate (that
given a line
`
and a point
P 6∈ `
, there exists a unique line
`
0
containing
P
that
does not intersect
`
). Instead of proving the parallel postulate, they managed to
create a new geometry where this is false, and this is hyperbolic geometry.
Unfortunately, hyperbolic geometry is much more complicated, since we
cannot directly visualize it as a subset of
R
3
. Instead, we need to develop the
machinery of a Riemannian metric in order to properly describe hyperbolic
geometry. In a nutshell, this allows us to take a subset of
R
2
and measure
distances in it in a funny way.
4.1 Review of derivatives and chain rule
We start by reviewing some facts about taking derivatives, and make explicit
the notation we will use.
Definition (Smooth function). Let
U ⊆ R
n
be open, and
f
= (
f
1
, ··· , f
m
) :
U → R
m
. We say
f
is smooth (or
C
∞
) if each
f
i
has continuous partial derivatives
of each order. In particular, a
C
∞
map is differentiable, with continuous first-
order partial derivatives.
Definition (Derivative). The derivative for a function
f
:
U → R
m
at a point
a ∈ U
is a linear map d
f
a
:
R
n
→ R
m
(also written as D
f
(
a
) or
f
0
(
a
)) such that
lim
h→0
kf(a + h) − f(a) −df
a
· hk
khk
→ 0,
where h ∈ R
n
.
If
m
= 1, then d
f
a
is expressed as
∂f
∂x
a
(a), ··· ,
∂f
∂x
n
(a)
, and the linear map
is given by
(h
1
, ··· , h
n
) 7→
n
X
i=1
∂f
∂x
i
(a)h
i
,
i.e. the dot product. For a general
m
, this vector becomes a matrix. The Jacobian
matrix is
J(f )
a
=
∂f
i
∂x
j
(a)
,
with the linear map given by matrix multiplication, namely
h 7→ J(f)
a
· h.
Example. Recall that a holomorphic (analytic) function of complex variables
f : U ⊆ C → C has a derivative f
0
, defined by
lim
|w|→0
|f(z + w) − f(z) −f
0
(z)w|
|w|
→ 0
We let f
0
(z) = a + ib and w = h
1
+ ih
2
. Then we have
f
0
(z)w = ah
1
− bh
2
+ i(ah
2
+ bh
1
).
We identify
R
2
=
C
. Then
f
:
U ⊆ R
2
→ R
2
has a derivative d
f
z
:
R
2
→ R
2
given by
a −b
b a
.
We’re also going to use the chain rule quite a lot. So we shall write it out
explicitly.
Proposition (Chain rule). Let
U ⊆ R
n
and
V ⊆ R
p
. Let
f
:
U → R
m
and
g : V → U be smooth. Then f ◦ g : V → R
m
is smooth and has a derivative
d(f ◦ g)
p
= (df)
g(p)
◦ (dg)
p
.
In terms of the Jacobian matrices, we get
J(f ◦ g)
p
= J(f )
g(p)
J(g)
p
.
4.2 Riemannian metrics
Finally, we get to the idea of a Riemannian metric. The basic idea of a Rieman-
nian metric is not too unfamiliar. Presumably, we have all seen maps of the
Earth, where we try to draw the spherical Earth on a piece of paper, i.e. a subset
of
R
2
. However, this does not behave like
R
2
. You cannot measure distances on
Earth by placing a ruler on the map, since distances are distorted. Instead, you
have to find the coordinates of the points (e.g. the longitude and latitude), and
then plug them into some complicated formula. Similarly, straight lines on the
map are not really straight (spherical) lines on Earth.
We really should not think of Earth a subset of
R
2
. All we have done was
to “force” Earth to live in
R
2
to get a convenient way of depicting the Earth, as
well as a convenient system of labelling points (in many map projections, the
x
and y axes are the longitude and latitude).
This is the idea of a Riemannian metric. To describe some complicated
surface, we take a subset
U
of
R
2
, and define a new way of measuring distances,
angles and areas on
U
. All these information are packed into an entity known
as the Riemannian metric.
Definition (Riemannian metric). We use coordinates (
u, v
)
∈ R
2
. We let
V ⊆ R
2
be open. Then a Riemannian metric on
V
is defined by giving
C
∞
functions E, F, G : V → R such that
E(P ) F (P )
F (P) G(P )
is a positive definite definite matrix for all P ∈ V .
Alternatively, this is a smooth function that gives a 2
×
2 symmetric positive
definite matrix, i.e. inner product
h·, ·i
P
, for each point in
V
. By definition, if
e
1
, e
2
are the standard basis, then
he
1
, e
1
i
P
= E(P )
he
1
, e
2
i
P
= F (P)
he
2
, e
2
i
P
= G(P ).
Example. We can pick
E
=
G
= 1 and
F
= 0. Then this is just the standard
Euclidean inner product.
As mentioned, we should not imagine
V
as a subset of
R
2
. Instead, we should
think of it as an abstract two-dimensional surface, with some coordinate system
given by a subset of
R
2
. However, this coordinate system is just a convenient way
of labelling points. They do not represent any notion of distance. For example,
(0, 1) need not be closer to (0, 2) than to (7, 0). These are just abstract labels.
With this in mind,
V
does not have any intrinsic notion of distances, angles
and areas. However, we do want these notions. We can certainly write down
things like the difference of two points, or even the compute the derivative of
a function. However, these numbers you get are not meaningful, since we can
easily use a different coordinate system (e.g. by scaling the axes) and get a
different number. They have to be interpreted with the Riemannian metric.
This tells us how to measure these things, via an inner product “that varies with
space”. This variation in space is not an oddity arising from us not being able to
make up our minds. This is since we have “forced” our space to lie in
R
2
. Inside
V
, going from (0
,
1) to (0
,
2) might be very different from going from (5
,
5) to
(6
,
5), since coordinates don’t mean anything. Hence our inner product needs
to measure “going from (0
,
1) to (0
,
2)” differently from “going from (5
,
5) to
(6, 5)”, and must vary with space.
We’ll soon come to defining how this inner product gives rise to the notion
of distance and similar stuff. Before that, we want to understand what we can
put into the inner product
h·, ·i
P
. Obviously these would be vectors in
R
2
, but
where do these vectors come from? What are they supposed to represent?
The answer is “directions” (more formally, tangent vectors). For example,
h
e
1
,
e
1
i
P
will tell us how far we actually are going if we move in the direction
of e
1
from
P
. Note that we say “move in the direction of e
1
”, not “move by
e
1
”. We really should read this as “if we move by
h
e
1
for some small
h
, then
the distance covered is
h
p
he
1
, e
1
i
P
”. This statement is to be interpreted along
the same lines as “if we vary
x
by some small
h
, then the value of
f
will vary
by
f
0
(
x
)
h
”. Notice how the inner product allows us to translate a length in
R
2
(namely khe
1
k
eucl
= h) into the actual length in V .
What we needed for this is just the norm induced by the inner product. Since
what we have is the whole inner product, we in fact can define more interesting
things such as areas and angles. We will formalize these ideas very soon, after
getting some more notation out of the way.
Often, instead of specifying the three functions separately, we write the metric
as
E du
2
+ 2F du dv + G dv
2
.
This notation has some mathematical meaning. We can view the coordinates
as smooth functions
u
:
V → R
,
v
:
U → R
. Since they are smooth, they have
derivatives. They are linear maps
du
P
: R
2
→ R dv
P
: R
2
→ R
(h
1
, h
2
) 7→ h
1
(h
1
, h
2
) 7→ h
2
.
These formula are valid for all
P ∈ V
. So we just write d
u
and d
v
instead.
Since they are maps
R
2
→ R
, we can view them as vectors in the dual space,
d
u,
d
v ∈
(
R
2
)
∗
. Moreover, they form a basis for the dual space. In particular,
they are the dual basis to the standard basis e
1
, e
2
of R
2
.
Then we can consider d
u
2
,
d
u
d
v
and d
v
2
as bilinear forms on
R
2
. For
example,
du
2
(h, k) = du(h)du(k)
du dv(h, k) =
1
2
(du(h)dv(k) + du(k)dv(h))
dv
2
(h, k) = dv(h)dv(k)
These have matrices
1 0
0 0
,
0
1
2
1
2
0
,
0 0
0 1
respectively. Then we indeed have
E du
2
+ 2F du dv + G dv
2
=
E F
F G
.
We can now start talking about what this is good for. In standard Euclidean
space, we have a notion of length and area. A Riemannian metric also gives a
notion of length and area.
Definition (Length). The length of a smooth curve
γ
= (
γ
1
, γ
2
) : [0
,
1]
→ V
is
defined as
Z
1
0
E ˙γ
2
1
+ 2F ˙γ
1
˙γ
2
+ G ˙γ
2
2
1
2
dt,
where E = E(γ
1
(t), γ
2
(t)) etc. We can also write this as
Z
1
0
h˙γ, ˙γi
1
2
γ(t)
dt.
Definition (Area). The area of a region W ⊆ V is defined as
Z
W
(EG −F
2
)
1
2
du dv
when this integral exists.
In the area formula, what we are integrating is just the determinant of the
metric. This is also known as the Gram determinant.
We define the distance between two points
P
and
Q
to be the infimum of
the lengths of all curves from
P
to
Q
. It is an exercise on the second example
sheet to prove that this is indeed a metric.
Example. We will not do this in full detail — the details are to be filled in in
the third example sheet.
Let V = R
2
, and define the Riemannian metric by
4(du
2
+ dv
2
)
(1 + u
2
+ v
2
)
2
.
This looks somewhat arbitrary, but we shall see this actually makes sense by
identifying
R
2
with the sphere by the stereographic projection
π
:
S
2
\{N } → R
2
.
For every point
P ∈ S
2
, the tangent plane to
S
2
at
P
is given by
{
x
∈ R
3
:
x
·
−−→
OP
= 0
}
. Note that we translated it so that
P
is the origin, so that we can
view it as a vector space (points on the tangent plane are points “from
P
”).
Now given any two tangent vectors x
1
,
x
2
⊥
−−→
OP
, we can take the inner product
x
1
· x
2
in R
3
.
We want to say this inner product is “the same as” the inner product provided
by the Riemannian metric on R
2
. We cannot just require
x
1
· x
2
= hx
1
, x
2
i
π(P)
,
since this makes no sense at all. Apart from the obvious problem that x
1
,
x
2
have three components but the Riemannian metric takes in vectors of two
components, we know that x
1
and x
2
are vectors tangent to
P ∈ S
2
, but to
apply the Riemannian metric, we need the corresponding tangent vector at
π(P) ∈ R
2
. To do so, we act by dπ
p
. So what we want is
x
1
· x
2
= hdπ
P
(x
1
), dπ
P
(x
2
)i
π(P)
.
Verification of this equality is left as an exercise on the third example sheet. It
is helpful to notice
π
−1
(u, v) =
(2u, 2v, u
2
+ v
2
− 1)
1 + u
2
+ v
2
.
In some sense, we say the surface
S
2
\ {N}
is “isometric” to
R
2
via the
stereographic projection
π
. We can define the notion of isometry between two
open sets with Riemannian metrics in general.
Definition (Isometry). Let
V,
˜
V ⊆ R
2
be open sets endowed with Riemannian
metrics, denoted as h·, ·i
P
and h·, ·i
∼
Q
for P ∈ V, Q ∈
˜
V respectively.
A diffeomorphism (i.e.
C
∞
map with
C
∞
inverse)
ϕ
:
V →
˜
V
is an isometry
if for every P ∈ V and x, y ∈ R
2
, we get
hx, yi
P
= hdϕ
P
(x), dϕ
P
(y)i
∼
ϕ(P )
.
Again, in the definition, x and y represent tangent vectors at
P ∈ V
, and
on the right of the equality, we need to apply d
ϕ
P
to get tangent vectors at
ϕ(P ) ∈
˜
V .
How are we sure this indeed is the right definition? We, at the very least,
would expect isometries to preserve lengths. Let’s see this is indeed the case. If
γ
: [0
,
1]
→ V
is a
C
∞
curve, the composition
˜γ
=
ϕ ◦ γ
: [0
,
1]
→
˜
V
is a path in
˜
V . We let P = γ(t), and hence ϕ(P) = ˜γ(t). Then
h˜γ
0
(t), ˜γ
0
(t)i
∼
˜γ(t)
= hdϕ
P
◦ γ
0
(t), dϕ
P
◦ γ
0
(t)i
∼
ϕ(P )
= hγ
0
(t), γ
0
(t)i
γ(t)=P
.
Integrating, we obtain
length(˜γ) = length(γ) =
Z
1
0
hγ
0
(t), γ
0
(t)i
γ(t)
dt.
4.3 Two models for the hyperbolic plane
That’s enough preparation. We can start talking about hyperbolic plane. We
will in fact provide two models of the hyperbolic plane. Each model has its own
strengths, and often proving something is significantly easier in one model than
the other.
We start with the disk model.
Definition (Poincar´e disk model). The (Poincar´e) disk model for the hyperbolic
plane is given by the unit disk
D ⊆ C
∼
=
R
2
, D = {ζ ∈ C : |ζ| < 1},
and a Riemannian metric on this disk given by
4(du
2
+ dv
2
)
(1 − u
2
− v
2
)
2
=
4|dζ|
2
(1 − |ζ|
2
)
2
, (∗)
where ζ = u + iv.
Note that this is similar to our previous metric for the sphere, but we have
1 − u
2
− v
2
instead of 1 + u
2
+ v
2
.
To interpret the term
|
d
ζ|
2
, we can either formally set
|
d
ζ|
2
= d
u
2
+ d
v
2
, or
interpret it as the derivative dζ = du + idv : C → C.
We see that (
∗
) is a scaling of the standard Riemannian metric by a factor
depending on the polar radius
r
=
|ζ|
2
. The distances are scaled by
2
1−r
2
, while
the areas are scaled by
4
(1−r
2
)
2
. Note, however, that the angles in the hyperbolic
disk are the same as that in
R
2
. This is in general true for metrics that are just
scaled versions of the Euclidean metric (exercise).
Alternatively, we can define the hyperbolic plane with the upper-half plane.
Definition (Upper half-plane). The upper half-plane is
H = {z ∈ C : Im(z) > 0}.
What is the appropriate Riemannian metric to put on the upper half plane?
We know D bijects to H via the M¨obius transformation
ϕ : ζ ∈ D 7→ i
1 + ζ
1 − ζ
∈ H.
This bijection is in fact a conformal equivalence, as defined in IB Complex
Analysis/Methods. The idea is to pick a metric on
H
such that this map is
an isometry. Then
H
together with this Riemannian metric will be the upper
half-plane model for the hyperbolic plane.
To avoid confusion, we reserve the letter
z
for points
z ∈ H
, with
z
=
x
+
iy
,
while we use ζ for points ζ ∈ D, and write ζ = u + iv. Then we have
z = i
1 + ζ
1 − ζ
, ζ =
z − i
z + i
.
Instead of trying to convert the Riemannian metric on
D
to
H
, which would be
a complete algebraic horror, we first try converting the Euclidean metric. The
Euclidean metric on R
2
= C is given by
hw
1
, w
2
i = Re(w
1
w
2
) =
1
2
(w
1
¯w
2
+ ¯w
1
w
2
).
So if
h·, ·i
eucl
is the Euclidean metric at
ζ
, then at
z
such that
ζ
=
z−i
z+i
, we
require (by definition of isometry)
hw, vi
z
=
dζ
dz
w,
dζ
dz
v
eucl
=
dζ
dz
2
Re(w¯v) =
dζ
dz
2
(w
1
v
1
+ w
2
v
2
),
where w = w
1
+ iw
2
, v = v
1
+ iv
2
.
Hence, on H, we obtain the Riemannian metric
dζ
dz
2
(dx
2
+ dy
2
).
We can compute
dζ
dz
=
1
z + i
−
z − i
(z + i)
2
=
2i
(z + i)
2
.
This is what we get if we started with a Euclidean metric. If we start with the
hyperbolic metric on
D
, we get an additional scaling factor. We can do some
computations to get
1 − |ζ|
2
= 1 −
|z − i|
2
|z + i|
2
,
and hence
1
1 − |ζ|
2
=
|z + i|
2
|z + i|
2
− |z − i|
2
=
|z + i|
2
4 Im z
.
Putting all these together, metric corresponding to
4|dζ|
2
(1−|ζ|
2
)
2
on D is
4 ·
4
|z + i|
4
·
|z + i|
2
4 Im z
2
· |dz|
2
=
|dz|
2
(Im z)
2
=
dx
2
+ dy
2
y
2
.
We now use all these ingredients to define the upper half-plane model.
Definition (Upper half-plane model). The upper half-plane model of the hyper-
bolic plane is the upper half-plane H with the Riemannian metric
dx
2
+ dy
2
y
2
.
The lengths on
H
are scaled (from the Euclidean one) by
1
y
, while the areas
are scaled by
1
y
2
. Again, the angles are the same.
Note that we did not have to go through so much mess in order to define
the sphere. This is since we can easily “embed” the surface of the sphere in
R
3
.
However, there is no easy surface in
R
3
that gives us the hyperbolic plane. As
we don’t have an actual prototype, we need to rely on the more abstract data of
a Riemannian metric in order to work with hyperbolic geometry.
We are next going to study the geometry of
H
, We claim that the following
group of M¨obius maps are isometries of H:
PSL(2, R) =
z 7→
az + b
cz + d
: a, b, c, d ∈ R, ad − bc = 1
.
Note that the coefficients have to be real, not complex.
Proposition. The elements of
PSL
(2
, R
) are isometries of
H
, and this preserves
the lengths of curves.
Proof. It is easy to check that PSL(2, R) is generated by
(i) Translations z 7→ z + a for a ∈ R
(ii) Dilations z 7→ az for a > 0
(iii) The single map z 7→ −
1
z
.
So it suffices to show each of these preserves the metric
|dz|
2
y
2
, where
z
=
x
+
iy
.
The first two are straightforward to see, by plugging it into formula and notice
the metric does not change.
We now look at the last one, given by z 7→ −
1
z
. The derivative at z is
f
0
(z) =
1
z
2
.
So we get
dz 7→ d
−
1
z
=
dz
z
2
.
So
d
−
1
z
2
=
|dz|
2
|z|
4
.
We also have
Im
−
1
z
= −
1
|z|
2
Im ¯z =
Im z
|z|
2
.
So
|d(−1/z)|
2
Im(−1/z)
2
=
|dz|
2
|z
4
|
(Im z)
2
|z|
4
=
|dz|
2
(Im z)
2
.
So this is an isometry, as required.
Note that each
z 7→ az
+
b
with
a >
0
, b ∈ R
is in
PSL
(2
, R
). Also, we can
use maps of this form to send any point to any other point. So
PSL
(2
, R
) acts
transitively on H. Moreover, everything in PSL(2, R) fixes R ∪ {∞}.
Recall also that each M¨obius transformation preserves circles and lines in the
complex plane, as well as angles between circles/lines. In particular, consider the
line
L
=
iR
, which meets
R
perpendicularly, and let
g ∈ PSL
(2
, R
). Then the
image is either a circle centered at a point in
R
, or a straight line perpendicular
to R.
We let
L
+
=
L ∩ H
=
{it
:
t >
0
}
. Then
g
(
L
+
) is either a vertical half-line
or a semi-circle that ends in R.
Definition (Hyperbolic lines). Hyperbolic lines in
H
are vertical half-lines or
semicircles ending in R.
We will now prove some lemmas to justify why we call these hyperbolic lines.
Lemma. Given any two distinct points
z
1
, z
2
∈ H
, there exists a unique
hyperbolic line through z
1
and z
2
.
Proof.
This is clear if
Re z
1
=
Re z
2
— we just pick the vertical half-line through
them, and it is clear this is the only possible choice.
Otherwise, if Re z
1
6= Re z
2
, then we can find the desired circle as follows:
R
z
1
z
2
It is also clear this is the only possible choice.
Lemma. PSL(2, R) acts transitively on the set of hyperbolic lines in H.
Proof.
It suffices to show that for each hyperbolic line
`
, there is some
g ∈
PSL
(2
, R
) such that
g
(
`
) =
L
+
. This is clear when
`
is a vertical half-line, since
we can just apply a horizontal translation.
If it is a semicircle, suppose it has end-points s < t ∈ R. Then consider
g(z) =
z − t
z − s
.
This has determinant
−s
+
t >
0. So
g ∈ PSL
(2
, R
). Then
g
(
t
) = 0 and
g
(
s
) =
∞
.
Then we must have
g
(
`
) =
L
+
, since
g
(
`
) is a hyperbolic line, and the only
hyperbolic lines passing through ∞ are the vertical half-lines. So done.
Moreover, we can achieve
g
(
s
) = 0 and
g
(
t
) =
∞
by composing with
−
1
z
.
Also, for any
P ∈ `
not on the endpoints, we can construct a
g
such that
g
(
P
) =
i ∈ L
+
, by composing with
z 7→ az
. So the isometries act transitively on
pairs (`, P ), where ` is a hyperbolic line and P ∈ `.
Definition (Hyperbolic distance). For points
z
1
, z
2
∈ H
, the hyperbolic distance
ρ
(
z
1
, z
2
) is the length of the segment [
z
1
, z
2
]
⊆ `
of the hyperbolic line through
z
1
, z
2
(parametrized monotonically).
Thus
PSL
(2
, R
) preserves hyperbolic distances. Similar to Euclidean space
and the sphere, we show these lines minimize distance.
Proposition. If
γ
: [0
,
1]
→ H
is a piecewise
C
1
-smooth curve with
γ
(0) =
z
1
, γ
(1) =
z
2
, then
length
(
γ
)
≥ ρ
(
z
1
, z
2
), with equality iff
γ
is a monotonic
parametrisation of [
z
1
, z
2
]
⊆ `
, where
`
is the hyperbolic line through
z
1
and
z
2
.
Proof.
We pick an isometry
g ∈ PSL
(2
, R
) so that
g
(
`
) =
L
+
. So without loss
of generality, we assume z
1
= iu and z
2
= iv, with u < v ∈ R.
We decompose the path as γ(t) = x(t) + iy(t). Then we have
length(γ) =
Z
1
0
1
y
p
˙x
2
+ ˙y
2
dt
≥
Z
1
0
|˙y|
y
dz
≥
Z
1
0
˙y
y
dt
= [log y(t)]
1
0
= log
v
u
This calculation also tells us that
ρ
(
z
1
, z
2
) =
log
v
u
. so
length
(
γ
)
≥ ρ
(
z
1
, z
2
)
with equality if and only if
x
(
t
) = 0 (hence
γ ⊆ L
+
) and
˙y ≥
0 (hence monotonic).
Corollary (Triangle inequality). Given three points z
1
, z
2
, z
3
∈ H, we have
ρ(z
1
, z
3
) ≤ ρ(z
1
, z
2
) + ρ(z
2
, z
3
),
with equality if and only if z
2
lies between z
1
and z
2
.
Hence, (H, ρ) is a metric space.
4.4 Geometry of the hyperbolic disk
So far, we have worked with the upper half-plane model. This is since the upper
half-plane model is more convenient for these calculations. However, sometimes
the disk model is more convenient. So we also want to understand that as well.
Recall that
ζ ∈ D 7→ z
=
i
1+ζ
1−ζ
∈ H
is an isometry, with an (isometric) inverse
z ∈ H 7→ ζ
=
z−i
z+i
∈ D
. Moreover, since these are M¨obius maps, circle-lines are
preserved, and angles between the lines are also preserved.
Hence, immediately from previous work on H, we know
(i) PSL(2, R)
∼
=
{M¨obius transformations sending D to itself} = G.
(ii)
Hyperbolic lines in
D
are circle segments meeting
|ζ|
= 1 orthogonally,
including diameters.
(iii) G
acts transitively on hyperbolic lines in
D
(and also on pairs consisting
of a line and a point on the line).
(iv)
The length-minimizing geodesics on
D
are a segments of hyperbolic lines
parametrized monotonically.
We write ρ for the (hyperbolic) distance in D.
Lemma. Let G be the set of isometries of the hyperbolic disk. Then
(i) Rotations z 7→ e
iθ
z (for θ ∈ R) are elements of G.
(ii) If a ∈ D, then g(z) =
z−a
1−¯az
is in G.
Proof.
(i)
This is clearly an isometry, since this is a linear map, preserves
|z|
and
|dz|, and hence also the metric
4|dz|
2
(1 − |z|
2
)
2
.
(ii)
First, we need to check this indeed maps
D
to itself. To do this, we first
make sure it sends {|z| = 1} to itself. If |z| = 1, then
|1 − ¯az| = |¯z(1 − ¯az)| = |¯z − ¯a| = |z −a|.
So
|g(z)| = 1.
Finally, it is easy to check
g
(
a
) = 0. By continuity,
G
must map
D
to itself.
We can then show it is an isometry by plugging it into the formula.
It is an exercise on the second example sheet to show all
g ∈ G
is of the form
g(z) = e
iθ
z − a
1 − ¯az
or
g(z) = e
iθ
¯z − a
1 − ¯a¯z
for some θ ∈ R and a ∈ D.
We shall now use the disk model to do something useful. We start by coming
up with an explicit formula for distances in the hyperbolic plane.
Proposition. If 0 ≤ r < 1, then
ρ(0, re
iθ
) = 2 tanh
−1
r.
In general, for z
1
, z
2
∈ D, we have
g(z
1
, z
2
) = 2 tanh
−1
z
1
− z
2
1 − ¯z
1
z
2
.
Proof.
By the lemma above, we can rotate the hyperbolic disk so that
re
iθ
is
rotated to r. So
ρ(0, re
iθ
) = ρ(0, r).
We can evaluate this by performing the integral
ρ(0, r) =
Z
r
0
2 dt
1 − t
2
= 2 tanh
−1
r.
For the general case, we apply the M¨obius transformation
g(z) =
z − z
1
1 − ¯z
1
z
.
Then we have
g(z
1
) = 0, g(z
2
) =
z
2
− z
1
1 − ¯z
1
z
2
=
z
1
− z
2
1 − ¯z
1
z
2
e
iθ
.
So
ρ(z
1
, z
2
) = ρ(g(z
1
), g(z
2
)) = 2 tanh
−1
z
1
− z
2
1 − ¯z
1
z
2
.
Again, we exploited the idea of performing the calculation in an easy case, and
then using isometries to move everything else to the easy case. In general, when
we have a “distinguished” point in the hyperbolic plane, it is often convenient to
use the disk model, move it to 0 by an isometry.
Proposition. For every point
P
and hyperbolic line
`
, with
P 6∈ `
, there is a
unique line
`
0
with
P ∈ `
0
such that
`
0
meets
`
orthogonally, say
` ∩ `
0
=
Q
, and
ρ(P, Q) ≤ ρ(P,
˜
Q) for all
˜
Q ∈ `.
This is a familiar fact from Euclidean geometry. To prove this, we again
apply the trick of letting P = 0.
Proof.
wlog, assume
P
= 0
∈ D
. Note that a line in
D
(that is not a diameter)
is a Euclidean circle. So it has a center, say C.
Since any line through
P
is a diameter, there is clearly only one line that
intersects
`
perpendicularly (recall angles in
D
is the same as the Euclidean
angle).
`
0
P
D
C
`
Q
It is also clear that
P Q
minimizes the Euclidean distance between
P
and
`
.
While this is not the same as the hyperbolic distance, since hyperbolic lines
through
P
are diameters, having a larger hyperbolic distance is equivalent to
having a higher Euclidean distance. So this indeed minimizes the distance.
How does reflection in hyperbolic lines work? This time, we work in the
upper half-plane model, since we have a favorite line L
+
.
Lemma (Hyperbolic reflection). Suppose
g
is an isometry of the hyperbolic
half-plane
H
and
g
fixes every point in
L
+
=
{iy
:
y ∈ R
+
}
. Then
G
is either
the identity or g(z) = −¯z, i.e. it is a reflection in the vertical axis L
+
.
Observe we have already proved a similar result in Euclidean geometry, and
the spherical version was proven in the first example sheet.
Proof.
For every
P ∈ H \ L
+
, there is a unique line
`
0
containing
P
such that
`
0
⊥ L
+
. Let Q = L
+
∩ `
0
.
L
+
`
0
P
Q
We see `
0
is a semicircle, and by definition of isometry, we must have
ρ(P, Q) = ρ(g(P), Q).
Now note that
g
(
`
0
) is also a line meeting
L
+
perpendicularly at
Q
, since
g
fixes
L
+
and preserves angles. So we must have
g
(
`
0
) =
`
0
. Then in particular
g
(
P
)
∈ `
0
. So we must have
g
(
P
) =
P
or
g
(
P
) =
P
0
, where
P
0
is the image
under reflection in L
+
.
Now it suffices to prove that if
g
(
P
) =
P
for any one
P
, then
g
(
P
) must be
the identity (if g(P ) = P
0
for all P , then g must be given by g(z) = −¯z).
Now suppose g(P ) = P , and let A ∈ H
+
, where H
+
= {z ∈ H : Re z > 0}.
L
+
P
P
0
A
A
0
B
Now if g(A) 6= A, then g(A) = A
0
. Then ρ(A
0
, P ) = ρ(A, P ). But
ρ(A
0
, P ) = ρ(A
0
, B) + ρ(B, P) = ρ(A, B) + ρ(B, P ) > ρ(A, P ),
by the triangle inequality, noting that
B 6∈
(
AP
). This is a contradiction. So
g
must fix everything.
Definition (Hyperbolic reflection). The map
R
:
z ∈ H 7→ −¯z ∈ H
is the
(hyperbolic) reflection in
L
+
. More generally, given any hyperbolic line
`
, let
T
be the isometry that sends ` to L
+
. Then the (hyperbolic) reflection in ` is
R
`
= T
−1
RT
Again, we already know how to reflect in
L
+
. So to reflect in another line
`
,
we move our plane such that ` becomes L
+
, do the reflection, and move back.
By the previous proposition,
R
`
is the unique isometry that is not identity
and fixes `.
4.5 Hyperb olic triangles
Definition (Hyperbolic triangle). A hyperbolic triangle
ABC
is the region
determined by three hyperbolic line segments
AB, BC
and
CA
, including extreme
cases where some vertices
A, B, C
are allowed to be “at infinity”. More precisely,
in the half-plane model, we allow them to lie in
R ∪ {∞}
; in the disk model we
allow them to lie on the unit circle |z| = 1.
We see that if A is “at infinity”, then the angle at A must be zero.
Recall for a region R ⊆ H, we can compute the area of R as
area(R) =
ZZ
R
dx dy
y
2
.
Similar to the sphere, we have
Theorem (Gauss-Bonnet theorem for hyperbolic triangles). For each hyperbolic
triangle ∆, say,
ABC
, with angles
α, β, γ ≥
0 (note that zero angle is possible),
we have
area(∆) = π − (α + β + γ).
Proof.
First do the case where
γ
= 0, so
C
is “at infinity”. Recall that we like
to use the disk model if we have a distinguished point in the hyperbolic plane.
If we have a distinguished point at infinity, it is often advantageous to use the
upper half plane model, since ∞ is a distinguished point at infinity.
So we use the upper-half plane model, and wlog
C
=
∞
(apply
PSL
(2
, R
))
if necessary. Then
AC
and
BC
are vertical half-lines. So
AB
is the arc of a
semi-circle. So AB is an arc of a semicircle.
C C
B
A
β
π − α
α
β
We use the transformation
z 7→ z
+
a
(with
a ∈ R
) to center the semi-circle at 0.
We then apply
z 7→ bz
(with
b >
0) to make the circle have radius 1. Thus wlog
AB ⊆ {x
2
+ y
2
= 1}.
Now we have
area(T ) =
Z
cos β
cos(π−α)
Z
∞
√
1−x
2
1
y
2
dy dx
=
Z
cos β
cos(π−α)
1
√
1 − x
2
dx
= [−cos
−1
(x)]
cos β
cos(π−α)
= π − α − β,
as required.
In general, we use
H
again, and we can arrange
AC
in a vertical half-line.
Also, we can move
AB
to
x
2
+
y
2
= 1, noting that this transformation keeps
AC vertical.
B
A
α
C
γ
δ
β
We consider ∆
1
= AB∞ and ∆
2
= CB∞. Then we can immediately write
area(∆
1
) = π − α − (β + δ)
area(∆
2
) = π − δ − (π − γ) = γ − δ.
So we have
area(T ) = area(∆
2
) − area(∆
1
) = π − α − β − γ,
as required.
Similar to the spherical case, we have some hyperbolic sine and cosine rules.
For example, we have
Theorem (Hyperbolic cosine rule). In a triangle with sides
a, b, c
and angles
α, β, γ, we have
cosh c = cosh a cosh b − sinh a sinh b cos γ.
Proof. See example sheet 2.
Recall that in
S
2
, any two lines meet (in two points). In the Euclidean plane
R
2
, any two lines meet (in one point) iff they are not parallel. Before we move
on to the hyperbolic case, we first make a definition.
Definition (Parallel lines). We use the disk model of the hyperbolic plane. Two
hyperbolic lines are parallel iff they meet only at the boundary of the disk (at
|z| = 1).
Definition (Ultraparallel lines). Two hyperbolic lines are ultraparallel if they
don’t meet anywhere in {|z| ≤ 1}.
In the Euclidean plane, we have the parallel axiom: given a line
`
and
P 6∈ `
,
there exists a unique line
`
0
containing
P
with
` ∩ `
0
=
∅
. This fails in both
S
2
and the hyperbolic plane — but for very different reasons! In
S
2
, there are no
such parallel lines. In the hyperbolic plane, there are many parallel lines. There
is a more deep reason for why this is the case, which we will come to at the very
end of the course.
4.6 Hyperb oloid model
Recall we said there is no way to view the hyperbolic plane as a subset of
R
3
,
and hence we need to mess with Riemannian metrics. However, it turns out we
can indeed embed the hyperbolic plane in R
3
, if we give R
3
a different metric!
Definition (Lorentzian inner product). The Lorentzian inner product on
R
3
has the matrix
1 0 0
0 1 0
0 0 −1
This is less arbitrary as it seems. Recall from IB Linear Algebra that we can
always pick a basis where a non-degenerate symmetric bilinear form has diagonal
made of 1 and
−
1. If we further identify
A
and
−A
as the “same” symmetric
bilinear form, then this is the only other possibility left.
Thus, we obtain the quadratic form given by
q(x) = hx, xi = x
2
+ y
2
− z
2
.
We now define the 2-sheet hyperboloid as
{x ∈ R
2
: q(x) = −1}.
This is given explicitly by the formula
x
2
+ y
2
= z
2
− 1.
We don’t actually need to two sheets. So we define
S
+
= S ∩ {z > 0}.
We let
π
:
S
+
→ D ⊆ C
=
R
2
be the stereographic projection from (0, 0, -1) by
π(x, y, z) =
x + iy
1 + z
= u + iv.
P
π(P )
We put r
2
= u
2
+ v
2
. Doing some calculations, we show that
(i) We always have r < 1, as promised.
(ii) The stereographic projection π is invertible with
σ(u, v) = π
−1
(u, v) =
1
1 − r
2
(2u, 2v, 1 + r
2
) ∈ S
+
.
(iii) The tangent plane to S
+
at P is spanned by
σ
u
=
∂σ
σu
, σ
v
=
∂σ
∂v
.
We can explicitly compute these to be
σ
u
=
2
(1 − r
2
)
2
(1 + u
2
− v
2
, 2uv, 2u),
σ
v
=
2
(1 − r
2
)
2
(2uv, 1 + v
2
− u
2
, 2v).
We restrict the inner product
h·, ·i
to the span of
σ
u
, σ
v
, and we get a
symmetric bilinear form assigned to each u, v ∈ D given by
E du
2
+ 2F du dv + G dv
2
,
where
E = hσ
u
, σ
u
i =
4
(1 − r
2
)
2
,
F = hσ
u
, σ
v
i = 0,
G = hσ
v
, σ
v
i =
4
(1 − r
2
)
2
.
We have thus recovered the Poincare disk model of the hyperbolic plane.
5 Smooth embedded surfaces (in R
3
)
5.1 Smooth embedded surfaces
So far, we have been studying some specific geometries, namely Euclidean, spheri-
cal and hyperbolic geometry. From now on, we go towards greater generality, and
study arbitrary surfaces. We will mostly work with surfaces that are smoothly
embedded as subsets of
R
3
, we can develop notions parallel to those we have had
before, such as Riemannian metrics and lengths. At the very end of the course,
we will move away from the needless restriction restriction that the surface is
embedded in R
3
, and study surfaces that just are.
Definition (Smooth embedded surface). A set
S ⊆ R
3
is a (parametrized)
smooth embedded surface if every point
P ∈ S
has an open neighbourhood
U ⊆ S
(with the subspace topology on
S ⊆ R
3
) and a map
σ
:
V → U
from an open
V ⊆ R
2
to U such that if we write σ(u, v) = (x(u, v), y(u, v), z(u, v)), then
(i) σ is a homeomorphism (i.e. a bijection with continuous inverse)
(ii) σ
is
C
∞
(smooth) on
V
(i.e. has continuous partial derivatives of all orders).
(iii)
For all
Q ∈ V
, the partial derivatives
σ
u
(
Q
) and
σ
v
(
Q
) are linearly inde-
pendent.
Recall that
σ
u
(Q) =
∂σ
∂u
(Q) =
∂x
∂u
∂y
∂u
∂z
∂u
(Q) = dσ
Q
(e
1
),
where e
1
, e
2
is the standard basis of R
2
. Similarly, we have
σ
v
(Q) = dσ
Q
(e
2
).
We define some terminology.
Definition (Smooth coordinates). We say (
u, v
) are smooth coordinates on
U ⊆ S.
Definition (Tangent space). The subspace of
R
3
spanned by
σ
u
(
Q
)
, σ
v
(
Q
) is
the tangent space T
P
S to S at P = σ(Q).
Definition (Smooth parametrisation). The function
σ
is a smooth parametrisa-
tion of U ⊆ S.
Definition (Chart). The function σ
−1
: U → V is a chart of U.
Proposition. Let
σ
:
V → U
and
˜σ
:
˜
V → U
be two
C
∞
parametrisations of a
surface. Then the homeomorphism
ϕ = σ
−1
◦ ˜σ :
˜
V → V
is in fact a diffeomorphism.
This proposition says any two parametrizations of the same surface are
compatible.
Proof.
Since differentiability is a local property, it suffices to consider
ϕ
on some
small neighbourhood of a point in
V
. Pick our favorite point (
v
0
, u
0
)
∈
˜
V
. We
know σ = σ(u, v) is differentiable. So it has a Jacobian matrix
x
u
x
v
y
u
y
v
z
y
z
v
.
By definition, this matrix has rank two at each point. wlog, we assume the first
two rows are linearly independent. So
det
x
u
x
v
y
u
y
v
6= 0
at (v
0
, u
0
) ∈
˜
V . We define a new function
F (u, v) =
x(u, v)
y(u, v)
.
Now the inverse function theorem applies. So
F
has a local
C
∞
inverse, i.e.
there are two open neighbourhoods (
u
0
, v
0
)
∈ N
and
F
(
u
0
, v
0
)
∈ N
0
⊆ R
2
such
that f : N → N
0
is a diffeomorphism.
Writing
π
:
˜σ → N
0
for the projection
π
(
x, y, z
) = (
x, y
) we can put these
things in a commutative diagram:
σ(N)
N N
0
π
F
σ
.
We now let
˜
N
=
˜σ
−1
(
σ
(
N
)) and
˜
F
=
π ◦ ˜σ
, which is yet again smooth. Then we
have the following larger commutative diagram.
σ(N)
N N
0
˜
N
π
F
σ
˜
F
˜σ
.
Then we have
ϕ = σ
−1
◦ ˜σ = σ
−1
◦ π
−1
◦ π ◦ ˜σ = F
−1
◦
˜
F ,
which is smooth, since
F
−1
and
˜
F
are. Hence
ϕ
is smooth everywhere. By
symmetry of the argument,
ϕ
−1
is smooth as well. So this is a diffeomorphism.
A more practical result is the following:
Corollary. The tangent plane T
Q
S is independent of parametrization.
Proof. We know
˜σ(˜u, ˜v) = σ(ϕ
1
(˜u, ˜v), ϕ
2
(˜u, ˜v)).
We can then compute the partial derivatives as
˜σ
˜u
= ϕ
1,˜u
σ
u
+ ϕ
2,˜u
σ
v
˜σ
˜v
= ϕ
1,˜v
σ
u
+ ϕ
2,˜v
σ
v
Here the transformation is related by the Jacobian matrix
ϕ
1,˜u
ϕ
1,˜v
ϕ
2,˜u
ϕ
2,˜v
= J(ϕ).
This is invertible since
ϕ
is a diffeomorphism. So (
σ
˜u
, σ
˜v
) and (
σ
u
, σ
v
) are
different basis of the same two-dimensional vector space. So done.
Note that we have
˜σ
˜u
× ˜σ
˜v
= det(J(ϕ))σ
u
× σ
v
.
So we can define
Definition (Unit normal). The unit normal to S at Q ∈ S is
N = N
Q
=
σ
u
× σ
v
kσ
u
× σ
v
k
,
which is well-defined up to a sign.
Often, instead of a parametrization
σ
:
V ⊆ R
2
→ U ⊆ S
, we want the
function the other way round. We call this a chart.
Definition (Chart). Let
S ⊆ R
3
be an embedded surface. The map
θ
=
σ
−1
:
U ⊆ S → V ⊆ R
2
is a chart.
Example. Let
S
2
⊆ R
3
be a sphere. The two stereographic projections from
±e
3
give two charts, whose domain together cover S
2
.
Similar to what we did to the sphere, given a chart
θ
:
U → V ⊆ R
2
, we can
induce a Riemannian metric on
V
. We first get an inner product on the tangent
space as follows:
Definition (First fundamental form). If
S ⊆ R
3
is an embedded surface, then
each
T
Q
S
for
Q ∈ S
has an inner product from
R
3
, i.e. we have a family of inner
products, one for each point. We call this family the first fundamental form.
This is a theoretical entity, and is more easily worked with when we have
a chart. Suppose we have a parametrization
σ
:
V → U ⊆ S
, a
,
b
∈ R
2
, and
P ∈ V . We can then define
ha, bi
P
= hdσ
P
(a), dσ
P
(b)i
R
3
.
With respect to the standard basis e
1
,
e
2
∈ R
2
, we can write the first fundamental
form as
E du
2
+ 2F du dv + G dv
2
,
where
E = hσ
u
, σ
u
i = he
1
, e
1
i
P
F = hσ
u
, σ
v
i = he
1
, e
2
i
P
G = hσ
v
, σ
v
i = he
2
, e
2
i
P
.
Thus, this induces a Riemannian metric on
V
. This is also called the first
fundamental form corresponding to
σ
. This is what we do in practical examples.
We will assume the following property, which we are not bothered to prove.
Proposition. If we have two parametrizations related by
˜σ
=
σ ◦ ϕ
:
˜
V → U
,
then ϕ :
˜
V → V is an isometry of Riemannian metrics (on V and
˜
V ).
Definition (Length and energy of curve). Given a smooth curve Γ : [
a, b
]
→
S ⊆ R
3
, the length of γ is
length(Γ) =
Z
b
a
kΓ
0
(t)k dt.
The energy of the curve is
energy(Γ) =
Z
b
a
kΓ
0
(t)k
2
dt.
We can think of the energy as something like the kinetic energy of a particle
along the path, except that we are missing the factor of
1
2
m
, because they are
annoying.
How does this work with parametrizations? For the sake of simplicity, we
assume Γ([
a, b
])
⊆ U
for some parametrization
σ
:
V → U
. Then we define the
new curve
γ = σ
−1
◦ Γ : [a, b] → V.
This curve has two components, say γ = (γ
1
, γ
2
). Then we have
Γ
0
(t) = (dσ)
γ(t)
( ˙γ
1
(t)e
1
+ ˙γ
2
(t)e
2
) = ˙γ
1
σ
u
+ ˙γ
2
σ
v
,
and thus
kΓ
0
(t)k = h˙γ, ˙γi
1
2
P
= (E ˙γ
2
1
+ 2F ˙γ
1
˙γ
2
+ G ˙γ
2
2
)
1
2
.
So we get
length Γ =
Z
b
a
(E ˙γ
2
1
+ 2F ˙γ
1
˙γ
2
+ G ˙γ
2
2
)
1
2
dt.
Similarly, the energy is given by
energy Γ =
Z
b
a
(E ˙γ
2
1
+ 2F ˙γ
1
˙γ
2
+ G ˙γ
2
2
) dt.
This agrees with what we’ve had for Riemannian metrics.
Definition (Area). Given a smooth
C
∞
parametrization
σ
:
V → U ⊆ S ⊆ R
3
,
and a region T ⊆ U, we define the area of T to be
area(T ) =
Z
θ(T )
p
EG −F
2
du dv,
whenever the integral exists (where θ = σ
−1
is a chart).
Proposition. The area of
T
is independent of the choice of parametrization.
So it extends to more general subsets
T ⊆ S
, not necessarily living in the image
of a parametrization.
Proof. Exercise!
Note that in examples,
σ
(
V
) =
U
often is a dense set in
S
. For example, if
we work with the sphere, we can easily parametrize everything but the poles. In
that case, it suffices to use just one parametrization σ for area(S).
Note also that areas are invariant under isometries.
5.2 Geodesics
We now come to the important idea of a geodesic. We will first define these for
Riemannian metrics, and then generalize it to general embedded surfaces.
Definition (Geodesic). Let
V ⊆ R
2
u,v
be open, and
E
d
u
2
+ 2
F
d
u
d
v
+
G
d
v
2
be a Riemannian metric on V . We let
γ = (γ
1
, γ
2
) : [a, b] → V
be a smooth curve. We say
γ
is a geodesic with respect to the Riemannian metric
if it satisfies
d
dt
(E ˙γ
1
+ F ˙γ
2
) =
1
2
(E
u
˙γ
2
1
+ 2F
u
˙γ
1
˙γ
2
+ G
u
˙γ
2
2
)
d
dt
(F ˙γ
1
+ G ˙γ
2
) =
1
2
(E
v
˙γ
2
1
+ 2F
v
˙γ
1
˙γ
2
+ G
v
˙γ
2
2
)
for all t ∈ [a, b]. These equations are known as the geodesic ODEs.
What exactly do these equations mean? We will soon show that these are
curves that minimize (more precisely, are stationary points of) energy. To do so,
we need to come up with a way of describing what it means for
γ
to minimize
energy among all possible curves.
Definition (Proper variation). Let
γ
: [
a, b
]
→ V
be a smooth curve, and let
γ(a) = p and γ(b) = q. A proper variation of γ is a C
∞
map
h : [a, b] ×(−ε, ε) ⊆ R
2
→ V
such that
h(t, 0) = γ(t) for all t ∈ [a, b],
and
h(a, τ ) = p, h(b, τ) = q for all |τ | < ε,
and that
γ
τ
= h( ·, τ) : [a, b] → V
is a C
∞
curve for all fixed τ ∈ (−ε, ε).
Proposition. A smooth curve
γ
satisfies the geodesic ODEs if and only if
γ
is
a stationary point of the energy function for all proper variation, i.e. if we define
the function
E(τ) = energy(γ
τ
) : (−ε, ε) → R,
then
dE
dτ
τ=0
= 0.
Proof. We let γ(t) = (u(t), v(t)). Then we have
energy(γ) =
Z
b
a
(E(u, v) ˙u
2
+ 2F (u, v) ˙u ˙v + G(u, v) ˙v
2
) dt =
Z
b
a
I(u, v, ˙u, ˙v) dt.
We consider this as a function of four variables
u, ˙u, v, ˙v
, which are not necessarily
related to one another. From the calculus of variations, we know
γ
is stationary
if and only if
d
dt
∂I
∂ ˙u
=
∂I
∂u
,
d
dt
∂I
∂ ˙v
=
∂I
∂v
.
The first equation gives us
d
dt
(2(E ˙u + F ˙v)) = E
u
˙u
2
+ 2F
u
˙u ˙v + G
u
˙v
2
,
which is exactly the geodesic ODE. Similarly, the second equation gives the other
geodesic ODE. So done.
Since the definition of a geodesic involves the derivative only, which is a local
property, we can easily generalize the definition to arbitrary embedded surfaces.
Definition (Geodesic on smooth embedded surface). Let
S ⊆ R
3
be an embed-
ded surface. Let Γ : [
a, b
]
→ S
be a smooth curve in
S
, and suppose there is a
parametrization
σ
:
V → U ⊆ S
such that
im
Γ
⊆ U
. We let
θ
=
σ
−1
be the
corresponding chart.
We define a new curve in V by
γ = θ ◦Γ : [a, b] → V.
Then we say Γ is a geodesic on
S
if and only if
γ
is a geodesic with respect to
the induced Riemannian metric.
For a general Γ : [
a, b
]
→ V
, we say Γ is a geodesic if for each point
t
0
∈
[
a, b
],
there is a neighbourhood
˜
V
of
t
0
such that
im
Γ
|
˜
V
lies in the domain of some
chart, and Γ|
˜
V
is a geodesic in the previous sense.
Corollary. If a curve Γ minimizes the energy among all curves from
P
= Γ(
a
)
to Q = Γ(b), then Γ is a geodesic.
Proof.
For any
a
1
, a
2
such that
a ≤ a
1
≤ b
1
≤ b
, we let Γ
1
= Γ
|
[a
1
,b
1
]
. Then Γ
1
also minimizes the energy between
a
1
and
b
1
for all curves between Γ(
a
1
) and
Γ(b
1
).
If we picked
a
1
, b
1
such that Γ([
a
1
, b
1
])
⊆ U
for some parametrized neigh-
bourhood
U
, then Γ
1
is a geodesic by the previous proposition. Since the
parametrized neighbourhoods cover
S
, at each point
t
0
∈
[
a, b
], we can find
a
1
, b
1
such that Γ([a
1
, b
1
]) ⊆ U. So done.
This is good, but we can do better. To do so, we need a lemma.
Lemma. Let
V ⊆ R
2
be an open set with a Riemannian metric, and let
P, Q ∈ V
.
Consider
C
∞
curves
γ
: [
a, b
]
→ V
such that
γ
(0) =
P, γ
(1) =
Q
. Then such a
γ
will minimize the energy (and therefore is a geodesic) if and only if
γ
minimizes
the length and has constant speed.
This means being a geodesic is almost the same as minimizing length. It’s
just that to be a geodesic, we have to parametrize it carefully.
Proof.
Recall the Cauchy-Schwartz inequality for continuous functions
f, g ∈
C[0, 1], which says
Z
1
0
f(x)g(x) dx
2
≤
Z
1
0
f(x)
2
dx
Z
1
0
g(x)
2
dx
,
with equality iff
g
=
λf
for some
λ ∈ R
, or
f
= 0, i.e.
g
and
f
are linearly
dependent.
We now put f = 1 and g = k˙γk. Then Cauchy-Schwartz says
(length γ)
2
≤ energy(γ),
with equality if and only if ˙γ is constant.
From this, we see that a curve of minimal energy must have constant speed.
Then it follows that minimizing energy is the same as minimizing length if we
move at constant speed.
Is the converse true? Are all geodesics length minimizing? The answer is
“almost”. We have to be careful with our conditions in order for it to be true.
Proposition. A curve Γ is a geodesic iff and only if it minimizes the energy
locally, and this happens if it minimizes the length locally and has constant
speed.
Here minimizing a quantity locally means for every
t ∈
[
a, b
], there is some
ε > 0 such that Γ|
[t−ε,t+ε]
minimizes the quantity.
We will not prove this. Local minimization is the best we can hope for, since
the definition of a geodesic involves differentiation, and derivatives are local
properties.
Proposition. In fact, the geodesic ODEs imply kΓ
0
(t)k is constant.
We will also not prove this, but in the special case of the hyperbolic plane,
we can check this directly. This is an exercise on the third example sheet.
A natural question to ask is that if we pick a point
P
and a tangent direction
a, can we find a geodesic through P whose tangent vector at P is a?
In the geodesic equations, if we expand out the derivative, we can write the
equation as
E F
F G
¨γ
1
¨γ
2
= something.
Since the Riemannian metric is positive definite, we can invert the matrix and
get an equation of the form
¨γ
1
¨γ
2
= H(γ
1
, γ
2
, ˙γ
1
,
˙
γ
2
)
for some function
H
. From the general theory of ODE’s in IB Analysis II, subject
to some sensible conditions, given any
P
= (
u
0
, v
0
)
∈ V
and a = (
p
0
, q
0
)
∈ R
2
,
there is a unique geodesic curve
γ
(
t
) defined for
|t| < ε
with
γ
(0) =
P
and
˙γ
(0) = a. In other words, we can choose a point, and a direction, and then there
is a geodesic going that way.
Note that we need the restriction that
γ
is defined only for
|t| < ε
since we
might run off to the boundary in finite time. So we need not be able to define it
for all t ∈ R.
How is this result useful? We can use the uniqueness part to find geodesics.
We can try to find some family of curves
C
that are length-minimizing. To prove
that we have found all of them, we can show that given any point
P ∈ V
and
direction a, there is some curve in C through P with direction a.
Example. Consider the sphere
S
2
. Recall that arcs of great circles are length-
minimizing, at least locally. So these are indeed geodesics. Are these all? We
know for any
P ∈ S
2
and any tangent direction, there exists a unique great
circle through
P
in this direction. So there cannot be any other geodesics on
S
2
,
by uniqueness.
Similarly, we find that hyperbolic line are precisely all the geodesics on a
hyperbolic plane.
We have defined these geodesics as solutions of certain ODEs. It is possible
to show that the solutions of these ODEs depend
C
∞
-smoothly on the initial
conditions. We shall use this to construct around each point
P ∈ S
in a surface
geodesic polar coordinates. The idea is that to specify a point near
P
, we can
just say “go in direction
θ
, and then move along the corresponding geodesic for
time r”.
We can make this (slightly) more precise, and provide a quick sketch of how
we can do this formally. We let
ψ
:
U → V
be some chart with
P ∈ U ⊆ S
. We
wlog
ψ
(
P
) = 0
∈ V ⊆ R
2
. We denote by
θ
the polar angle (coordinate), defined
on V \ {0}.
θ
Then for any given
θ
, there is a unique geodesic
γ
θ
: (
−ε, ε
)
→ V
such that
γ
θ
(0) = 0, and ˙γ
θ
(0) is the unit vector in the θ direction.
We define
σ(r, θ) = γ
θ
(r)
whenever this is defined. It is possible to check that
σ
is
C
∞
-smooth. While
we would like to say that
σ
gives us a parametrization, this is not exactly true,
since we cannot define
θ
continuously. Instead, for each
θ
0
, we define the region
W
θ
0
= {(r, θ) : 0 < r < ε, θ
0
< θ < θ
0
+ 2π} ⊆ R
2
.
Writing V
0
for the image of W
θ
0
under σ, the composition
W
θ
0
V
0
U
0
⊆ S
σ
ψ
−1
is a valid parametrization. Thus σ
−1
◦ ψ is a valid chart.
The image (
r, θ
) of this chart are the geodesic polar coordinates. We have
the following lemma:
Lemma (Gauss’ lemma). The geodesic circles
{r
=
r
0
} ⊆ W
are orthogonal to
their radii, i.e. to
γ
θ
, and the Riemannian metric (first fundamental form) on
W
is
dr
2
+ G(r, θ) dθ
2
.
This is why we like geodesic polar coordinates. Using these, we can put the
Riemannian metric into a very simple form.
Of course, this is just a sketch of what really happens, and there are many
holes to fill in. For more details, go to IID Differential Geometry.
Definition (Atlas). An atlas is a collection of charts covering the whole surface.
The collection of all geodesic polars about all points give us an example.
Other interesting atlases are left as an exercise on example sheet 3.
5.3 Surfaces of revolution
So far, we do not have many examples of surfaces. We now describe a nice way
of obtaining surfaces — we obtain a surface
S
by rotating a plane curve
η
around
a line
`
. We may wlog assume that coordinates a chosen so that
`
is the
z
-axis,
and η lies in the x − z plane.
More precisely, we let η : (a, b) → R
3
, and write
η(u) = (f(u), 0, g(u)).
Note that it is possible that a = −∞ and/or b = ∞.
We require
kη
0
(
u
)
k
= 1 for all
u
. This is sometimes known as parametrization
by arclength. We also require
f
(
u
)
>
0 for all
u >
0, or else things won’t make
sense.
Finally, we require that
η
is a homeomorphism to its image. This is more
than requiring η to be injective. This is to eliminate things like
Then S is the image of the following map:
σ(u, v) = (f(u) cos v, f(u) sin v, g(u))
for
a < u < b
and 0
≤ v ≤
2
π
. This is not exactly a parametrization, since it is
not injective (
v
= 0 and
v
= 2
π
give the same points). To rectify this, for each
α ∈ R, we define
σ
α
: (a, b) ×(α, α + 2π) → S,
given by the same formula, and this is a homeomorphism onto the image. The
proof of this is left as an exercise for the reader.
Assuming this, we now show that this is indeed a parametrization. It is
evidently smooth, since
f
and
g
both are. To show this is a parametrization,
we need to show that the partial derivatives are linearly independent. We can
compute the partial derivatives and show that they are non-zero. We have
σ
u
= (f
0
cos v, f
0
sin v, g
0
)
σ
v
= (−f sin v, f cos v, 0).
We then compute the cross product as
σ
u
× σ
v
= (−fg
0
cos v, −fg
0
sin v, ff
0
).
So we have
kσ
u
× σ
v
k = f
2
(g
02
+ f
02
) = f
2
6= 0.
Thus every σ
α
is a valid parametrization, and S is a valid embedded surface.
More generally, we can allow
S
to be covered by several families of parametriza-
tions of type
σ
α
, i.e. we can consider more than one curve or more than one axis
of rotation. This allows us to obtain, say,
S
2
or the embedded torus (in the old
sense, we cannot view
S
2
as a surface of revolution in the obvious way, since we
will be missing the poles).
Definition (Parallels). On a surface of revolution, parallels are curves of the
form
γ(t) = σ(u
0
, t) for fixed u
0
.
Meridians are curves of the form
γ(t) = σ(t, v
0
) for fixed v
0
.
These are generalizations of the notions of longitude and latitude (in some
order) on Earth.
In a general surface of revolution, we can compute the first fundamental form
with respect to σ as
E = kσ
u
k
2
= f
02
+ g
02
= 1,
F = σ
u
· σ
v
= 0
G = kσ
v
k
2
= f
2
.
So its first fundamental form is also of the simple form, like the geodesic polar
coordinates.
Putting these explicit expressions into the geodesic formula, we find that the
geodesic equations are
¨u = f
df
du
˙v
2
d
dt
(f
2
˙v) = 0.
Proposition. We assume k˙γk = 1, i.e. ˙u
2
+ f
2
(u) ˙v
2
= 1.
(i) Every unit speed meridians is a geodesic.
(ii) A (unit speed) parallel will be a geodesic if and only if
df
du
(u
0
) = 0,
i.e. u
0
is a critical point for f.
Proof.
(i)
In a meridian,
v
=
v
0
is constant. So the second equation holds. Also, we
know k˙γk = |˙u| = 1. So ¨u = 0. So the first geodesic equation is satisfied.
(ii) Since o = o
u
, we know f (u
0
)
2
˙v
2
= 1. So
˙v = ±
1
f(u
0
)
.
So the second equation holds. Since
˙v
and
f
are non-zero, the first equation
is satisfied if and only if
df
du
= 0.
5.4 Gaussian curvature
We will next consider the notion of curvature. Intuitively, Euclidean space is
“flat”, while the sphere is “curved”. In this section, we will define a quantity
known as the curvature that characterizes how curved a surface is.
The definition itself is not too intuitive. So what we will do is that we first
study the curvature of curves, which is something we already know from, say, IA
Vector Calculus. Afterwards, we will make an analogous definition for surfaces.
Definition (Curvature of curve). We let
η
: [0
, `
]
→ R
2
be a curve parametrized
with unit speed, i.e.
kη
0
k
= 1. The curvature
κ
at the point
η
(
s
) is determined
by
η
00
= κn,
where n is the unit normal, chosen so that κ is non-negative.
If
f
: [
c, d
]
→
[0
, `
] is a smooth function and
f
0
(
t
)
>
0 for all
t
, then we can
reparametrize our curve to get
γ(t) = η(f(t)).
We can then find
˙γ(t) =
df
dt
η
0
(f(t)).
So we have
k˙γk
2
=
df
dt
2
.
We also have by definition
η
00
(f(t)) = κn,
where κ is the curvature at γ(t).
On the other hand, Taylor’s theorem tells us
γ(t + ∆t) −γ(t) =
df
dt
η
0
(f(t))∆t
+
1
2
"
d
2
f
dt
2
η
0
(f(t)) +
df
dt
2
η
00
(f(t))
#
+ higher order terms.
Now we know by assumption that
η
0
· η
0
= 1.
Differentiating thus give s
η
0
· η
00
= 0.
Hence we get
η
0
· n = 0.
We now take the dot product of the Taylor expansion with n, killing off all the
η
0
terms. Then we get
(γ(t + ∆t) −γ(t)) · n =
1
2
κk˙γk
2
(∆t)
2
+ ··· , (∗)
where κ is the curvature. This is the distance denoted below:
γ(t)
γ(t + ∆t)
(γ(t + ∆t) −γ(t)) · n
We can also compute
kγ(t + ∆t) −γ(t)k
2
= k˙γk
2
(∆t)
2
. (†)
So we find that
1
2
κ
is the ratio of the leading (quadratic) terms of (
∗
) and (
†
),
and is independent of the choice of parametrization.
We now try to apply this thinking to embedded surfaces. We let
σ
:
V →
U ⊆ S
be a parametrization of a surface
S
(with
V ⊆ R
2
open). We apply
Taylor’s theorem to σ to get
σ(u + ∆u, v + ∆v) − σ(u, v) = σ
u
∆u + σ
v
∆v
+
1
2
(σ
uu
(∆u
2
) + 2σ
uv
∆u∆v + σ
vv
(∆v)
2
) + ··· .
We now measure the deviation from the tangent plane, i.e.
(σ(u + ∆u, v + ∆v) − σ(u, v)) · N =
1
2
(L(∆u)
2
+ 2M ∆u∆v + N (∆v)
2
) + ··· ,
where
L = σ
uu
· N,
M = σ
uv
· N,
N = σ
vv
· N.
Note that N and
N
are different things. N is the unit normal, while
N
is the
expression given above.
We can also compute
kσ(u + ∆u, v + ∆v) − σ(u, v)k
2
= E(∆u)
2
+ 2F ∆u∆v + G(∆v)
2
+ ··· .
We now define the second fundamental form as
Definition (Second fundamental form). The second fundamental form on
V
with σ : V → U ⊆ S for S is
L du
2
+ 2M du dv + N dv
2
,
where
L = σ
uu
· N
M = σ
uv
· N
N = σ
vv
· N.
Definition (Gaussian curvature). The Gaussian curvature
K
of a surface of
S
at P ∈ S is the ratio of the determinants of the two fundamental forms, i.e.
K =
LN − M
2
EG −F
2
.
This is valid since the first fundamental form is positive-definite and in particular
has non-zero derivative.
We can imagine that
K
is a capital
κ
, but it looks just like a normal capital
K.
Note that
K >
0 means the second fundamental form is definite (i.e. either
positive definite or negative definite). If
K <
0, then the second fundamental
form is indefinite. If
K
= 0, then the second fundamental form is semi-definite
(but not definite).
Example. Consider the unit sphere
S
2
⊆ R
3
. This has
K >
0 at each point.
We can compute this directly, or we can, for the moment, pretend that
M
= 0.
Then by symmetry, N = M. So K > 0.
On the other hand, we can imagine a Pringle crisp (also known as a hyperbolic
paraboloid), and this has
K <
0. More examples are left on the third example
sheet. For example we will see that the embedded torus in
R
3
has points at
which K > 0, some where K < 0, and others where K = 0.
It can be deduced, similar to the curves, that
K
is independent of parametriza-
tion.
Recall that around each point, we can get some nice coordinates where the
first fundamental form looks simple. We might expect the second fundamental
form to look simple as well. That is indeed true, but we need to do some
preparation first.
Proposition. We let
N =
σ
u
× σ
v
kσ
u
× σ
v
k
be our unit normal for a surface patch. Then at each point, we have
N
u
= aσ
u
+ bσ
v
,
N
v
= cσ
u
+ dσ
v
,
where
−
L M
M N
=
a b
c d
E F
F G
.
In particular,
K = ad − bc.
Proof. Note that
N · N = 1.
Differentiating gives
N · N
u
= 0 = N · N
v
.
Since
σ
u
, σ
v
and N for an orthogonal basis, at least there are some
a, b, c, d
such
that
N
u
= aσ
u
+ bσ
v
N
v
= cσ
u
+ dσ
v
.
By definition of σ
u
, we have
N · σ
u
= 0.
So differentiating gives
N
u
· σ
u
+ N · σ
uu
= 0.
So we know
N
u
· σ
u
= −L.
Similarly, we find
N
u
= σ
v
= −M = N
v
· σ
u
, N
v
· σ
v
= −N.
We dot our original definition of N
u
,
N
v
in terms of
a, b, c, d
with
σ
u
and
σ
v
to
obtain
−L = aE + bF −M = aF + bG
−M = cE + dF −N = cF + dG.
Taking determinants, we get the formula for the curvature.
If we have nice coordinates on
S
, then we get a nice formula for the Gaussian
curvature K.
Theorem. Suppose for a parametrization
σ
:
V → U ⊆ S ⊆ R
3
, the first
fundamental form is given by
du
2
+ G(u, v) dv
2
for some G ∈ C
∞
(V ). Then the Gaussian curvature is given by
K =
−(
√
G)
uu
√
G
.
In particular, we do not need to compute the second fundamental form of the
surface.
This is purely a technical result.
Proof. We set
e = σ
u
, f =
σ
v
√
G
.
Then e and f are unit and orthogonal. We also let N = e
×
f be a third unit
vector orthogonal to e and f so that they form a basis of R
3
.
Using the notation of the previous proposition, we have
N
u
× N
v
= (aσ
u
+ bσ
v
) × (cσ
u
+ dσ
v
)
= (ad − bc)σ
u
× σ
v
= Kσ
u
× σ
v
= K
√
Ge × f
= K
√
GN.
Thus we know
K
√
G = (N
u
× N
v
) · N
= (N
u
× N
v
) · (e × f)
= (N
u
· e)(N
v
· f ) − (N
u
· f )(N
v
· e).
Since N · e = 0, we know
N
u
· e + N · e
u
= 0.
Hence to evaluate the expression above, it suffices to compute N
·
e
u
instead of
N
u
· e.
Since e · e = 1, we know
e · e
u
= 0 = e · e
v
.
So we can write
e
u
= αf + λ
1
N
e
v
= βf + λ
2
N.
Similarly, we have
f
u
= −˜αe + µ
1
N
f
v
= −
˜
βe + µ
2
N.
Our objective now is to find the coefficients µ
i
, λ
i
, and then
K
√
G = λ
1
µ
2
− λ
2
µ
1
.
Since we know e · f = 0, differentiating gives
e
u
· f + e · f
u
= 0
e
v
· f + e · f
v
= 0.
Thus we get
˜α = α,
˜
β = β.
But we have
α = e
u
· f = σ
uu
·
σ
v
√
G
=
(σ
u
· σ
v
)
u
−
1
2
(σ
u
· σ
u
)
v
1
√
G
= 0,
since σ
u
· σ
v
= 0, σ
u
· σ
u
= 1. So α vanishes.
Also, we have
β = e
v
· f = σ
uv
·
σ
v
√
G
=
1
2
G
u
√
G
= (
√
G)
u
.
Finally, we can use our equations again to find
λ
1
µ
1
− λ
2
µ
1
= e
u
· f
v
− e
v
· f
u
= (e · f
v
)
u
− (e · f
u
)
v
= −
˜
β
u
− (−˜α)
u
= −(
√
G)
uu
.
So we have
K
√
G = −(
√
G)
uu
,
as required. Phew.
Observe, for
σ
as in the previous theorem,
K
depends only on the first
fundamental from, not on the second fundamental form. When Gauss discovered
this, he was so impressed that he called it the Theorema Egregium, which means
Corollary (Theorema Egregium). If
S
1
and
S
2
have locally isometric charts,
then K is locally the same.
Proof.
We know that this corollary is valid under the assumption of the previous
theorem, i.e. the existence of a parametrization
σ
of the surface
S
such that the
first fundamental form is
du
2
+ G(u, v) dv
2
.
Suitable
σ
includes, for each point
P ∈ S
, the geodesic polars (
ρ, θ
). However,
P
itself is not in the chart, i.e.
P 6∈ σ
(
U
), and there is no guarantee that there
will be some geodesic polar that covers
P
. To solve this problem, we notice that
K
is a
C
∞
function of
S
, and in particular continuous. So we can determine the
curvature at P as
K(P ) = lim
ρ→0
K(ρ, σ).
So done.
Note also that every surface of revolution has such a suitable parametrization,
as we have previously explicitly seen.
6 Abstract smooth surfaces
While embedded surfaces are quite general surfaces, they do not include, say,
the hyperbolic plane. We can generalize our notions by considering surfaces
“without embedding in R
3
”. These are known as abstract surfaces.
Definition (Abstract smooth surface). An abstract smooth surface
S
is a metric
space (or Hausdorff (and second-countable) topological space) equipped with
homeomorphisms
θ
i
:
U
i
→ V
i
, where
U
i
⊆ S
and
V
i
⊆ R
2
are open sets such
that
(i) S =
S
i
U
i
(ii) For any i, j, the transition map
φ
ij
= θ
j
◦ θ
−1
i
: θ
j
(U
i
∩ U
j
) → θ
i
(U
i
∩ U
j
)
is a diffeomorphism. Note that
θ
j
(
U
i
∩ U
j
) and
θ
i
(
U
i
∩ U
j
) are open
sets in
R
2
. So it makes sense to talk about whether the function is a
diffeomorphism.
Like for embedded surfaces, the maps
θ
i
are called charts, and the collection
of θ
i
’s satisfying our conditions is an atlas etc.
Definition (Riemannian metric on abstract surface). A Riemannian metric on
an abstract surface is given by Riemannian metrics on each
V
i
=
θ
i
(
U
i
) subject to
the compatibility condition that for all
i, j
, the transition map
φ
ij
is an isometry,
i.e.
hdϕ
P
(a), dϕ
P
(b)i
ϕ(P )
= ha, bi
P
Note that on the left, we are computing the Riemannian metric on
V
i
, while on
the left, we are computing it on V
j
.
Then we can define lengths, areas, energies on an abstract surface S.
It is clear that every embedded surface is an abstract surface, by forgetting
that it is embedded in R
3
.
Example. The three classical geometries are all abstract surfaces.
(i) The Euclidean space R
2
with dx
2
+ dy
2
is an abstract surface.
(ii)
The sphere
S
2
⊆ R
2
, being an embedded surface, is an abstract surface
with metric
4(dx
2
+ dy
2
)
(1 + x
2
+ y
2
)
2
.
(iii) The hyperbolic disc D ⊆ R
2
is an abstract surface with metric
4(dx
2
+ dy
2
)
(1 − x
2
− y
2
)
2
.
and this is isometric to the upper half plane H with metric
dx
2
+ dy
2
y
2
Note that in the first and last example, it was sufficient to use just one chart
to cover every point of the surface, but not for the sphere. Also, in the case of the
hyperbolic plane, we can have many different charts, and they are compatible.
Finally, we notice that we really need the notion of abstract surface for the
hyperbolic plane, since it cannot be realized as an embedded surface in
R
3
. The
proof is not obvious at all, and is a theorem of Hilbert.
One important thing we can do is to study the curvature of surfaces.
Given a
P ∈ S
, the Riemannian metric (on a chart) around
P
determines
a “reparametrization” by geodesics, similar to embedded surfaces. Then the
metric takes the form
dρ
2
+ G(ρ, θ) dθ
2
.
We then define the curvature as
K =
−(
√
G)
ρρ
√
G
.
Note that for embedded surfaces, we obtained this formula as a theorem. For
abstract surfaces, we take this as a definition.
We can check how this works in some familiar examples.
Example.
(i) In R
2
, we use the usual polar coordinates (ρ, θ), and the metric becomes
dρ
2
+ ρ
2
dθ
2
,
where x = ρ cos θ and y = ρ sin θ. So the curvature is
−(
√
G)
ρρ
√
G
=
−(ρ)
ρρ
ρ
= 0.
So the Euclidean space has zero curvature.
(ii)
For the sphere
S
, we use the spherical coordinates, fixing the radius to be
1. So we specify each point by
σ(ρ, θ) = (sin ρ cos θ, sin ρ sin θ, cos ρ).
Note that
ρ
is not really the radius in spherical coordinates, but just one
of the angle coordinates. We then have the metric
dρ
2
+ sin
2
ρ dθ
2
.
Then we get
√
G = sin ρ,
and K = 1.
(iii)
For the hyperbolic plane, we use the disk model
D
, and we first express
our original metric in polar coordinates of the Euclidean plane to get
2
1 − r
2
2
(dr
2
+ r
2
dθ
2
).
This is not geodesic polar coordinates, since
r
is given by the Euclidean
distance, not hyperbolic distance. We will need to put
ρ = 2 tanh
−1
r, dρ =
2
1 − r
2
dr.
Then we have
r = tanh
ρ
2
,
which gives
4r
2
(1 − r
2
)
2
= sinh
2
ρ.
So we finally get
√
G = sinh ρ,
with
K = −1.
We see that the three classic geometries are characterized by having constant
0, 1 and −1 curvatures.
We are almost able to state the Gauss-Bonnet theorem. Before that, we need
the notion of triangulations. We notice that our old definition makes sense for
(compact) abstract surfaces
S
. So we just use the same definition. We then
define the Euler number of an abstract surface as
e(S) = F − E + V,
as before. Assuming that the Euler number is independent of triangulations, we
know that this is invariant under homeomorphisms.
Theorem (Gauss-Bonnet theorem). If the sides of a triangle
ABC ⊆ S
are
geodesic segments, then
Z
ABC
K dA = (α + β + γ) − π,
where
α, β, γ
are the angles of the triangle, and d
A
is the “area element” given
by
dA =
p
EG −F
2
du dv,
on each domain
U ⊆ S
of a chart, with
E, F, G
as in the respective first
fundamental form.
Moreover, if S is a compact surface, then
Z
S
K dA = 2πe(S).
We will not prove this theorem, but we will make some remarks. Note that
we can deduce the second part from the first part. The basic idea is to take a
triangulation of
S
, and then use things like each edge belongs to two triangles
and each triangle has three edges.
This is a genuine generalization of what we previously had for the sphere
and hyperbolic plane, as one can easily see.
Using the Gauss-Bonnet theorem, we can define the curvature
K
(
P
) for a
point
P ∈ S
alternatively by considering triangles containing
P
, and then taking
the limit
lim
area→0
(α + β + γ) − π
area
= K(P ).
Finally, we note how this relates to the problem of the parallel postulate we have
mentioned previously. The parallel postulate, in some form, states that given a
line and a point not on it, there is a unique line through the point and parallel
to the line. This holds in Euclidean geometry, but not hyperbolic and spherical
geometry.
It is a fact that this is equivalent to the axiom that the angles of a triangle
sum to
π
. Thus, the Gauss-Bonnet theorem tells us the parallel postulate is
captured by the fact that the curvature of the Euclidean plane is zero everywhere.