Chapter 1

The First Fundamental

Form

The first thing to do in geometry if you are confined to a surface in R3 is to

measure the distance between two points of the surface. This will usually be
different from the distance between these points as measured in the ambient
three dimensional space, since the straight line segment which furnishes the
shortest path between the points of R3 will generally not be contained in the
surface. The object that allows one to compute lengths on a surface, and also
angles and areas, is the first fundamental form (also called the metric in the
physicist’s usage) of the surface.

1.1 Lengths of curves on surfaces

A person walking along a curve γ on a surface S, travels a distance
Z
||γ̇|| dt.

To compute this we have to find the length of tangent vectors to the surface,
which can be computed in the following way.

Definition 1.1 Let p be a point of a surface S. The first fundamental form

of S at p associates to tangent vectors v, w ∈ Tp S the scalar

< v, w >p,S = v · w.

Thus, < v, w >p,S is the dot product, restricted to tangent vectors to S at p.

We shall usually omit one or both of the subscripts unless there is some danger
of confusion as to which point or surface is intended.

1
2 CHAPTER 1. THE FIRST FUNDAMENTAL FORM

The first fundamental form < , > is an example of an inner product: this
follows immediately from the fact that the dot product defines an inner product
on R3 .
Suppose that σ(u, v) is a surface patch of S. Then, any tangent vector to S
at a point p in the image of σ can be expressed uniquely as a linear combination
of σ u and σ v . Define maps du : Tp S → R and dv : Tp S → R by

du(v) = λ, dv(v) = µ if v = λσu + µσv ,

for some λ, µ ∈ R. du and dv are linear maps. Then, using the fact that < , >
is a symmetric bilinear form, we have

< v, v >= λ2 < σ u , σ u > +2λµ < σ u , σ v > +µ2 < σ v , σ v > .

Writing
2 2
E = ||σ u || , F = σ u · σ v , G = ||σ v || ,
this becomes

< v, v >= Eλ2 + 2F λµ + Gµ2 = Edu(v)2 + 2F du(v)dv(v) + Gdv(v)2 .

The expession
Edu2 + 2F dudv + Gdv 2
is called the first fundamental form of the surface patch σ(u, v). Note that the
coefficients E, F, G and the linear maps du, dv depend on the choice of surface
patch for S, but the first fundamental form itself depends only on S and p.

If γ is a curve lying in the image of a surface patch σ(u, v). Note that the
coefficients E, F, G and the linear maps du, dv depend on the choice of surface
patch for S, but the first fundamental form itself depends only on S and p.
If γ is a curve lying in the image of a surface patch σ,we have

γ(t) = σ(u(t), v(t))

for some smooth functions u(t) and v(t). Then, denoting d/dt by a dot, we have
γ̇ = u̇σu + v̇σv by the chain rule, so

< γ̇, γ̇ >= E u̇2 + 2F u̇v̇ + Gv̇ 2 ,

and the length of γ is given by

Z
(E u̇2 + 2F u̇v̇ + Gv̇ 2 )1/2 dt.

Example 1.2 For the plane

σ(u, v) = a + up + vq
1.1. LENGTHS OF CURVES ON SURFACES 3

(see Example 2.2) with p and q being perpendicular unit vectors, we have
2 2
σ u = p, σ v = q, so E = ||σ u || = ||p|| = 1, F = σ u · σ v = u · v = 0, G =
2 2
||σ v || = ||q|| = 1, and the first fundamental form is simply

du2 + dv 2 .

Example 1.3 Consider a surface of revolution

σ(u, v) = (f (u) cos v, f (u) sin v, g(u)).

Recall from Example 2.25 that we can assume that f (u) > 0 for all values of u
and that the profile curve u → (f (u), 0, g(u)) is unit speed, i.e., f˙2 + ġ 2 = 1 (a
dot denoting d/du). Then

σ u = (f˙ cos u, ġ sin v, ġ), σ v = (−f sin v, f cos v, 0),

2 2
E = ||σ u || = f˙2 + ġ 2 = 1, F = σ u · σ v = 0, G = ||σ v || = f 2 .

So the first fundamental form is

du2 + f (u)2 dv 2 .

A special case is the unit sphere S 2 in latitude-longitude coordinates (Exam-

ple 2.4). We take u = θ, v = ϕ, f (θ) = cos θ, g(θ) = sin θ, giving the first
fundamental form of S 2 as
dθ2 + cos θ2 dϕ2 .

Example 1.4 We consider a generalized cylinder

σ(u, v) = σ(u) + va

defined in Example ??.

As we saw in Example 2.25, we can assume that γ is unit-speed, a is a unit

vector, and γ is contained in a plane perpendicular to a. Then, denoting d/du
2 2
by a dot, σ u = γ̇, σ v = a, so E = ||σ u || = ||σ̇|| = 1, F = σ u · σ v = γ̇ · a =
2 2
0, G = ||σ v || = a = 1, and the first fundamental form of σ is

du2 + dv 2 .

Note that this is the same as the first fundamental form of the plane (Example
3.2). The geometrical reason for this coincidence will be discussed in the next
section.
4 CHAPTER 1. THE FIRST FUNDAMENTAL FORM

Suppose that a surface patch σ̃(ũ, ṽ) is a reparametrization of asurface patch

σ(u, v), and let

Ẽdũ2 + 2F̃ dũdṽ + G̃dṽ 2 and Edu2 + 2F dudv + Gdv 2

∂u ∂u ∂v ∂v
be their first (i)] du = ∂ ũ du + ∂ ṽ dṽ, dv = ∂ ũ dũ + ∂ ṽ dṽ. If
 ∂u ∂u

J= ∂ ũ ∂ ṽ
∂v ∂v
∂ ũ ∂ ṽ

is the Jacobian matrix of the reparametrization map (u, v) → (ũ, ṽ), and J t is
the transpose of J, then
   
Ẽ F̃ E F
= Jt J.
F̃ G̃ f G

By the chain rule,

∂u ∂v ∂u ∂v
σ̃ ũ = σ u + σ v , σ̃ ṽ = σ u + σv ,
∂ ũ ∂ ũ ∂ṽ ∂ṽ
which gives
 2  2
∂u ∂u ∂v ∂v
Ẽ = σ̃ ũ · σ̃ ũ = E + 2F +G .
∂ ũ ∂ ũ ∂ ũ ∂ ũ

Similar expressions for F and G can be found; multiplying out the matrices
shows that these formulas are equivalent to the matrix equation in the question.
Following the procedure given,

Edu2 + 2F dudv + Gdv 2

 2   
∂u ∂u ∂u ∂u ∂v ∂v
=E dũ + dṽ + 2F dũ + dṽ dũ + dṽ
∂ ũ ∂ṽ ∂ ũ ∂ṽ ∂ ũ ∂ṽ
 2
∂v ∂v
+G dũ + dṽ . (1.1)
∂ ũ ∂ṽ

The coefficient of dũ2 is

 2  2
∂u ∂u ∂v ∂v
E + 2F +G ,
∂ ũ ∂ ũ ∂ ũ ∂ ũ

which agrees with the expression for Ẽ found above. Similarly for F̃ and G̃.

1.2 Isometries of surfaces

We have seen in Example 2.4 that a plane and a generalized cylinder, when
suitably parametrized, have the same first fundametal form. The geometric
1.2. ISOMETRIES OF SURFACES 5

reason for this is not hard to see. A plane piece of paper can be ‘wrapped’ on
a cylinder in an obvious way without crumpling the paper (see Example 2.15).
If we draw a curve on a plane, then after wrapping it becomes a curve on the
cylinder. Because there is no crumpling, the lengths of these two curves will be
the same. Since the lengths are computed as the integral of (the square root of)
the first fundamental form, it is plausible that the first fundamental forms of
the two surfaces should be the same. Experiment suggests, on the other hand,
that it is impossible to wrap a plane sheet of paper around a sphere without
crumpling. Thus, we expect that a plane and a sphere do not have the same
fundamental form.
This following definition makes precise what it means to wrap one surface
onto another without crumpling.

Definition 1.5 If S1 and S2 are surfaces, a smooth map f : S1 → S2 is called

a local isometry if it takes any curve in S1 to a curve of the same length in
S2 . If a local isometry f : S1 → S2 exists, we say that S1 and S2 are locally
isometric.

We shall see that every local isometry is a local diffeomorphism; a local

isometry that is a diffeomorphism is called an isometry. It is obvious that any
composite of local isometries is a local isometry, and that the inverse of any
isometry is an isometry.
To express the condition for a local isometry in a more useful form, we need
the following construction. Let f : S1 → S2 be a smooth map and let p ∈ S1 .
For v, w ∈ TS1 , define

f ∗< v, w >p =< Dp f (v), Dp f (w) >f (p)

Then, f ∗< , >p is a symmetric bilinear form on Tp S1 . Indeed,, the symmetry

is obvious and if λ, λ0 ∈ R, v, v0 , w ∈ Tp ,

f ∗< λv + λ0 v0 , w >p =< Dp f (λv + λ0 v0 ), Dp f (w) >f (p)

=< λDp f (v) + λ0 Dp f (v0 ), Dp f (w) >f (p)
= λ < Dp f (v), Dp f (w) >f (p)
+ λ0 < Dp f (v0 ), Dp f (w) >f (p)
= λf < v, w >p +λ0 f ∗< v0 , w >p .
∗

Theorem 1.6 A smooth map f : S1 → S2 is a local isometry if and only if

the symmetric bilinear forms < , >p and f ∗< , >p on Tp S1 are equal for all
p ∈ S1 .
6 CHAPTER 1. THE FIRST FUNDAMENTAL FORM

Proof If γ 1 is a curve on S1 , the length of the part of γ 1 with endpoints γ 1 (t0 )

and γ 1 (t1 ) is
Z t1
< γ˙1 , γ˙1 >1/2 dt. (1.2)
t0

The length of the corresponding part of the curve γ 2 = f ◦ γ 1 on S2 is

Z t1 Z t1 Z t1
< γ̇ 2 , γ̇ 2 >1/2 dt = < Df (γ̇ 1 ), Df (γ̇ 1 ) >1/2 dt = f ∗< γ̇ 1 , γ̇ 1 >1/2 dt.
t0 t0 t0
(1.3)
It is now obvious that, if the two symmetric bilinear forms in the statement of
the theorem are equal, the curves γ 1 and f ◦ γ 1 have the same length.
Conversely, suppose that the integrals in (1.2) and in (1.3) are equal for all
curves γ on S1 . Then, the integrals must be the same for all γ:

< γ̇, γ̇ >= f ∗< γ̇, γ̇ > .

Since any tangent vector v to S1 is the tangent vector of a curve on S1 , it follows

that
< v, v >= f ∗< v, v > for all v. (1.4)
Since < , > and f ∗< , > are symmetric bilinear forms, it follows from (1.4)
that they are equal.
Thus, f is a local isometry if and only if

< Dp f (v), Dv f (w) >f (p) =< v, w >p

for all p ∈ S1 and v, w ∈ Tp S1 . This means that the linear map Df : Tp S1 →

Tp S1 is an isometry. In short, f is a local isometry if and only if Dp f is an
isometry for all p ∈ S1 .
It follows from this theorem that every local isometry is a local diffeomor-
phism. Indeed, let f : S1 → S2 be a local isometry and let p ∈ S1 . If Dp f is not
invertible, there is a non-zero tangent vector v ∈ Tp S1 such that Dp f (v) = 0.
But this gives a contradiction: since f is a local isometry,

0 6=< v, v >p =< Dp f, Dp f >f (p) =< 0, 0 >= 0.

Hence, Dp f is invertible,and so f is a local diffeomorphism (Proposition 2.20).

It will be useful to express Theorem 3.6 in terms of surface patches.

Corollary 1.7 A local diffeomorphism f : S1 → S2 is a local isometry if and

only if, for any surface patch σ 1 of S1 , the patches σ 1 and f ◦ σ 1 of S1 and S2 ,
respectively, have the same first fundamental form.

Proof In view of the throrem, we have to show that the patches σ 1 and
f ◦ σ 1 = σ 2 , say, have the same first fundamental form if and only if the
1.2. ISOMETRIES OF SURFACES 7

symmetric bilinear forms < , >p and f ∗< , >p are equal for all p in the image
of σ 1 .
The first fundamental form of σ i (i=1,2) is Ei du2 + 2Fi dudv + Gi dv 2 , where
Ei =<(σ i )u , (σ i )u>, Fi =<(σ i )u , (σ i )v>, Gi =<(σ i )v , (σ i )v>. We compute

<(σ 2 )u , (σ 2 )u>=<Df ((σ 1 )u ), Df ((σ 1 )u )>= f ∗<(σ 1 )u , (σ 1 )u> .

Thus, if < >= f ∗< >, then E1 = E2 , and simlarly F1 = F2 and G1 =

G2 . Conversely, if these last three equations hold, then <v, w>= f ∗<v, w>
whenever the tangent vectors v, w are of the form (σ 1 )u or (σ 1 )v. The blinearity
property then implies that <v, w>= f ∗<v, w> for all v, w.
This proof actually shows that, if p ∈ S1 is in the image of a surface patch
σ 1 , then σ 1 and f ◦ σ 1 have the same first fundamental form at p if and only
if Dp f is an isometry: it follows that, if p is in the image of another surface
patch σ 2 , then σ 1 and f ◦ σ 1 have the same first fundamental form at p if and
only if the same is true of σ 2 and f ◦ σ 2 .

Example 3.9 The map f from the yz-plane to the unit cylinder defined in Ex-
ample 2.24 is a local isometry. For, if we use the surface patch σ 1 (u, v) = (0, u, v)
for the plane and σ 2 (u, v) = (cos u, sin u, v) for the cylinder, then f (σ 1 (u, v)) =
σ 2 (u, v), and by Example 3.4 σ 1 and σ 2 have the same first fundamental form.

Conformal mappings of surfaces

Suppose that two curves γ and γ̃ on a surface S intersect at a point p. The
angle of intersection of γ and γ̃ at p is defined to be the angle between the
tangent vectors σ̇ and γ̃˙ (evaluated at t = t0 and t = t̃0 , respectively). Using
the dot product formula for the angle between vectors, we see that θ is given by

γ̇ · γ̃˙ < γ̇, γ̃˙ >

cos θ =   = (1.5)
|| γ̇||  γ̃˙  < γ̇, γ̇ >1/2 < γ̃,˙ γ̃˙ >1/2

As usual, it will be useful to have an expression for this in terms of a surface

patch. Suppose then that γ and γ̃ lie in a surface patch σ of S, so that
γ(t) = σ(u(t), v(t)) and γ̃(t) = σ(ũ(t), ṽ(t)) for some smooth functions u, v, ũ
and ṽ. If Edu2 + 2F dudv + Gdv 2 is the first fundamental form of σ, then by
(1.5) we have

E u̇ũ˙ + 2F (u̇ṽ˙ + ũ˙ v̇) + Gv̇ ṽ˙

cos θ = (1.6)
(E u̇2 + 2F u̇v̇ + Gv̇ 2 )1/2 (E ũ˙ 2 + 2F ũ˙ ṽ˙ + Gṽ)
˙ 1/2

Example 3.10 The parameter curves on a surface patch σ(u, v) can be parametrized
by
γ(t) = σ(u0 , t), γ̃(t) = σ(t, v0 ),
8 CHAPTER 1. THE FIRST FUNDAMENTAL FORM

respectively, where u0 is the constant value of u and v0 is the constant value of

v in the two cases. Thus

u(t) = u0 , v(t) = t, ũ = t, ṽ(t) = v0 ,

u̇ = 0, v̇ = 1, ũ˙ = 1, ṽ˙ = 0.

These parameter curves intersect at the point σ(u0 , v0 ) of the surface. By Eq.
1.6, their angle of intersection θ is given by
F
cos θ = √ ,
EG
where E, F and G are evaluated at (u0 , v0 ). In particular, the parameter curves
are orthogonal if and only if F = 0.

Corresponding to the Definition 3.5 of a local isometry, we have the following

definition.

Definition 1.8 If S1 and S2 are surfaces, a conformal map f : S1 → S2

is a local diffeomorphism such that, if γ and γ̃ are any two curves on S1 that
intersect, say at a point p ∈ S1 , and if γ 2 and γ̃ are their images under f ,
theangle of intersection of γ 1 and γ̃ 1 at p is equal to the angle of intersection
of γ 2 and γ̃ 2 at f (p).

In short, f is conformal if and only if it preserves angles.

It is obvious that any composite of conformal maps is conformal, and that

the inverse of any conformal diffeomorphism is conformal.
As a special case, if σ : U → R3 is a surface, then σ may be viewed as a map
from an open subset of the plane (namely U ), parametrized by (u, v) in the usual
way, and the image S of σ, and we say that σ is a conformal paramerization or
a conformal surface patch of S if this map between surfaces is conformal.

Theorem 1.9 A local diffeomorphism f : S1 → S2 is conformal if and only if

there is a function λ : S1 → R such that

f ∗ < v, w >= λ(p) < v, w > for all p ∈ S1 and v, w ∈ Tp S1 .

It is not hard to see that the function λ, if it exists, is necessarily smooth.

Proof Let γ and γ̃ be two curves on S1 that intersect at a point p ∈ S1 . The

angle θ of intersection of the curves is given by Eq. 1.5. The corresponding
angle of intersection of the curves f ◦ γ and f ◦ γ̃ on S2 is obtained from the
1.2. ISOMETRIES OF SURFACES 9

expression on the right-hand side of Eq. 1.5 by replacing γ̇ and γ̃˙ with (f ∗ γ).
and (f ◦ γ̃)., respectively. Now

< (f ◦ γ)., (f ◦ γ̃).> =< Dp f (γ̇), Dp f (γ̃)

˙ >f (p) = f ∗< γ̇, γ̃˙ >p ,

with similar expressions for < (f ◦ γ)., (f ◦ γ).>f (p) and < (f ◦ γ̃)., (f ◦ γ̃).>f (p) .
Thus, to compute the angle of intersection of the curves f ◦ γ and f ◦ γ̃ on S2 ,
we must replace < > in the numerator and denominator of the expression of
the right-hand side of Eq. 1.5 by f ∗< , >= λ < , >, this replacement leaves
the expression in Eq. 1.5 unchanged (since the factor λ cancels out) and so f
is conformal.
For the converse, we must show that if

< γ̇, γ̃˙ > f ∗< γ̇, γ̃˙ >

= ∗ (1.7)
< γ̇, γ̇ >1/2 < γ̃,˙ γ̃˙ >1/2 f < γ̇, γ̇ >1/2 f ∗< γ̃, ˙ γ̃˙ >1/2

for all pairs of intersecting curves γ and γ̃ on S1 , then f ∗< , > is proportional
to < , >. Since every tangent vector to S1 is the tangent vector of a curve on
S1 , Eq. 1.7 implies that

< v, w > f ∗< v, w >

= (1.8)
< v, v >1/2 < w, w >1/2 f ∗< v, v >1/2 f ∗< w, w >1/2

for all tangent vectors v, w to S1 .

Choose an orthonormal basis {v1 , v2 } of the tangent plane to S1 with respect
to its first fundamental form < , >. Let

λ = f ∗< v1 , v1 >, µ = f ∗< v1 , v2 >, ν = f ∗< v 2 , v 2 > .

We apply Eq. 1.8 with v = v1 and w = cos θv1 + sin θv2 , where θ ∈ R. This
gives
λ cos θ + µ sin θ
cos θ = q .
λ(λ cos2 θ + 2µ sin θ cos θ + ν sin2 θ)

Taking θ = π/2 gives µ = 0, which implies that

λ = λ cos2 θ + ν sin2 θ for all θ ∈ R.

Hence λ = ν. This implies that f ∗< v, w >= λ < v, w > whenever v and w are
basis vectors. Since both sides are bilinear forms, it follows that f ∗< , >=< , >.

Reinterpreting this result in terms of surface patches gives

Corollary 1.10 A local diffeomorphism f : S1 → S2 is conformal if and only

if, for any surface patch σ of S1 , the first fundamental form of the patches σ of
S1 and f ◦ σ of S2 are proportional.
10 CHAPTER 1. THE FIRST FUNDAMENTAL FORM

In particular, a surface patch σ(u, v) is conformal if and only if its first

fundamental form is λ(du2 + dv 2 ) for some smooth function λ(u, v).

Example 1.11 We consider the unit sphere S 2 .

If q is any point of S 2 other than the north pole n = (0, 0, 1), the straight line
joining n and q intersects the xy-plane at some point p, say. The map that
takes q to p is called stereographic projection from S 2 to the plane, and we
denote it by Π. We are going to show that Π is conformal.
Let p = (u, v, 0), q = (x, y, z). Since p, q, n lie on a straight line, there is a
scalar ρ such that
q − n = ρ(p − n),
and hence

(x, y.z) = (0, 0, 1) + ρ((u, v, 0) − (0, 0, 1)) = (ρu, ρv, 1 − ρ). (1.9)

Hence, ρ = 1 − z, u = x/(1 − z), v = y/(1 − z) and we have

 
x y
Π(x, y, z) = , ,0 .
1−z 1−z

On the other hand, from Eq. 1.9 and x2 + y 2 + z 2 = 1 we get ρ = 2/(u2 +

2
v + 1), and hence

u2 + v 2 − 1
 
2u 2v
q= , , .
u2 + v 2 + 1 u2 + v 2 + 1 u2 + v 2 + 1

If we denote the right-hand side by σ 1 (u, v), then σ 1 is a parametrization

of S 2 with the north pole removed. Parametrizing the xy-plane by σ 2 (u, v) =
(u, v, 0), we have
Π(σ 1 (u, v)) = σ 2 (u, v).
1.2. ISOMETRIES OF SURFACES 11

According to Corollary 3.12, to show that Π is conformal, we have to show

that the first fundamental forms of σ 1 and σ 2 are proportional. The first
fundamental form of σ 2 is du2 + dv 2 . As to σ 1 , we get
2(v 2 − u2 + 1)
 
−4uv 4u
(σ 1 )u = , , ,
(u2 + v 2 + 1)2 (u2 + v 2 + 1)2 (u2 + v 2 + 1)2
2(u2 − v 2 + 1)
 
−4uv 4v
(σ 1 )v = , , (1.10)
(u2 + v 2 + 1)2 (u2 + v 2 + 1)2 (u2 + v 2 + 1)2
This gives
4(v 2 − u2 + 1)2 + 16u2 v 2 + 16u2 4
E1 = (σ 1 )u · (σ 1 )u = = 2 .
(u2 + v 2 + 1)4 (u + v 2 + 1)2

Similarly, F1 = 0, G1 = 4/(u2 + v 2 + 1)2 . Thus, the first fundamental form of

σ 2 is λ times that of σ 1 , where λ = 41 (u2 + v 2 + 1)2 .
It is often useful to think of Π as a map to the complex numbers C rather
than to the xy-plane, by identifying u + iv ∈ C with (u, v, 0). Moreover, we can
parametrize S 2 itself in a partly complex way by identifying (x, y, z) ∈ S 2 with
(x + iy, z). Then, S 2 becomes the set of pairs (w, z) where w ∈ C, z ∈ R and
|w|2 + z 2 = 1. Stereographic projection then takes the simple form
w
Π(w, z) = ,
1−z
and the surface patch σ 1 is given by
||w||2 − 1
 
2w
σ 1 (w) = , .
|w|2 + 1 |w|2 + 1
The inconvenience of having to exclude the north pole from the domain of
definition of Π be overcome by introducing a ‘point at infinity’ ∞ and defining
the ‘extended complex plane’ C∞ = R ∩ ∞. If we agree that Π maps the north
pole to ∞, it defines a bijection Π : S 2 → C∞ .
Returning now to the general case, it is natural to ask when there is a
conformal map between two surfaces. The surprising answer is that this is alway
the case locally: if p1 and p2 are points of two surfaces S1 and S2 , respectively,
there are open subsets O1 of S1 containing p1 and O2 of S2 containing p2 and a
conformal diffeomorphism O1 → O2 . This follows from the following theorem:

Theorem 1.12 Every surface has an atlas consisting of conformal surface

patches.

Equiareal maps and a theorem of Archimedes

Suppose that σ : U → R3 is a surface patch on a surface S. The image of
σ is covered by the two families of parameter curves obtained by setting u =
12 CHAPTER 1. THE FIRST FUNDAMENTAL FORM

constant and v = constant, respectively. Fix (u0 , v0 ) ∈ U ; since the change in

σ(u, v) corresponding to a small change ∆u in u is approximatly σ u ∆u and
that corresponding to a small change ∆v in v is approximately σ ∆ v, the part
of the surface contained by the parameter curves on the surface corresponding
to u = u0 , u = u0 + ∆u, v = v0 , v = v0 + ∆v is approximately a parallelogram
in the plane with sides given by σ u ∆u and ∆v (the derivatives being evaluated
at (u0 , v0 )).
Recalling that the area of a parallelogram in the plane with sides a and b is
||a×b||, we see that the area of the parallelogram on the surface is approximately
||σ u δu × σ v ∆v || = ||σ u × σ v ||∆u∆v.
This suggests the following definition.

Definition 3.16 The area Aσ (R) of the part σ(R) of a surface patch
σ : U → R3 corresponding to a region R ⊆ U is
Z
Aσ (R) = ||σ u × σ v || dudv.
R

The quantity ||σ u × σ v || that appears in the definition of area is easily computed
in terms of the first fundamental form Edu2 + 2F dudv + Gdv 2 of σ:

Proposition 1.13 ||σ u × σ v || = (EG − F 2 )1/2 .

Proof We use a result from vector algebra in R3 :

(a × b) · (c × d) = (a · d)(c · d) − (a · d)(b · c).
Applying this to ||σ u × σ v ||2 = (σ u × σ v ) · (σ u × σ v ), we get
||σ u × σ v ||2 = (σ u · σ u )(σ v · σ v ) − (σ u · σ v )2 = EG − F 2 .
Thus, our definition of area is
Z
Aσ (R) = (EG − F 2 )1/2 dudv.
R

We still have to check that this definition is unchanged if σ is reparametrized.

Proposition 1.14 The area of a surface patch is unchanged by reparametriza-

tion.

Proof Let σ : U → R3 be a surface patch and let σ̃ : Ũ → R3 be a

reparametrization of σ, with reparametrization map Φ : Ũ → U . Thus,if
Φ(ũ, ṽ) = (u, v), we have
σ̃(ũ, ṽ) = σ(u, v).
1.2. ISOMETRIES OF SURFACES 13

Let R̃ ⊆ Ũ be a region,and let R = Φ(R) ⊆ U . We have to prove that

Z Z
||σ u × σ v || dudv = ||σ̃ ũ × σ̃ ṽ || dũdṽ.
R R̃

We showed in the proof of Proposition 2.13 that

σ̃ ũ × σ̃ ṽ = det(J(Φ))σ u × σ v ,
where J(Φ) is the Jacobian matrix of Φ. Hence
Z Z
||σ̃ ũ × σ̃ ṽ || dũdṽ = |detJ(Φ)| ||σ u × σ v || dũdṽ.
R̃ R̃

By the change of variables formula for double integrals, the right-hand side of
this equation is exactly Z
||σ u × σ v || dudv.
R

Definition 1.15 Let S1 and S2 be two surfaces. A local diffeomorphism f :

S1 → S2 is said to be equiareal if it takes any region in S1 to a region of the
same area in S2 .

Theorem 1.16 A local diffeomorphism f : S1 → S2 is equiareal if and only

if, for any surface patch σ(u, v) on S1 , the first fundamental forms
E1 du2 + 2F1 dudv + G1 dv 2 and E2 du2 + 2F2 dudv + G2 dv 2
of the patches σ on S1 and f ◦ σ on S2 satisfy
E1 G1 − F12 = E2 G2 − F22 . (1.11)

Proof Let σ : U → R3 . The statement that f is equiareal is equivalent to

Z Z
(E1 G1 − F12 )1/2 dudv = (E2 G2 − F22 ) dudv
R R

for all regions R ⊆ U . This holds if and only if

E1 G1 − F12 = E2 G2 − F22 .

As with isometries and conformal maps, it is obvious that the composite if

equiareal diffeomorphisms is equiareal, and that the inverse any any equiareal
diffeomorphism is equiareal.
14 CHAPTER 1. THE FIRST FUNDAMENTAL FORM

p
q

One of the most famous examples of an equiareal map was found by

Archimedes. Consider the unit sphere x2 + y 2 + z 2 = 1 and the unit cylin-
der x2 + y 2 = 1. The sphere is contained inside the cylinder, and the two
surfaces touch each other along the circle x2 + y 2 = 1 in the xy-plane. For
each point p in S 2 other than the poles (0, 0, ±1), there is a unique straight line
parallel to the xy-plane and passing through the point p and the z-axis. The
line intersects the cylinder in two points, one of which, say q, is closest to p.
Let f be the map from S 2 (with the two poles removed) to the cylinder that
takes p to q.
To find a formula for f , let (x, y, z) be the Cartesian coordinates of p, and
(X, Y, Z) those of q. Since the line through p and q is parallel to the xy-plane,
we have Z = z and (X, Y ) = λ(x, y) for some scalar λ. Since (X, Y, Z) is on the
cylinder,
1 = X 2 + Y 2 = λ2 (x2 + y 2 ),
λ = ±(x2 + y 2 )−1/2 . (1.12)
Taking the + sign gives the point q, so we get
 
x y
f (x, y, z) = , , z .
(x2 + y 2 )1/2 (x2 + y 2 )1/2

Theorem 1.17 (Archimedes’ Theorem) The map f is an equiareal diffeomor-

phism.
1.2. ISOMETRIES OF SURFACES 15

Proof We take the atlas for the surface S1 consisting of the sphere minus the
north and south poles with two patches, both given by the formula
σ 1 (θ, ϕ) = (cos θ cos ϕ, cos θ sin ϕ, sin ϕ),
and defined on the open sets
{−π/2 < θ < π/2, 0 < ϕ < 2π} and {−π/2 < θ < π/2, −π < ϕ < π}.
The image of σ 1 (θ, ϕ) under the map is the point
σ 2 (θ, ϕ) = (cos ϕ, sin ϕ, sin θ) (1.13)
of the cylinder. It is easy to check that this gives an atlas for the surface S2 ,
consisting of the part of the cylinder between the planes z = 1 and z = −1,
with two patches, both given by Eq. 1.15 and defined on the same two open
sets as σ 1 . We have to show that Eq. 1.11 holds.
We computed the coefficients E1 , F1 and G1 of the first fundamental form
of σ 1 in Example 3.3:
E1 = 1, F1 = 0, G1 = cos2 θ.
For σ 2 , we get (σ 2 )θ = (0, 0, cos θ), (γ 2 )ϕ = (− sin ϕ, cosϕ, 0), and so
E2 = cos2 θ, F2 = 0, G2 = 1.
It is now clear that Eq. 1.11 holds.
Note that, since f corresponds to simply to the identity map (θ, ϕ) 7→ (θ, ϕ)
in terms of the parametrizations σ 1 and σ 2 of the unit sphere and cylinder,
respectively, it follows that f is a diffeomorphism.

Theorem 1.18 The area of a spherical triangle on the unit sphere S 2 with
internal angles α, β and γ is
α + β + γ − π.

Proof We begin by using Archimedes’ Theorem 3.20 to compute the area of a

‘lune’, i.e., the area enclosed between two great circles:

θ
16 CHAPTER 1. THE FIRST FUNDAMENTAL FORM

We can assume that the great circles intersect at the poles, since this can
be achieved by applying a rotation of S 2 , and this does not change areas. If θ
is the angle between them, the image of the lune under the map f is a curved
rectangle on the cylinder of width θ and height 2.

If we now apply the isometry which unwraps the cylinder on the plane, this
curved rectangle on the cylinder will map to a genuine rectangle on the plane,
with width θ and height 2. By Archimedes’ theorem, the lune has the same
area as the curved rectangle on the cylinder, and since every isometry is an
eqauiareal map, this has the same area as the genuine rectangle in the plane,
namely 2θ. Note that this correctly gives the area of the whole sphere to be 4π.

Turning now to the proof of the theorem,let A, B and C be the vertices of

the triangle (so that α is the angle at A,etc.). The three great circles, of which
the sides of the triangles are arcs, divide S 2 into eight triangles, as shown in the
following diagram (in which A0 is the antipodal point of A, etc.).

B
β
C0
α
γ
A
C A0

B0
1.2. ISOMETRIES OF SURFACES 17

Note that the two triangles with vertices A, B, C and A0 , B 0 , C 0 together

form a lune with angle α, etc. Hence, denoting the tiangle with vertices A, B, C
by ABC and its area by A(ABC), etc., we have, by the preceeding calculation,

A(ABC) + A(A0 BC) = 2α,

A(ABC) + A(AB 0 C) = 2β,
A(ABC) + A(ABC 0 ) = 2γ,
(1.14)

Adding these equations, we get

2A(ABC) + {A(ABC) + A(A0 BC) + A(AB 0 C) + A(ABC 0 )} = 2α + 2β + 2γ.

(1.15)
Now, the triangles ABC, AB 0 C, AB 0 C and ABC 0 together make a hemisphere
(namely, the hemishere containing the vertex A with boudary the great circle
passing through B and C), so

A(ABC) + A(A0 BC) + A(AB 0 C) + A(ABC 0 ) = 2π. (1.16)

Finally, since the map that takes each point of S 2 to its antipodal point is an
isometry, and hence equiareal, we have

A(A0 BC) = A(AB 0 C 0 ).

Inserting this into Eq. 1.16, we see that the term in {} on the left-hand side of
Eq. 1.15 is equal to 2π. Rearranging now gives the result.
This theorem is a special case of the Gauss-Bonnet Theorem. A more general
formulation is
KA = α + β + γ − π,
where K is the Gaussian curvature. In the case at hand, K = 1/R2 , where R
is the radius of the sphere. For the unit sphere we then have K = 1, and thus
the theorem appears in the form given in Theorem 3.22. In the application to
Euclidean geometry, where the triangle has sides which are segments of straight
lines, we have K = 0, so the theorem reduces to the well-known result that
the sum of the angles in a triangle is π. This is the motivation for designating
this quantity as the ‘angular defect’. We see that ‘small’ triangles, where A
tends to zero, become in a good approximation ‘Euclidean’. We will verify
these statements when we have at our disposal the definition of the Gaussian
curvature.