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MATH329

Geometry of Curves and Surfaces

Copyright c© Alexander C. R. Belton 2015

Hyperlinked edition

All rights reserved

The right of Alexander Belton to be identified as the authorof this work has been asserted by him in accordance with

the Copyright, Designs and Patents Act 1988.

Contents

Contents i

Introduction iiiPreface to the expanded version . . . . . . . . . . . . . . . . . . . . . . iiiPreface to the original version . . . . . . . . . . . . . . . . . . . . . . . iiiConventions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv

1 Curves 1

1.1 Review: vectors in R3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1The scalar product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2Orthonormal bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3The vector product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3Lines and planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6Spheres and circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.2 Parameterised curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7Rectifiable curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8Some calculus for vector-valued functions . . . . . . . . . . . . . . . . . 11Best linear approximation . . . . . . . . . . . . . . . . . . . . . . . . . 11Space-filling curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.3 Unit-speed curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13The tangent vector field . . . . . . . . . . . . . . . . . . . . . . . . . . 15Smooth curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16Curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16The normal vector field . . . . . . . . . . . . . . . . . . . . . . . . . . . 17Best circular approximation . . . . . . . . . . . . . . . . . . . . . . . . 18The binormal vector field . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2 Surfaces – local theory 21Partial derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.1 The first fundamental form . . . . . . . . . . . . . . . . . . . . . . . . . . 27Calculating area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30Surface reparameterisations . . . . . . . . . . . . . . . . . . . . . . . . 32Orthogonal reparameterisation . . . . . . . . . . . . . . . . . . . . . . . 35

i

Contents

Isometric surface patches . . . . . . . . . . . . . . . . . . . . . . . . . . 35Isometries and reparameterisations . . . . . . . . . . . . . . . . . . . . 37

2.2 Geodesics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38Normal and geodesic curvature . . . . . . . . . . . . . . . . . . . . . . . 38Geodesics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39The geodesic equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 41Geodesics and surfaces of revolution . . . . . . . . . . . . . . . . . . . . 44Meridians and parallels . . . . . . . . . . . . . . . . . . . . . . . . . . . 45Geodesics and isometries . . . . . . . . . . . . . . . . . . . . . . . . . . 46Curves of extremal length . . . . . . . . . . . . . . . . . . . . . . . . . 47

2.3 The second fundamental form . . . . . . . . . . . . . . . . . . . . . . . . 49Principal curvatures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

2.4 Gaussian curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57The Gauss map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58Theorema Egregium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

3 Surfaces – global theory 63

3.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

3.2 Plane curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

3.3 Simple closed curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69Simple closed curves in surfaces . . . . . . . . . . . . . . . . . . . . . . 72Curvilinear polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

3.4 Global surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77Charts and atlases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77Subdivisions and the Euler characteristic . . . . . . . . . . . . . . . . . 82The global Gauss–Bonnet theorem . . . . . . . . . . . . . . . . . . . . . 83

A Auxiliary results 85

A.1 Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

A.2 Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

B Solutions to selected exercises 89

Bibliography 99

Index 101

ii

Introduction

Preface to the expanded version

This version of these notes contains a new chapter, on the global theory of surfaces astypified by three variations on the Gauss–Bonnet theorem. Some minor amendmentshave been made to the previous text; for example, the generating curve for a surfaceof revolution is taken to lie in the x-z plane, so as to be consistent with the usuallongitude-latitude parameterisation of the sphere.

Alexander C. R. BeltonLancaster, 6th January 2015

Preface to the original version

These notes are intended as a gentle introduction to the differential geometry of curvesand surfaces. Much effort has been expended to keep technicalities to a minimum, butcertain prerequisites are unavoidable; some of the necessary analytical and algebraicresults are collecting in two appendices. The focus is on local properties and we workin R3 throughout. Unusually for a text on geometry, there are no figures: the reader isinvited to provide their own in the many gaps left for this purpose.

The book closest to the presentation given here is the excellent text of Andrew Pressley[6], which contains far more material and goes into much greater detail. (Numbersin square brackets refer to items in the bibliography.) It also differs in some respectsfrom the conventions and notation adopted here, so the reader should take care whenswitching from one to the other. The Oxford University lecture notes of Graeme Segal[8] were invaluable for the production of the second chapter of these notes, on surfaces.John Roe’s book [7] is a pleasant exposition of geometry with a different emphasis (andsome overlap) with ours; a venerable but still excellent treatment of differential geometry(both local and global) is [12].

The author happily acknowledges his debt to all those who tried to teach him differentialgeometry, particularly Professors R. L. Hudson and N. J.Hitchen, and Dr P. J. Braam.

This document was typeset using LATEX2ε with Peter Wilson’s memoir class and theAMS-LATEX packages. The index was produced with the aid of the MakeIndex program.

iii

Introduction

Conventions

The notation “P := Q” means that the quantity P is defined to equal Q. For example,if x ∈ R then

|x| :={

x if x > 0,

−x if x < 0.

If f : A → B is a function and b ∈ B then f ≡ b means that f is the constant functionwith value b, i.e., f(a) = b for all a ∈ A.

iv

One Curves

1.1 Review: vectors in R3

Recall that R3 is the three-dimensional vector space with scalar field R: elements of R3

are triples of real numbers, often written as column vectors, with vector-space operations(addition and scalar multiplication) defined “coordinate-wise”:

v1v2v3

+

w1

w2

w3

=

v1 + w1

v2 + w2

v3 + w3

and λ

v1v2v3

=

λv1λv2λv3

for all v1, v2, v3, w1, w2, w3, λ ∈ R. For convenience, we will also denote vectors in R3

as row vectors,v = (v1, v2, v3), w = (w1, w2, w3)

and so on, and allow scalars to act on the right as well as the left:

vλ = (λv1, λv2, λv3) = λu.

The zero vector 0 = (0, 0, 0).

We regard R3 as a mathematical model for the space of everyday experience.

The standard basis of R3 consists of the unit vectors

i :=

100

, j :=

010

and k :=

001

and every vector v in R3 has a unique representation of the form

v = v1i+ v2j+ v3k,

where v1, v2, v3 ∈ R.

1

1. Curves

The scalar product

Recall that R3 has a natural inner product, known as the scalar product or dot product :algebraically, this is

v ·w = v1w1 + v2w2 + v3w3;

geometrically, we havev ·w = |v| |w| cosθ,

where

|v| =√v21 + v22 + v23

is the magnitude of the vector v (and similarly for |w|) and θ is the angle betweenthe vectors v and w. (The sense in which this angle is measured does not matter,since cosine is an even function.) Two vectors are orthogonal (or perpendicular) if theirscalar product is zero (so the angle between them equals π/2). The magnitude |v −w|corresponds to the distance between v and w.

It is simple to check that the dot product is commutative (or symmetric),

v ·w = w · v for all v, w ∈ R3,

and bilinear (linear in each argument):

(u+ λv) ·w = (u ·w) + λ(v ·w) and u · (v + λw) = (u · v) + λ(u ·w)

for all u, v, w ∈ R3 and λ ∈ R.

Furthermore,v · v = |v|2 for all v ∈ R

3

and the Cauchy–Schwarz inequality holds:

|v ·w| 6 |v| |w| for all v, w ∈ R3,

with equality if and only if the vectors v and w are linearly dependent. Recall that

|v +w| 6 |v|+ |w| and∣∣ |v| − |w|

∣∣ 6 |v −w| for all v, w ∈ R3. (1.1)

Exercise 1.1. Prove the second of the two inequalities (1.1). [Hint: write v as(v −w) +w and apply the first inequality.]

2

1.1. Review: vectors in R3

Orthonormal bases

An orthonormal basis for R3 is a set {u,v,w} consisting of three unit vectors, so that

|u| = |v| = |w| = 1,

which are mutually orthogonal :

u · v = v ·w = w · u = 0.

Given such an orthonormal basis {u,v,w}, any vector a ∈ R3 can be written uniquely

as a linear combination of these vectors:

a = (u · a)u+ (v · a)v + (w · a)w. (1.2)

Furthermore, if a = λu+ µv + νw then Pythagoras’s theorem holds:

|a|2 = λ2 + µ2 + ν2. (1.3)

The standard basis {i, j,k} is an orthonormal basis, and there are many more.

Exercise 1.2. Prove (1.2) and (1.3). [Hint for the latter: recall that |a|2 = a · a.]The vector product

Recall that R3 has a vector product , which is also known as the cross product . This is abilinear map

× : R3 × R3 → R

3; (v,w) 7→ v×w,

where

v ×w :=

∣∣∣∣∣∣

i j kv1 v2 v3w1 w2 w3

∣∣∣∣∣∣= (v2w3 − v3w2)i+ (v3w1 − v1w3)j + (v1w2 − v2w1)k

=(v2w3 − v3w2, v3w1 − v1w3, v1w2 − v2w1

)

and the determinant is calculated by expansion along the first row. (The notation v∧wis used by some authors for the vector product of v with w.)

Note that the vector product is anti-commutative,

v ×w = −w × v for all v, w ∈ R3,

but is not associative: in general,

(u× v)×w 6= u× (v ×w). (1.4)

3

1. Curves

The vector product has the following geometrical interpretation: the magnitude of v×wis equal to

|v| |w| sin θ,where θ ∈ [0, π] is the angle between v and w, and the direction of v×w is perpendicularto both v and w and such that (v, w, v × w) is a right-handed triple. Physically, acorkscrew turned from v to w will move in the direction of v ×w.

Exercise 1.3. If v and w are linearly independent then they span a plane, so thereare only two directions which are orthogonal to both vectors. However, if v and w arelinearly dependent then they lie on a straight line (they are co-linear) and there areinfinitely many directions which are perpendicular to this line: why is there no problemwith the geometrical interpretation in this case? [Hint: look at the magnitude of v×w.]

The product v×w is a vector which is orthogonal to both v and w. In particular,

i× j = k, j× k = i and k× i = j. (1.5)

If u and v are orthogonal unit vectors then (u, v, u × v) is a right-handed triple oforthogonal unit vectors or orthonormal triple: this is a ordered collection of orthonormalvectors, (u, v, w), such that

u× v = w, v×w = u and w × u = v.

By convention, our orthonormal triples will be right-handed unless otherwise specified.

Exercise 1.4. Use the vector-product identities (1.5) satisfied by i, j and k to findvectors u, v and w which demonstrate the non-associativity of ×; that is, find u, vand w such that (1.4) holds.

Exercise 1.5. Let v and w be two vectors in R3. To what geometrical quantity doesthe number |v| |w| sin θ correspond? [Hint: it is the product of two lengths, so is anarea: of what?]

4

1.1. Review: vectors in R3

Exercise 1.6. Use the answer to Exercise 1.5 to give a geometrical interpretation ofthe scalar triple product

[u,v,w] := u · (v ×w),

where u, v and w are vectors in R3. [Hint: up to a choice of sign, this is the volume ofsomething.]

Show also that

[u,v,w] =

∣∣∣∣∣∣

u1 u2 u3v1 v2 v3w1 w2 w3

∣∣∣∣∣∣.

Deduce that the scalar triple product is unchanged by cyclically permuting its arguments,but changes sign if two of its arguments are transposed.

Exercise 1.7. Let {u,v,w} be an orthonormal basis. Show that if [u, v, w] = 1 then(u, v, w) is a right-handed triple. What other values can [u, v, w] take, and whathappens then?

Exercise 1.8. Prove the following identity:

|v ×w|2 = |v|2|w|2 − (v ·w)2 for all v, w ∈ R3.

Exercise 1.9. Let u, v, w, z ∈ R3. By writing both sides in terms of coordinates, showthat

u× (v ×w) = (u ·w)v− (u · v)w.Using this, deduce that

(u× v) · (w × z) = (u ·w)(v · z)− (u · z)(v ·w) =

∣∣∣∣∣u ·w u · zv ·w v · z

∣∣∣∣∣ .

[Hint: for this deduction, consider the scalar triple product [u× v,w, z].]

5

1. Curves

Lines and planes

Given distinct points a and b in R3, the straight line passing through them is the set ofpoints

{λa+ (1− λ)b : λ ∈ R},with the subset

{λa+ (1− λ)b : 0 6 λ 6 1}being the straight-line segment with end points a and b.

Alternatively, a straight line L may be specified by giving a direction, that is, a unitvector u, and a point a through which the line L passes. Then

L = {a+ λu : λ ∈ R}.

(More generally, the direction u can be any non-zero vector which is parallel to L.)

A plane P may be specified by giving two linearly independent vectors v and w whichare parallel to P and a point a which lies in P ; in this case

P = {a+ λv + µw : λ, µ ∈ R}.

Alternatively, P may be specified by giving a unit vector c to which P is orthogonal andthe perpendicular distance d from the origin to P , measured in the direction c; in termsof these quantities,

P = {r ∈ R3 : r · c = d}.

Exercise 1.10. Show that the different ways of specifying straight lines are equivalent,by explaining how to go from each one to the other. Do the same thing for planes. [Hintfor the latter: given any unit vector u, there exist vectors v and w such that {u,v,w}is an orthonormal basis.]

Exercise 1.11. Prove that a straight line L may be written as

{r ∈ R3 : r× c = d}

where c and d are two unit vectors. How does this representation relate to the othertwo ways of describing L?

6

1.2. Parameterised curves

Spheres and circles

A sphere is the collection of all points in R3 equidistant from its centre, this distance beingcalled the radius. If d = (a, b, c) is the centre and r > 0 the radius then r = (x, y, z)lies on the sphere if and only if

|r− d| = r ⇐⇒ |r− d|2 = r2 ⇐⇒ (x− a)2 + (y − b)2 + (z − c)2 = r2.

A circle C is the collection of all points in a plane P equidistant from its centre, a pointin P . If d ∈ P is the centre and the circle has radius r > 0 then

C = {r ∈ R3 : |r− d| = r} ∩ {d+ λv + µw : λ, µ ∈ R}

= {d+ λv + µw : |λv + µw| = r}.

where the vectors v and w are linearly independent and parallel to P . If v and w areorthogonal unit vectors then

C ={d+ λv + µw : λ2 + µ2 = r2

}=

{d+ (r cos θ)v + (r sin θ)w : θ ∈ (−π, π]

}.

1.2 Parameterised curvesYou are probably used to thinking of a curve as the graph of some function (for example,the parabola y = x2 or the sinusoidal wave y = sin x) or, more generally, as the set ofpoints satisfying an equation, such as the circle

{(x, y) ∈ R2 : x2 + y2 = 1}.

In these notes, we will prefer a dynamic perspective: we are not just interested in thepoints which make up the curve, but how those points are traced out.

Another difference is that we will consider curves which live in R3 and not R2; it should be

clear that this is a straightforward generalisation – we can always restrict our attentionto planar curves (those which lie in a plane) if we wish. (We will study certain propertiespeculiar to planar curves in Chapter 3.)

7

1. Curves

Definition 1.1. A parameterised curve is a continuous function

γ : (a, b) → R3; t 7→ γ(t) =

(γ1(t), γ2(t), γ3(t)

),

where the open interval

(a, b) := {x ∈ R : a < x < b}for a ∈ {−∞}∪R and b ∈ R∪ {∞} such that a < b; more concisely, −∞ 6 a < b 6 ∞.We include intervals such as (−∞, 0), (2,∞) and (−∞,∞) = R.

As t runs over the interval from a to b, we imagine the point γ(t) tracing out the curvein R3; the parameter t can be thought of as time. The requirement of continuity ensuresthat the curve is connected : informally, this means that the curve is made up of just onepiece, without any jumps or gaps as the parameter t moves through the interval (a, b).

It may seem more natural for the parameter interval to be closed, in other words, toinclude the end points a and b (as long as these are finite). However, we will be interestedin curves which are differentiable functions, and excluding the end points simplifiesmatters a little: there is no need to worry about one-sided derivatives.

Rectifiable curves

Suppose we have a parameterised curve γ : (a, b) → R3. A natural question to askis, how long is it? If γ is part of a straight line, the answer is easy; if it is madeup of straight-lines segments (is piecewise linear or polygonal) then the answer is alsostraightforward. This suggests a way to approach the general case: by using polygonal

approximation.

We divide the interval (a, b) into subintervals, by taking t0, t1, . . . , tn such that

a < t0 < t1 < · · · < tn < b.

We then sum the lengths of each straight-line segment between consecutive points γ(ti−1)and γ(ti) as i runs from 1 to n. As we include more and more points in the subdivisionof (a, b), we expect this approximation to become closer and closer to the “true” lengthof the curve γ.

8


The approximate length of γ given by this procedure is

L = |γ(t1)− γ(t0)|+ |γ(t2)− γ(t1)|+ · · ·+ |γ(tn)− γ(tn−1)| =n∑

i=1

|γ(ti)− γ(ti−1)|.

If the set of all such approximations

S ={ n∑

i=1

|γ(ti)− γ(ti−1)| : a < t0 < · · · < tn < b, n > 1}

is bounded above then the curve is said to be rectifiable and its length is defined to equalthe supremum (the least upper bound) of S.

It is not easy to see how, in practice, these approximations could be efficiently calculatedfor all but the simplest of examples. However, for a large class of curves, there is a way.The trick is to re-write L in the following manner:

L =

n∑

i=1

∣∣∣∣γ(ti)− γ(ti−1)

ti − ti−1

∣∣∣∣ (ti − ti−1).

If the points ti−1 and ti are close together then the ratio

γ(ti)− γ(ti−1)

ti − ti−1

≈ γ′(ti−1),

the derivative of γ at ti−1. (Don’t worry too much about whether this makes sense forthe moment; if necessary, pretend that the function γ is scalar valued.) Hence

L ≈n∑

i=1

|γ ′(ti−1)|(ti − ti−1) ≈∫ b

a

|γ′(t)| dt.

This gives a way to calculate the length of γ, provided that it is “sufficiently smooth”.

Definition 1.2. A curve γ : (a, b) → R3 is continuously differentiable if

γ′(t) := lim

h→0

γ(t+ h)− γ(t)

h

exists for all t ∈ (a, b) and the derivative γ′ : (a, b) → R3 is a continuous function;

thinking dynamically, the vector γ ′(t) is the velocity of the curve at time t and |γ ′(t)| isits speed . If γ is continuously differentiable then it has length

ℓ(γ) :=

∫ b

a

|γ′(t)| dt.

It may be shown that this definition of length agrees with the previous one. We will notworry about this, as we will be dealing only with continuously differentiable curves (andso can use this definition).

9

1. Curves

An immediate question presents itself: what do we mean by the limit of a function withvalues in R3? If

v : (a, b) → R3; t 7→

(v1(t), v2(t), v3(t)

)

is a vector-valued function and s ∈ (a, b) then the limit

limt→s

v(t)

exists if and only if the limits of the coordinate functions do: that is,

limt→s

v(t) =(limt→s

v1(t), limt→s

v2(t), limt→s

v3(t)),

in the sense that the left-hand side exists if and only if each of the limits on the right-handside do, in which case the two sides are equal.

(If you know about metric spaces, you should convince yourself that this is equivalentto regarding R

3 as a metric space, where the metric d is such that d(v,w) = |v−w| forall v, w ∈ R3.)

In particular, the curve

γ : (a, b) → R3; t 7→

(γ1(t), γ2(t), γ3(t)

)

is continuously differentiable if and only if the real-valued functions γ1, γ2 and γ3 aredifferentiable on (a, b) and the derivatives γ′1, γ

′

2 and γ′3 are continuous, in which case,

γ′(t) =

(γ′1(t), γ

′

2(t), γ′

3(t))

for all t ∈ (a, b).

Example 1.3. If r > 0 is fixed and γ(t) := (r cos t, r sin t, 0) for all t ∈ (−π, π) thenthis curve describes a circle in the x-y plane with radius r and centre (0, 0, 0), minus thepoint (−r, 0, 0).

Sinceγ′(t) = (−r sin t, r cos t, 0) for all t ∈ (−π, π),

the curve γ has length

ℓ(γ) =

∫ π

−π

√r2 sin2 t+ r2 cos2 t+ 02 dt =

∫ π

−π

r dt = 2πr,

as we would expect.

10


Exercise 1.12. Show that the curve

γ : (−1, 1) → R3; t 7→

(√1 + t2, 0, 2

)

is continuously differentiable and find its length. Suggest a better parameterisation forthis curve.

Some calculus for vector-valued functions

Exercise 1.13. Let v : (a, b) → R3 and w : (a, b) → R3 be differentiable at t ∈ (a, b).Show that v ·w is differentiable at t, with

(v ·w)′(t) = v′(t) ·w(t) + v(t) ·w′(t).

Show also that v × w is differentiable at t and give a formula for (v × w)′(t). Finally,show that if v(t) 6= 0 then |v| is differentiable at t, with |v|′(t) = (v · v′)(t)/|v|(t).

Exercise 1.14. Use the usual (scalar-valued) form of the chain rule to prove the followingversion: if f : (c, d) → (a, b) is differentiable at t ∈ (c, d) and v : (a, b) → R

3 isdifferentiable at f(t) then

v ◦ f : (c, d) → R3; x 7→ v

(f(x)

)

is differentiable at t, with (v ◦ f)′(t) = v′(f(t)

)f ′(t).

Best linear approximation

Suppose the curve γ : (a, b) → R3 is differentiable at the point t ∈ (a, b) and let

v : R → R3; λ 7→ γ(t) + λd

be a straight-line segment through γ(t) in the direction of the vector d ∈ R3 \ {0}.

If h is small thenγ(t + h) = γ(t) + hγ ′(t) + o(h),

simply by the definition of γ ′(t) and o(h) (see Notation A.1). Hence

|γ(t+ h)− v(h)| = |γ(t) + hγ ′(t) + o(h)− γ(t)− hd| = h|γ ′(t)− d|+ o(h)

and the “error” between γ(t + h) and v(h) is smallest if and only if d = γ′(t). Thus,

as long as γ ′(t) 6= 0, the straight line λ 7→ γ(t) + λγ ′(t) is the best linear approximation

to γ at t.

11

1. Curves

Space-filling curves

A continuously differentiable curve cannot take up much room, as the following pair ofexercises demonstrates.

Exercise 1.15. Let γ : (a, b) → R3 be continuously differentiable and let c, d ∈ (a, b)be such that c < d. Explain why

∫ d

c

|γ ′(t)| dt > |γ(d)− γ(c)|. (1.6)

[Hint: think geometrically.] Given n > 1, prove there exist points

c = t0 < t1 < · · · < tn = d

such that s(ti)− s(ti−1) = s(d)/n for i = 1, . . . , n, where

s(t) :=

∫ t

c

|γ ′(r)| dr for all t ∈ [c, d].

[Hint: use the intermediate-value theorem, Theorem A.2.] Deduce that the set

γ([c, d]

):= {γ(t) : c 6 t 6 d}

can be covered by n spheres, each of radius s(d)/n. [Hint: centre the spheres at γ(ti)for i = 1, . . . , n.]

Exercise 1.16. Prove that a continuously differentiable curve γ cannot “fill space”.[Hint: suppose there exist c, d ∈ (a, b) such that c < d and γ

([c, d]

)contains a cube of

positive volume. Use Exercise 1.15 to obtain a contradiction.]

However, if the smoothness condition is dropped then more pathological behaviour maybe obtained. In fact, there exists a continuous curve γ : [0, 1] → R3 which has image thewhole of the unit cube [0, 1]3. (This was first proved by Peano in an article [5] publishedin 1890.) Such space-filling curves were quite shocking when they first appeared butthey are now well understood. A good explanation of how to construct one can be foundin [4, Chapter 7, §44].

12

1.3. Unit-speed curves

1.3 Unit-speed curvesIf two curves γ1 : (a, b) → R3 and γ2 : (c, d) → R3 have the same image, so that

{γ1(t) : t ∈ (a, b)} = {γ2(t) : t ∈ (c, d)},then each is a different parameterisation of that image. For a given curve, length asmeasured along it, called arc length, can be used to obtain a particularly well-behavedparameterisation.

Definition 1.4. A continuously differentiable curve γ : (a, b) → R3 is regular if γ ′(t) 6= 0for all t ∈ (a, b). Geometrically, a regular curve may be approximated to first order atevery point by a straight line, as shown above.

Exercise 1.17. Show that the continuously differentiable curve

γ : (−1, 1) → R3; t 7→

(t3, t6, t9

)

is not regular. Find a regular curve with the same image as γ.

Definition 1.5. Suppose the continuously differentiable curve γ is regular and fix apoint m ∈ (a, b). If

s(t) :=

∫ t

m

|γ ′(r)| dr for all t ∈ (a, b)

then s is the arc-length function for γ with starting point m. This function has a strictlypositive, continuous derivative, so is strictly increasing and thus a bijection from (a, b)onto (c, d), where

c := −∫ m

a

|γ′(r)| dr and d :=

∫ b

m

|γ ′(r)| dr.

(Different choices of starting point give rise to arc-length functions which differ only bya constant.)

By Theorem A.3, the inverse s−1 : (c, d) → (a, b) is continuously differentiable; let

γ := γ ◦ s−1 : (c, d) → R3; x 7→ γ

(s−1(x)

).

Then γ and γ have the same image, but γ is a unit-speed parameterisation of this image:if x ∈ (c, d) then Exercise 1.14 and Theorem A.3 imply that

|γ ′(x)| =∣∣γ ′

(s−1(x)

)∣∣ |(s−1)′(x)| =∣∣γ ′

(s−1(x)

)∣∣ |s′(s−1(x)

)∣∣−1= 1,

since s′(t) = |γ ′(t)| for all t ∈ (a, b). This will be our preferred choice of parameterisation:it makes many formulae more simple.

Note that if n ∈ (c, d) then the arc-length function s for γ with starting point n is suchthat

s(x) =

∫ x

n

|γ ′(r)| dr =∫ x

n

dr = x− n for all x ∈ (c, d) :

the curve γ is parameterised by arc length.

13

1. Curves

Exercise 1.18. Prove the converse statement, that a curve parameterised by arc lengthhas unit speed.

Example 1.6. The curve

γ : R → R3; t 7→

(3 cosh t, 4 sinh t, 3t

)

has tangent vectorγ′(t) =

(3 sinh t, 4 cosh t, 3

)

with magnitude

|γ′(t)| =√9 sinh2 t+ 16 cosh2 t+ 9 =

√25 cosh2 t = 5 cosh t,

for all t ∈ R. Hence the arc-length function for γ with starting point 0 is

s : R → R; t 7→∫ t

0

5 cosh r dr = 5 sinh t,

which has inverses−1 : R → R; x 7→ sinh−1

(x5

).

The reparameterised curve

γ := γ ◦ s−1 : R → R3; x 7→

(3

5

√25 + x2,

4

5x, 3 sinh−1

(x5

))

has unit speed, which may be verified directly: if x ∈ R then, asd

dzsinh−1 z =

1√1 + z2

,

|γ ′(x)|2 = 9

25

x2

25 + x2+

16

25+

9

25

25

25 + x2

=9x2 + 16(25 + x2) + 225

25(25 + x2)

=25x2 + 625

25x2 + 625

= 1.

This example shows that reparameterisation by arc length may make calculations moredifficult in specific cases, although it makes the general theory more straightforward.

Exercise 1.19. Let γ : (a, b) → R3 be a regular curve. Show that its length is unchangedby reparameterisation: if f : (c, d) → (a, b) has strictly positive, continuous derivativethen

γ := γ ◦ f : (c, d) → R3; x 7→ γ

(f(x)

)

is a regular curve with the same length as γ. Prove the same holds if f : (c, d) → (a, b)has strictly negative, continuous derivative.

14


The tangent vector field

If the regular curve γ : (a, b) → R3 is parameterised by arc length then the unit tangentvector at s ∈ (a, b) is

t(s) := γ′(s);

note that t(s) is a unit vector because γ has unit speed. The function

t : (a, b) → R3; s 7→ t(s)

is the unit tangent vector field to the curve γ.

Given a curve γ : (a, b) → R3, any related vector-valued function v : (a, b) → R3 definedon the same parameter interval is called a vector field , and v(t) is the value of the vectorfield at the point t ∈ (a, b). (We say “the point t”, rather than “the point γ(t)”, sincethe curve γ need not be simple: distinct points s and t in (a, b) may correspond to thesame point γ(t) = γ(s). To put it another way, points on the curve are specified bythe parameter t rather than their location in R3.) We will also be interested in scalar

fields : these are simply maps of the form f : (a, b) → R, where f depends on γ in somemanner.

Lemma 1.7. Suppose v : (a, b) → R3 has constant magnitude: there exists c ∈ R such

that |v(t)| = c for all t ∈ (a, b). If v is differentiable at t ∈ (a, b) then v(t) · v′(t) = 0.

Proof. As v · v = |v|2 ≡ c2, it follows that

0 = (v · v)′(t) = v′(t) · v(t) + v(t) · v′(t) = 2v(t) · v′(t).

15

1. Curves

Smooth curves

It will be convenient to strengthen the differentiability condition: a curve γ : (a, b) → R3

is smooth if it is infinitely differentiable; that is, the nth derivative γ(n)(t) exists for

all t ∈ (a, b) and n > 1, where γ(0) := γ and

γ(n)(t) := lim

h→0

γ(n−1)(t+ h)− γ

(n−1)(t)

hfor all t ∈ (a, b).

In terms of coordinates, the curve γ is smooth if and only if each of its coordinatefunctions is infinitely differentiable.

Exercise 1.20. Prove that if γ is smooth then so is any arc-length function for γ.Deduce that a regular smooth curve has a smooth unit-speed reparameterisation.

Henceforth, curves will be taken to be smooth unless it is stated otherwise.

Curvature

Let γ : (a, b) → R3 be a unit-speed curve. It follows from Lemma 1.7 that t′ is everywhereorthogonal to t, the unit tangent vector field. The curvature κ, a non-negative scalarfield, is defined by setting

κ(s) := |t′(s)| = |γ ′′(s)| for all s ∈ (a, b) :

the curvature of a unit-speed curve is the magnitude of the derivative of the tangentvector field t. (More generally, curvature can be defined for any regular smooth curve:see Exercise 1.26).

Exercise 1.21. Fix r > 0 and let

γ : (−π, π) → R3; t 7→ (r cos t, r sin t, 0)

be the circle in the x-y plane with radius r and centre (0, 0, 0) (minus one point).Show that γ is regular. Find an arc-length function for γ and so obtain a unit-speedreparameterisation γ. Prove that γ has constant curvature 1/r.

Exercise 1.22. Prove that a unit-speed curve γ with zero curvature lies on a straightline.

16


The normal vector field

If γ : (a, b) → R3 is a unit-speed curve with curvature κ(s) 6= 0 at some point s ∈ (a, b)then n(s) := t′(s)/κ(s) is a unit vector, the unit normal vector . This is such that

t(s) · n(s) = 0 and t′(s) = κ(s)n(s).

If the curvature scalar field κ is strictly positive then the unit normal vector field

n : (a, b) → R3; s 7→ n(s) := t′(s)/κ(s)

is well defined, orthogonal to t and such that t′ = κn.

Exercise 1.23. Suppose the unit-speed curve γ : (a, b) → R3 has strictly positivecurvature and is planar: there exists a constant unit vector u and a constant d ∈ R suchthat γ(s) · u = d for all s ∈ (a, b). Show that {t(s), n(s), u} is an orthonormal basisof R3, for all s ∈ (a, b). [Hint: differentiate γ · u twice.] Use this to prove that, in thiscase, n′ = −κt. [Hint: let n′ = λt+ µn+ νu. Find λ, µ and ν.]

Exercise 1.24. Find the normal vector field n for the unit-speed circle γ found inExercise 1.21.

Example 1.8. If γ is a planar unit-speed curve with constant curvature κ > 0 then γ

lies on a circle.

To see this, suppose γ lies in the plane

P := {r ∈ R3 : r · u = d},

where the unit vector u and the scalar d are constant. By Exercise 1.23, if c := γ+κ−1nthen

c · u = γ · u+ κ−1n · u ≡ d+ 0 = d

and c lies in P . Furthermore, again by Exercise 1.23,

c′ = (γ + κ−1n)′ = t+ κ−1n′ = t− t ≡ 0,

and therefore c is constant. Finally,

|γ − c| = |κ−1n| ≡ κ−1,

so γ lies on the circle in the plane P with centre c and radius κ−1.

17

1. Curves

Best circular approximation

Given a unit-speed curve γ : (a, b) → R3, fix s ∈ (a, b). For sufficiently small h, Taylor’stheorem (Theorem A.4) implies that

γ(s+ h) = γ(s) + hγ ′(s) + 12h2γ ′′(s) + o(h2). (1.7)

If κ(s) = 0 then γ′′(s) = t′(s) = 0 and

|γ(s+ h)− γ(s)− ht(s)| = o(h2),

so the best linear approximation to γ has second-order contact at s (not just the usualfirst-order contact). Otherwise, κ(s) > 0 and n(s) is well defined, so

γ(s+ h) = γ(s) + ht(s) + 12h2κ(s)n(s) + o(h2)

for small h. In the first-order situation we saw that the line through γ(s) in the directionof t(s) was the best straight-line approximation to γ at s. Now we will show that thecircle C passing through γ(s) with centre γ(s) + rn(s), lying in the plane

P := {γ(s) + λt(s) + µn(s) : λ, µ ∈ R},

gives the best circular approximation to γ when the radius r = κ(s)−1.

The point γ(s) + ht(s) + kn(s) in P will be close to γ(s + h) if k is chosen suitably;by (1.7), the error

η := |γ(s+ h)− γ(s)− ht(s)− kn(s)| = |12κ(s)h2 − k|+ o(h2).

Furthermore, γ(s) + ht(s) + kn(s) lies on the circle

C = {v ∈ P : |v− γ(s)− rn(s)|2 = r2} = {γ(s) + λt(s) + µn(s) : λ2 + (µ− r)2 = r2}

if and only ifh2 + (k − r)2 = r2 ⇐⇒ k = r ±

√r2 − h2.

18


The negative square root gives a point on C close to γ(s+ h); the positive choice givesthe point at the other end of the chord parallel to n(s). With the negative choice, abinomial expansion gives that

k = r(1− (1− r−2h2)1/2

)= 1

2r−1h2 + o(h2)

and therefore the error

η = |12κ(s)h2 − k|+ o(h2) = 1

2|κ(s)− r−1|h2 + o(h2).

Hence C has second-order contact at γ(s) if and only if r = κ(s)−1; the error η is o(h2)(smaller than second order) in this case

The circle in the osculating plane P which has centre γ(s)+κ(s)−1n(s) and radius κ(s)−1,called the osculating circle at s, is the best circular approximation to γ at the point s.

The binormal vector field

Suppose the unit-speed curve γ has strictly positive curvature, so that n is well defined.The cross product b := t × n is the binormal vector field associated with γ. At everypoint s ∈ (a, b), the triple (

t(s), n(s), b(s))

is a (right-handed) orthonormal triple which evolves with the curve.

In particular, any vector field v can be decomposed into components in the t, n and bdirections:

v = (t · v)t+ (n · v)n+ (b · v)b.

We will use this observation to analyse how n and b evolve. Note that

b′ = (t× n)′ = t′ × n+ t× n′ = κn× n+ t× n′ = t× n′;

in particular, t · b′ = [t, t,n′] ≡ 0. By Lemma 1.7, b · b′ ≡ 0, and therefore

b′ = (n · b′)n = [n, t,n′]n = [n′,n, t]n = −[n′, t,n]n = −τn,

where τ := n′ · b is the torsion of γ; this scalar field τ measures how much the curvedeviates from the t-n plane.

Furthermore, n = b× t, so

n′ = (b× t)′ = b′ × t+ b× t′ = −τn× t+ b× κn = τb− κt.

In summary, we have the Serret–Frenet equations [1, 9, 10].

19

1. Curves

Theorem 1.9. (Serret–Frenet) If the smooth curve γ has unit speed and strictlypositive curvature then the following equations hold.

t′ = κn

n′ = −κt +τb

b′ = −τn(S–F)

In the above, t, n and b are the unit tangent, normal and binormal vector fields for γ,whereas κ and τ are the curvature and torsion scalar fields for γ.

Exercise 1.25. Let γ : (a, b) → R3 be a unit-speed curve which has strictly positivecurvature and zero torsion: τ ≡ 0. Show that γ lies in a plane: there exist a constantunit vector u and a constant scalar d such that γ(s) · u = d for all s ∈ (a, b). [Hint: usethe third Serret–Frenet equation.]

Example 1.10. If the unit-speed curve γ has strictly positive curvature and lies in aplane then γ has zero torsion.

To see this, suppose γ · u ≡ d, where the unit vector u and the scalar d are constant.It follows from Exercise 1.23 that u is orthogonal to t and n. Hence u is proportionalto b; as both are unit vectors, b = ±u. Since b is continuous, b · u = ±u · u = ±1 isconstant (by the intermediate-value theorem, Theorem A.2) and therefore b is constant(because b = (b · u)u). By the third Serret–Frenet equation, τ ≡ 0.

Exercise 1.26. Let γ : (a, b) → R3 be a regular smooth curve with an arc-lengthfunction s : (a, b) → (c, d). Suppose the unit-speed curve γ = γ ◦ s−1 : (c, d) → R3

has unit tangent vector field t, curvature scalar field κ (which is strictly positive), unit

normal vector field n, unit binormal vector field b and torsion scalar field τ . Show that

γ′ = |γ ′|(t ◦ s), (t ◦ s)′ = |γ ′|

((κn) ◦ s

)and γ

′ × γ′′ = |γ′|3

((κb) ◦ s

).

Deduce that

κ ◦ s = |γ ′ × γ′′|

|γ ′|3 .

Show further that

[γ ′,γ ′′,γ ′′′] = |γ ′|6((κ2τ) ◦ s

)and τ ◦ s = [γ ′,γ′′,γ ′′′]

|γ ′ × γ ′′|2 .

Exercise 1.27. Let the unit-speed curve γ : (a, b) → R3 have constant curvature κ > 0and constant torsion τ . Prove that γ describes a helix. [Hint: find n′′.]

20

Two Surfaces – local theory

In this chapter, we will begin an investigation into the local theory of surfaces. Theobjects of interest are surface patches : these are sufficiently differentiable, injective mapsfrom open subsets of R2 to R3, and we think of them as equipping a region of space witha two-dimensional coordinate system. In Chapter 3 we will consider the global theoryof how these patches fit together to form more complicated surfaces.

A parameterised curve is a map γ : I → R3, where I is an open subinterval of R. Fora surface, this parameter interval needs to be replaced by a suitable subset of R2, onewhich is open. This is for the same reason we choose to define curves on open intervals– those without endpoints; the key idea is that we must have room to define derivatives.

Recall that the open disc with centre (x, y) ∈ R2 and radius r > 0 is the set

D(x, y; r) :={(u, v) ∈ R

2 : (u− x)2 + (v − y)2 < r2}.

A set U ⊆ R2 is open if every point in U is contained in some open disc which liescompletely within U . Formally, U is open if for every (x, y) ∈ U there exists r > 0 suchthat D(x, y; r) ⊆ U . (Note that r can depend on x and y.) If a function is defined on anopen set then we can easily talk about the limit of that function at any point in the set.

Example 2.1. The empty set ∅ is open, since it satisfies the definition vacuously. Thewhole of R2 is also an open set: given (x, y) ∈ R2, the disc D(x, y; 1) ⊆ R2 (and 1 canbe replaced by any other choice of r > 0). Any open disc is open – draw a picture toconvince yourself of this – as is any open rectangle, a set of the form

(a, b)× (c, d) = {(x, y) ∈ R2 : a < x < b, c < y < d},

where −∞ 6 a < b 6 ∞ and −∞ 6 c < d 6 ∞.

21

2. Surfaces – local theory

We shall also require that the parameter set U ⊆ R2 is connected, so that every pair ofpoints in U can be joined by a path which lies within U . This requirement ensures thata surface is also connected, and thus composed of only one piece.

Formally, the open set U ⊆ R2 is connected if, for all x, y ∈ U , there exists a continuousfunction f : [0, 1] → U such that f(0) = x and f(1) = y. All the sets in Example 2.1are connected. (For comparison, the connected subsets of R are precisely the intervals.)

Partial derivatives

Given a vector-valued map v : U → R3, where U is an open set in R

2, the partial

derivatives ∂1v and ∂2v are defined by setting

∂1v(x, y) :=∂v

∂x(x, y) = lim

h→0

v(x+ h, y)− v(x, y)

h

and

∂2v(x, y) :=∂v

∂y(x, y) = lim

h→0

v(x, y + h)− v(x, y)

h

for all (x, y) ∈ U . Similarly to the one-dimensional case, if v = (v1, v2, v3) then

∂1v(x, y) =∂v

∂x(x, y) =

(∂v1∂x

(x, y),∂v2∂x

(x, y),∂v3∂x

(x, y)),

in the sense that the left-hand side exists if and only if each derivative on the right-handside does, and then they are equal. Of course, the same holds for the partial derivativewith respect to the second variable.

A map v : U → R3 is smooth if it has partial derivatives of all orders, so that

∂11v := ∂1(∂1v), ∂12v := ∂1(∂2v), ∂21v := ∂2(∂1v) and ∂22v := ∂2(∂2v)

exist, and similarly for the higher derivatives. It is a standard result that if the mixedsecond-order partial derivatives are continuous, as they must be for a smooth map, thenthey are equal: ∂12v = ∂21v. A similar result holds for all higher derivatives: mixedpartial derivatives of a smooth map may be calculated in any order.

22

Definition 2.2. A surface patch is a smooth injection r : U → R3, where U is a non-empty connected open set in R2. We think of the map (x, y) 7→ r(x, y) as introducing asystem of coordinates on the image r(U).

A surface patch r : U → R3 is regular if the tangent vector fields ∂1r and ∂2r are linearlyindependent at every point of U .

(Given a surface patch r : U → R3, any function v : U → R3 which depends on r is avector field , as in the one-dimensional case, and similarly for a scalar field.)

Let r : U → R3 be a surface patch. The tangent vector fields ∂1r and ∂2r are linear

approximations to the surface in the u and v directions; if they are linearly independentat the point (x, y) ∈ U then they span a plane, the tangent plane

T(x,y)r := {λ ∂1r(x, y) + µ ∂2r(x, y) : λ, µ ∈ R}.

The plane r(x, y)+T(x,y)r is the best linear approximation to the regular surface patch rat the point (x, y), in the same way that the tangent line {γ(t) + λγ ′(t) : λ ∈ R} isthe best linear approximation to the regular curve γ at the point t. A proof of this factrequires the two-variable version of Taylor’s theorem, so we omit it.

Exercise 2.1. Explain why the surface patch r is regular if and only if |∂1r× ∂2r| > 0everywhere.

23


Example 2.3. Let γ : I → R3 be a regular curve which is simple (if s, t ∈ I are suchthat γ(s) = γ(t) then s = t) and planar (there exist a unit vector c and a scalar d whichare constant and such that γ · c ≡ d). If

r : I × R → R3; (x, y) 7→ γ(x) + y c

then r is a surface patch, called a generalised cylinder or prism.

To see that r is injective, note that

r(u, v) = r(x, y) =⇒ r(u, v) · c = r(x, y) · c =⇒ v = y

and then γ(u) = γ(x) so, because γ is simple, u = x. Furthermore,

∂1r(x, y) = γ′(x) and ∂2r(x, y) = c.

Since γ(x) · c = d, for all x ∈ I, differentiating both sides of this equality gives that

0 = γ′(x) · c+ γ(x) · c′ = γ

′(x) · c,

so γ′(x) and c are orthogonal. Hence if (x, y) ∈ I × R then

|∂1r× ∂2r|(x, y) = |γ ′(x)× c| = |γ ′(x)| |c| = |γ ′(x)| > 0,

as γ is regular, and the surface patch r is regular as well.

24

Exercise 2.2. Let γ : I → R3 be a regular curve which is simple and which lies ina plane that does not contain the origin: there exists a unit vector c and a non-zeroscalar d such that γ · c ≡ d. Prove that

r : I × (0,∞) → R3; (x, y) 7→ y γ(x)

is a regular surface patch. This type of surface is called a generalised cone.

Definition 2.4. If r : U → R3 is a regular surface patch then

N :=∂1r× ∂2r

|∂1r× ∂2r|

is the unit normal vector field , made up of unit vectors which are orthogonal to thesurface. (The capital letter N is chosen to distinguish it from n, the unit normal to acurve.)

Exercise 2.3. Show that the surface patch

r : R× (−π, π) → R3; (z, φ) 7→ (z cosφ, z sinφ, z)

is not regular. Describe the image r(R× (−π, π)

).

25


Example 2.5. The surface patch

r : (−π/2, π/2)× (−π, π) → R3; (θ, φ) 7→ (cos θ cosφ, cos θ sin φ, sin θ)

is a parameterisation of the unit sphere S2 = {r ∈ R3 : |r| = 1}, minus the semicirculararc from the north pole (0, 0, 1) to the south pole (0, 0, −1) which passes through(−1, 0, 0); the coordinate θ corresponds to latitude (or elevation from the x-y plane)and φ to longitude (or azimuth).

Since

∂1r(θ, φ) = (− sin θ cos φ, − sin θ sinφ, cos θ)

and ∂2r(θ, φ) = (− cos θ sin φ, cos θ cosφ, 0),

it follows that

(∂1r× ∂2r)(θ, φ) =

∣∣∣∣∣∣

i j k− sin θ cosφ − sin θ sinφ cos θ− cos θ sinφ cos θ cosφ 0

∣∣∣∣∣∣

= (− cos2 θ cos φ, − cos2 θ sinφ, − sin θ cos θ).

As |∂1r× ∂2r|(θ, φ) = cos θ, we obtain the unit normal

N(θ, φ) = (− cos θ cosφ, − cos θ sinφ, − sin θ) = −r(θ, φ).

26

2.1. The first fundamental form

2.1 The first fundamental formSuppose r : U → R3 is a regular surface patch which contains the curve γ : I → R3,where I is an open interval:

γ(t) ∈ r(U) = {r(u, v) : (u, v) ∈ U} for all t ∈ I.

For each t ∈ I, there exists(u(t), v(t)

)∈ U such that γ(t) = r

(u(t), v(t)

). In other

words, we can (and will) regard such a curve as a map of the form

r ◦ (u, v) : t 7→ r(u(t), v(t)

),

where the pair(u, v) : I → U ; t 7→

(u(t), v(t)

).

Example 2.6. The curve

γ : R → R3; t 7→ (cosh t, sinh t, 1)

lies in the hyperbolic paraboloid

r : R× R → R3; (x, y) 7→ (x, y, x2 − y2)

and has the form r ◦ (u, v), where

u : R → R; t 7→ cosh t and v : R → R; t 7→ sinh t.

(The first claim follows from the identity cosh2 t− sinh2 t = 1, valid for all t ∈ R.)

27


Recall that a curve is said to be smooth if it is infinitely differentiable. Since r haspartial derivatives of all orders, in order to show that γ = r◦ (u, v) is smooth, it suffices,by the chain rule, to verify that the functions u : I → R and v : I → R are infinitelydifferentiable; we call (u, v) a smooth pair if this is the case.

Exercise 2.4. Let r : U → R3 be a regular surface patch and let (u, v) : I → U be asmooth pair. Use the identity

γ′(t) = ∂1r(u(t), v(t)) u

′(t) + ∂2r(u(t), v(t)) v′(t) for all t ∈ I (2.1)

to prove that γ = r ◦ (u, v) is regular if and only if u′(t)2 + v′(t)2 > 0 for all t ∈ I.

We shall insist henceforth that every smooth pair (u, v) satisfies this regularity condition,so that the resulting curve r ◦ (u, v) is always regular.It follows from (2.1) that

γ′ · γ ′ = ∂1r(u, v) · ∂1r(u, v)(u′)2 + ∂1r(u, v) · ∂2r(u, v)u′v′

+ ∂2r(u, v) · ∂1r(u, v)u′v′ + ∂2r(u, v) · ∂2r(u, v)(v′)2

= E(u, v)(u′)2 + 2F (u, v)u′v′ +G(u, v)(v′)2,

where the parameter t is omitted for clarity and the scalar fields

E := ∂1r · ∂1r,F := ∂1r · ∂2r = ∂2r · ∂1r

and G := ∂2r · ∂2r.

(Here and in what follows, if v : U → R3 is a vector field and (u, v) : I → U is a smoothpair then, for convenience and clarity, we will write v ◦ (u, v) as v(u, v) et cetera, andsimilarly for scalar fields. Hence we have the function

v(u, v) : I → R3; t 7→ v

(u(t), v(t)

).

It is essential to remember that v(u, v) is a function on I which depends on the curveparameter t, and is not the value of v at a point (u, v) ∈ R3.)

Thus if I = (a, b) then the curve γ has length

ℓ(γ) :=

∫ b

a

|γ ′(t)| dt

=

∫ b

a

√E(u, v)(u′)2 + 2F (u, v)u′v′ +G(u, v)(v′)2(t) dt

=

∫ b

a

√E(u(t), v(t)

)u′(t)2 + 2F

(u(t), v(t)

)u′(t)v′(t) +G

(u(t), v(t)

)v′(t)2 dt.

(2.2)

28


The scalar fields E, F and G are the coefficients of the first fundamental form,

E du2 + 2F du dv +G dv2. (2.3)

For convenience, we will sometimes call the triple (E, F,G) the first fundamental form.

We shall regard (2.3) as a purely formal expression. However, the following calculationis suggestive: if ds is an infinitesimal piece of arc length (whatever that may mean) then

ds2 = |dγ|2 = dγ · dγ = (∂1r du+ ∂2r dv) · (∂1r du+ ∂2r dv)

= ∂1r · ∂1r du2 + (∂1r · ∂2r+ ∂2r · ∂1r) du dv + ∂2r · ∂2r dv2

= E du2 + 2F du dv +G dv2,

from which it follows that

s =

∫ds =

∫ √E du2 + 2F du dv +G dv2 =

∫ √E(u′)2 + 2Fu′v′ +G(v′)2 dt.

Example 2.7. The surface patch

r : R× R → R3; (x, y) 7→ (x, y, x2 + y2)

is an elliptic paraboloid . This has tangent vectors

∂1r(x, y) = (1, 0, 2x) and ∂2r(x, y) = (0, 1, 2y),

so is regular, and the coefficients of the first fundamental form are

E(x, y) = 1 + 4x2, F (x, y) = 4xy and G(x, y) = 1 + 4y2.

Example 2.8. If r > 0 is fixed and

γ : (−π, π) → R3; t 7→ (r cos t, r sin t, r2)

is a curve in the elliptic paraboloid r of Example 2.7 then γ = r ◦ (u, v), whereu : (−π, π) → R; t 7→ r cos t and v : (−π, π) → R; t 7→ r sin t.

Hence u′(t) = −r sin t, v′(t) = r cos t and

E(u(t), v(t)

)= 1+4r2 cos2 t, F

(u(t), v(t)

)= 4r2 cos t sin t, G

(u(t), v(t)

)= 1+4r2 sin2 t.

This curve has length

ℓ(γ) =

∫ π

−π

√(1 + 4r2 cos2 t)r2 sin2 t− 8r4 cos2 t sin2 t+ (1 + 4r2 sin2 t)r2 cos2 tdt

=

∫ π

−π

√r2 sin2 t+ r2 cos2 t dt

= 2πr;

since γ is a parameterisation of the circle x2 + y2 = r2, except for one point, this is asexpected.

29


Exercise 2.5. Let r : U → R3 be a regular surface patch. Express |∂1r× ∂2r| in termsof the first fundamental form. [Hint: Exercise 1.8 of Section 1.1 may be useful.]

Exercise 2.6. Let r : U → R3 be a regular surface patch and let (x, y) ∈ U . Showthat the angle θ between the tangent vectors ∂1r(x, y) and ∂2r(x, y) is such that

cos θ =F√EG

(x, y). (2.4)

An orthogonal parameterisation is one such that F ≡ 0; use (2.4) to explain why thisterminology is appropriate.

Calculating area

As well as determining the length of curves, the first fundamental form can be used tocalculate the area of part of a surface patch. The key to this is the following interpretationof the vector product.

Consider the parallelogram PQRS, with vertices labelled clockwise starting with thebottom left corner. The line QS divides this figure into two congruent triangles, PQSand QRS, so the parallelogram has area

APQRS =(12×∣∣∣−→

PS∣∣∣× vertical height

)+(12×

∣∣∣−→

QR∣∣∣× vertical height

).

The vertical height equals∣∣∣−→

PQ∣∣∣ sin θ, where θ is the angle between the vectors

−→

PS

and−→

PQ, and−→

PS =−→

QR, so

APQRS =∣∣∣−→

PS∣∣∣∣∣∣−→

PQ∣∣∣ sin θ =

∣∣∣−→

PS×−→

PQ∣∣∣,

the magnitude of the vector product of−→

PS and−→

PQ. Furthermore, by Exercise 1.8 ofSection 1.1,

APQRS =∣∣∣−→

PS×−→

PQ∣∣∣ =

√∣∣∣−→

PS∣∣∣2∣∣∣

−→

PQ∣∣∣2

−(

−→

PS ·−→

PQ)2

.

30


Hence an infinitesimally small parallelogram with sides dx = ∂1r dx and dy = ∂2r dywill have area

dAr =√|dx|2|dy|2 − (dx · dy)2 =

√∂1r · ∂1r dx2 ∂2r · ∂2r dy2 − (∂1r · ∂2r dx dy)2.

Thus if V ⊆ U then the area of r(V ) equals

Ar(V ) =

∫ ∫

V

dAr :=

∫ ∫

V

√EG− F 2(x, y) dx dy. (2.5)

Example 2.9. If

r : (−π/2, π/2)× (−π, π) → R3; (θ, φ) 7→ (cos θ cosφ, cos θ sin φ, sin θ)

is the parameterisation of the unit sphere S2 from Example 2.5 then

∂1r(θ, φ) = (− sin θ cos φ, − sin θ sinφ, cos θ)

and ∂2r(θ, φ) = (− cos θ sin φ, cos θ cosφ, 0),

so the first fundamental form has coefficients

E(θ, φ) = 1, F (θ, φ) = 0 and G(θ, φ) = cos2 θ.

Hence the sphere has area

Ar

((−π/2, π/2)× (−π, π)

)=

∫ π

−π

∫ π/2

−π/2

cos θ dθ dφ = 2π

[sin θ

]π/2

−π/2

= 4π,

as expected.

Example 2.10. Consider a (circular) cylinder which exactly encloses the sphere of theExample 2.5: this has a natural coordinate system obtained from the sphere by projectionparallel to the x-y plane. The corresponding surface patch is

r : (−π/2, π/2)× (−π, π) → R3; (θ, φ) 7→ (cos φ, sinφ, sin θ),

for which

∂1r(θ, φ) = (0, 0, cos θ) and ∂2r(θ, φ) = (− sinφ, cosφ, 0).

It follows that

E(θ, φ) = cos2 θ, F (θ, φ) = 0 and G(θ, φ) = 1;

in particular, (EG− F 2)(θ, φ) = cos2 θ, the same as for the sphere.

(This is not the “usual” parameterisation of the cylinder; vertical height is measuredthrough the angle of elevation θ rather than directly.)

31


Surface reparameterisations

Let r : U → R3 be a surface patch. A reparameterisation of this surface patch is abijection Φ : U → U , where U is a non-empty connected open set in R2, such that Φand Φ−1 are smooth (that is, have partial derivatives of all orders). Informally, this isjust a change of coordinate system; the composite map r ◦ Φ is a regular surface patchwhich describes the same region of R3 as r does.

The Jacobian of

Φ := (φ, ψ) : U → U ; (z, w) 7→(φ(z, w), ψ(z, w)

)

is the 2× 2-matrix-valued function

JΦ : (z, w) 7→[∂1φ(z, w) ∂2φ(z, w)

∂1ψ(z, w) ∂2ψ(z, w)

],

where ∂1φ(z, w) =∂φ

∂z(z, w) et cetera. (See Definition A.5 for more about Jacobians.)

Exercise 2.7. Explain why the Jacobian determinant, det JΦ, is either everywherestrictly positive or everywhere strictly negative. [Hint: explain why JΦ is invertible,then apply the intermediate-value theorem, Theorem A.2.]

A reparameterisation Φ : U → U such that det JΦ(u) > 0 for all u ∈ U is orientation

preserving ; one with det JΦ(u) < 0 for all u ∈ U is orientation reversing. (Exercise 2.8explains this terminology.)

If r := r ◦ Φ, where the reparameterisation Φ = (φ, ψ), then

∂1r = ∂1r(φ, ψ) ∂1φ+ ∂2r(φ, ψ) ∂1ψ and ∂2r = ∂1r(φ, ψ) ∂2φ+ ∂2r(φ, ψ) ∂2ψ,

so[∂1r∂2r

]=

[∂1φ ∂1ψ∂2φ ∂2ψ

] [∂1r(φ, ψ)∂2r(φ, ψ)

]= (JΦ)t

[∂1r ◦ Φ∂2r ◦ Φ

], (2.6)

where At denotes the transpose of the matrix A.

Exercise 2.8. Prove that

N :=∂1r× ∂2r

|∂1r× ∂2r|=

{N ◦ Φ if Φ is orientation preserving,

−N ◦ Φ if Φ is orientation reversing.

(In other words, the unit normal vector field N is unchanged by a orientation-preservingreparameterisation, and changes sign, so is reflected in the tangent plane, after anorientation-reversing reparameterisation.)

It follows from (2.6) that[E F

F G

]=

[∂1r · ∂1r ∂1r · ∂2r∂2r · ∂1r ∂2r · ∂2r

]= (JΦ)t

[E ◦ Φ F ◦ ΦF ◦ Φ G ◦ Φ

]JΦ. (2.7)

32


In particular,

EG− F 2 = (det JΦ)2((EG− F 2) ◦ Φ

). (2.8)

Exercise 2.9. Use the identity (2.8) to prove that if the surface patch r is regular thenso is the reparameterised patch r := r ◦ Φ.

Another consequence of (2.8) is that the area

Ar(V ) =

∫ ∫

V

√EG− F 2 dz dw =

∫ ∫

V

(√EG− F 2 ◦ Φ) | detJΦ| dz dw

=

∫ ∫

Φ(V )

√EG− F 2 dx dy

=(Ar ◦ Φ

)(V ),

where the penultimate identity follows from the change-of-variables formula for doubleintegrals (Theorem A.6). Thus area, like length, is unchanged by reparameterisation.

Example 2.11. The circular cylinder of Example 2.10 admits the reparameterisation

Φ : (−1, 1)× (−π, π) → (−π/2, π/2)× (−π, π); (z, φ) 7→ (sin−1 z, φ).

This coordinate system corresponds to our normal notion of distance on the cylinder:note that

r := r ◦ Φ : (−1, 1)× (−π, π) → R3; (z, φ) 7→ (cosφ, sinφ, z)

and a short calculation (Exercise) shows that

E ≡ 1, F ≡ 0 and G ≡ 1,

the same first fundamental form as the plane. We saw above that area is invariant underreparameterisation, so combining this observation with Example 2.10 gives a celebratedresult of Archimedes: the area of a sphere is equal to area of its circumscribing cylinder.Archimedes was so proud of this theorem that he had its statement inscribed upon histomb.

In fact, we have shown a stronger result: projection (parallel to the x-y plane) from thesphere onto the cylinder preserves area. This gives a way of producing a map of theEarth which represents area correctly; many atlases contain maps obtained by such acylindrical projection.

33


Exercise 2.10. A great circle C is the curve described by the intersection of the unitsphere S2 with a plane through the origin; in vector terms,

C = {r ∈ R3 : |r|2 = 1} ∩ {r ∈ R

3 : r · u = 0},

where u is a unit vector. Show that two distinct great circles meet at exactly two pointswhich are antipodal : they lie on a straight line through the origin. [Hint: let

C1 = {r ∈ R3 : |r|2 = 1, r · u1 = 0} and C2 = {r ∈ R

3 : |r|2 = 1, r · u2 = 0},

where u1 and u2 are linearly independent. Let v be a unit vector which is orthogonal toboth u1 and u2, and show that if r ∈ C1 ∩ C2 then r = νv for some ν ∈ R.] A lune isa region of the unit sphere bounded by two semicircles from distinct great circles; anytwo great circles divide the sphere into four lunes. Explain why a lune has area 2α,where α is the interior angle of the lune made by the semicircles at their intersection(the antipodes). Verify that this is consistent with Example 2.9.

Exercise 2.11. A spherical triangle is a region on the unit sphere S2 bounded by(parts of) three great circles. By using the result of Exercise 2.10 on the area of lunes,prove that a spherical triangle has area

∆ := α + β + γ − π,

where α, β and γ are the interior angles of the triangle. [Hint: let A, B and Γ be thevertices of the triangle, with interior angles α, β and γ respectively, and let A′, B′ and Γ′

be the corresponding antipodal points. Show that

A(ABΓ) +A(AB′Γ) = 2β

and3∆ +A(A′BΓ) +A(AB′Γ) +A(ABΓ′) = 2(α + β + γ),

where A(ABΓ) denotes the area of the triangle ABΓ et cetera. Now consider the regioncovered by the triangles ABΓ, A′BΓ, A′B′Γ and AB′Γ.]

34


Orthogonal reparameterisation

As seen in Section 1.3, any regular curve may be reparameterised by arc length, and thissimplifies many formulae. It is natural to wonder if something similar can be done forsurfaces.

Given a regular surface patch r : U → R3 and a point (x, y) ∈ U , it is possible toobtain a “good” reparameterisation about this point, although it may be necessary toshrink U . More formally, there exist non-empty connected open sets V and V , with(x, y) ∈ V ⊆ U , and a reparameterisation Φ : V → V which makes r := r ◦ Φ an

orthogonal parameterisation: the new first fundamental form (E, F , G) has F ≡ 0. Inother words, we can find a coordinate system for a neighbourhood of r(x, y) such that,in this coordinate system, the tangent vector fields are orthogonal there.

To establish this result, however, would take us too far afield and we will have no usefor it in what follows. Some authors make use of it (for example, [7, §§12.3–4]) but thesimplifications it brings are rather nullified by the complications of its proof.

Isometric surface patches

Informally, two surface patches r1 : U1 → R3 and r2 : U2 → R3 are isometric if the notionof distance measured along curves in them is the same, so that curves corresponding tothe same smooth pair have the same length. More formally, for this to make sense thedomains U1 and U2 of the patches must be identical, U1 = U2 = U , and then we say thepatches are isometric if, for any smooth pair

(u, v) : I → U ; t 7→(u(t), v(t)

)

the curves γ1 : I → R3; t 7→ r1(u(t), v(t)

)and γ2 : I → R3; t 7→ r2

(u(t), v(t)

)have the

same length, i.e., ∫

I

|(γ1)′(t)| dt =

∫

I

|(γ2)′(t)| dt.

Since this must hold for any smooth pair (u, v), it follows that the first fundamentalforms are equal:

E1(x, y) = E2(x, y), F1(x, y) = F2(x, y) and G1(x, y) = G2(x, y)

at all points (x, y) ∈ U . (The proof of this is Exercise 2.12.) The converse is immediate– if the first fundamental forms are the same then ℓ(γ1) = ℓ(γ2), by (2.2) – and so wehave the following theorem.

Theorem 2.12. Two regular surface patches r1 : U → R3 and r2 : U → R3 are isometricif and only if their first fundamental forms are equal.

35


Exercise 2.12. Let r1 : U → R3 and r2 : U → R3 be regular surface patches with firstfundamental forms (E1, F1, G1) and (E2, F2, G2) respectively. Prove that if

∫

I

√E1(u, v)(u′)2 + 2F1(u, v)u′v′ +G1(u, v)(v′)2 dt

=

∫

I

√E2(u, v)(u′)2 + 2F2(u, v)u′v′ +G2(u, v)(v′)2 dt

for any open interval I ⊆ R and for any smooth pair (u, v) : I → U then (E1, F1, G1)and (E2, F2, G2) are equal everywhere on U . [Hint: let (a, b) ∈ U and choose ε > 0 suchthat the square [a − ε, a + ε] × [b − ε, b + ε] ⊆ U ; explain why this is possible. Thenconsider (u, v) : (−ε, ε) → U ; t 7→ (a+ t, b) and two similar smooth pairs.]

Example 2.13. Since a cylinder can be obtained by rolling up a piece of paper, theplane and the cylinder should be isometric surfaces: the act of rolling up does not distortdistances. Let

rS : R× (−π, π) → R3; (x, y) 7→ (x, y, 0)

be a strip in the x-y plane and let

rC : R× (−π, π) → R3; (z, φ) 7→ (cosφ, sinφ, z)

be the cylinder x2 + y2 = 1, minus the line {(−1, 0, z) : z ∈ R}. Then

∂1rS ≡ (1, 0, 0), ∂2rS ≡ (0, 1, 0) and (ES, FS, GS) ≡ (1, 0, 1),

whereas

∂1rC ≡ (0, 0, 1), ∂2rC(z, φ) = (− sinφ, cosφ, 0) and (EC , FC , GC) ≡ (1, 0, 1).

Thus rS and rC are isometric, as expected.

36


Example 2.14. Now consider the surface of revolution obtained by rotating about thez axis the circle (x− 2)2 + z2 = 1 with centre (2, 0, 0) and unit radius in the x-z plane:this is a torus, with parameterisation

r : (−π, π)× (−π, π) → R3; (θ, φ) 7→

((2 + cos θ) cosφ, (2 + cos θ) sinφ, sin θ

).

(The circles x2 + y2 = 1 and (x + 2)2 + z2 = 1 are omitted.) The coordinate θ givesthe angle on the generating circle, measured from the positive x axis, and φ correspondsto the angle through which this generating circle has been rotated about the z axis,measured in the positive sense (anticlockwise) from the x-z plane.

It is straightforward to verify that the tangent vectors

∂1r(θ, φ) = (− sin θ cos φ, − sin θ sinφ, cos θ),

∂2r(θ, φ) = (−(2 + cos θ) sinφ, (2 + cos θ) cosφ, 0)

and the first fundamental form

(E, F,G)(θ, φ) =(1, 0, (2 + cos θ)2

).

Hence the torus is not isometric to the open square (−π, π) × (−π, π): if you try toconstruct one by taking a piece of paper and rolling it into a cylinder then bending thecylinder into a torus, you will find the second step impossible without squashing andtrying to stretch the paper.

Isometries and reparameterisations

Do not confuse the idea of being isometric with that of reparameterisation. The latterjust corresponds to a change of coordinate system, and all the geometrical features areunchanged; the former preserves some geometrical quantities, such as length and area,but other features may be completely different. For example, the cylinder and the planeare isometric, but a cylinder contain geodesics (see below) which intersect themselves,and this does not occur in the plane.

37


2.2 GeodesicsNormal and geodesic curvature

Let the regular surface patch r : U → R3 contain the smooth curve

γ = r ◦ (u, v) : I → R3; t 7→ r

(u(t), v(t)

).

Suppose γ has unit speed; recall that this is always possible, by reparameterising thecurve using its arc length.

We can use the tools developed in Chapter 1 to analyse this curve, ignoring the surfacein which it lies, but we can also use properties of the surface to give a more detailedpicture. For example, the curve γ has tangent vector

t(s) = γ′(s) = ∂1r

(u(s), v(s)

)u′(s) + ∂2r

(u(s), v(s)

)v′(s) ∈ T(u(s),v(s))r,

the tangent plane to the surface. Since N, the unit normal to the surface, is orthogonalto the tangent plane, the vector fields t and N(u, v) are orthogonal and

(t(s), N

(u(s), v(s)

), B(s) := t(s)×N

(u(s), v(s)

) )

is a right-handed orthonormal basis of R3 at every point s ∈ I. Using this basis toexpress the derivative of the tangent vector field, and recalling that t′ ·t ≡ 0, we see that

t′ = (t′ · t)t+(t′ ·N(u, v)

)N(u, v) + (t′ ·B)B = κnN(u, v) + κgB, (2.9)

where the scalar fieldκn := t′ ·N(u, v)

is the normal curvature of the curve γ and the scalar field

κg := t′ ·B = [t′, t,N(u, v)]

is the geodesic curvature of γ. Note that, by Pythagoras’s theorem,

κ2 = κ2n + κ2g,

where κ := |t′| is the curvature of γ.

Warning. Pressley defines geodesic curvature [6, p.127] with the opposite sign to us.For some reason, the most common convention in the literature is to work with the left-handed orthonormal triple

(t, B, N(u, v)

)rather than the right-handed choice made

above, which we prefer for its consistency with the Serret–Frenet apparatus.

38

2.2. Geodesics

Exercise 2.13. Suppose γ = r ◦ (u, v) : I → R3 is a unit-speed curve with strictlypositive curvature, which lies in the regular surface patch r : U → R3. Let ψ(s) denotethe angle from n(s) to N

(u(s), v(s)

)for all s ∈ I, so that

N(u, v) = n cosψ + b sinψ.

(The existence of such a smooth function ψ follows from Lemma 3.6.) Show that thenormal curvature κn = κ cosψ, the surface binormal

B = −n sinψ + b cosψ

and the geodesic curvature κg = −κ sinψ. Deduce that

t′ = κnN(u, v) + κgB,

N(u, v)′ = −κnt + τgB

and B′ = −κgt − τgN(u, v)

where τg := τ + ψ′ is the geodesic torsion. (This is an analogue of the Serret–Frenetequations for the orthonormal triple of vector fields

(t,N(u, v),B

).)

Geodesics

A particle which is free to move in R3 and which has zero acceleration will either remainat rest or travel with constant velocity in a straight line: this is Newton’s first law ofmotion. However, if the particle is required to move upon some surface patch, then itmay be necessary for the particle to accelerate in order that it stays in contact with thesurface. As long as there is no component of acceleration in the tangent plane (so that,infinitesimally, the particle does not appear to change velocity at all) then the particleis accelerating as little as possible, subject to the constraint of remaining on the surface.The curve traced out by such a particle is called a geodesic.

Formally, the curve γ = r ◦ (u, v) : I → R3 in the surface patch r is a geodesic if andonly if

γ′′(t) · ∂1r

(u(t), v(t)

)= γ

′′(t) · ∂2r(u(t), v(t)

)= 0 for all t ∈ I.

Equivalently, γ is a geodesic if and only if γ ′′ is a scalar multiple of N everywhere.

Exercise 2.14. Suppose the unit-speed curve γ : I → R3 lies on a straight line. Provethat γ is a geodesic.

39


Example 2.15. The parameterisation of the unit sphere S2 given in Example 2.5,

r : (−π/2, π/2)× (−π, π) → R3; (θ, φ) 7→ (cos θ cos φ, cos θ sinφ, sin θ),

has tangent vectors

∂1r(θ, φ) = (− sin θ cosφ, − sin θ sin φ, cos θ)

and ∂2r(θ, φ) = (− cos θ sinφ, cos θ cosφ, 0).

If γ : (−π, π) → R3; s 7→ r(0, s) is the equator, so that u ≡ 0 and v(s) = s, then

γ(s) = (cos s, sin s, 0),

γ′(s) = (− sin s, cos s, 0)

and γ′′(s) = (− cos s, − sin s, 0) = −γ(s)

for all s ∈ (−π, π). Hence

γ′′(s) · ∂1r(0, s) = (− cos s, − sin s, 0) · (0, 0, 1) = 0

andγ′′(s) · ∂2r(0, s) = (− cos s, − sin s, 0) · (− sin s, cos s, 0) = 0,

which shows that γ is a geodesic.

There is nothing particularly special about the equator; any great circle, the intersectionof the sphere with a plane through its centre, is a geodesic when parameterised by arclength. The converse is also true; see the following exercise.

Exercise 2.15. Let r : U → R3 be a regular surface patch contained in the unitsphere S2. Prove that the unit normal vector field N is such that N = ±r. Deduce thata unit-speed curve γ = r ◦ (u, v) : I → S2 is a geodesic if and only if γ ′′ + γ ≡ 0, and so

γ(s) = a cos s+ b sin s

for all s ∈ I, where a and b are orthogonal unit vectors. [Thus γ is part of the greatcircle in the plane spanned by a and b.]

Proposition 2.16. Any geodesic has constant speed and so a very simple unit-speedreparameterisation.

Proof. If γ = r ◦ (u, v) : I → R3 is a geodesic then

d

dt|γ ′(t)|2 = d

dt

(γ′(t) · γ ′(t)

)= 2γ ′(t) · γ ′′(t) = 0 for all t ∈ I,

since γ′ lies in the tangent plane. Hence |γ′| is constant.

40

2.2. Geodesics

Suppose γ = r ◦ (u, v) is a unit-speed geodesic. Then γ′′ = t′ is proportional to the unit

normal to the surface N(u, v), hence its geodesic curvature

κg = t′ ·B = [t′, t,N(u, v)] ≡ 0.

The converse is clear from (2.9) and so we have the following proposition, which explainswhy κg bears the name it does.

Proposition 2.17. A unit-speed curve in a surface is a geodesic if and only if it haszero geodesic curvature everywhere: κg ≡ 0.

Exercise 2.16. Show that the torsion τ of a unit-speed geodesic γ with strictly positivecurvature is equal to its geodesic torsion τg, defined in Exercise 2.13.

The geodesic equations

The smooth curve γ = r ◦ (u, v) is a geodesic if and only if

γ′′ · ∂1r(u, v) = γ

′′ · ∂2r(u, v) = 0

everywhere on γ. Now, (v ·w)′ = v′ ·w+v ·w′ for any smooth vector fields v and w, so

(γ′ · ∂1r(u, v)

)′

= γ′′ · ∂1r(u, v) + γ

′ ·(∂1r(u, v)

)′

and rearranging this shows that

γ′′ · ∂1r(u, v) =

(γ′ · ∂1r(u, v)

)′ − γ

′ ·(∂1r(u, v)

)′

.

Since γ′ = ∂1r(u, v) u

′ + ∂2r(u, v) v′, it follows that

γ′′ · ∂1r(u, v)

=((∂1r · ∂1r)(u, v) u′ + (∂2r · ∂1r)(u, v) v′

)′

− (∂1r(u, v) u′ + ∂2r(u, v) v

′) · (∂11r(u, v) u′ + ∂21r(u, v) v′)

= (E(u, v) u′ + F (u, v) v′)′

−((∂1r · ∂11r)(u, v)(u′)2 + (∂2r · ∂11r+ ∂1r · ∂21r)(u, v)u′v′ + (∂2r · ∂21r)(u, v)(v′)2

),

(2.10)

where ∂11r := ∂1(∂1r) et cetera:

∂11r(x, y) =∂2r

∂x2(x, y), ∂21r(x, y) =

∂r

∂y∂x(x, y) and ∂22r(x, y) =

∂2r

∂y2(x, y).

Moreover,

∂1E = ∂1(∂1r · ∂1r) = ∂11r · ∂1r+ ∂1r · ∂11r = 2∂1r · ∂11r,∂1F = ∂1(∂1r · ∂2r) = ∂11r · ∂2r+ ∂1r · ∂12r = ∂2r · ∂11r+ ∂1r · ∂21r

and ∂1G = ∂1(∂2r · ∂2r) = ∂12r · ∂2r+ ∂2r · ∂12r = 2∂2r · ∂21r.

41


Applying these to (2.10) yields the identity

γ′′ · ∂1r(u, v)= (E(u, v) u′ + F (u, v) v′)′ − 1

2(∂1E(u, v)(u

′)2 + 2∂1F (u, v)u′v′ + ∂1G(u, v)(v

′)2);

similar working gives the equation

γ′′ · ∂2r(u, v)= (F (u, v) u′ +G(u, v) v′)′ − 1

2(∂2E(u, v)(u

′)2 + 2∂2F (u, v)u′v′ + ∂2G(u, v)(v

′)2).

Hence the smooth curve γ = r ◦ (u, v) is a geodesic if and only if the geodesic equations

are satisfied:

(E(u, v)u′ + F (u, v)v′)′ = 12(∂1E(u, v)(u

′)2 + 2∂1F (u, v)u′v′ + ∂1G(u, v)(v

′)2) (2.11)

and

(F (u, v)u′ +G(u, v)v′)′ = 12(∂2E(u, v)(u

′)2 + 2∂2F (u, v)u′v′ + ∂2G(u, v)(v

′)2). (2.12)

These equations are usually very hard to solve explicitly.

Example 2.18. For the plane

r : R2 → R3; (x, y) 7→ (x, y, 0),

the first fundamental form has coefficients E = G ≡ 1 and F ≡ 0. The geodesicequations become

u′′ = 0 and v′′ = 0,

which have the general solutions u = at + b and v = ct + d for constants a, b, c, d ∈ R.Hence γ = r ◦ (u, v) has the form

γ : t 7→ vt+w,

where v = (a, c, 0) and w = (b, d, 0). In other words, for the plane, geodesics andstraight lines are the same thing.

42

2.2. Geodesics

Example 2.19. The cylinder

r : R× (−π, π) → R3; (z, φ) 7→ (cosφ, sin φ, z)

has first fundamental form dz2 + dφ2, the same as for the plane. As in Example 2.18above, the geodesic equations become u′′ = 0 and v′′ = 0, so u(t) = at+b and v(t) = ct+dfor constants a, b, c and d. In this case, the geodesic γ is such that

γ(t) = r(u(t), v(t)

)= r(at+ b, ct + d) = (cos(ct + d), sin(ct + d), at+ b)

and the geodesics on the cylinder fall into three classes:

(i) if a = 0 and c 6= 0 then γ lies on the parallel r(b, ·) : φ 7→ r(b, φ), a circle parallelto the x-y plane;

(ii) if c = 0 and a 6= 0 then γ lies on the meridian r(·, d) : z 7→ r(z, d), a straight lineparallel to the z axis;

(iii) if a 6= 0 and c 6= 0 then γ describes a helix .

(If a = c = 0 then γ is not a curve.)

43


Geodesics and surfaces of revolution

Letγ : I → R

3; s 7→(f(s), 0, g(s)

)

be a simple unit-speed curve in the x-z plane, such that f(s) > 0 for all s ∈ I. As γ hasunit speed, if s ∈ I then

1 = |γ ′(s)|2 = |(f ′(s), 0, g′(s)

)|2 = f ′(s)2 + g′(s)2.

A surface of revolution is formed by rotating this curve about the z axis; it may beparameterised in the following fashion:

r : I × (−π, π) → R3; (s, φ) 7→

(f(s) cosφ, f(s) sinφ, g(s)

). (2.13)

The coordinate s corresponds to the point γ(s) on the curve γ, and the coordinate φdescribes the angle through which this point has been rotated.

Exercise 2.17. Verify that the map r is injective.

Exercise 2.18. Verify that the surface of revolution r given by (2.13) is regular.

Since∂1r(s, φ) =

(f ′(s) cosφ, f ′(s) sinφ, g′(s)

)

and∂2r(s, φ) =

(−f(s) sinφ, f(s) cosφ, 0

),

the coefficients of the first fundamental form are

E(s, φ) = f ′(s)2 + g′(s)2 = 1, F (s, φ) = 0 and G(s, φ) = f(s)2.

44

2.2. Geodesics

Exercise 2.19. Show that the geodesic equations for the surface of revolution r givenby (2.13) have the form

u′′ = f(u)f ′(u)(v′)2 and(f(u)2v′)′ = 0. (2.14)

Meridians and parallels

A meridian on the surface of revolution r is a curve of the form

r(·, φ0) : I → R3; s 7→ r(s, φ0) =

(f(s) cosφ0, f(s) sinφ0, g(s)

),

where φ0 ∈ (−π, π) is constant; it is a copy of the curve γ which generates r, rotatedthrough an angle of φ0 radians about the z axis.

Similarly, a parallel of the surface of revolution r has the form

r(s0, ·) : (−π, π) 7→ R3; φ 7→ r(s0, φ) =

(f(s0) cosφ, f(s0) sinφ, g(s0)

),

where s0 ∈ I is constant. This is a circle (minus one point) parallel to the x-y plane,formed by rotating the point γ(s0) through 2π radians.

Exercise 2.20. Use Exercise 2.19 to prove that the meridian r(·, φ0) is a geodesic forany φ0 ∈ (−π, π), but a parallel r(s0, ·) is a geodesic if and only if f ′(s0) = 0.

45


Geodesics and isometries

A unit-speed curve γ = r ◦ (u, v) in the surface patch r is a geodesic if and only if itsgeodesic curvature κg := t′ ·B = 0. However,

B := t×N(u, v)

= (∂1r(u, v) u′ + ∂2r(u, v) v

′)× ∂1r× ∂2r

|∂1r× ∂2r|(u, v)

=

((∂1r · ∂2r)∂1r− (∂1r · ∂1r)∂2r

)(u, v) u′ +

((∂2r · ∂2r)∂1r− (∂2r · ∂1r)∂2r

)(u, v) v′√

|∂1r|2|∂2r|2 − (∂1r · ∂2r)2(u, v)

=(F (u, v)u′ +G(u, v)v′)∂1r(u, v)− (E(u, v)u′ + F (u, v)v′)∂2r(u, v)√

EG− F 2(u, v),

since, by Exercises 1.9 and 1.8 in Section 1.1,

u× (v×w) = (u ·w)v− (u · v)w for all u, v, w ∈ R3

and|u× v|2 = |u|2|v|2 − (u · v)2 for all u, v ∈ R

3.

Moreover, the identity t = ∂1r(u, v) u′ + ∂2r(u, v) v

′ implies that

t′ = ∂11r(u, v)(u′)2 + 2∂12r(u, v) u

′v′ + ∂22r(u, v)(v′)2 + ∂1r(u, v) u

′′ + ∂2r(u, v) v′′

and therefore√EG− F 2(u, v) κg equals

(F (u, v)u′ +G(u, v)v′)

×((∂1r · ∂11r)(u, v)(u′)2 + 2(∂1r · ∂12r)(u, v) u′v′ + (∂1r · ∂22r)(u, v)(v′)2

+ E(u, v) u′′ + F (u, v) v′′)

− (E(u, v) u′ + F (u, v) v′)

×((∂2r · ∂11r)(u, v)(u′)2 + 2(∂2r · ∂12r)(u, v) u′v′ + (∂2r · ∂22r)(u, v)(v′)2

+ F (u, v) u′′ +G(u, v) v′′)

= (u′)3(F ∂1r · ∂11r−E ∂2r · ∂11r)(u, v)+ (u′)2v′(G∂1r · ∂11r+ 2F ∂1r · ∂12r− 2E ∂2r · ∂12r− F ∂2r · ∂11r)(u, v)+ u′(v′)2(F ∂1r · ∂22r+ 2G∂1r · ∂12r− 2F ∂2r · ∂12r−E ∂2r · ∂22r)(u, v)+ (v′)3(G∂1r · ∂22 − F ∂2r · ∂22r)(u, v)+ (EG− F 2)(u, v)(u′′v′ − u′v′′).

Since

∂1E = 2∂1r · ∂11r, ∂1F = ∂1r · ∂12r+ ∂2r · ∂11r, ∂1G = 2∂2r · ∂12r (2.15)

and

∂2E = 2∂1r · ∂12r, ∂2F = ∂1r · ∂22r+ ∂2r · ∂12r, ∂2G = 2∂2r · ∂22r, (2.16)

46

2.2. Geodesics

it follows that

κg = A(u, v)(u′)3 +B(u, v)(u′)2v′ + C(u, v)u′(v′)2 +D(u, v)(v′)3

+√EG− F 2(u, v)(u′′v′ − u′v′′),

where

A =12F∂1E − E∂1F + 1

2E∂2E√

EG− F 2, B =

12G∂1E + 3

2F∂2E − E∂1G− F∂1F√EG− F 2

,

C =F∂2F − 3

2F∂1G+G∂2E − 1

2E∂2G√

EG− F 2and D =

G∂2F − 12G∂1G− 1

2F∂2G√

EG− F 2.

The precise form of A, B, C and D is unimportant; the key thing to observe is thatgeodesic curvature depends only on the coefficients of the first fundamental form andtheir partial derivatives. Consequently, geodesics are preserved by isometry.

Theorem 2.20. If the surface patches r1 : U → R3 and r2 : U → R3 are isometric then,for any smooth pair (u, v) : I → U , the curve γ1 := r1 ◦ (u, v) is a geodesic in r1 if andonly if γ2 := r2 ◦ (u, v) is a geodesic in r2.

Exercise 2.21. Use the geodesic equations to give another proof of Theorem 2.20, thatthat geodesics are preserved by isometry.

This leads us to consider another characterisation of geodesics, in terms of length.

Curves of extremal length

Let a and b be distinct points in the regular surface patch r and suppose γ : I → R3 isa regular curve in r which passes through a and b. There there exist a, b ∈ I such thatγ(a) = a and γ(b) = b; without loss of generality, suppose a < b. The curve γ|(a,b) is ageodesic if and only if its length is an extremum of the lengths of all curves between thepoints a and b, in the following sense.

Letγ·: (−ε, ε)× I → R

3; (c, t) 7→ γc(t) := r(uc(t), vc(t)

)

be a family of smooth curves in r : U → R3 which depends smoothly on the parameter c,i.e.,

u· : (−ε, ε)× I → U ; (c, t) 7→ uc(t) and v· : (−ε, ε)× I → U ; (c, t) 7→ vc(t)

have partial derivatives of all orders. Suppose γc(a) = a and γc(b) = b for all c ∈ (−ε, ε),and suppose also that γ0 = γ.

47


The length of γc is

l(c) =

∫ b

a

√E(uc, vc)(u′c)

2 + 2F (uc, vc)u′cv′

c +G(uc, vc)(v′c)2 dt

for all c ∈ (−ε, ε), by (2.2).

Theorem 2.21. The curve γ|(a,b) is a geodesic if and only if l′(0) = 0 for every familyof smooth curves γ

·as above.

Proof. This is a little involved so we omit it; details can be found in [6, Chapter 8].

If γ is the shortest path between a and b then it is a geodesic, since l′(0) = 0 in thiscase. The converse does not hold: a geodesic must be a stationary point for length, butit need not be a minimum. For example, if a and b are points on a sphere then the greatcircle on which they lie (the intersection of the sphere with the plane through a, b andthe origin) gives two paths from a to b; unless these points are antipodal (that is, theylie on a straight line in R3 through the origin) one of these paths will be longer than theother, but both are geodesics.

Furthermore, there may not exist a shortest path between two points. To see this,consider the punctured plane P∗ = {(x, y, 0) : x, y ∈ R} \ {(0, 0, 0)}. If a = (−1, 0, 0)and b = (1, 0, 0) then any path from a to b has length at least 2, and there is no pathof length 2 between a and b in P∗. (Think of the origin as a deep hole which must beavoided by anyone walking from a to b.)

Exercise 2.22. Show that any regular curve from a = (−1, 0, 0) to b = (1, 0, 0) whichlies in P∗ = (R2 \ {(0, 0)})× {0} has length strictly greater than two, but if ε > 0 thenthere exists a smooth curve from a to b of length less than 2 + ε.

48

2.3. The second fundamental form

2.3 The second fundamental formIf γ = r ◦ (u, v) is a unit-speed curve in the regular surface patch r then the tangentvector field

t := γ′ = ∂1r(u, v) u

′ + ∂2r(u, v) v′ (2.17)

and

t′ = ∂11r(u, v)(u′)2 + ∂21r(u, v) u

′v′ + ∂1r(u, v) u′′

+ ∂12r(u, v) u′v′ + ∂22r(u, v)(v

′)2 + ∂2r(u, v) v′′

= ∂11r(u, v)(u′)2 + 2∂12r(u, v) u

′v′ + ∂22r(u, v)(v′)2 + ∂1r(u, v) u

′′ + ∂2r(u, v) v′′.

It follows that

κn := t′ ·N(u, v) = (∂11r ·N)(u, v)(u′)2 + 2(∂12r ·N)(u, v)u′v′ + (∂22r ·N)(u, v)(v′)2

= L(u, v)(u′)2 + 2M(u, v)u′v′ +N(u, v)(v′)2, (2.18)

where L, M and N are the coefficients of the second fundamental form,

L(u, v) du2 + 2M(u, v) du dv +N(u, v) dv2, (2.19)

with

L := ∂11r ·N, M := ∂12r ·N and N := ∂22r ·N. (2.20)

The normal curvature κn measures how much the tangent vector t is moving out ofthe tangent plane as we move along γ. More generally, the second fundamental formgives a measure of how much the surface r deviates from its tangent planes. (The firstfundamental form governs the local intrinsic geometry of the surface: what can be seento first order, by looking at the tangent plane. The second fundamental form, on theother hand, is extrinsic: it depends on how the surface sits inside R3. The plane andthe cylinder have the same first fundamental form, but their second fundamental formsdiffer.)

49


Example 2.22. Let r : U → R3 be a regular surface patch with vanishing secondfundamental form: L = M = N ≡ 0. Show that ∂1N = ∂2N ≡ 0. Deduce that if U isan open rectangle then N is constant and r lies in a plane.

Since ∂1r ·N ≡ 0 and L ≡ 0, it follows that

0 ≡ ∂1(∂1r ·N) = ∂11r ·N+ ∂1r · ∂1N = L+ ∂1r · ∂1N = ∂1r · ∂1N.

Similarly, the facts that ∂2r ·N ≡ 0 and M ≡ 0 imply that ∂2r · ∂1N ≡ 0. Thus ∂1N iseverywhere orthogonal to the tangent plane, so is a scalar multiple of the unit normal N.However, N is a unit vector, so

0 ≡ ∂1(N ·N) = ∂1N ·N+N · ∂1N = 2N · ∂1N

and therefore ∂1N ≡ 0. Similar working shows that ∂2N ≡ 0.

As ∂1N = ∂2N ≡ 0 and any two points in U can be connected by a path made up of ahorizontal line segment and a vertical line segment, the function N is constant. (This isthe two-variable version of a very familiar result: a function which has zero derivativeon some interval is constant on that interval.) Furthermore,

∂1(r ·N) = ∂1r ·N+ r · ∂1N ≡ 0 and ∂2(r ·N) = ∂2r ·N+ r · ∂2N ≡ 0,

so r ·N is also constant (for the same reason); let r ·N ≡ d. This working shows that rlies in the plane {v ∈ R

3 : v ·N = d}.

In Example 2.22, the restriction that U is an open rectangle may be dropped; the resultholds for any connected open set U .

Exercise 2.23. Let r = r ◦ Φ, where Φ is a reparameterisation of the regular surfacepatch r. Show that

[L M

M N

]= ±(JΦ)t

[L ◦ Φ M ◦ ΦM ◦ Φ N ◦ Φ

]JΦ,

where the positive sign is taken if Φ preserves orientation and the negative sign if Φ isorientation reversing, and JΦ is the Jacobian of Φ: see p.32.

50


Principal curvatures

Let r : U → R3 be a regular surface patch; throughout this section, for clarity, we supposethe point (x, y) ∈ U to be fixed and omit mentioning it explicitly: ∂1r = ∂1r(x, y) et

cetera.

If

t1 = λ1 ∂1r+ µ1 ∂2r and t2 = λ2 ∂1r+ µ2 ∂2r (2.21)

are any two tangent vectors then their scalar product t1 · t2 equals

λ1λ2∂1r ·∂1r+λ1µ2∂1r ·∂2r+µ1λ2∂2r ·∂1r+µ1µ2∂2r ·∂2r =[λ1 µ1

][E F

F G

] [λ2µ2

];

with respect to the ordered basis (∂1r, ∂2r), the matrix comprising the coefficients of thefirst fundamental form corresponds to the inner product on the tangent plane T(x,y)r.Hence

QI : T(x,y)r → R; λ ∂1r+ µ ∂2r =

[λµ

]7→

[λ µ

][E F

F G

] [λµ

]

is a quadratic form on the tangent plane which gives the squared magnitude of a tangentvector. Similarly, the quadratic form

QII : T(x,y)r → R; λ ∂1r+ µ ∂2r =

[λµ

]7→

[λ µ

][L M

M N

] [λµ

]

gives the normal curvature κn of a unit-speed curve through r(x, y) which has the tangentvector λ ∂1r(x, y)+µ ∂2r(x, y) at that point, by (2.17) and (2.18). Using diagonalisation,we will find the maximum and minimum values of κn over all such curves, that is, theextreme values of QII under the constraint that QI ≡ 1.

51


Suppose the tangent vectors t1 and t2 from (2.21) are chosen so that {t1, t2} is anorthonormal basis of the tangent plane T(x,y)r. If

A :=

[λ1 λ2µ1 µ2

]

then

At

[E F

F G

]A =

[t1·t1 t1·t2t2·t1 t2·t2

]=

[1 0

0 1

]= I; (2.22)

in particular, the matrix A is invertible. Since(At

[L M

M N

]A

)t

= At

[L M

M N

]t

(At)t = At

[L M

M N

]A,

the principal-axis theorem (Theorem A.8) yields a real 2×2 matrix B which is orthogonal(that is, BtB = BBt = I) and such that

BtAt

[L M

M N

]AB =

[κ1 0

0 κ2

], (2.23)

where κ1 and κ2 are real numbers with κ1 6 κ2.

Furthermore, by (2.22), it follows that

BtAt

[E F

F G

]AB = BtIB = BtB = I; (2.24)

the matrix C := AB simultaneously diagonalises the matrices corresponding to the firstand the second fundamental forms. Let the linear transformation

TC : T(x,y)r → T(x,y)r; λ ∂1r+ µ ∂2r =

[λµ

]7→ C

[λµ

]= λC ∂1r+ µC ∂2r

and note that TC is invertible because the matrix C is. If s = λ ∂1r+µ ∂2r is any tangentvector then

|TCs|2 = (TCs) · (TCs) =[λ µ

]Ct

[E F

F G

]C

[λµ

]=

[λ µ

] [ λµ

]= λ2 + µ2,

so{unit vectors in T(x,y)r} = {TCs : λ2 + µ2 = 1}.

Furthermore, the value of the second fundamental form at TCs is

QII(TCs) =[λ µ

]Ct

[L M

M N

]C

[λµ

]=

[λ µ

][κ1 0

0 κ2

] [λµ

]= κ1λ

2 + κ2µ2.

Hence

{QII(t) : t is a unit tangent vector} = {κ1λ2 + κ2µ2 : λ2 + µ2 = 1}

= {κ1x+ κ2(1− x) : 0 6 x 6 1} = [κ1, κ2]

and κ1 and κ2 are the principal curvatures, scalar fields on r giving the minimum andmaximum values of the normal curvature κn.

52


The corresponding principal directions are unit tangent vectors fields, t1 and t2, on rcorresponding to the directions in which the minimum and maximum values of thenormal curvature, κ1 and κ2, are attained, so that QII(t1) = κ1 and QII(t2) = κ2. Apoint where κ1 = κ2 is called an umbilic, and every unit tangent vector is principal atsuch a point.

Suppose the principal curvatures are distinct: κ1 6= κ2. If t = TCs is a unit tangentvector, where s = λ ∂1r+ µ ∂2r, then λ

2 + µ2 = QI(t) = 1 and

QII(t) = κ1 ⇐⇒ κ1λ2 + κ2µ

2 = κ1

⇐⇒ κ1λ2 + κ2(1− λ2) = κ1

⇐⇒ (κ1 − κ2)λ2 = κ1 − κ2

⇐⇒ λ2 = 1.

Hence t is a principal direction corresponding to κ1 if and only if t = ±TC∂1r, andsimilar working shows that t is a principal direction corresponding to κ2 if and onlyif t = ±TC∂2r. In particular, up to a choice of sign, the principal directions are uniquewherever the principal curvatures are distinct.

Exercise 2.24. Suppose the principal curvatures κ1 and κ2 are distinct at some point.Show that the principal directions t1 and t2 are orthogonal there.

Since the matrix C was obtained non-constructively, with the assistance of the principal-axis theorem, there remains the question of how to calculate in practice the principalcurvatures and principal directions. We proceed as follows.

By (2.23), the principal curvatures κ1 and κ2 are the eigenvalues of Ct

[L M

M N

]C.

Moreover, from (2.24) and the fact that C is invertible,

det

(κI − Ct

[L M

M N

]C

)= 0 ⇐⇒ det

(Ct

(κ

[E F

F G

]−[L M

M N

])C

)= 0

⇐⇒ det

(κ

[E F

F G

]−[L M

M N

])= 0.

Thus the principal curvatures κ1 and κ2 may be calculated in a straightforward manner,as the roots of the quadratic equation

det

[κE − L κF −M

κF −M κG−N

]= 0. (2.25)

53


To find the principal directions, note that if

s = λ ∂1r+ µ ∂2r and t = TCs = λC ∂1r+ µC ∂2r

then[κE − L κF −M

κF −M κG−N

][λCµC

]=

[00

]

⇐⇒ Ct

(κ

[E F

F G

]−[L M

M N

])C

[λµ

]=

[00

]

⇐⇒[κ−κ1 0

0 κ−κ2

] [λµ

]=

[00

]

⇐⇒[(κ1 − κ)λ(κ2 − κ)µ

]=

[00

].

Recall that t = TCs is a principal direction corresponding to κ1 if and only if λ = ±1and µ = 0. From the previous working, this is equivalent to

[κ1E − L κ1F −Mκ1F −M κ1G−N

] [λCµC

]=

[00

],

and a similar result holds for κ2.

To summarise, the tangent vector t = z ∂1r+w ∂2r is a principal direction correspondingto κi (where i = 1 or i = 2) if and only if

[κiE − L κiF −M

κiF −M κiG−N

][zw

]=

[00

], (2.26)

and the principal directions may be found by solving this equation, with the constraintthat Ez2 + 2Fzw +Gw2 = 1.

Exercise 2.25. Let r = r ◦ Φ, where Φ is a reparameterisation of the regular surfacepatch r. Prove that the principal curvatures κ1 and κ2 of r are equal to κ1◦Φ and κ2◦Φ,where κ1 and κ2 are the principal curvatures of r.

54


Exercise 2.26. Let E, F , G and L, M , N be constants. Use the method of Lagrangemultipliers to show that the extreme values of

κ(x, y) := Lx2 + 2Mxy +Ny2,

subject to the constraint

f(x, y) := Ex2 + 2Fxy +Gy2 = 1,

satisfy

det

[κE − L κF −MκF −M κG−N

]= 0.

[Hint: letF (x, y) := κ(x, y)− λ

(f(x, y)− 1

)

and show first that λ = κ if F is stationary, so that∂F

∂x=∂F

∂y= 0, and f(x, y) = 1.]

Example 2.23. The sphere of radius a > 0, minus the poles and one meridian, has theparameterisation

r : (−π/2, π/2)× (−π, π) → R3; (θ, φ) 7→ (a cos θ cosφ, a cos θ sinφ, a sin θ).

(When a = 1, this is the parameterisation of Example 2.5.) This surface has firstfundamental form a2 dθ2+a2 cos2 θ dφ2 and second fundamental form a dθ2+a cos2 θ dφ2,so the principal curvatures are the roots of the equation

0 = det

[κa2 − a 0

0 κa2 cos2 θ − a cos2 θ

]= a2(κa− 1)2 cos2 θ.

Hence the principal curvatures are equal and have the constant value 1/a. In particular,every point is an umbilic and every tangent vector is a principal direction, as we shouldexpect: at each point, the sphere curves equally in every direction.

55


Example 2.24. Fix a > 0. The cylinder

r : R× (−π, π) → R3; (z, φ) 7→ (a cosφ, a sin φ, z)

has first fundamental form dz2 + a2 dφ2 and second fundamental form a dφ2. Hence theprincipal curvatures are the roots of

det

[κ 0

0 a2κ− a

]= 0 ⇐⇒ κ(a2κ− a) = 0,

so κ1 ≡ 0 and κ2 ≡ 1/a. If we visualise the situation, the minimum deviation from thetangent plane should occur along a meridian (as this is a straight line in the cylinder, solies in the tangent plane) and the maximum along a parallel.

To see if this conjecture holds, we solve the equations

[0 0

0 −a

][λ1µ1

]=

[00

]and

[1/a 0

0 0

] [λ2µ2

]=

[00

]

subject to the constraints λ2i + a2µ2i = 1 for i = 1 and i = 2.

The first yields (λ1, µ1) = (±1, 0) and the second yields (λ2, µ2) = (0,±1/a), so theprincipal directions are

t1(z, φ) = ±∂1r(z, φ) = (0, 0, ±1)

and

t2(z, φ) = ±1

a∂2r(z, φ) = ±(− sin φ, cos φ, 0).

It is an exercise to verify that the meridian and the parallel through r(z, φ) have tangentvectors which are proportional to these, and so show that our conjecture is correct.

Exercise 2.27. Do isometric surfaces have the same principal curvatures? [Hint:consider a plane and a cylinder.]

56

2.4. Gaussian curvature

2.4 Gaussian curvature

Definition 2.25. The Gaussian curvature of a regular surface patch r : U → R3 is thescalar field K := κ1κ2, the product of the principal curvatures.

If the surface r : U → R3 has Gaussian curvature K(x, y) > 0 at some point (x, y) ∈ Uthen the principal curvatures κ1(x, y) and κ2(x, y) have the same sign, so the surfacenear to r(x, y) is convex or concave, lying on one side of the tangent plane T(x,y)r.If K(x, y) < 0 then the principal curvatures have different signs and r(x, y) is a saddle

point , where the surface crosses the tangent plane.

Theorem 2.26. The Gaussian curvature

K =LN −M2

EG− F 2, (2.27)

where E, F , G and L, M , N are the coefficients of the first and second fundamentalforms, respectively.

Proof. If b and c are the roots of a quadratic polynomial then

a(x− b)(x− c) = ax2 − a(b+ c)x+ abc = a2x2 + a1x+ a0,

so their product bc equals a0/a2. Since the principal curvatures are roots of the quadratic

det

(κ

[E F

F G

]−[L M

M N

])= det

[κE − L κF −M

κF −M κG−N

]

= (κE − L)(κG−N)− (κF −M)2

= (EG− F 2)κ2 − (EN + LG− 2FM)κ+ LN −M2,

it follows thatK = κ1κ2 = (LN −M2)/(EG− F 2).

Exercise 2.28. Why is EG− F 2 > 0 at all points in a regular surface patch?

57


Example 2.27. For the sphere with radius a > 0, parameterised as in Example 2.23,the formula (2.27) becomes

K(θ, φ) =a2 cos2 θ

a4 cos2 θ=

1

a2,

which agrees with the values of the principal curvatures found in that example.

An important property of Gaussian curvature is its invariance under reparameterisation.

Exercise 2.29. Let r = r ◦ Φ, where Φ is a reparameterisation of the regular surfacepatch r. Prove that the Gaussian curvature K of r is equal to K ◦ Φ, where K is theGaussian curvature of r.

The Gauss map

If r : U → R3 is a regular surface patch then the unit normal vector field N is a smoothmap from U to the unit sphere S2. When thought of in this way, N is called the Gauss

map. As the point (x, y) varies over U , the vector r(x, y) traces out the surface r(U)and N(x, y) traces out a region on the sphere.

Exercise 2.30. Explain why the vectors ∂1N(x, y) and ∂2N(x, y) lie in the tangentplane T(x,y)r for all (x, y) ∈ U , so that

∂1N = A∂1r+B ∂2r and ∂2N = C ∂1r+D ∂2r,

where A, B, C and D are scalar fields.

By considering the partial derivatives of ∂1r ·N and ∂2r ·N, prove that

∂1r · ∂1N = −L, ∂1r · ∂2N = ∂2r · ∂1N = −M and ∂2r · ∂2N = −N.

Deduce that the Weingarten matrix

[−A −C−B −D

]=

[E FF G

]−1 [

L MM N

].

Hence show that ∂1N× ∂2N = K ∂1r× ∂2r, where K is the Gaussian curvature.

58


Exercise 2.31. Explain why [N, ∂1N, ∂2N] = ±|∂1N× ∂2N|.

Exercise 2.32. Let Dε := D(z, w; ε) ⊆ U be the open disc of radius ε and centre (z, w).Explain why the area traced out on the sphere by N(x, y) as (x, y) varies over Dε is

AN(Dε) =

∫ ∫

Dε

|(∂1N× ∂2N)(x, y)| dx dy.

We let the signed area

sAN(Dε) =

∫ ∫

Dε

[N, ∂1N, ∂2N](x, y) dx dy;

Exercise 2.31 shows why this is a sensible definition. By using the fact that

limε→0

1

πε2

∫ ∫

Dε

f(x, y) dx dy = f(z, w)

for any continuous function f : U → R, show that

limε→0

sAN(Dε)

Ar(Dε)

exists and equals K(z, w), where Ar(Dε) is the area of the region on r with coordinatesin Dε, as defined by (2.5). [Use the usual “rules for limits”.]

Theorema Egregium

Although it appears from Theorem 2.26 that Gaussian curvature depends on both thefirst and the second fundamental forms, this is not in fact the case: Gaussian curvaturedepends only on the first fundamental form and its derivatives. This result is Gauss’sTheorema Egregium (the Latin for “remarkable theorem”).

Theorem 2.28. (Theorema Egregium) If the regular surface patches r1 : U → R3

and r2 : U → R3 are isometric then their Gaussian curvatures are equal.

The key step in the proof of Theorem 2.28 is the following lemma, which is also a vitalcomponent in the proof of the Gauss–Bonnet Theorem: see Theorem 3.13.

Lemma 2.29. Let (e, f , N) be a orthonormal triple of vector fields on the regularsurface patch r : U → R3, where N is, as usual, the unit normal to the surface. Then

∂1e · ∂2f − ∂2e · ∂1f =LN −M2

√EG− F 2

,

where E, F , G and L, M , N are the coefficients of the first and second fundamentalforms, respectively.

Proof. By Exercise 2.30, we have that ∂1N× ∂2N = K ∂1r× ∂2r. Therefore

(∂1N× ∂2N) ·N = K (∂1r× ∂2r) ·N = K |∂1r× ∂2r|N ·N =LN −M2

√EG− F 2

,

by Theorem 2.26 and the fact that |∂1r× ∂2r| =√EG− F 2.

59


Now,

(∂1N× ∂2N) ·N = (∂1N× ∂2N) · (e× f) = (∂1N · e)(∂2N · f)− (∂2N · e)(∂1N · f),

by Exercise 1.9. Furthermore,

N · e ≡ 0 =⇒ 0 ≡ ∂i(N · e) = ∂iN · e +N · ∂ie =⇒ ∂iN · e = −N · ∂ie,

for i = 1 and i = 2, and the same holds with e replaced by f . Hence

LN −M2

√EG− F 2

= (N · ∂1e)(N · ∂2f)− (N · ∂2e)(N · ∂1f).

Finally, note that when we write the vector fields ∂ie and ∂if in terms of the orthonormaltriple (e, f , N), certain terms do not appear: as e · e ≡ 1, so e · ∂ie ≡ 0, and similarlyfor ∂if . Hence

∂1e = (f · ∂1e)f + (N · ∂1e)N,∂2e = (f · ∂2e)f + (N · ∂2e)N,∂1f = (e · ∂1f)e + (N · ∂1f)N

and ∂2f = (e · ∂2f)e + (N · ∂2f)N,

so

∂1e · ∂2f − ∂2e · ∂1f = (N · ∂1e)(N · ∂2f)− (N · ∂2e)(N · ∂1f) =LN −M2

√EG− F 2

,

as claimed.

Proof of Theorem 2.28. It follows from Lemma 2.29 and Theorem 2.26 that

K =∂1e · ∂2f − ∂2e · ∂1f√

EG− F 2.

Thus it suffices to show that ∂1e · ∂2f − ∂2e · ∂1f depends only on the coefficients of thefirst fundamental form and their partial derivatives, when e and f are suitably chosen.

Let

e :=∂1r

|∂1r|and f :=

∂2r− (e · ∂2r)e|∂2r− (e · ∂2r)e|

;

since {∂1r, ∂2r} is linearly independent everywhere, the vector fields e and f are welldefined. (They are the result of applying the Gram-Schmidt orthogonalisation procedureto this set.)

By construction, {e, f , N} is an orthonormal basis everywhere; to see that (e, f , N) isa right-handed triple, it suffices to verify that [e, f , N] = [N, e, f ] = N · (e× f) ≡ 1, byExercise 1.7.

60


For this, note that

|∂1r× ∂2r| = (EG− F 2)1/2,

|∂1r| = E1/2

and |∂2r− (e · ∂2r)e| = |∂2r− |∂1r|−2(∂1r · ∂2r)∂1r|= |∂2r−E−1F ∂1r|= (G− 2E−1F 2 + E−2F 2E)1/2

= E−1/2(EG− F 2)1/2.

As ∂1r× e ≡ 0, it follows that

N · (e× f) =(∂1r× ∂2r) · (∂1r× ∂2r)

(EG− F 2)1/2E1/2 E−1/2(EG− F 2)1/2≡ 1,

as required.

Now,

∂2(∂1e · f)− ∂1(∂2e · f) = ∂21e · f + ∂1e · ∂2f − ∂12e · f − ∂2e · ∂1f= ∂1e · ∂2f − ∂2e · ∂1f ,

so it suffices to use (2.15) and (2.16) to note that

∂1e · f =(∂1E ∂1r

2E3/2+∂11r

E1/2

)· E

1/2(∂2r− E−1F∂1r)

(EG− F 2)1/2

=∂11r · ∂2r

(EG− F 2)1/2− F ∂11r · ∂1rE(EG− F 2)1/2

=2∂1F − ∂2E

2(EG− F 2)1/2− F∂1E

2E(EG− F 2)1/2

and ∂2e · f =(∂2E ∂1r

2E3/2+∂21r

E1/2

)· E

1/2(∂2r− E−1F∂1r)

(EG− F 2)1/2

=∂21r · ∂2r

(EG− F 2)1/2− F ∂21r · ∂1rE(EG− F 2)1/2

=∂1G

2(EG− F 2)1/2− F ∂2E

2E(EG− F 2)1/2,

so they depend only on E, F , G and their partial derivatives.

Corollary 2.30. If the parameterisation of the regular surface patch r : U → R3 isorthogonal, so that the coefficients (E, F,G) of the first fundamental form are suchthat F ≡ 0, then the Gaussian curvature

K =−1

2√EG

(∂2

( ∂2E√EG

)+ ∂1

( ∂1G√EG

)).

61


Proof. If F ≡ 0 then, following the working in the proof of Theorem 2.28,

K =1√EG

(∂1e · ∂2f − ∂2e · ∂1f

)

=1√EG

(∂2(∂1e · f)− ∂1(∂2e · f)

)

=1√EG

(∂2

( −∂2E2√EG

)− ∂1

( ∂1G√EG

)),

as claimed.

Example 2.31. The sphere and the plane have different Gaussian curvature, which isinvariant under reparameterisation. Hence there is no isometric map from the sphere tothe plane, no matter how the plane is parameterised. In particular, any flat map of theEarth must distort distances.

62

Three Surfaces – global theory

3.1 OverviewIn this chapter we present three closely related theorems, named after Carl Gauss andPierre Bonnet. The first is the Gauss–Bonnet theorem for smooth simple closed curves .

Theorem 3.1. Let γ be a unit-speed simple closed curve in the regular surface patch r,and suppose γ is positively oriented. Then

∫ ∫

intγ

K dAr −∫

γ

κg ds = 2π,

where int γ is the interior of the curve γ, K is the Gaussian curvature of r, dAr is thearea element of r and κg is the geodesic curvature of γ.

Informally, a simple closed curve is one which starts at some point and travels back tothat point without crossing itself; such a curve γ lies in the surface patch r as long asthe whole of γ and its interior are contained in r, and the curve is positively oriented ifits interior lies on the left of the curve as we proceed around it.

It is not straightforward to make sense of these seemingly simple concepts in a rigorousfashion. For example, given the equator on the unit sphere, is the northern or southernhemisphere its interior?

63

3. Surfaces – global theory

The second theorem is a generalisation of the first, called the Gauss–Bonnet theorem for

curvilinear polygons .

Theorem 3.2. Let γ be a unit-speed curvilinear polygon in the regular surface patch r,and suppose γ is positively oriented and has n vertices. Then

∫ ∫

intγ

K dAr −∫

γ

κg ds = (2− n)π +n∑

i=1

αi,

where intγ, K, dAr and κg are as in Theorem 3.1 and α1, . . . , αn are the interior anglesat the vertices of γ.

A curvilinear polygon is a simple closed curve which is smooth apart from a finite numberof corners, or vertices, where the curve changes direction sharply (so is not differentiablethere). A polygon in the plane, made up of a finite number of straight-line segments,provides a familiar example of such a curve. Note that taking n = 0 reduces Theorem 3.2to Theorem 3.1, so this is a generalisation of the former result.

The third and final theorem is the Gauss–Bonnet theorem for closed surfaces .

Theorem 3.3. If S is a closed surface then∫ ∫

S

K dA = 2πχ, (3.1)

where χ is the Euler characteristic of S.

A surface S is a subset of R3 made up of one or more regular surface patches; it is closedif S consists of only one piece (so is connected) and is compact : given any collection ofopen sets which covers S, a finite number of them can be selected which still covers S.The plane

{(x, y, 0) : x, y ∈ R}and the cylinder

{(x, y, z) : x2 + y2 = 1, z ∈ R}are connected but non-compact surfaces, as is the open unit disc

{(x, y, 0) : x2 + y2 < 1}.

64

3.1. Overview

The unit sphere{(x, y, z) : x2 + y2 + z2 = 1}

and the torus {(x, y, z) :

(√x2 + y2 − 2

)2+ z2 = 1

}

are closed surfaces.

[It is difficult to appreciate the definition of compactness at first, so an alternativecharacterisation may be helpful: a set S ⊆ R3 is compact if and only if it is bounded (Slies in some ball centred at the origin) and contains its limit points (a point x ∈ R3 isa limit point of S if, for all ε > 0, there exists p ∈ S such that |x − p| < ε; hence Scontains its limit points if every point not in S lies a positive distance away from S).]

The Euler characteristic of a closed surface is defined by the formula

χ = V − E + F,

where V is the number of vertices, E is the number of edges and F is the number offaces in a subdivision of S. This is a way of splitting up the surface into a finite numberof curvilinear polygons, the interiors of which form the faces, which are either disjointor share common edges or vertices. The precise choice does not matter; any subdivisionof a given surface will produce the same value for χ.

If a surface is orientable (so there is a consistent choice of an outward direction from thesurface and hence the surface has two sides, like the sphere and torus, but unlike theMobius strip) then

χ = 2− 2g,

where g is the genus of S, the number of holes in it: the unit sphere has genus 0 andthe torus has genus 1. Every closed surface is orientable and homeomorphic to a toruswith g holes, for some g > 0: two surfaces are homeomorphic if there is a bijectionbetween them which is continuous and has continuous inverse.

The most remarkable thing about Theorem 3.3 is the different character of the twosides of (3.1). The left-hand side is analytical and geometrical, defined using area andcurvature; the right-hand side is topological and combinatorial, defined by countingvertices, edges and faces or holes. The fact that these two coincide is the source ofinspiration for a significant amount of recent mathematical research.

65


3.2 Plane curves

Definition 3.4. A plane curve is a map

c : (a, b) → R2; t 7→

(u(t), v(t)

),

where −∞ 6 a < b 6 ∞, such that the coordinate functions

u : (a, b) → R and v : (a, b) → R

are continuous. The plane curve c is smooth if the coordinate functions u and v havederivatives of all orders (i.e., are infinitely differentiable). [Thus a smooth plane curveis the same thing as a smooth pair, considered earlier.]

A smooth plane curve is regular if the tangent vector field

c′ : (a, b) → R2; t 7→ (u′(t), v′(t)

)

is nowhere zero; equivalently,

|c′(t)|2 = u′(t)2 + v′(t)2 > 0 for all t ∈ (a, b),

where |x| denotes the magnitude of the vector x ∈ R2.

If c is a regular plane curve and m ∈ (a, b) then the bijection

ℓ : (a, b) → (c, d); t 7→∫ t

m

|c′(x)| dx

is the arc-length function for c with starting point m, where

c :=

∫ a

m

|c′(t)| dt = −∫ m

a

|c′(t)| dt and d :=

∫ b

m

|c′(t)| dt;

note that c could equal −∞ and d could equal ∞. The arc-length function ℓ is smoothand has strictly positive derivative, so the inverse

ℓ−1 : (c, d) → (a, b)

is also a smooth bijection. The curve c := c ◦ ℓ−1 is a reparameterisation of c which hasunit speed, i.e.,

|c′(s)| = 1 for all s ∈ (c, d).

66

3.2. Plane curves

Exercise 3.1. Prove that an arc-length function ℓ for a regular plane curve c = (u, v) isinfinitely differentiable. Prove further that the inverse ℓ−1 is also infinitely differentiable.

For a unit-speed plane curve

c : (a, b) → R2; s 7→

(u(s), v(s)

),

the unit normal vector field is

n : (a, b) → R2; s 7→

(−v′(s), u′(s)

).

Since the map [xy

]7→

[cos θ − sin θ

sin θ cos θ

] [xy

]

rotates the point (x, y) ∈ R2 about the origin through θ radians in the positive (anti-clockwise) sense, geometrically the normal vector

n(s) =(−v′(s), u′(s)

)

is obtained by rotating the tangent vector

t(s) =(u′(s), v′(s)

)

through π/2 radians anticlockwise.

Note thatt(s) · n(s) = −u′(s)v′(s) + v′(s)u′(s) = 0

and1 = |t(s)|2 = u′(s)2 + v′(s) = |n(s)|,

so, for all s ∈ (a, b) the set {t(s), n(s)} is an orthonormal basis of R2, analogous to theSerret–Frenet basis {t(s), n(s), b(s)} attached to a unit-speed curve in R3.

67


This orthonormal basis may be used to decompose the vector t′(s):

t′(s) = (t(s) · t′(s))t(s) + (n(s) · t′(s))n(s).

However, as t has unit magnitude,

0 =d

ds|t(s)|2 = d

ds

(t(s) · t(s)

)= t′(s) · t(s) + t(s) · t′(s) = 2t(s) · t′(s),

sot′(s) = κs(s)n(s) for all s ∈ (a, b),

where the scalar field

κs : (a, b) → R; s 7→ κs(s) := n(s) · t′(s)

is the signed curvature of the curve c. Note that

|t′| = |κs n| = |κs|,

so the absolute value |κs| is the magnitude of the derivative of the tangent vector (thesame as for curves in R3).

If the scalar curvature is positive then the curve is moving in the direction of the normalvector; if negative, the curve is bending away from the normal vector. The followingtheorem makes this idea more precise.

Theorem 3.5. Let u ∈ R2 be a fixed unit vector. If c is a unit-speed plane curve and ψ

is the angle from u to t, measured in the positive sense, then

κs =dψ

ds.

Proof. Let v be the unit vector obtained by rotating u through π/2 radians in thepositive sense and note that {u,v} is an orthonormal basis, so

t = (u · t)u+ (v · t)v = cosψ u+ sinψ v.

Thent′ = (− sinψ u+ cosψ v)ψ′

andn = cosψ v − sinψ u

(as rotating u and v through π/2 radians gives v and −u, respectively) so

κs = n · t′ = (sin2 ψ + cos2 ψ)ψ′ = ψ′.

68

3.3. Simple closed curves

Those of a cautious frame of mind may wonder if the angle ψ in Theorem 3.5 is welldefined. The following lemma should reassure them.

Lemma 3.6. Letx : (a, b) → R and y : (a, b) → R

be smooth functions such that x(s)2 + y(s)2 = 1 for all s ∈ (a, b). If s0 ∈ (a, b) is fixedand ψ0 ∈ R is chosen so that

(cosψ0, sinψ0) =(x(s0), y(s0)

)

then

ψ : (a, b) → R; s 7→ ψ0 +

∫ s

s0

x(t)y′(t)− x′(t)y(t) dt

is a smooth function such that

(cosψ(s), sinψ(s)

)=

(x(s), y(s)

)for all s ∈ (a, b).

Proof. Consider the function

F : (a, b) → R; s 7→(x(s)− cosψ(s)

)2+(y(s)− sinψ(s)

)2;

since F (s0) = 0, the result holds if F ′ ≡ 0. However, as x2 + y2 ≡ 1, it follows bydifferentiating that xx′ + yy′ ≡ 0 and therefore

12F ′ = (x− cosψ)(x′ + ψ′ sinψ) + (y − sinψ)(y′ − ψ′ cosψ)

= xx′ + yy′ − (x′ cosψ + y′ sinψ) + ψ′(x sinψ − y cosψ)

= 0− (x2 + y2)(x′ cosψ + y′ sinψ) + (xy′ − x′y)(x sinψ − y cosψ)

= −x(xx′ + yy′) cosψ − y(xx′ + yy′) sinψ

≡ 0,

as required. That ψ is smooth follows from the fundamental theorem of calculus andthe smoothness of x and y.

3.3 Simple closed curves

Definition 3.7. Let c : R → R2 be a plane curve. If there exists a > 0 such that

(i) c|[0,a) is injective, so that c(s) = c(t) if and only if s = t, for all s, t ∈ [0, a)

and

(ii) c(t) = c(t+ a) for all t ∈ R

then c is a simple closed plane curve with period a.

69


Theorem 3.8. A simple closed plane curve c partitions the plane R2 into three disjointnon-empty sets, the image of c, c(R), the interior of c, int c, and the exterior of c, ext c;the latter two sets are open and have c(R) as their boundary. [The boundary of a set S isthe set of limit points of S which do not lie in S.] Each of these three sets is connected:any two points in the set may be joined by a path which lies within that set. The interiorof c is bounded and the exterior of c is unbounded: there exists r > 0 such that

int c ⊆ {(x, y) ∈ R2 : x2 + y2 < r}

and, for all R > 0,ext c ∩ {(x, y) ∈ R

2 : x2 + y2 > R} 6= ∅.

Proof. This is the Jordan curve theorem. A proof is surprisingly difficult, so we omit it.The concerned reader may look at Theorem 63.4 in [4]; a simpler proof, valid only forpiecewise-smooth curves, is contained in [2, Section VIII.5].

A unit-speed simple closed plane curve is positively oriented if the normal vector fieldpoints to the interior of the curve at every point, and negatively oriented if the normalvector field points to the exterior of the curve at every point. The normal vector fieldcannot flip from interior to exterior, or vice versa, for reasons of continuity; hence everyunit-speed simple closed plane curve has an orientation, either positive or negative.

We extend the notation of orientation to any regular simple closed plane curve, by sayingsuch a curve is positively oriented if a unit-speed reparameterisation is, and similarly fornegatively oriented.

Exercise 3.2. Prove that any unit-speed reparameterisation of a regular simple closedplane curve c with period a is a regular simple closed plane curve with period

ℓ(c) :=

∫ a

0

|c′(t)| dt.

Exercise 3.3. Suppose c is a regular simple closed plane curve and let c(t) := c(−t)for all t ∈ R. Show that c is also a regular simple closed plane curve, with the sameperiod as c. Show further that c and c have opposite orientations.

70


The following result is known as Green’s theorem.

Theorem 3.9. Let c = (u, v) : R → R2 be a regular simple closed curve with period a,which is positively oriented. Suppose f : U → R and g : U → R are smooth functionsdefined on the open set U ⊆ R2 which contains c and its interior. Then

∫

c

f du+ g dv =

∫ ∫

int c

∂1g − ∂2f dx dy,

where the notation for the first integral is shorthand for

∫ a

0

f(u(t), v(t)

)u′(t) + g

(u(t), v(t)

)v′(t) dt.

Proof. The strategy is to prove the theorem for particularly nice choices of c, then toshow that any c can be reduced to this case by suitable subdivision of the region int c;this is somewhat similar to the proof of Cauchy’s theorem in MATH215. Details can befound in [3, Section 5.2].

Corollary 3.10. The area A(int c) enclosed by the positively oriented regular simpleclosed curve c = (u, v) : R → R2 with period a equals

1

2

∫ a

0

u(t)v′(t)− u′(t)v(t) dt.

Proof. Let f(x, y) = −y and g(x, y) = x for all x, y ∈ R. Then, by Green’s theorem,

∫ a

0

−v(t)u′(t) + u(t)v′(t) dt =

∫ ∫

int c

1 + 1 dx dy = 2A(int c).

Example 3.11. The simple closed curve

c : R → R2; t 7→ (r cos t, r sin t)

is positively oriented and has period 2π; it describes in an anticlockwise motion the circleabout the origin with radius r > 0.

By Corollary 3.10, the interior of this circle has area

1

2

∫ 2π

0

r2 cos2 t + r2 sin t dt = πr2,

as expected.

71


Simple closed curves in surfaces

Definition 3.12. Let r : U → R3 be a regular surface patch and let

c : R → R2; t 7→

(u(t), v(t)

)

be a regular simple closed plane curve with period a. If the open set U contains c andits interior then

γ := r ◦ c : R → R3; t 7→ r

(u(t), v(t)

)

is a simple closed curve in r with period a and interior intγ := r(int c). By definition,the curve γ has the same orientation as c.

Exercise 3.4. Prove that a simple closed curve γ in r with period a is a regular smoothcurve such that γ|[0,a) is injective and γ(t) = γ(t+ a) for all t ∈ R.

Exercise 3.5. Let γ be a simple closed curve in r with period a, let m ∈ R and let

ℓ : R → R; t 7→∫ t

m

|γ ′(x)| dx.

Prove that ℓ is a smooth bijection with smooth inverse. Deduce that γ := γ ◦ ℓ−1 is aunit-speed smooth curve in r with period

ℓ(γ) :=

∫ a

0

|γ ′(x)| dx.

72


We can now state the first Gauss–Bonnet theorem. The minus sign on the left-handside below appears as we have defined the geodesic curvature in the opposite sense toPressley: compare this with [6, Theorem 11.1].

Theorem 3.13. Let γ := r◦c be a unit-speed simple closed curve in the regular surfacepatch r : U → R3 and suppose γ is positively oriented. Then

∫ ∫

intγ

K dAr −∫

γ

κg ds = 2π,

where K is the Gaussian curvature of r, dAr is the area element of r and κg is thegeodesic curvature of γ. The integrals are defined as follows:

∫ ∫

intγ

K dAr :=

∫ ∫

int c

K(u, v)√EG− F 2(u, v) du dv

and

∫

γ

κg ds :=

∫ a

0

κg(s) ds,

where E, F and G are the coefficients of the first fundamental form of r and a is theperiod of γ.

Proof. Let (e, f , N) be the orthonormal triple obtained by applying the Gram–Schmidtprocedure to {∂1r, ∂2r, N}, as in the proof of Theorem 2.28. The proof here consists ofcalculating

I :=

∫

γ

e · f ′ ds =∫ a

0

e(u, v) · f(u, v)′ ds

in two different ways, where

c = (u, v) : R → R2; s 7→

(u(s), v(s)

).

First, note that∫ a

0

e(u, v) · f(u, v)′ ds =∫ a

0

e(u, v) · (∂1f(u, v)u′ + ∂2f(u, v)v′) ds

=

∫

c

e · ∂1f du+ e · ∂2f dv.

By Green’s theorem, Theorem 3.9, this equals∫ ∫

int c

∂1(e · ∂2f)− ∂2(e · ∂1f) du dv =∫ ∫

int c

∂1e · ∂2f − ∂2e · ∂1f du dv

and, from Lemma 2.29, it follows that

I =

∫ ∫

int c

LN −M2

√EG− F 2

du dv =

∫ ∫

int c

K√EG− F 2 du dv =

∫ ∫

intγ

K dAr.

73


Now use Lemma 3.6 to let θ(s) be the angle between t(s) := γ′(s) and e

(u(s), v(s)

), so

thatt = e(u, v) cos θ + f(u, v) sin θ.

As (e, f , N) is an orthonormal triple of vector fields,

t×N(u, v) = −f(u, v) cos θ + e(u, v) sin θ

and

t′ = e(u, v)′ cos θ − e(u, v) θ′ sin θ + f(u, v)′ sin θ + f(u, v) θ′ cos θ

= e(u, v)′ cos θ + f(u, v)′ sin θ − t×N(u, v) θ′.

It follows that

κg := t′ ·(t×N(u, v)

)= −θ′ + (−e(u, v)′ · f(u, v) cos2 θ + f(u, v)′ · e(u, v) sin2 θ)= −θ′ + e(u, v) · f(u, v)′,

using the facts that t×N(u, v) is a unit vector and

e(u, v) · e(u, v) ≡ 1 =⇒ e(u, v)′ · e(u, v) ≡ 0,

f(u, v) · f(u, v) ≡ 1 =⇒ f(u, v)′ · f(u, v) ≡ 0

and e(u, v) · f(u, v) ≡ 0 =⇒ e(u, v)′ · f(u, v) = −e(u, v) · f(u, v)′.

The result now follows, since

−∫ a

0

κg ds =

∫ a

0

θ′ ds−∫ a

0

e(u, v) · f(u, v)′ ds = 2π − I

because∫γθ′ ds :=

∫ a

0θ′ ds = θ(a)− θ(0) = 2π.

For this final step, we argue as follows. It may be shown that the interior of any simpleclosed curve is simply connected: it has no holes. Consequently, the curve γ may becontinuously deformed until it is a small circle γ0. As 1

2π

∫γθ′ ds is integer valued and

depends on γ in a continuous manner, it must be the case that∫γθ′ ds =

∫γ

0

θ′ ds. The

curve γ0 is very small, so e is almost constant on γ0 and θ is almost the angle made bythe tangent vector γ ′

0 with a fixed direction. Finally, as the tangent vector travels onceround a circle anticlockwise, it turns through an angle of 2π.

74


Curvilinear polygons

If the simple closed curve γ is only piecewise smooth then a modified version of the Gauss–Bonnet theorem still holds. Most of the working above in the proof of Theorem 3.13remains valid; Green’s theorem still applies, but the final claim, that

∫γθ′ ds = 2π, must

be modified.

Definition 3.14. The simple closed curve c : R → R2 with period a is a curvilinear

polygon with vertices 0 = t1 < · · · < tn < a if

(i) for i = 1, . . . , n, the edge c|(ti,ti+1) is a regular smooth curve , where tn+1 := a, andthe limit tangent vectors

c′(ti+) := limt→ti+

c′(t) and c′(ti+1−) := limt→ti+1−

c′(t)

at the start and the end of the edge are well defined

and

(ii) the vectors c′(ti+) and c′(ti−) are linearly independent for i = 1, . . . , n, wherec′(0−) := c′(a−).

Since a curvilinear polygon is a simple closed curve, it has an interior. A curvilinearpolygon c has unit speed if each of its edges do, and is positively oriented if the unitnormal to each edge, reparameterised by arc length if necessary, points to the interiorof c.

Let r : U → R3 be a regular surface patch. If c : R → R2 is a curvilinear polygon suchthat U contains c and its interior then γ := r ◦ c is a curvilinear polygon in r with thesame orientation, the same period a and the same vertices 0 = t1 < t2 < · · · < tn as c,and with edges γ|(ti,ti+1) for i = 1, . . . , n.

75


Consider∫γθ′ ds, where γ is a unit-speed curvilinear polygon in r. At each vertex ti

the value of θ jumps by the exterior angle δi, where cos δi = γ′(ti−) · γ ′(ti+).

Hence the integral

∫

γ

θ′ ds = 2π −n∑

i=1

δi = 2π −n∑

i=1

(π − αi),

where αi := π − δi is the interior angle at the vertex ti. (If γ is a polygon in the planethen θ′ ≡ 0 on each side and we have the familiar result that the sum of the interiorangles

∑ni=1 αi equals (n−2)π.) This extension of the previous result gives the following

variant of the Gauss–Bonnet theorem.

Theorem 3.15. Let γ be a positively oriented unit-speed curvilinear polygon in theregular surface patch r. If γ has interior angles α1, . . . , αn at its n vertices then

∫ ∫

intγ

K dAr −∫

γ

κg ds = (2− n)π +n∑

i=1

αi.

Example 3.16. A curvilinear triangle is a curvilinear polygon with three vertices; ageodesic triangle is a curvilinear triangle with edges which are geodesics. For a unit-speed geodesic triangle γ with interior angles α1, α2 and α3, the Gauss–Bonnet theoremgives that ∫ ∫

intγ

K dAr = −π + (α1 + α2 + α3).

For the plane, where K ≡ 0, the edges of a geodesic triangle are straight lines and wehave the familiar result that

α1 + α2 + α3 = π;

the sum of the interior angles of a plane triangle is π.

For the unit sphere, where K ≡ 1, the edges of a geodesic triangle are great circles andthe Gauss–Bonnet theorem gives that

A(intγ) = α1 + α2 + α3 − π.

In particular, α1 + α2 + α3 > π: the sum of the interior angles of a spherical triangleexceed π.

76

3.4. Global surfaces

3.4 Global surfacesCharts and atlases

The reader may wonder why the regular surface patches we have considered often havepieces missing; the sphere of Example 2.5 is lacking a line of longitude, for example.This is not due to carelessness or lack of effort; topological methods may be used toshow that there is no continuous bijection r : U → S2, where U is an open set in R2 and

S2 := {(x, y, z) ∈ R3 : x2 + y2 + z2 = 1}

is the unit sphere.

If we want to deal with surfaces such as S2 as a whole, it is necessary to build on thefoundations from Chapter 2; the idea is to cover the surface with patches, which introducesystems of local coordinates. The systems, or charts, are required to be compatible wherethey overlap, so that the change of coordinates is a smooth function.

Formally, we proceed as follows. Let S be a subset of R3, and suppose there exists afamily of regular surface patches {ri : Ui → S | i ∈ I} which covers S, i.e., the union∪i∈iSi = S, where Si := ri(Ui).

Definition 3.17. Since a surface patch ri : Ui → Si := ri(Ui) ⊆ S is injective, thecoordinate chart

φi : Si → Ui; p 7→ r−1i (p) = (φ1

i (p), φ2i (p)

)

is well defined, and the real numbers φ1i (p) and φ

2i (p) are the coordinates of the point p

with respect to this chart.

It is required that these charts be compatible: if Si ∩Sj 6= ∅ then the transition function

φj ◦ φ−1i : φi(Si ∩ Sj) → φj(Si ∩ Sj)

must be smooth, i.e., have partial derivatives of all orders, for all i, j ∈ I. For this tomake sense, the domain of the transition function must be an open subset of R2.

77


Definition 3.18. A set U ⊆ R3 is open if every point in U is contained in some openball lying within U : for all u ∈ U there exists r > 0 such that the ball

B(u; r) := {v ∈ R3 : |u− v| < r} ⊆ U.

[This is the same as the definition for R2, but with “disc” replaced by “ball”.]

Let S be a subset of R3. A set V ⊆ S is said to be relatively open in S if there exists anopen set W ⊆ R3 such that W ∩ S = V .

[If you know about such things, you should verify that the open sets in R3 are thosewhich come from the standard Euclidean metric, and that S has the relative topologyobtained by restricting this metric.]

Exercise 3.6. [For those who know some topology.] Prove that if V is relatively openin S then r−1

i (V ) is open in R2. Deduce that if Si is relatively open in S then φi(Si∩Sj)is open in R2 for all j ∈ I.

This leads us to the following definition; in practice, it turns out to be slightly moreconvenient to work with the coordinate charts than their inverses, the surface patches.

Definition 3.19. A surface (S,A) is a set S ⊆ R3 with an atlas

A := {φi : Si → Ui | i ∈ I}of coordinate charts; for all i ∈ I, the map φi is a continuous bijection from the set Si,which is relatively open in S, to the set Ui, which is open in R

2, and its inverse

ri := φ−1i : Ui → Si ⊆ R

3

is a regular surface patch. Moreover, these charts cover S, so that ∪i∈ISi = S.

The following theorem, which states that transition functions are automatically smooth,is an application of the inverse-function theorem, Theorem A.7. [The proof may safelybe skipped by those without the necessary knowledge of topology.]

Theorem 3.20. Let φ1 : S1 → U1 and φ2 : S2 → U2 be coordinate charts. Then thetransition function

φ1 ◦ φ−12 : φ2(S1 ∩ S2) → φ1(S1 ∩ S2)

is smooth.

Proof. It suffices to show that the transition function is infinitely differentiable at anarbitrary point (x0, y0) ∈ φ2(S1 ∩ S2). Fix this point and note that

(z0, w0) := r−11

(r2(x0, y0)

)∈ φ1(S1 ∩ S2) ⊆ U1

and, by regularity, the 2× 3 matrix[∂1r1(z0, w0)

∂2r1(z0, w0)

]

has linearly independent rows.

78


As row rank and column rank are equal, this matrix must have (at least) two linearlyindependent columns; let these be the ith and jth columns, where 1 6 i < j 6 3. If

π : R3 → R2; (u1, u2, u3) 7→ (ui, uj)

then the 2× 2 matrix

A =

[π(∂1r1(z0, w0)

)

π(∂2r1(z0, w0)

)]

is invertible and the mapΦ := π ◦ r1 : U1 → R

2

has Jacobian JΦ(z0, w0) = A. By the inverse-function theorem, Theorem A.7, there

exist open sets V ⊆ U1 and V ⊆ R2 such that (z0, w0) ∈ V and Φ∣∣V

: V → V is asmooth bijection with smooth inverse.

Now note thatr−11

∣∣r1(V )

= Φ∣∣−1

V◦ π

∣∣r1(V )

,

since if v ∈ V and p = r1(v) then Φ(v) ∈ V and

r−11 (p) = v = Φ

∣∣−1

V

(Φ(v)

)= Φ

∣∣−1

V

(π(r1(v)

))= Φ

∣∣−1

V

(π(p)

).

Letting W := r−12

(r1(V )

), it follows that W is open (since V is open and both r−1

1 = φ1

and r2 are continuous) and

φ1 ◦ φ−12

∣∣W

= r−11 ◦ r2

∣∣W

= r1∣∣−1

r1(V )◦ r2

∣∣W

= Φ∣∣−1

V◦ π

∣∣r1(V )

◦ r2∣∣W;

the right-hand side is the composition of three smooth maps, so is smooth. Finally,

(x0, y0) = r−12

(r1(z0, w0)

)∈ r−1

2

(r1(V )

)= W

so the result follows.

Exercise 3.7. Explain why, by increasing the number of charts in a given atlas ifnecessary, one may assume that each codomain Ui is an open disc in R2. Explain furtherwhy, with a little more work, this may be taken to be the open unit disc

D := {(u, v) ∈ R2 : u2 + v2 < 1}.

We will only consider embedded surfaces in these notes: those which exist as subsetsof R3. There is a more general theory, where a surface is an abstract topological space,and which considers curves, surfaces and higher-dimensional spaces on the same footing:this is the theory of differentiable manifolds .

79


Example 3.21. The sphere S2 has an atlas consisting of six coordinate charts,

φz+ : {(x, y, z) ∈ S2 : z > 0} → D; (x, y, z) 7→ (x, y),

φz− : {(x, y, z) ∈ S2 : z < 0} → D; (x, y, z) 7→ (x, y),

φy+ : {(x, y, z) ∈ S2 : y > 0} → D; (x, y, z) 7→ (x, z),

φy− : {(x, y, z) ∈ S2 : y < 0} → D; (x, y, z) 7→ (x, z),

φx+ : {(x, y, z) ∈ S2 : x > 0} → D; (x, y, z) 7→ (y, z)

and φx− : {(x, y, z) ∈ S2 : x < 0} → D; (x, y, z) 7→ (y, z),

where the common codomain

D := D(0, 0; 1) = {(u, v) ∈ R2 : u2 + v2 < 1}

is the open unit disc; the domain of each chart is a hemisphere, and is relatively openin S2 because the sets

{(x, y, z) ∈ R3 : x > 0}, {(x, y, z) ∈ R

3 : x < 0},{(x, y, z) ∈ R

3 : y > 0}, {(x, y, z) ∈ R3 : y < 0},

{(x, y, z) ∈ R3 : z > 0} and {(x, y, z) ∈ R

3 : z < 0}

are all open in R3.

To compute the transition functions, consider φy− ◦ φ−1z+; the other cases are similar.

Letting Sy− denote the domain of φy−, and so on, it follows that

Sy− ∩ Sz+ = {(x, y, z) ∈ S2 : y < 0 and z > 0}

and the regular surface patch inverse to φz+ is

rz+ := φ−1z+ : D → Sz+; (u, v) 7→

(u, v,

√1− u2 − v2

).

Hence the transition function

φy− ◦ φ−1z+ : {(u, v) ∈ D : v < 0} → {(u, v) ∈ D : v > 0}; (u, v) 7→

(u,

√1− u2 − v2

),

which is smooth.

80


Exercise 3.8. Use stereographic projection to obtain an atlas for the sphere S2 whichconsists of only two charts.

Exercise 3.9. From its realisation as a surface of revolution, it may seem that thecircular cylinder

S1 × R := {(x, y, z) ∈ R3 : x2 + y2 = 1, z ∈ R}

requires an atlas with at least two coordinate charts. Find a single chart which suffices.

[Hint: S1 × R can be seen as C \ {0}; how?]

A surface S is connected if, for every pair of points x, y ∈ S there exists a path from xto y, i.e., a continuous function f : [0, 1] → S such that f(0) = x and f(1) = y.

A surface S is closed if it is connected and compact : if S is a family of relatively opensets in S which covers S then there exist S1, . . . , Sn ∈ S such that S = S1 ∪ · · · ∪Sn. Inparticular, any atlas for S may be reduced to an atlas containing only a finite number ofcharts. [In R

n, compact sets are those which are bounded and contain their limit points,as mentioned earlier.]

Exercise 3.10. Show that the plane

{(x, y, 0) ∈ R3 : x, y ∈ R}

is not compact. [Hint: consider the atlas consisting of the charts

φn : {(x, y, 0) ∈ R3 : x2 + y2 < n} → {(x, y) : x2 + y2 < n}; (x, y, 0) 7→ (x, y),

for all n > 1.]

A surface is orientable if there is a compatible atlas such that the transition maps betweencharts are always orientation preserving; it is non-orientable otherwise. The sphere andtorus are orientable; the Mobius strip is not. In fact, every closed surface is orientable.

Surfaces S1 and S2 are homeomorphic if there exists a continuous bijection f : S1 → S2

with continuous inverse f−1 : S2 → S1.

Theorem 3.22. Every closed surface is homoeomorphic to a torus with g holes, forsome g > 0. The number g is the genus of the surface.

Proof. This is another big theorem, with a complicated proof which we shall omit.

81


There is a similar classification theorem for non-orientable surfaces; see [11, Section 5].Non-orientable closed surfaces do exist, but not as subsets of R3. A Klein bottle is theclosed surface obtained by identifying pairs of opposite points on the edges of a Mobiusstrip; if realised in three dimensions, it must intersect itself. The abstract theory ofmanifolds avoids issues like this.

Subdivisions and the Euler characteristic

Definition 3.23. Let S be a closed surface. A subdivision of S is a finite collection ofcurvilinear polygons such that

(i) each face, the image of a curvilinear polygon and its interior, lies in a regularsurface patch with inverse contained in the atlas for S,

(ii) every point in S belongs to a face,

(iii) faces are disjoint, or meet at a common vertex or a common edge

and

(iv) each edge belongs to exactly two faces.

Given such a subdivision, the Euler characteristic

χ := V −E + F,

where V is the number of distinct vertices, E is the number of distinct edges and F isthe number of faces in the subdivision.

82


Theorem 3.24. Any subdivision of a closed surface S has Euler characteristic χ = 2−2g,where g is the genus of S.

Proof. See [11, Theorem 5.1].

Hence the Euler characteristic depends only on the surface, and not a particular choiceof subdivision; hence we speak of the Euler characteristic of a surface. Furthermore, onemay determine the genus of a surface intrinsically, by constructing a subdivision andcomputing the Euler characteristic.

Theorem 3.25. Every closed surface possesses a triangulation: a subdivision whereevery polygon has three vertices.

Proof. See [11, Section 4].

It is even possible to show that a geodesic triangulation exists for any closed surface, i.e.,a subdivision where every edge is a geodesic and every face is a geodesic triangle.

The global Gauss–Bonnet theorem

Theorem 3.26. If S is a closed surface then∫ ∫

S

K dA = 2πχ,

where K is the Gaussian curvature and χ is the Euler characteristic of S.

This is quite an astonishing result; the quantity on the right has an analytical nature,depending on area and curvature, whereas the left is a topological quantity, dependingonly on the number of holes in the surface.

83


Proof of Theorem 3.26. Suppose S has a geodesic triangulation, and let γ be a geodesictriangle bounding a face in this triangulation. By the Gauss–Bonnet theorem for curvi-linear polygons, ∫ ∫

intγ

K dAr = α1 + α2 + α3 − π,

where r is a regular surface patch containing γ and its interior; it follows from earlierresults that this quantity is independent of the parameterisation of intγ. Summing overall faces gives ∫ ∫

S

K dA = 2πV − πF,

where V is the number of vertices and F the number of faces in the subdivision, sincethe sum of interior angles at a vertex is 2π. Furthermore, each edge in the subdivisionis the boundary of two faces, so

2E = 3F

and therefore∫ ∫

S

K dA = 2π(V − 12F ) = 2π(V − 3

2F + F ) = 2π(V − E + F ) = 2πχ,

as required.

84

A Auxiliary results

A.1 Analysis

Notation A.1. (Landau) Let f and g be functions defined on an open interval whichcontains 0. We write f(h) = o

(g(h)

)if limh→0 f(h)/g(h) = 0. For example, if the

function f is differentiable at x, with f ′(x) = c, then

f(x+ h)− f(x)

h− c→ 0 as h→ 0,

so f(x+h)− f(x)−hc = o(h). We can re-arrange this to give the following proposition,a simple re-writing of the definition of differentiability: a function f is differentiable at xwith derivative c if and only if f(x+ h) = f(x) + hc + o(h).

Theorem A.2. (Intermediate-value theorem) If f : [a, b] → R is continuous and ylies between f(a) and f(b), so either f(a) 6 y 6 f(b) or f(b) 6 y 6 f(a), then thereexists x ∈ [a, b] such that f(x) = y.

Proof. MATH210.

Theorem A.3. If f : I → J is a differentiable bijection between open subintervals of Rwith f ′(x) > 0 for all x ∈ I then f−1 : J → I is also differentiable, with

(f−1)′(y) = 1/f ′(f−1(y)

)for all y ∈ J.

If f ′ is continuous then so is (f−1)′.

Proof. MATH210. For the final comment, note that if f ′ is continuous then so is f ′◦f−1,and f ′(x) > 0 for all x ∈ I.

Theorem A.4. (Taylor’s theorem) Let f : (a − h, a + h) → R be n + 1-timesdifferentiable, where h > 0. If x ∈ (a − h, a + h) then there exists c between x and asuch that

f(x) = f(a)+ f ′(a)(x− a)+f ′′(a)

2(x− a)2+ · · ·+ f (n)(a)

n!(x− a)n+

f (n+1)(c)

(n + 1)!(x− a)n+1.

85

A. Auxiliary results

Definition A.5. (The Jacobian) Suppose Φ = (φ, ψ) : U → R2 has continuous partialderivatives, where U ⊆ R2 is open. The Jacobian of Φ is the matrix-valued function JΦdefined on U by setting

JΦ(u, v) =

[∂1φ(u, v) ∂2φ(u, v)

∂1ψ(u, v) ∂2ψ(u, v)

]for all (u, v) ∈ U,

where ∂1φ(u, v) :=∂φ

∂u(u, v) and so on. If u ∈ U then it may be shown that

Φ(u+ h) = Φ(u) + JΦ(u)h+ o(|h|),with the magnitude of vectors in R

2 defined in the usual way, in the sense that

limh→0

|Φ(u+ h)− Φ(u)− JΦ(u)h||h| = 0.

(Note that this makes sense: JΦ(u)h is the product of the matrix JΦ(u) with thecolumn vector h.) Hence the Jacobian is the appropriate generalisation, for functions oftwo variables, of the derivative considered as the best linear approximation.

If Ψ : V → U and Φ : U → R2 have continuous partial derivatives, where U and V are

open subsets of R2, then it may be shown that

J(Ψ ◦ Φ)(v) = JΨ(Φ(v)

)JΦ(v) for all v ∈ V.

Compare this with the one-dimensional version: if f : I → R and g : J → I aredifferentiable on the open intervals I and J , respectively, then

(f ◦ g)′(x) = f ′(g(x)

)g′(x) for all x ∈ J.

Hence if Ψ : V → U is invertible and Φ = Ψ−1 then[1 0

0 1

]= JΨ

(Ψ−1(v)

)JΨ−1(v),

so JΨ(Ψ−1(v)

)is invertible and

JΨ−1(v) =(JΨ

(Ψ−1(v)

))−1

for all v ∈ V.

(This is an exact analogue of the formula

(f−1)′(y) = 1/f ′(f−1(y)

),

valid for any invertible function f which has non-zero derivative at f−1(y).)

Theorem A.6. If V ⊆ R2 is open and Φ : V → R

2 has continuous partial derivativesand Jacobian JΦ such that det JΦ(v) 6= 0 for all v ∈ V then

∫ ∫

Φ(V )

f(x, y) dx dy =

∫ ∫

V

f(Φ(u, v)

)| det JΦ(u, v)| du dv

for any sufficiently nice function f : Φ(V ) → R.

Proof. MATH113. “Sufficiently nice” includes all the functions in these notes.

86

A.2. Algebra

Theorem A.7. (Inverse-function theorem) Let Φ : U → R2 be smooth, where theset U ⊆ R2 is open, and suppose v ∈ U is such that JΦ(v) is invertible. There exist

open sets V and V such that v ∈ V ⊆ U , Φ(v) ∈ V and Φ|V : V → V is invertible withsmooth inverse.

A.2 Algebra

Theorem A.8. (Principal-axis theorem) Let C be a real 2 × 2 matrix. If C issymmetric (that is, the transpose Ct = C) then there exists a real 2× 2 matrix D whichis orthogonal (i.e., DtD = DDt = I) and such that

DtCD =

[λ1 0

0 λ2

],

where λ1 6 λ2.

87

B Solutions to selected exercises

Solution to the first part of Exercise 1.9

Let u = (u1, u2, u3) et cetera. Then u× (v ×w) equals

u×

∣∣∣∣∣∣

i j kv1 v2 v3w1 w2 w3

∣∣∣∣∣∣=

∣∣∣∣∣∣

i j ku1 u2 u3

v2w3 − v3w2 v3w1 − v1w3 v1w2 − v2w1

∣∣∣∣∣∣

= (u2v1w2 − u2v2w1 − u3v3w1 + u3v1w3,

u1v2w1 − u1v1w2 − u3v3w2 + u3v2w3,

u1v3w1 − u1v1w3 − u2v2w3 + u2v3w2)

=((u1w1 + u2w2 + u3w3)v1 − (u1v1 + u2v2 + u3v3)w1,

(u1w1 + u2w2 + u3w3)v2 − (u1v1 + u2v2 + u3v3)w2,

(u1w1 + u2w2 + u3w3)v3 − (u1v1 + u2v2 + u3v3)w3))

= (u ·w)v − (u · v)w,

as claimed.

Solution to the first part of Exercise 1.15

Since |γ(d)− γ(c)| is the straight-line distance from γ(c) to γ(d), whereas

∫ d

c

|γ ′(t)| dtis the distance measured along the curve γ, the inequality

∫ d

c

|γ ′(t)| dt > |γ(d)− γ(c)|

holds because a straight line gives the shortest distance in R3 between two points. Fora rigorous proof of this fact, we proceed as follows.

Let u ∈ R3 be a unit vector such that

(γ(d)− γ(c)

)· u = |γ(d)− γ(c)|

and note thatd

dt

(γ(t) · u

)= γ

′(t) · u+ γ(t) · u′ = γ′(t) · u.

89

B. Solutions to selected exercises

By the fundamental theorem of calculus,

|γ(d)− γ(c)| = γ(d) · u− γ(c) · u =

∫ d

c

d

dt

(γ(t) · u

)dt =

∫ d

c

γ′(t) · u dt

and the Cauchy–Schwarz inequality implies that

γ′(t) · u 6 |γ ′(t) · u| 6 |γ ′(t)| |u| = |γ′(t)|.

Thus

|γ(d)− γ(c)| =∫ d

c

γ′(t) · u dt 6

∫ d

c

|γ′(t)| dt.

Solution to Exercise 1.18

If γ : (a, b) → R3 is parameterised by arc length then there exists m ∈ (a, b) such that

s(t) :=

∫ t

m

|γ′(r)| dr = t

for all t ∈ (a, b). Hence |γ ′(t)| = s′(t) = 1 for all t ∈ (a, b) and γ has unit speed.


If κ ≡ 0 then t′ ≡ 0 and t ≡ c for some constant vector c. It follows that γ ′ = t ≡ cand γ(s) = s c+ d for all s, where d is a constant vector: γ lies on a straight line.


Since t and n are orthogonal, it suffices to prove that each of these vector fields isorthogonal to u. Differentiating,

d ≡ γ · u =⇒ 0 ≡ γ′ · u+ γ · u′ =⇒ 0 ≡ t · u

and t is orthogonal to u. Differentiating once more,

0 ≡ t · u =⇒ 0 ≡ t′ · u+ t · u′ =⇒ 0 ≡ κn · u;as κ(s) 6= 0 for all s, it follows that n and u are orthogonal, as required.

For the final claim, note that

n · n ≡ 1 =⇒ n′ · n+ n · n′ ≡ 0 =⇒ n′ · n ≡ 0

andn · u ≡ 0 =⇒ n′ · u+ n · u′ ≡ 0 =⇒ n′ · u ≡ 0,

son′ = (t · n′)t+ (n · n′)n+ (u · n′)u = (t · n′)t.

Furthermore, t · n ≡ 0 and therefore

0 ≡ t′ · n+ t · n′ = κn · n+ t · n′ =⇒ −κ = t · n′,

from which it follows that n′ = −κt.

90


Note that b is constant, by the third Serret–Frenet equation, and

(γ · b)′ = γ′ · b+ γ · b′ = t · b+ γ · 0 ≡ 0.

Hence γ · b ≡ c for some constant c and the result follows.


The chain rule gives that

(v ◦ s)′ = (v′ ◦ s)s′ = |γ ′|(v′ ◦ s)

for any vector field v : (a, b) → R3. Taking v = γ = γ ◦ s−1, we have

γ′ = (γ ◦ s)′ = |γ ′|(γ ′ ◦ s) = |γ ′|(t ◦ s); (B.1)

taking v = t shows that

(t ◦ s)′ = |γ ′|(t′ ◦ s) = |γ ′|((κn) ◦ s

), (B.2)

by the definition of n. It follows from differentiating (B.1) that

γ′′ = |γ ′|′(t ◦ s) + |γ ′|(t ◦ s)′

= |γ ′|′(t ◦ s) + |γ ′|2((κn) ◦ s

), (B.3)

where the last equality holds by (B.2). Hence

γ′ × γ

′′ = |γ ′|(t ◦ s)× γ′′ = |γ′|3

((κb) ◦ s

), (B.4)

as required. The deduction is immediate, since b has unit magnitude.

For the next part, it follows by differentiating (B.4) that

γ′ × γ

′′′ = γ′′ × γ

′′ + γ′ × γ

′′′

= (γ ′ × γ′′)′

=(|γ ′|3(κ ◦ s

))′(b ◦ s) + |γ ′|3(κ ◦ s)(b ◦ s)′

=(|γ ′|3(κ ◦ s)

)′

(b ◦ s)− |γ′|4((κτn) ◦ s

),

by the initial observation and the third Serret–Frenet equation. Thus

[γ ′,γ ′′,γ ′′′] = −γ′′ · (γ ′ × γ

′′′) = |γ ′|6((κ2τ) ◦ s

),

by (B.3), as claimed. The final result follows.

91



For injectivity, note that

v γ(u) = xγ(y) =⇒ vd = xd =⇒ v = x,

where the first implication follows from taking the scalar product of each side with c,and then γ(u) = γ(y), whence u = y.

For regularity, note that

∂1r(u, v) = v γ ′(u) and ∂2r(u, v) = γ(u),

so (∂1r×∂2r)(u, v) = v γ ′(u)×γ(u), which has positive magnitude unless γ ′(u) and γ(u)are linearly dependent. However, γ(u) · c = d and γ

′(u) · c = 0; if aγ(u) + bγ ′(u) = 0then

0 = 0 · c = aγ(u) · c+ bγ ′(u) · c = ad,

so a = 0, but then |bγ ′(u)| = |0| = 0 and (since γ is regular) b = 0.


As∂1r(z, φ) = (cosφ, sin φ, 1) and ∂2r(z, φ) = (−z sin φ, z cosφ, 0),

it follows that

(∂1r× ∂2r)(z, φ) =

∣∣∣∣∣∣

i j kcosφ sinφ 1

−z sin φ z cosφ 0

∣∣∣∣∣∣= (−z cosφ, −z sinφ, z)

and

|∂1r× ∂2r|(z, φ) =√z2 cos2 φ+ z2 sin2 φ+ z2 =

√2 |z|.

Hence ∂1r(0, φ) and ∂2r(0, φ) are linearly dependent (which may also be seen directly,since ∂2r(0, φ) is the zero vector) and r is not regular.

This surface is the double cone x2 + y2 = z2, minus the half lines {(−z, 0, z) : z 6= 0};it fails to be regular at (0, 0, 0), where the apices of the two cones meet.


The first Serret–Frenet equation and the fact that (t, n, b) is an orthonormal triple ofvector fields imply that

κn := t′ ·N(u, v) = κn · (n cosψ + b sinψ) = κ cosψ.

Furthermore,

B := t×N(u, v) = t× (n cosψ + b sinψ) = b cosψ − n sinψ,

as required; it follows from this that

κg := t′ ·B = κn · (b cosψ − n sinψ) = −κ sinψ.

92

For the last part, note that the first equation holds by definition. For the others, thesecond and third Serret–Frenet equations and the previous working give that

N(u, v)′ = n′ cosψ − ψ′ n sinψ + b′ sinψ + ψ′ b cosψ

= (−κt+ τb) cosψ + ψ′ B− τn sinψ

= −κ cosψ t+ (τ + ψ′)B

= −κnt+ τgB

and

B′ = −n′ sinψ − ψ′ n cosψ + b′ cosψ − ψ′ b sinψ

= (κt− τb) sinψ − ψ′ N(u, v)− τn cosψ

= κ sinψ t− (τ + ψ′)N(u, v)

= −κgt− τgN(u, v).


For the first part, note that

r · r ≡ 1 =⇒ ∂1r · r = ∂2r · r ≡ 0,

so the unit vector field r is orthogonal to the tangent plane everywhere, and thusN = ±r.

The unit-speed curve γ = r ◦ (u, v) is a geodesic if and only if the component of γ ′′ inthe tangent plane is zero:

0 ≡ γ′′ − (N(u, v) · γ ′′)N(u, v) = γ

′′ − (γ · γ ′′)γ,

by the first part. However,

γ · γ ≡ 1 =⇒ γ′ · γ ≡ 0 =⇒ γ

′′ · γ + γ′ · γ ′ ≡ 0

and γ′ · γ ′ ≡ 1, so γ is a geodesic if and only if γ ′′ + γ ≡ 0, as claimed.

Solving for each component separately,

γ(s) =(a1 cos s+ b1 sin s, a2 cos s+ b2 sin s, a3 cos s+ b3 sin s) = a cos s+ b sin s

for all s ∈ I. If 0 ∈ I then

|a| = |γ(0)| = 1, |b| = |γ′(0)| = 1 and a · b = (γ · γ ′)(0) = 0,

so a and b are orthogonal unit vectors. For the general case, fix s0 ∈ I and note that

γ = r ◦ (u, v) : I − s0 → S2; s 7→ γ(s+ s0) = r(u(s+ s0), v(s+ s0)

)

is a geodesic if and only if γ is, because

γ′′(s) = γ

′′(s+ s0) and N(u(s+ s0), v(s+ s0)

)= N

(u(s), v(s)

).

93


Hence, by the previous working, there exist orthogonal unit vectors a and b such that

γ(s) = γ(s− s0) = a cos(s− s0) + b sin(s− s0) = a cos s+ b sin s

for all s ∈ I, where

a := a cos s0 − b sin s0 and b := a sin s0 + b cos s0;

since a and b are orthogonal unit vectors, so are a and b.


If γ = r ◦ (u, v) is a geodesic then κg ≡ 0, therefore

κn = t′ = κnN(u, v) + κgB = κnN(u, v).

Hence n = ±N(u, v) (with the sign fixed everywhere on γ) and

b = t× n = ±t×N(u, v) = ±B

(with the same sign as for n). By the third Serret–Frenet equation and its analogue fromExercise 2.13,

∓τn = ±b′ = B′ = −κgt− τgN(u, v) = ∓τgN(u, v)

and thus τg = τ , as required.


The surface of revolution r given by (2.13) is such that

(∂1r× ∂2r)(s, φ) =

∣∣∣∣∣∣

i j kf ′(s) cosφ f ′(s) sinφ g′(s)−f(s) sinφ f(s) cosφ 0

∣∣∣∣∣∣

=(−f(s)g′(s) cosφ, −f(s)g′(s) sinφ, f(s)f ′(s)

).

Hence the squared magnitude

|∂1r× ∂2r|2(s, φ) = f(s)2(g′(s)2 + f ′(s)2) = f(s)2 > 0

for all (s, φ) ∈ I × (−π, π) and therefore r is regular.


For a meridian, u′ ≡ 1 and v′ ≡ 0, so both of the geodesic equations (2.14) are satisfied.

For a parallel, u ≡ s0 (so u′ ≡ 0) and v′ ≡ 1; the equations become

0 = f(s0)f′(s0) and 2f(s0)f

′(s0)u′ = 0.

The latter is always satisfied and, because f(s0) > 0, the former is satisfied if and onlyif f ′(s0) = 0.

94


Let γ : I → P∗ be a regular curve such that

γ(a) = a = (−1, 0, 0) and γ(b) = b = (1, 0, 0)

for some a, b ∈ I with a < b. It follows from (1.6) that

ℓ(γ|(a,b)) =∫ b

a

|γ′(t)| dt > |γ(b)− γ(a)| = 2;

suppose for contradiction that we have equality.

Letting γ(t) =(x(t), y(t), z(t)

), we see that

2 = x(b)− x(a) =

∫ b

a

x′(t) dt 6

∫ b

a

√x′(t)2 + y′(t)2 + z′(t)2 dt =

∫ b

a

|γ′(t)| dt = 2,

so equality holds throughout. Hence

∫ b

a

√x′(t)2 + y′(t)2 + z′(t)2 − x′(t) dt = 0,

and as the integrand is non-negative, it must be zero everywhere on [a, b]. In particular,y′(t) = z′(t) = 0 for all t ∈ (a, b), so y(t) = y(a) = 0 and z(t) = z(a) = 0 for allt ∈ [a, b]. Since x(a) < 0 and x(b) > 0, continuity and the intermediate-value theorem(Theorem A.2) give the existence of c ∈ (a, b) such that x(c) = 0, but then it followsthat γ(c) =

(x(c), y(c), z(c)

)= 0 6∈ P∗. This is the desired contradiction.

For the second part, let δ > 0, define the smooth curve

γ : R → P∗; t 7→(t, 1

2δ(1− t2), 0

)

and note thatγ′(t) = (1, −δt, 0) for all t ∈ R.

Then γ(−1) = a = (−1, 0, 0), γ(1) = b = (1, 0, 0) and

ℓ := ℓ(γ|(−1,1)) =

∫ 1

−1

√1 + δ2t2 dt = 2

∫ 1

0

√1 + δ2t2 dt. (B.5)

If α := sinh−1 δ > 0 and x := sinh−1(δt) then

ℓ =2

δ

∫ α

0

cosh2 x dx =1

δ

∫ α

0

1 + cosh 2x dx =1

δ

[x+ 1

2sinh 2x

]α0=

α

sinhα+ coshα,

since 2 cosh2 x = 1 + cosh 2x and sinh 2x = 2 sinh x cosh x for all x ∈ R. It is clear from(B.5) that ℓ is an increasing function of δ, and as δ → 0+, so α→ 0+ and ℓ→ 2. Hencefor all ε > 0 there exists δ > 0 such that ℓ < 2 + ε, as required.

95



As the principal directions t1 = ±TC∂1r and t2 = ±TC∂2r, it follow that

t1 · t2 =[±1 0

]Ct

[E F

F G

]C

[0±1

]=

[±1 0

][1 0

0 1

] [0±1

]= 0,

as claimed.


If F is stationary at (x, y) then

0 =∂F

∂x= 2Lx+ 2My − λ(2Ex+ 2Fy) (B.6)

and 0 =∂F

∂y= 2Mx+ 2Ny − λ(2Fx+ 2Gy). (B.7)

Adding x times the first to y times the second gives that

0 = 2(Lx2 + 2Mxy +Ny2)− 2λ(Ex2 + 2Fxy +Gy2) ⇐⇒ λ = κ(x, y)

if f(x, y) = 1. Hence the equations (B.6) and (B.7) may be written as the matrixequation

2

[L− κE M − κF

M − κF N − κG

][xy

]=

[00

].

Note that x and y cannot both be zero, for then f(x, y) = 0. Hence the matrix on theleft-hand side above must be singular and so

det

[L− κE M − κF

M − κF N − κG

]= 0,

as required.


Note that

K =LN − M2

EG− F 2=

(det JΦ)2((LN −M2) ◦ Φ

)

(det JΦ)2((EG− F 2) ◦ Φ

) = K ◦ Φ,

by Theorem 2.26, Exercise 2.23 and (2.8).


Since ∂1N and ∂2N lie in the tangent plane, their cross product is orthogonal to thetangent plane, so is a multiple of the unit normal N. Hence

∂1N× ∂2N = ±|∂1N× ∂2N|N.

The claim follows.

96


Note that

sAN(Dε) =

∫ ∫

Dε

[N, ∂1N, ∂2N](x, y) dx dy

and

Ar(Dε) =

∫ ∫

Dε

|∂1r× ∂2r|(x, y) dx dy.

Hence, as ε→ 0,

sAN(Dε)

Ar(Dε)=

1

πε2

∫ ∫

Dε

[N, ∂1N, ∂2N](x, y) dx dy

1

πε2

∫ ∫

Dε

|∂1r× ∂2r|(x, y) dx dy

→ [N, ∂1N, ∂2N]

|∂1r× ∂2r|(z, w)

=N · (K ∂1r× ∂2r)

|∂1r× ∂2r|(z, w) (by Exercise 2.30)

= (KN ·N)(z, w)

= K(z, w).


If V = W ∩ S, where W ⊆ R3 is open, then r−1

i (V ) = r−1i (W ∩ S) = r−1

i (W ), which isopen because ri is continuous.

Hence φi(Si∩Sj) = r−1i (Si∩Sj) = r−1

i (Sj) is open in R2 if Sj is relatively open in S.

97

Bibliography

[1] F. Frenet, Sur les courbes a double courbure, J. de Math. Pures et Appliquees 17(1852), 437–447. [Cited on page 19.]

[2] T. W. Gamelin, Complex analysis, Springer, New York, 2001. [Cited on page 70.]

[3] P. C. Matthews, Vector calculus, Springer, London, 1998. [Cited on page 71.]

[4] J. R. Munkres, Topology, second edition, Prentice-Hall, New Jersey, 2000. [Citedon pages 12 and 70.]

[5] G. Peano, Sur une courbe, qui remplit toute une aire plane, Math. Ann. 36 (1890),157–160. [Cited on page 12.]

[6] A. Pressley, Elementary differential geometry, Springer, London, 2001. [Cited onpages iii, 38, 48 and 73.]

[7] J. Roe, Elementary geometry, Oxford University Press, Oxford, 1993. [Cited onpages iii and 35.]

[8] G. B. Segal, Geometry of surfaces, Mathematical Institute lecture notes, Univer-sity of Oxford, 1986. [Cited on page iii.]

[9] J.-A. Serret, Sur quelques formules relatives a la theorie des courbes a double

courbure, J. de Math. Pures et Appliquees 16 (1851), 193–207. [Cited on page 19.]

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99

Index

Numbers set in italic type refer to the page where the term is first defined.

A

anti-commutative . . . . . . . . . . . . . . . . . . 3

antipodal . . . . . . . . . . . . . . . . . . . . . . . 34

antipodes . . . . . . . . . . . . . . . . . . . . . . . 34

approximation

best circular . . . . . . . . . . . . . . . . . . . 19

best linear . . . . . . . . . . . . . . . . . 11, 23

polygonal . . . . . . . . . . . . . . . . . . . . . . 8

arc length . . . . . . . . . . . . . . . . . . . . 13, 16

arc-length function . . . . . . . . . . . . . 13, 16

Archimedes . . . . . . . . . . . . . . . . . . . . . . 33

area . . . . . . . . . . . . . . . . . . . . . . . . 31, 33

signed . . . . . . . . . . . . . . . . . . . . . . . 59

atlas . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

azimuth . . . . . . . . . . . . . . . . . . . . . . . . 26

B

basis

orthonormal . . . . . . . . . . . . . . . . . . . . 3

standard . . . . . . . . . . . . . . . . . . . . . . 1

best circular approximation . . . . . . . . . . 19

bilinear . . . . . . . . . . . . . . . . . . . . . . . . . 2

binormal vector field . . . . . . . . . . . . . . . 19

bottle

Klein . . . . . . . . . . . . . . . . . . . . . . . . 82

boundary . . . . . . . . . . . . . . . . . . . . . . . 70

bounded . . . . . . . . . . . . . . . . . . . . . . . . 65

C

Cauchy–Schwarz inequality . . . . . . . . . . . 2

chain rule

vector-valued . . . . . . . . . . . . . . . . . . 11

characteristic

Euler . . . . . . . . . . . . . . . . . . . . . 65, 82

chartcoordinate . . . . . . . . . . . . . . . . . . . . 77

circle . . . . . . . . . . . . . . . . . . . . . 7, 10, 16great . . . . . . . . . . . . . . . . . . . . . 34, 40osculating . . . . . . . . . . . . . . . . . . . . 19

closed surface . . . . . . . . . . . . . . . . . 64, 81co-linear . . . . . . . . . . . . . . . . . . . . . . . . . 4

compact . . . . . . . . . . . . . . . . . . . . . 64, 81cone

generalised . . . . . . . . . . . . . . . . . . . . 25

connected . . . . . . . . . . . . . . . . . . . . . . . 22

connected surface . . . . . . . . . . . . . . . . . 81

continuously differentiable . . . . . . . . . 9, 10coordinate chart . . . . . . . . . . . . . . . . . . 77

cross product . . . . . . . . . . . . . . . . . . . . . 3

curvature . . . . . . . . . . . . . . . . . . . . . . . 16

Gaussian . . . . . . . . . . . . . . . . . . . . . 57

geodesic . . . . . . . . . . . . . . . . 38, 41, 46

normal . . . . . . . . . . . . . . . . . . . . 38, 51principal . . . . . . . . . . . . . . . . . . . . . 52

signed . . . . . . . . . . . . . . . . . . . . . . . 68

zero . . . . . . . . . . . . . . . . . . . . . . . . . 16curve

in a surface . . . . . . . . . . . . . . . . 27, 38parameterised . . . . . . . . . . . . . . . . . . 8

planar . . . . . . . . . . . . . . . . . . 7, 17, 24plane . . . . . . . . . . . . . . . . . . . . . . . . 66

rectifiable . . . . . . . . . . . . . . . . . . . . . . 9

regular . . . . . . . . . . . . . . . . . . . . . . . 13simple . . . . . . . . . . . . . . . . . . . . . . . 24

simple closed . . . . . . . . . . . . . . . . . . 63smooth . . . . . . . . . . . . . . . . . . . . . . 16space-filling . . . . . . . . . . . . . . . . . . . 12

curvilinear polygon . . . . . . . . . . . . . 64, 75

101

Index

curvilinear triangle . . . . . . . . . . . . . . . . 76

cylinder . . . . . . . . . . . . . 31, 33, 36, 43, 56generalised . . . . . . . . . . . . . . . . . . . . 24

cylindrical projection . . . . . . . . . . . . . . 33

D

derivativeof a vector-valued function . . . . . . . . . 9

determinantJacobian . . . . . . . . . . . . . . . . . . . . . 32

differentiablecontinuously . . . . . . . . . . . . . . . . . 9, 10

differentiable manifold . . . . . . . . . . . . . . 79direction

principal . . . . . . . . . . . . . . . . . . . . . 53

discopen . . . . . . . . . . . . . . . . . . . . . . . . 21

dot product . . . . . . . . . . . . . . . . . . . . . . 2

E

elevation . . . . . . . . . . . . . . . . . . . . . . . . 26

elliptic paraboloid . . . . . . . . . . . . . . . . . 29

equationsgeodesic . . . . . . . . . . . . . . . . . . . 42, 45Serret–Frenet . . . . . . . . . . . . . . . . . . 19

Euler characteristic . . . . . . . . . . . . . 65, 82extrinsic geometry . . . . . . . . . . . . . . . . . 49

F

fieldscalar . . . . . . . . . . . . . . . . . . . . . . . 23vector . . . . . . . . . . . . . . . . . . . . . . . 23

first fundamental form . . . . . . . . . . 29, 51form

first fundamental . . . . . . . . . . . . 29, 51quadratic . . . . . . . . . . . . . . . . . . . . . 51second fundamental . . . . . . . . . . . . . 49

functionarc-length . . . . . . . . . . . . . . . . . 13, 16transition . . . . . . . . . . . . . . . . . . 77, 78

G

Gauss map . . . . . . . . . . . . . . . . . . . . . . 58

Gauss–Bonnet theoremfor closed surfaces . . . . . . . . . . . . . . 64for curvilinear polygons . . . . . . . . . . 64for smooth simple closed curves . . . . . 63

Gaussian curvature . . . . . . . . . . . . . . . . 57

generalised cone . . . . . . . . . . . . . . . . . . 25

generalised cylinder . . . . . . . . . . . . . . . . 24

genus . . . . . . . . . . . . . . . . . . . . . . . 65, 81

geodesic . . . . . . . . . . . . . . . . . . . . . 39, 47geodesic curvature . . . . . . . . . . . 38, 41, 46

geodesic equations . . . . . . . . . . . . . . 42, 45geodesic torsion . . . . . . . . . . . . . . . 39, 41geodesic triangle . . . . . . . . . . . . . . . . . . 76

geodesic triangulation . . . . . . . . . . . . . . 83

geometry

extrinsic . . . . . . . . . . . . . . . . . . . . . 49intrinsic . . . . . . . . . . . . . . . . . . . . . . 49

great circle . . . . . . . . . . . . . . . . 34, 40, 48

Green’s theorem . . . . . . . . . . . . . . . . . . 71

H

helix . . . . . . . . . . . . . . . . . . . . . . . . . . 43

homeomorphic . . . . . . . . . . . . . . . . 65, 81hyperbolic paraboloid . . . . . . . . . . . . . . 27

I

inequality

Cauchy–Schwarz . . . . . . . . . . . . . . . . . 2

Intermediate-value theorem . . . . . . . . . . 85

intervalopen . . . . . . . . . . . . . . . . . . . . . . . . . 8

intrinsic geometry . . . . . . . . . . . . . . . . . 49

Inverse-function theorem . . . . . . . . . . . . 87isometric . . . . . . . . . . . . . . . . . 35, 37, 47

J

Jacobian . . . . . . . . . . . . . . . . . . . . . 32, 86Jacobian determinant . . . . . . . . . . . . . . 32

Jordan curve theorem . . . . . . . . . . . . . . 70

K

K (Gaussian curvature) . . . . . . . . . . . . 57

Klein bottle . . . . . . . . . . . . . . . . . . . . . 82

L

Landau notation . . . . . . . . . . . . . . . . . . 85latitude . . . . . . . . . . . . . . . . . . . . . . . . 26

length . . . . . . . . . . . . . . . . . . . . . . 14, 47approximate . . . . . . . . . . . . . . . . . . . . 9

of a rectifiable curve . . . . . . . . . . . . . . 9

of a smooth curve . . . . . . . . . . . . . 9, 28limit

of a vector-valued function . . . . . . . . 10

limit point . . . . . . . . . . . . . . . . . . . . . . 65

line . . . . . . . . . . . . . . . . . . . . . . . . . 6, 39

102

longitude . . . . . . . . . . . . . . . . . . . . . . . 26

lune . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

M

magnitude . . . . . . . . . . . . . . . . . . . . . . . 2

constant . . . . . . . . . . . . . . . . . . . . . 15manifold

differentiable . . . . . . . . . . . . . . . . . . 79map

Gauss . . . . . . . . . . . . . . . . . . . . . . . 58

smooth . . . . . . . . . . . . . . . . . . . . . . 22

matrixWeingarten . . . . . . . . . . . . . . . . . . . 58

meridian . . . . . . . . . . . . . . . . . . . . . 43, 45mutually orthogonal . . . . . . . . . . . . . . . . 3

N

negatively oriented . . . . . . . . . . . . . . . . 70

Newton . . . . . . . . . . . . . . . . . . . . . . . . 39normal curvature . . . . . . . . . . . . . . 38, 51

O

open disc . . . . . . . . . . . . . . . . . . . . . . . 21

open interval . . . . . . . . . . . . . . . . . . . . . 8

open rectangle . . . . . . . . . . . . . . . . . . . 21

open set . . . . . . . . . . . . . . . . . . . . . 21, 78orientable . . . . . . . . . . . . . . . . . . . . . . . 65orientable surface . . . . . . . . . . . . . . . . . 81

orthogonal . . . . . . . . . . . . . . . . . . . . . . . 2

orthogonal parameterisation . . . . . . 30, 35orthonormal basis . . . . . . . . . . . . . . . . . . 3

orthonormal triple . . . . . . . . . . . . . . . 4, 19osculating circle . . . . . . . . . . . . . . . . . . 19

osculating plane . . . . . . . . . . . . . . . . . . 19

P

pairsmooth . . . . . . . . . . . . . . . . . . . . . . 28

paraboloidelliptic . . . . . . . . . . . . . . . . . . . . . . . 29

hyperbolic . . . . . . . . . . . . . . . . . . . . 27

parallel . . . . . . . . . . . . . . . . . . . . . . 43, 45parameterisation

orthogonal . . . . . . . . . . . . . . . . . 30, 35parameterised curve . . . . . . . . . . . . . . . . 8

patchsurface . . . . . . . . . . . . . . . . . . . . . . . 23

period . . . . . . . . . . . . . . . . . . . . . . . . . 69

perpendicular . . . . . . . . . . . . . . . . . . . . . 2

piecewise linear . . . . . . . . . . . . . . . . . . . . 8

planar curve . . . . . . . . . . . . . . . . 7, 17, 24

plane . . . . . . . . . . . . . . . . . . 6, 36, 42, 62

osculating . . . . . . . . . . . . . . . . . . . . 19

tangent . . . . . . . . . . . . . . . . . . . . . . 23

plane curve . . . . . . . . . . . . . . . . . . . . . . 66

simple closed . . . . . . . . . . . . . . . . . . 69

point

limit . . . . . . . . . . . . . . . . . . . . . . . . 65

saddle . . . . . . . . . . . . . . . . . . . . . . . 57

polygon

curvilinear . . . . . . . . . . . . . . . . . 64, 75

polygonal . . . . . . . . . . . . . . . . . . . . . . . . 8

polygonal approximation . . . . . . . . . . . . . 8

positively oriented . . . . . . . . . . . . . . 63, 70

principal curvature . . . . . . . . . . . . . . . . 52

principal direction . . . . . . . . . . . . . . . . . 53

principal-axis theorem . . . . . . . . . . . . . . 52

prism . . . . . . . . . . . . . . . . . . . . . . . . . . 24

product

cross . . . . . . . . . . . . . . . . . . . . . . . . . 3

dot . . . . . . . . . . . . . . . . . . . . . . . . . . 2

scalar . . . . . . . . . . . . . . . . . . . . . . . . 2

scalar triple . . . . . . . . . . . . . . . . . . . . 5

vector . . . . . . . . . . . . . . . . . . . . . . . . 3

product rule . . . . . . . . . . . . . . . . . . . . . 11

projection

cylindrical . . . . . . . . . . . . . . . . . . . . 33

Pythagoras’s theorem . . . . . . . . . . . . . . . 3

Q

quadratic form . . . . . . . . . . . . . . . . . . . 51

R

R3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

rectangle

open . . . . . . . . . . . . . . . . . . . . . . . . 21

rectifiable curve . . . . . . . . . . . . . . . . . . . 9

regular curve . . . . . . . . . . . . . . . . . . . . 13

regular surface patch . . . . . . . . . . . . 23, 33

relatively open set . . . . . . . . . . . . . . . . . 78

reparameterisation . . 14, 32, 33, 37, 40, 50

orientation-preserving . . . . . . . . . . . . 32

orientation-reversing . . . . . . . . . . . . . 32

smooth . . . . . . . . . . . . . . . . . . . . . . 16

right-handed triple . . . . . . . . . . . . . . . . . 4

rule

product . . . . . . . . . . . . . . . . . . . . . . 11

103

Index

S

saddle point . . . . . . . . . . . . . . . . . . . . . 57

scalar field . . . . . . . . . . . . . . . . . . . 15, 23scalar product . . . . . . . . . . . . . . . . . . . . . 2

scalar triple product . . . . . . . . . . . . . . . . 5

second fundamental form . . . . . . . . . . . . 49

Serret–Frenet equations . . . . . . . . . . 19, 39set

open . . . . . . . . . . . . . . . . . . . . . 21, 78relatively open . . . . . . . . . . . . . . . . . 78

signed area . . . . . . . . . . . . . . . . . . . . . . 59

signed curvature . . . . . . . . . . . . . . . . . . 68

simple closed curve . . . . . . . . . . . . . . . . 63simple curve . . . . . . . . . . . . . . . . . . . . . 24

smooth curve . . . . . . . . . . . . . . . . . . . . 16smooth map . . . . . . . . . . . . . . . . . . . . . 22

smooth pair . . . . . . . . . . . . . . . . . . . . . 28

space-filling curve . . . . . . . . . . . . . . . . . 12speed . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

constant . . . . . . . . . . . . . . . . . . . . . 40sphere . . . . . . . . . . . . 7, 31, 40, 55, 62, 80

unit . . . . . . . . . . . . . . . . . . . . . . . . . 26

spherical triangle . . . . . . . . . . . . . . . . . 34

standard basis . . . . . . . . . . . . . . . . . . . . 1

subdivision . . . . . . . . . . . . . . . . . . . 65, 82surface . . . . . . . . . . . . . . . . . . . . . . . . . 78

closed . . . . . . . . . . . . . . . . . . . . 64, 81connected . . . . . . . . . . . . . . . . . . . . 81

of revolution . . . . . . . . . . . . . . . . . . 44orientable . . . . . . . . . . . . . . . . . . . . 81

surface patch . . . . . . . . . . . . . . . . . . . . 23

regular . . . . . . . . . . . . . . . . . . . . 23, 33

T

tangent plane . . . . . . . . . . . . . . . . . . . . 23

tangent vector fields . . . . . . . . . . . . . . . 23

Taylor’s theorem . . . . . . . . . . . . . . . . . . 85theorem

Green’s . . . . . . . . . . . . . . . . . . . . . . 71Jordan curve . . . . . . . . . . . . . . . . . . 70principal-axis . . . . . . . . . . . . . . . . . . 52Pythagoras’s . . . . . . . . . . . . . . . . . . . 3

Theorema Egregium . . . . . . . . . . . . . . . 59

tomb . . . . . . . . . . . . . . . . . . . . . . . . . . 33

torsion . . . . . . . . . . . . . . . . . . . . . . . . . 19

geodesic . . . . . . . . . . . . . . . . . . . 39, 41

zero . . . . . . . . . . . . . . . . . . . . . . . . . 20

torus . . . . . . . . . . . . . . . . . . . . . . . . . . 37

transition function . . . . . . . . . . . . . 77, 78

transpose . . . . . . . . . . . . . . . . . . . . . . . 32

triangle

curvilinear . . . . . . . . . . . . . . . . . . . . 76

geodesic . . . . . . . . . . . . . . . . . . . . . . 76

spherical . . . . . . . . . . . . . . . . . . . . . 34

triangulation . . . . . . . . . . . . . . . . . . . . 83

geodesic . . . . . . . . . . . . . . . . . . . . . . 83

triple

orthonormal . . . . . . . . . . . . . . . . . 4, 19

right-handed . . . . . . . . . . . . . . . . . . . 4

U

umbilic . . . . . . . . . . . . . . . . . . . . . . 53, 55

unit normal vector . . . . . . . . . . . . . . . . 17

unit normal vector field . . . . . . . . . . 17, 32

unit sphere . . . . . . . . . . . . . . . . . . . . . . 26

unit tangent vector . . . . . . . . . . . . . . . . 15

unit tangent vector field . . . . . . . . . . . . 15

V

vector

unit normal . . . . . . . . . . . . . . . . . . . 17

unit tangent . . . . . . . . . . . . . . . . . . . 15

zero . . . . . . . . . . . . . . . . . . . . . . . . . . 1

vector field . . . . . . . . . . . . . . . . . . . 15, 23

binormal . . . . . . . . . . . . . . . . . . . . . 19

unit normal . . . . . . . . . . . . . 17, 25, 32

unit tangent . . . . . . . . . . . . . . . . . . . 15

vector product . . . . . . . . . . . . . . . . . . . . 3

velocity . . . . . . . . . . . . . . . . . . . . . . . . . 9

W

Weingarten matrix . . . . . . . . . . . . . . . . 58

Z

zero torsion . . . . . . . . . . . . . . . . . . . . . 20

zero vector . . . . . . . . . . . . . . . . . . . . . . . 1

104

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