MATH 165 DIFFERENTIAL GEOMETRYzimmer.csufresno.edu/~sdelcroix/165notes.pdfMATH 165 DIFFERENTIAL GEOMETRY ... We normally omit the multiplication symbol · and use the empty notation.

California State University, Fresno

MATH 165

DIFFERENTIAL GEOMETRY

Fall 2003 Instructor : Stefaan Delcroix

Chapter 1

Preliminaries

1.1 Vector Spaces

Definition : A real vector space is a set V , whose elements are called vectors, together with two binary operations+ : V ×V �→ V and · : R×V �→ V , called addition and scalar multiplication, which satisfy the following nine axioms :

(a) u + v = v + u for all u,v ∈ V

(b) (u + v) + w = u + (v + w) for all u,v,w ∈ V

(c) there exists 0 ∈ V such that 0 + v = v + 0 for all v ∈ V

(d) for all v ∈ V , there exists −v ∈ V such that v + (−v) = 0

(e) (rs) · u = r · (s · u) for all u ∈ V and all r, s ∈ R

(f) (r + s) · u = r · u + s · u for all u ∈ V and all r, s ∈ R

(g) r · (u + v) = r · u + r · v for all u,v ∈ V and all r ∈ R

(h) 0 · u = 0 for all u ∈ V

(i) 1 · u = u for all u ∈ V

We normally omit the multiplication symbol · and use the empty notation.

Definitions :

(1) Let W = {vi | i ∈ I} be a subset of V . We say that W is linearly independent if whenever a finite linearcombination

∑i∈I aivi is zero then ai = 0 for all i ∈ I. We say that W is linearly dependent if there exists a

finite linear combination∑

i∈I aivi = 0 with some ak �= 0.

(2) A subset W of V spans V if for each v ∈ V there exist a finite subset {v1,v2, . . . ,vn} of W and real numbersa1, . . . an such that v =

∑ni=1 aivi.

(3) A basis of a vector space V is a linearly independent spanning set.

Theorem 1.1 Every vector space has a basis. Any two bases have the same number of elements, or all have infinitelymany elements.

Definitions :

(1) The number of elements of a basis for a vector space V is called the dimension of V .

1

(2) Let β = {vi | i ∈ I} be a basis for V . Then every v ∈ V can be written uniquely as a finite sum v =∑

I∈I aivi.The numbers ai are called components of v with respect to β.

(3) An inner product on a vector space V is a function 〈 , 〉 : V × V �→ R such that

(a) 〈u,v〉 = 〈v,u〉 for all u,v ∈ V

(b) 〈u, rv + sw〉 = r 〈u,v〉 + s 〈u,w〉 for all u,v,w ∈ V and all r, s ∈ R

(c) 〈u,u〉 ≥ 0 and 〈u,u〉 = 0 ⇐⇒ u = 0 for all u ∈ V

(4) If V has an inner product then for all v ∈ V , the length or norm of v is ‖v‖ =√〈v,v〉

Remark : In R3 the ordinary dot product is defined by 〈(a1, a2, a3), (b1, b2, b3)〉 = a1b1 + a2b2 + a3b3

Lemma 1.2 (Cauchy-Schwarz Inequality) For all u,v ∈ V , we have that | 〈u,v〉 | ≤ ‖u‖‖v‖ and we have equalityif and only if u and v are linearly dependent.

Definitions : Let u,v ∈ V .

(1) If u �= 0 �= v, then we define the angle θ between u and v as θ = cos−1

( 〈u,v〉‖u‖‖v‖

)(2) u is orthogonal (or perpendicular) to v if 〈u,v〉 = 0.

Theorem 1.3 If V has dimension n and an inner product, then there exists a basis {v1,v2, . . . ,vn} such that ‖vi‖ = 1for i = 1, . . . , n and vi is perpendicular to vj for all 1 ≤ i, j ≤ n with i �= j.

Remark : Such a basis is called orthonormal and can be obtained by a process called the Gram-Schmidt orthogonal-ization.

1.2 Linear Transformations and Eigenvectors

Definitions :

(1) Let V and W be vector spaces. A linear transformation from V to W is a function T : V �→ W such thatT (au + bv) = aT (u) + bT (v) for all u,v ∈ V and all a, b ∈ R.

(2) A linear transformation is an isomorphism if it is one-to-one and onto.

(3) Let T : V �→ V be a linear transformation. A real number λ is an eigenvalue of T if there exists a nonzero vectorv ∈ V such that T (v) = λv. Such a vector v is called an eigenvector of T corresponding to λ.

1.3 Orientation and Cross Products

Definitions :

(1) Let {u1,u2, . . . ,un} and {v1,v2, . . . ,vn} be two ordered bases of V and define a matrix A = (aij)1≤i,j≤n byvj =

∑ni=1 aijui. We say that the two bases give the same orientation to V if det(A) > 0 and give opposite

orientations to V if det(A) < 0.

(2) We denote the basis {(1, 0, 0), (0, 1, 0), (0, 0, 1)} of R3 by {e1, e2, e3}. Its orientation will be called right handed.

2

(3) If u =∑3

i=1 aiei and v =∑3

i=1 biei are vectors in R3, the cross product of u and v is the vector

u × v = (a2b3 − a3b2)e1 + (a3b1 − a1b3)e2 + (a1b2 − a2b1)e3

By abuse of notation, this may be written as u× v =

∣∣∣∣∣∣e1 e2 e3

a1 a2 a3

b1 b2 b3

∣∣∣∣∣∣Lemma 1.4 Let u,v,w ∈ R

3 and r ∈ R. Then the following holds :

(a) u × v = −v × u

(b) (ru) × v = r(u × v)

(c) u × v = 0 if and only if u and v are linearly dependent.

(d) (u + v) × w = (u × w) + (v × w)

(e) u × v is perpendicular to both u and v under the usual dot product of R3.

(f) ‖u× v‖ = ‖u‖‖v‖ sin(θ) where θ is the angle between u and v (under the usual dot product)

(g) {u,v,u × v} gives a right handed orientation to R3 if {u,v} is linearly independent.

(h) 〈u × v,w〉 = 〈u,v × w〉

Definition : The mixed (or triple) scalar product of u,v,w is [u,v,w] = 〈u × v,w〉

1.4 Lines, Planes and Spheres

We mention the following vector equations :

(1) Let x0,v ∈ R3 with v �= 0. The equation of the line through x0 and parallel to v is

α(t) = x0 + tv where t ∈ R

(2) Let x1,x2 ∈ R3 with x1 �= x2. The equation of the line through x1 and x2 is

α(t) = x1 + t(x2 − x1) where t ∈ R

(3) Let x0,n ∈ R3 with n �= 0. The equation of the plane through x0 and perpendicular to n is

〈x − x0,n〉 = 0

(4) Let x0,u,v ∈ R3 with {u,v} linearly independent. The equation of the plane through x0 and parallel to both

u and v is〈x − x0,u× v〉 = 0

(5) Let m ∈ R3 and r > 0. The equation of the sphere with center m and radius r is

〈x − m,x− m〉 = r2

3

1.5 Vector Calculus

Let V be a real vector space, {v1,v2, . . . ,vn} a basis for V and f : R �→ V a function. Then we can write f =∑ni=1 fi(t)vi. If the component functions fi(t) are differentiable or integrable, we may differentiate or integrate f

componentswise :dfdt

=n∑

i=1

(dfi(t)

dt

)vi and

∫ b

a

f(t) dt =n∑

i=1

(∫ b

a

fi(t) dt

)vi

Note that these definitions do not depend on the choice of basis for V .Similarly, if f is a vector-valued function of several variables, we may take partial derivatives or multiple integrals.

Lemma 1.5 Let f ,g : R �→ V . Then the following holds :

(a)d

dt(f × g) =

dfdt

× g + f × dgdt

.

(b) Suppose that V has an inner product 〈 , 〉. Thend

dt〈f ,g〉 =

⟨dfdt

, g

⟩+⟨f ,

dgdt

⟩. In particular, If ‖f‖ is

constant, thendfdt

is perpendicular to f .

Definitions :

(1) A function f : R �→ R is of class Ck if all derivatives up through order k exist and are continuous.

(2) A function f : Rn �→ R is of class Ck if all its (mixed) partial derivatives of order k or less exist and are

continuous.

(3) A vector-valued function is of class Ck if all of its components with respect to a given basis are of class Ck.

Theorem 1.6 (chain rule) Let x be a function of the variables u1, u2, . . . , un which are functions of the variablesv1, v2, . . . , vm. Then

∂x∂vk

=n∑

i=1

∂x∂ui

∂ui

∂vkfor k = 1, 2, . . . , m

4

Chapter 2

Local Curve Theory

2.1 Basic Definitions

Definition : A regular curve in R3 is a function α : (a, b) �→ R

3 which is of class Ck for some k ≥ 1 and for whichdα

dt�= 0 for all t ∈ (a, b).

Let α(t) be a regular curve in R3. We will mostly assume that α(t) is of class C3.

We can view a curve as the path of a particle moving in 3-space : the position at time t of the particle is given byα(t).

Definitions :

(1) A quantity is called a geometric quantity if it only depends on the image set of α (in R3) and not on the particular

parametrization.

(2) The velocity vector at t = t0 is the derivativedα

dt(t0)

(3) The velocity vector field is the vector-valued functiondα

dt

(4) The speed at t = t0 is∥∥∥∥dα

dt(t0)∥∥∥∥

(5) The tangent vector field is the vector-valued function T(t) =

dα

dt∥∥∥∥dα

dt

∥∥∥∥(6) The tangent line at the point t = t0 is given by

l = {w ∈ R3 |w = α(t0) + λT(t0) , λ ∈ R} = {w ∈ R

3 |w = α(t0) + µdα

dt(t0) , µ ∈ R}

(7) A reparametrization of a curve α : (a, b) �→ R3 is a bijection g : (c, d) �→ (a, b) such that both g and g−1 are of

class Ck for some k ≥ 1.

Theorem 2.1 Let g : (c, d) �→ (a, b) be a reparametrization of the curve α : (a, b) �→ R3. Put β = α ◦ g. Then β is a

regular curve.

5

Proof : Clearly β is at least of class C1. Let r denote the variable on (c, d). Using the chain rule, we get that

dβ

dr=

dα

dt

dg

dr

Since α is regular, we have thatdα

dt�= 0. Note that g(g−1(t)) = t for all t ∈ (a, b). Since g and g−1 are of class C1,

we can derive this with respect to t. Using the chain rule, we get thatdg

dr

dg−1

dt= 1 for all t ∈ (a, b). In particular,

dg

dr�= 0 for all r ∈ (c, d). So

dβ

dr�= 0 and β is regular. �

Remark : If g : (c, d) �→ (a, b) is a reparametrization, thendg

dr�= 0 for all r ∈ (c, d). Hence we have that either

dg

dris

always positive or always negative for all r ∈ (c, d).

Theorem 2.2 Let α : (a, b) �→ R3 be a regular curve and g : (c, d) �→ (a, b) a reparametrization. Put β = α ◦ g. Let

T (resp. S) be the tangent vector field of α (resp. β). Then S = ±T.

Proof : Using the chain rule, we get that

S =

dβ

dr∥∥∥∥dβ

dr

∥∥∥∥ =

dα

dt

dg

dr∥∥∥∥dα

dt

∥∥∥∥ ∣∣∣∣dg

dr

∣∣∣∣ =

dα

dt∥∥∥∥dα

dt

∥∥∥∥dg

dr∣∣∣∣dg

dr

∣∣∣∣ =

dg

dr∣∣∣∣dg

dr

∣∣∣∣T

Since eitherdg

dr> 0 or

dg

dr< 0 for all r ∈ (c, d), we get that either

dg

dr∣∣∣∣dg

dr

∣∣∣∣ = 1 or

dg

dr∣∣∣∣dg

dr

∣∣∣∣ = −1. So S = ±T . �

2.2 Arc Length

Definition : Let α �→ R3 be a regular curve and [a1, b1] ⊂ (a, b). The length of the curve segment α : [a1, b1] �→ R

3 is∫ b1

a1

∥∥∥∥dα

dt

∥∥∥∥ dt

Theorem 2.3 Let α : (a, b) �→ R3 be a regular curve and g : (c, d) �→ (a, b) a reparametrization of α. Put β =

α ◦ g. Let [c1, d1] ⊂ [c, d] and [a1, b1] ⊂ (a, b) such that g([c1, d1]) = [a1, b1]. Then the length of the curve segmentsα : [a1, b1] �→ R

3 and β : [c1, d1] �→ R3 are the same.

Proof : Using the chain rule, we get that the length of the curve segment β : [c1, d1] �→ R3 is∫ d1

c1

∥∥∥∥dβ

dr

∥∥∥∥ dr =∫ d1

c1

∥∥∥∥dα

dt

dg

dr

∥∥∥∥ dr =∫ d1

c1

∥∥∥∥dα

dt

∥∥∥∥ ∣∣∣∣dg

dr

∣∣∣∣ dr

Suppose first thatdg

dr> 0 for all r ∈ (c, d). Then g(c1) = a1, g(d1) = b1 and dt =

dg

drdr. Hence

∫ d1

c1

∥∥∥∥dα

dt

∥∥∥∥ dg

drdr =

∫ b1

a1

∥∥∥∥dα

dt

∥∥∥∥ dt

6

which is the length of the curve segment α : [a1, b1] �→ R3.

The case ”dg

dr< 0 for all r ∈ (c, d)” is done similarly. �

Definition : Let α : (a, b) �→ R3 be a regular curve and t0 ∈ (a, b). Put

h(t) =∫ t

t0

∥∥∥∥dα

dt

∥∥∥∥ dt for all t ∈ (a, b)

Then s = h(t) is called the arc length along α.

Theorem 2.4 Let α : (a, b) �→ R3 be a regular curve and s = h(t) its arc length. Then the following holds :

(a)ds

dt=∥∥∥∥dα

dt

∥∥∥∥ for all t ∈ (a, b)

(b) s = h(t) is a bijection from (a, b) onto some interval (c, d) and h−1(s) is a reparametrization.

(c) Put β = α ◦ h−1. Let T be the tangent vector field for β(s). Then T =dβ

ds.

Proof : : (a),(b) We easily get thatds

dt=

dh

dt=∥∥∥∥dα

dt

∥∥∥∥ > 0 for all t ∈ (a, b). So h(t) is injective and hence a

bijection from (a, b) onto some interval (c, d). Also, if α is of class Ck, then so is h. Since h−1 is the inverse of h, we

get thatdh−1

ds=

1dhdt

for all s ∈ (c, d). Note thatdh

dt�= 0 for all t ∈ (a, b). So h and h−1 are both of the same class

and h−1 is a reparametrization.(c) Using the chain rule and (a), we get that

dβ

ds=

dα

dt

dh−1

ds=

dα

dt

1dh

dt

=

dα

dt∥∥∥∥dα

dt

∥∥∥∥In particular,

∥∥∥∥dβ

ds

∥∥∥∥ = 1 and T =

dβ

ds∥∥∥∥dβ

ds

∥∥∥∥ =dβ

ds. �

Remarks :

(1) The variable s is the arc length.

(2) 0 ∈ (c, d)

(3) Practically, it may be impossible to reparametrize a curve by its arc length : we may not be able to calculates = h(t) or h−1(s).

2.3 Curvature and The Frenet-Serret Apparatus

Definition : A curve α : (a, b) �→ R3 is a unit speed curve if

∥∥∥∥dα

dt

∥∥∥∥ = 1 for all t ∈ (a, b).

Remark : Suppose that α : (a, b) �→ R3 is a unit speed curve. Then the arc length s is given by

s =∫ t

t0

∥∥∥∥dα

dt

∥∥∥∥ dt =∫ t

t0

1 dt = t − t0

7

We will assume that 0 ∈ (a, b) and that t0 = 0. Hence s = t and we will write α(s) : the curve is parametrized by its

arc length. Then T =dα

ds.

Definitions : Let α(s) be a unit speed curve.

(1) The curvature of α(s) is κ(s) = ‖T′(s)‖

(2) The principal normal vector field to α(s) is the vector field N(s) =T′(s)κ(s)

(3) The binormal vector field to α(s) is the vector field B(s) = T(s) × N(s)

(4) The torsion of α(s) is the real-valued function τ(s) = −〈B′(s),N(s)〉(5) The Frenet-Serret apparatus of α(s) is the set {κ(s), τ(s),T(s),N(s),B(s)}

Lemma 2.5 Let α(s) be a unit speed curve. Then for all s such that κ(s) �= 0, the set {T(s),N(s),B(s)} is anorthonormal set.

Proof : Pick s such that κ(s) �= 0. Clearly, ‖T‖ = 1. Since κ(s) = ‖T′‖, ‖N‖ = 1. By Lemma 1.5(b), 〈T,T′〉 = 0and so 〈T,N〉 = 0 . Let θ be the angle between T and N. Then θ = 90◦. Since B = T × N, we have that〈B,T〉 = 〈B,N〉 = 0 by Lemma 1.4(e) and ‖B‖ = ‖T‖‖N‖ sin(θ) = 1 by Lemma 1.4(f). �

Proposition 2.6 Let α(s) : (a, b) �→ R3 be a unit speed curve such that κ(s) = 0 for all s in [c, d] ⊂ (a, b). Then the

curve segment α : [c, d] �→ R3 is a straight line.

Proof : Since ‖T′(s)‖ = κ(s) = 0 for all s ∈ [c, d], there exists v ∈ R3 such that T(s) = v for all s ∈ [c, d]. By

Theorem 2.4(c),dα

ds= T. Hence

α(s) − α(c) =∫ s

c

dα

dσdσ =

∫ s

c

T(σ) dσ =∫ s

c

v dσ = (s − c)v for all s ∈ [c, d]

This is the equation of the line through α(c) and parallel to v. �

2.4 The Frenet-Serret Theorem

Lemma 2.7 Let V be an n-dimensional vector space with an inner product and E = {e1, e2, . . . , en} an orthonormalset of n elements of V . Then the following holds :

(a) E is a basis for V

(b) v =n∑

i=1

〈ei,v〉 ei for all v ∈ V .

Proof : (a) Since the number of elements of E is equal to the dimension of V , it suffices to prove that E is linearlyindependent. Let c1, c2, . . . , cn be real numbers such that

∑ni=1 ciei = 0. Then

0 =

⟨n∑

i=1

ciei, ej

⟩=

n∑i=1

ci 〈ei, ej〉 =n∑

i=1

ciδij = cj for j = 1, 2, . . . , n

Hence E is linearly independent.(b) Pick v ∈ V . Since E is a basis for V , there exist real numbers d1, d2, . . . , dn such that v =

∑ni=1 diei. Hence

〈ej,v〉 =

⟨ej ,

n∑i=1

diei

⟩=

n∑i=1

di 〈ej , ei〉 =n∑

i=1

diδij = dj for j = 1, 2, . . . , n

So (b) holds. �

8

Theorem 2.8 (Frenet-Serret) Let α(s) : (a, b) �→ R3 be a unit speed curve with κ(s) �= 0 for all s ∈ (a, b) and

{κ, τ,T,N,B} its Frenet-Serret apparatus. Then the following holds :

(a) T′(s) = κ(s)N(s)

(b) N′(s) = −κ(s)T(s) + τ(s)B(s)

(c) B′(s) = −τ(s)N(s)

Proof : By Lemma 2.5, {T(s),N(s),B(s)} is an orthonormal set. By Lemma 2.7(b), we have that

v = 〈T,v〉T + 〈N,v〉N + 〈B,v〉B for all v ∈ R3 (∗)

(a) This follows from the definition of κ and T.(b) Differentiating 0 = 〈T,N〉, we get that 0 = 〈T′,N〉 + 〈T,N′〉 = 〈κN,N〉 + 〈T,N′〉 = κ + 〈T,N′〉. Hence〈T,N′〉 = −κ.Since N is a unit vector, 〈N,N′〉 = 0 by Lemma 1.5(b).Differentiating 0 = 〈B,N〉, we get that 0 = 〈B′,N〉 + 〈B,N′〉 = −τ + 〈B,N〉. Hence 〈B,N′〉 = τ .Using (*) with v = N′, we get that

N′ = 〈T,N′〉T + 〈N,N′〉N + 〈B,N′〉B = −κT + τB

(c) Differentiating 0 = 〈T,B〉, we get that 0 = 〈T′,B〉 + 〈T,B′〉 = 〈κN,B〉 + 〈T,B′〉 = 〈T,B′〉. Hence 〈T,B′〉 = 0.Since B is a unit vector, 〈B,B′〉 = 0 by Lemma 1.5(b).Using (*) with v = B′, we get that

B′ = 〈T,B′〉T + 〈N,B′〉N + 〈B,B′〉B = −τN

by definition of τ . �

Theorem 2.9 Let α(s) : (a, b) �→ R3 be a unit speed curve with κ(s) �= 0 for all s ∈ (a, b). Then the following are

equivalent :

(a) The image of α lies in a plane (or α is a plane curve).

(b) B is a constant vector.

(c) τ(s) = 0 for all s ∈ (a, b).

Proof : Note that (b) and (c) are equivalent by Theorem 2.8(c).Suppose first that (a) holds. By making an appropriate choice of coordinates in R

3, we may assume that α lies in theXY -plane. So α(s) = (x(s), y(s), 0). Then we get that

T(s) = (x′(s), y′(s), 0) and T′(s) = (x′′(s), y′′(s), 0)

Since N =T′

‖T′‖ , we see that both T(s) and N(s) are vectors in the XY -plane. Since B(s) is a unit vector that is

ortogonal to both T(s) and N(s), we get that B(s) = (0, 0,±1) for all s ∈ (a, b). Hence τ(s) = −〈B′(s),N(s)〉 = 0.So (b) holds.Suppose next that (b) holds. Let x0 be an element on α, say x0 = α(0). Put n = B(s) for all s ∈ (a, b). Then we getthat

〈α(s) − x0,n〉′ = 〈α′(s),n〉 = 〈T(s),B(s)〉 = 0 for all s ∈ (a, b)

So 〈α(s) − x0,n〉 is a constant. Hence 〈α(s) − x0,n〉 = 〈α(0) − x0,n〉 = 0 for all s ∈ (a, b) and α lies in the planethrough x0 and perpendicular to n. So (a) holds. �

Definitions : Let α : (a, b) �→ R3 be a unit speed curve.

(1) The osculating plane of α(s) is the plane through α(s) and perpendicular to B(s).

9

(2) The normal plane of α(s) is the plane through α(s) and perpendicular to T(s).

(3) The rectifying plane of α(s) is the plane through α(s) and perpendicular to N(s).

Proposition 2.10 Let α(s) : (a, b) �→ R3 be a unit speed curve and x0 ∈ R

3 such that every normal plane to α(s)goes through x0 for all s ∈ (a, b). Then the image of α lies on a sphere.

Proof : Since the normal plane to α(s) is orthogonal to T(s), we have that 〈α(s) − x0,T(s)〉 = 0 for all s ∈ (a, b).Hence 〈α(s) − x0, α(s) − x0〉′ = 2 〈α(s) − x0, α

′(s)〉 = 2 〈α(s) − x0,T(s)〉 = 0 for all s ∈ (a, b). So there exists c ∈ R

with 〈α(s) − x0, α(s) − x0〉 = c for all s ∈ (a, b). Note that c ≥ 0. If c = 0, then α(s) = x0 for all s ∈ (a, b) and α isnot regular, a contradiction. So c > 0 and the image of α lies on the sphere with center x0 and radius

√c. �

2.5 The Fundamental Existence and Uniqueness Theorem for Curves

Theorem 2.11 (Fundamental Theorem of Curves) Let (a, b) be an interval with a < 0 < b, κ(s) > 0 a C1

function on (a, b), τ (s) a continuous function on (a, b), x0 a fixed point of R3 and {D,E,F} a fixed right handed

orthonormal basis of R3. Then there exists a C3 regular curve α(s) : (a, b) �→ R

3 such that

(a) The parameter s is the arc length from α(0)

(b) α(0) = x0, T(0) = D, N(0) = E and B(0) = F

(c) κ(s) = κ(s) and τ(s) = τ (s)

Moreover, α(s) is unique with these properties.

Proof : �

2.6 Non-Unit Speed Curves

Let α(t) be a regular curve and s(t) its arc length. The next theorem will tell us how to determine the Frenet-Serret

apparatus of α in terms of t. We use the prime notation for derivatives with respect to s :dα

ds= α′,

d2α

ds2= α′′, etc.

We denote derivatives with respect to t by dots :ds

dt= s,

dα

dt= α,

d2α

dt2= α, etc.

Theorem 2.12 Let α(t) be a regular curve in R3. Then the following holds :

(a) T =α

‖α‖

(b) B =α × α

‖α × α‖(c) N = B× T

(d) κ =‖α × α‖‖α‖3

(e) τ =[α, α,

...α]

‖α × α‖2

10

Proof : Using the chain rule, we get that

α = α′s = sT (∗)

Note that s = ‖α‖. So (a) follows.Deriving (*) with respect to t and using the chain rule and the Frenet-Serret equations, we get that

α = sT + sT = sT + sT′s = sT + s2κN (∗∗)

Taking the cross product of (*) with (**), we find that

α × α = (sT) × (sT + s2κN) = s3κT × N = s3κB (∗ ∗ ∗)

Taking the norm on both sides of (***), we get that

‖α × α‖ = ‖s3κB‖ = |s3κ|‖B‖ = s3κ = ‖α‖3κ

which proves (d).Suppose that κ �= 0. Then (b) follows immediately from (***). Since {T,N,B} is a right handed orthonormal basis,(c) holds.Deriving (**) with respect to t and using the chain rule and the Frenet Serret equations, we get that

...α =

...s T + sT + (s2κ)˙N + s2κN =

...s T + sT′s + (s2κ)˙N + s2κN′s =

...s T + ssκN + (s2κ)˙N− s3κ2T + s3κτB

Hence ...α = (

...s − s3κ2)T + (ssκ + (s2κ)˙)N + s3κτB and

[α, α, α] = 〈α × α,...α〉 =

⟨s3κB,

...α⟩

=⟨s3κB, s3κτB

⟩= s6κ2τ = τ(s3κ)2 = τ‖α × α‖2

which proves (e). �

Theorem 2.13 Let α(t) be a regular curve in R3. Then the following holds :

(a) T = κ‖α‖N(b) N = −κ‖α‖T + τ‖α‖B(c) B = −τ‖α‖N

Proof : (a) Using the chain rule and the Frenet-Serret equations, we get that T = T′s = κN‖α‖, which proves (a).(b) and (c) are done similarly. �

11

Chapter 3

Global Theory of Plane Curves

In this chapter, we will study plane curves (so τ ≡ 0). We will assume that a suitable choice of coordinates has beenmade in R

3 so that the curve lies in the XY -plane. We will disregard the Z-coordinate an write the curve as if it liesin R

2.

3.1 The Rotation Index of a Plane Curve

Definitions : Let α(s) : (a, b) �→ R2 be a unit speed C2 plane curve.

(a) The tangent vector field t(s) to α is t(s) = α′(s) for all s ∈ (a, b).

(b) The normal vector field n(s) to α is the unique unit vector field n(s) such that {t(s),n(s)} gives a right handedorthonormal basis of R

2 for each s ∈ (a, b).

(c) The plane curvature k(s) of α is given by k(s) = 〈t′(s),n(s)〉

Theorem 3.1 Let α(s) : (a, b) �→ R2 be a unit speed plane curve. Then the following holds :

(a) t′(s) = k(s)n(s) for all s ∈ (a, b).

(b) If α(s) = (x(s), y(s)) for all s ∈ (a, b) and x, y are of class C2, then

t(s) = (x′(s), y′(s)) , n(s) = (−y′(s), x′(s)) and k(s) = x′(s)y′′(s) − y′(s)x′′(s) for all s ∈ (a, b)

(c) t(s) = T(s) for all s ∈ (a, b)

(d) Suppose that N(s) exists for all s ∈ (a, b). Then n(s) = ±N(s), κ(s) = |k(s)|, n(s) is differentiable andn′(s) = −k(s)t(s) for all s ∈ (a, b).

Proof : (a) Since t(s) is a unit vector, we have that 〈t(s), t′(s)〉 = 0 for all s ∈ (a, b) by Lemma 1.5(b). Hencet′(s) = 〈t′(s), t(s)〉 t(s) + 〈t′(s),n(s)〉n(s) = k(s)n(s) by Lemma 2.7(b).(b) Clearly, t(s) = (x′(s), y′(s)). Put m(s) = (−y′(s), x′(s)) for all s ∈ (a, b). Pick s ∈ (a, b). Then 〈t(s),m(s)〉 = 0and ‖m(s)‖ =

√(−y′(s))2 + (x′(s))2 = ‖t(s)‖ = 1. Hence {t(s),m(s)} is an orthonormal basis for R

2. The change of

basis matrix A to change from the standard basis {(1, 0), (0, 1)} to {t(s),m(s)} is[

x′(s) −y′(s)y′(s) x′(s)

]. Since det(A) =

(x′(s))2 + (y′(s))2 = ‖t(s)‖2 = 1, {t(s),m(s)} is a right handed basis. So m(s) = n(s).Finally, k(s) = 〈t′(s),n(s)〉 = 〈(x′′(s), y′′(s)), (−y′(s), x′(s))〉 = x′(s)y′′(s) − y′(s)x′′(s).(c) This is obvious.(d) Suppose that N(s) exists for all s ∈ (a, b). Since N(s) is a unit vector in the XY -plane, orthogonal to T(s) = t(s),we have that n(s) = ±N(s). So (x′′, y′′) = T′ = κN = ±κn = ±κ(−y′, x′). Suppose that (x′′, y′′) = κ(−y′, x′).Then k = x′y′′ − y′x′′ = κ(x′)2 + κ(y′)2 = κ‖t‖2 = κ and n′ = (−y′′, x′′) = (−κx′,−κy′) = −kt. Similarly for(x′′, y′′) = −κ(−y′, x′). �

12

Theorem 3.2 Let α(t) : (a, b) �→ R2 : t �→ (x(t), y(t)) be a regular plane curve. Then

t =α

‖α‖ =(x, y)√x2 + y2

, n =(−y, x)√x2 + y2

and k =x y − y x

(x2 + y2)32

Proof : Let s(t) be the arc length along α. Then s = ‖α‖. We use the prime notation for derivatives with repect

to s. Using the chain rule, we get that α = α′s = ts. Hence t =α

s=

α

‖α‖ =(x, y)√x2 + y2

. By Theorem 3.1(b),

n =(−y, x)√x2 + y2

. Using the chain rule and Theorem 3.1(a), we get that α = ts + ts = t′s2 + ts = ks2n + st. So

α× α = (st)× (ks2n+ st) = ks3t×n = ks3e3 since {t,n} is a right handed orthonormal basis of R2. But α = (x, y)

and α = (x, y) and so

α × α =

∣∣∣∣∣∣e1 e2 e3

x y 0x y 0

∣∣∣∣∣∣ = (x y − y x)e3

Hence ks3 = x y − y x and so k =x y − y x

s3=

x y − y x

(x2 + y2)32. �

Definitions :

(a) A function f : R �→ V is periodic if there exists a constant a > 0 such that f(t) = f(a + t) for all t ∈ R. Theperiod of f is the smallest such number a.

(b) A regular curve α : R �→ R3 is closed if α is periodic.

(c) A regular curve α(t) is simple if one of the following holds :

(1) α(t) is one-to-one

(2) α(t) is a closed curve with period a such that α(t1) �= α(t2) for all t1, t2 ∈ [0, a) with t1 �= t2

(d) Let α(s) be a regular closed unit speed plane curve with period L. We define θ(s) =∫ s

0 k(σ) dσ for all s ∈ [0, L].

Lemma 3.3 Let α(s) be a regular closed unit speed plane curve with period L and θ(s) =∫ s

0k(σ) dσ for all s ∈ [0, L].

If t(0) = (1, 0) then t(s) = (cos(θ(s)), sin(θ(s)) for all s ∈ [0, L].

Proof : Define a continuous function ϕ(s) by t(s) = (cos(ϕ(s)), sin(ϕ(s))) for all s ∈ [0, L]. Since t(0) = (1, 0), wecan choose ϕ(0) = 0. By Lemma 3.1(b), we get that

t′(s) = ϕ′(s)(− sin(ϕ(s)), cos(ϕ(s))) = ϕ′(s)n(s) = k(s)n(s) for all s ∈ [0, L]

Hence ϕ′(s) = k(s) = θ′(s) for all s ∈ [0, L]. So there exists a constant c with θ(s) = ϕ(s) + c for all s ∈ [0, L]. Sinceθ(0) = 0 = ϕ(0), we have that θ(s) = ϕ(s) and t(s) = (cos(θ(s)), sin(θ(s))) for all s ∈ [0, L]. �

Definition : Let α(s) be a regular closed unit speed plane curve with period L. The rotation index of α(s) is theinteger

iα =θ(L) − θ(0)

2π=

θ(L)2π

Theorem 3.4 (Rotation Index Theorem) The rotation index of a regular simple closed unit speed plane curve is±1.

Proof : �

13

Definitions :

(a) A piecewise Ck regular curve is a continuous function α(s) : [a, b] �→ R3 and a finite set of points a = s0 < s1 <

· · · < sn−1 < sn = b such that α : (si, si+1) �→ R3 is a regular Ck curve for i = 0, 1, . . . , n − 1.

Suppose that α(s) : [a, b] �→ R2 is a piecewise regular plane curve. Let δθi be the angle through which t

rotates on the segment α : (si, si+1) �→ R2 for i = 0, 1, . . . , n − 1. Put t−(si) = lim

s�→s−i

t(s) for i = 1, 2, . . . , n and

t+(si) = lims�→s+

i

t(s) for i = 0, 1, . . . , n − 1. Let ∆θi ∈ [−π, π] be the angle between t−(si) and t+(si) (measured

form t−(si) to t+(si)) for i = 1, 2, . . . , n − 1 (if α is closed, put ∆θ0 = ∆θn the angle between t−(sn) andt+(s0)). If |∆θi| = π, we can figure out the sign of ∆θi by finding the sign of the angle between t(si − ε) andt(si + ε) for some small ε.

(b) The rotation index of a piecewise regular closed plane curve α is the integer

iα =

n−1∑i=0

(δθi + ∆θi)

2π

Theorem 3.5 The rotation index of a piecewise regular simple closed plane curve is ±1.

Proof : �

3.2 The Isoperimetric Inequality

Theorem 3.6 (Green) Let C be a closed plane C2 curve, which bounds a region R and is traversed counterclockwise.

Then∫C(f dx + g dy) =

∫∫R

(∂g

∂x− ∂f

∂y

)dx dy for all differentiable functions f and g defined on R.

Proof : �

Lemma 3.7 Let α(t) : [a, b] �→ R2 be a simple closed plane curve which is traversed counterclockwise and whose image

bounds a region R. Then the area of R is given by∫ b

a

x(t)y′(d) dt = −∫ b

a

y(t)x′(t) dt

Proof : By definition, the area of R is given by∫∫

R1 dx dy. Using Green’s Theorem with f(x, y) = 0 and g(x, y) = x,

we get that ∫α

xdy =∫α

(f dx + g dy) =∫∫

R

(∂g

∂x− ∂f

∂y

)dx dy =

∫∫R

1 dx dy = area of R

Since α(t) = (x(t), y(t)) for all t ∈ [a, b], we get that∫α

xdy =∫ b

a

x(t)dy(t) =∫ b

a

x(t)y′(t) dt.

Similarly, by using Green’s Theorem with f(x, y) = −y and g(x, y) = 0 we get that the area of R is given by

−∫ b

a

y(t)x′(t) dt. �

Lemma 3.8 Let a and b be positive real numbers. Then√

ab ≤ a + b

2. Moreover, we have equality if and only if a = b.

Proof : We have that

√ab ≤ a + b

2⇐⇒ 2

√ab ≤ a+b ⇐⇒ 4ab = (2

√ab)2 ≤ (a+b)2 = a2+2ab+b2 ⇐⇒ 0 ≤ a2−2ab+b2 ⇐⇒ 0 ≤ (a−b)2

This proves the lemma. �

14

Theorem 3.9 (Isoperimetric Inequality) Let α be a simple closed regular plane curve of length (perimeter) L.Let A be the area of the region bounded by α. Then L2 ≥ 4πA. Moreover, we have equality if and only if α is a circle.

Proof : Let l1, l2 be two parallel lines tangent to α with α bounded between them. Let β be a circle tangent tol1 and l2 which does not intersect α. Let r be the radius of that circle (note that for now, r depends on l1). Wechoose a coordinate system with the origin at the center of the circle and the Y -axis parallel to l1. Let s ∈ [0, L] bethe arc length on α. Put α(s) = (x(s), y(s)). Then β(s) = (x(s), y(s)). Note that s is not the arc length on β. By

Lemma 3.7, the area of the region bounded by α is A =∫ L

0

x(s)y′(s) ds and the area of the region bounded by β is

πr2 = −∫ L

0

y(s)x′(s) ds. Hence

A + πr2 =∫ L

0

(x(s)y′(s) − y(s)x′(s)) ds =∫ L

0

〈(x′, y′), (−y, x)〉 ds ≤∫ L

0

| 〈(x′, y′), (−y, x)〉 | ds

By the Cauchy-Schwarz Inequality (Lemma 1.2), we get that

| 〈(x′, y′), (−y, x)〉 | ≤ ‖(x′, y′)‖‖(−y, x)‖ =√

(x′)2 + (y′)2√

(y)2 + x2 = ‖α′(s)‖‖β(s)‖ = 1 · r = r

Hence we have that

A + πr2 =∫ L

0

〈(x′, y′), (−y, x)〉 ds ≤∫ L

0

| 〈(x′, y′), (−y, x)〉 | ds ≤∫ L

0

r ds = rL (∗)

By Lemma 3.8 (with a = A and b = πr2), we get that

√Aπr2 ≤ A + πr2

2≤ rL

2(∗∗)

Hence Aπr2 ≤ r2L2

4and so L2 ≥ 4πA.

Suppose now that L2 = 4πA. So we have equalities in (**). By Lemma 3.8, A = πr2. Moreover, we have equalitiesin (**). So 〈(x′, y′), (−y, x)〉 = r. By Lemma 1.2, (−y, x) and (x′, y′) are linearly dependent. So (−y, x) = c(x′, y′)for some c ∈ R. Then r = 〈(x′, y′), (−y, x)〉 = 〈(x′, y′), c(x′, y′)〉 = c((x′)2 + (y′)2) = c‖α′‖2 = c. Hence we have thatx(s) = ry′(s) for all s ∈ (0, L).Since A = πr2, we see that r only depends on A, not on l1. So we can go over the same construction and calculationswith lines l3 and l4 which are perpendicular to l1. Denote the coordinates in the new coordinate system (related to l3and l4) by (x, y). Then we have again that x(s) = ry′(s) for all s ∈ (0, L). Note that the x-axis (resp. y-axis) is paralellto and pointing in the same (resp. opposite) direction of the y-axis (resp. x-axis). Hence there are constants a and bsuch that x(s) = y(s) − b and y(s) = a − x(s) for all s ∈ (0, L). So y(s) − b = x(s) = ry′(s) = r(a − x(s))′ = −rx′(s)for all s ∈ (0, L). Hence

x2(s) + (y(s) − b)2 = (ry′(s))2 + (−rx′(s))2 = r2((x′(s)2 + (y′(s))2) = r2‖α′(s)‖2 = r2 for all s ∈ (0, L)

So α is a circle with radius r and center (0, b) in the xy-coordinate system. �

3.3 Convex Curves and The Four-Vertex Theorem

Definition : A regular curve is convex if it lies on one side of each tangent line.

Theorem 3.10 A simple closed regular plane curve α(s) is convex if and only if k(s) has constant sign.

Proof : �

15

Remark : If α is a regular plane curve, P is a point on α and L is a line through P such that α lies on one side ofL, then L is the tangent line to α at P .

Definition : A vertex of a regular plane curve is a point where the plane curvature k has a relative minimum ormaximum.

Lemma 3.11 Let α(s) : [0, L] �→ R2 : s �→ (x(s), y(s)) be a regular unit speed closed plane curve and A, B, C ∈ R.

Then∫ L

0

(Ax(s) + By(s) + C)k′(s) ds = 0.

Proof : By Theorem 3.1(a)(b), we have that t(s) = (x′(s), y′(s)) and (x′′(s), y′′(s)) = t′(s) = k(s)n(s) =k(s)(−y′(s), x′(s)). Hence x′′(s) = −k(s)y′(s) and y′′(s) = k(s)x′(s). Note that x(0) = x(L), y(0) = y(L),

x′(0) = x′(L), y′(0) = y′(L) and k(0) = k(L). Hence∫ L

0

k′(s) ds = k(L) − k(0) = 0. Using integration by parts,

we get that ∫ L

0

x(s)k′(s) ds = [x(s)k(s)]L0 −∫ L

0

k(s)x′(s) ds = −∫ L

0

y′′(s) ds = −(y′(L) − y′(0)) = 0

Similalry, we get that∫ L

0

y(s)k′(s) ds = 0. Hence∫ L

0

(Ax(s) + By(s) + C)k′(s) ds = 0. �

Theorem 3.12 (Four-Vertex Theorem) A regular simple closed convex curve has at least four vertices.

Proof : Let α(s) : [0, L] �→ R2 be the parametrization of α using the arc length.

Suppose that α has at most 3 vertices. Since k(s) is a continuous function on [0, L], k reaches a maximum and aminimum on [0, L]. Hence α has at least two vertices, say p and q. Note that p �= q. By changing the ’starting point’on α, we may assume that p = α(0) and q = α(s1) for some 0 < s1 < L.Suppose that the tangent line to α at p is the same as the tangent line to α at q. Call this line L1. Then there existss ∈ (0, s1) such that the tangent line to α at α(s) is parallel to L1. Call this tangent line L2. Put r = α(s). Assumefirst that L2 = L1. Consider the intermediate point of the points p, q and r, say r. Let L3 be the tangent line to α ata point very close to r. Since α is convex, p and q must lie on the same side of L3. This is only possible if L3 = L1.So α is a line segment around the point p. But then k(s) is constant on some interval, a contradiction. Assume nextthat L2 �= L1. Since α is convex, α lies between the parallel lines L1 and L2, a contradiction since α is simple andclosed.So the tangent line to α at p is not the same as the tangent line to α at q.Let L be the line through p and q and β (resp. γ) be the ’arc’ of α on [0, s1] (resp. [s1, L]).Suppose β does not lie on one side of L. Then the curve α meets L is a point r with p �= r �= q. Consider theintermediate point of p, q and r, say p. If L is not the tangent line to α at p, then q and r are on different sides ofthat tangent line, a contradiction since α is convex. So L is the tangent line to α at p. Then L is a line that meets αin q (resp. r) such that α lies on one side of L. Hence L is also the tangent line to α at q (resp. r), a contradictionsince p and q have different tangent lines.So β lies on one side of L. Similarly, we get that γ lies on one side of L. Suppose that β and γ lie on the same side ofL. Then L is a line going through p (resp. q) such that α lies on one side of L. Hence L is the tangent line to α at pand q, again a contradiction since p and q have different tangent lines.So β and γ lie on different sides of L. Let Ax + By + C = 0 be the equation of L. Then we may assume thatAx(s) + By(s) + C ≥ 0 for all s ∈ (0, s1) and Ax(s) + By(s) + C ≤ 0 for all s ∈ (s1, L). Moreover, Ax(s) + By(s) + Cis not identically zero on (0, L) (otherwise α contains a line segment and k(s) would be constant on some interval).Suppose that α has only two vertices. We may assume that k reaches a minimum at p and a maximum at q. Thenk′(s) ≥ 0 for all s ∈ (0, s1) and k′(s) ≤ 0 for all s ∈ (s1, L). Hence (Ax(s) + By(s) + C)k′(s) ≥ 0 for all s ∈ (0, L) andis not identically zero, a contradiction to Lemma 3.11.So α has a third vertex, say on β. Hence k′(s) changes signs on β, say at s2 with 0 < s2 < s1. Then k′(s) > 0 for0 < s < s2, k′(s) < 0 for s2 < s < s1 and k′(s) > 0 for s1 < s < L, a contradiction since k′(s) has a minimum ats = 0.Hence α has at least four vertices. �

16

Chapter 4

Local Surface Theory

4.1 Basic Definitions

Definitions

(a) A subset U of R2 is open if for every point (a, b) ∈ U , there exists some real number ε > 0 such that (x, y) ∈ U

whenever (x − a)2 + (y − b)2 < ε2.

(b) Let U be an open subset R2 (with variables (u1, u2)) and r : U �→ R

3 a function. Then r is a Ck coordinate patch

(or simple surface) if r is one-to-one, of class Ck and∂r∂u1

× ∂r∂u2

�= 0 on U .

(c) Let U and V be open subsets of R2 (with variables (v1, v2) on V and variables (u1, u2) on U) and f : V �→ U :

(v1, v2) �→ (f1(v1, v2), f2(v1, v2)) a function. Then the Jacobian of f (notation : J(f)) is the matrix

J(f) =

∂f1

∂v1

∂f1

∂v2

∂f2

∂v1

∂f2

∂v2

(d) Let U and V be open subsets of R

2 and f : V �→ U a function. Then f is a Ck coordinate transformation if f isa bijection of class Ck and f−1 is also of class Ck.

Theorem 4.1 Let U and V be open subsets of R2 and f : V �→ U a bijection of class Ck. Then f is a coordinate

transformation if and only if det(J(f)) �= 0 in V.

Proof : Suppose first that det(J(f)) �= 0 on V . Then by the Inverse Function Theorem (see any advanced calculuscourse), f−1 is of class Ck. So f is a coordinate transformation.Suppose next that f is a coordinate transformation. Let (v1, v2) be the coordinates on V and (u1, u2) the coordinateson U . Let g : U �→ V : (u1, u2) �→ (g1(u1, u2), g2(u1, u2)) be the inverse of f . Then f ◦ g is the identity function on U .So

f1(g1(u1, u2), g2(u1, u2)) = u1 and f2(g1(u1, u2), g2(u1, u2)) = u2 for all (u1, u2) ∈ UWe use the chain rule to derive these equations with respect to u1 and u2 :

∂f1

∂v1

∂g1

∂u1+

∂f1

∂v2

∂g2

∂u1= 1 ,

∂f1

∂v1

∂g1

∂u2+

∂f1

∂v2

∂g2

∂u2= 0

∂f2

∂v1

∂g1

∂u1+

∂f2

∂v2

∂g2

∂u1= 0 ,

∂f2

∂v1

∂g1

∂u2+

∂f2

∂v2

∂g2

∂u2= 1

17

We can rewrite these equations in matrix form :∂f1

∂v1

∂f1

∂v2

∂f2

∂v1

∂f2

∂v2

∂g1

∂u1

∂g1

∂u2

∂g2

∂u1

∂g2

∂u2

=[

1 00 1

]or J(f)J(g) =

[1 00 1

]

So the matrix J(f) has an inverse. Hence det(J(f)) �= 0 for all (v1, v2) ∈ V . �

Theorem 4.2 Let U and V be open subsets of R2, g : U �→ R a continuous function and f : V �→ U a C1 coordinate

transformation. Then ∫∫U

g(u1, u2) du1 du2 =∫∫

Vg(f1(v1, v2), f2(v1, v2)) | det(J(f))| dv1 dv2

Proof : �

Lemma 4.3 Let r : U �→ R3 be a simple surface of class Ck and f : V �→ U a Ck coordinate transformation. Then

r ◦ f : V �→ R3 is a simple surface of class Ck.

Proof : Note that r ◦ f is one-to-one and of class Ck. Also, r(v1, v2) = r(f1(v1, v2), f2(v1, v2)). Using the chain rule,we get that

∂r∂v1

=∂r∂u1

∂f1

∂v1+

∂r∂u2

∂f2

∂v1and

∂r∂v2

=∂r∂u1

∂f1

∂v2+

∂r∂u2

∂f2

∂v2

Recall that u× u = 0 and u× v = −v × u for all u,v ∈ R3. Hence we get that

∂r∂v1

× ∂r∂v2

=(

∂r∂u1

∂f1

∂v1+

∂r∂u2

∂f2

∂v1

)×(

∂r∂u1

∂f1

∂v2+

∂r∂u2

∂f2

∂v2

)

=∂f1

∂v1

∂f1

∂v2

∂r∂u1

× ∂r∂u1

+∂f1

∂v1

∂f2

∂v2

∂r∂u1

× ∂r∂u2

+∂f2

∂v1

∂f1

∂v2

∂r∂u2

× ∂r∂u1

+∂f2

∂v1

∂f2

∂v2

∂r∂u2

× ∂r∂u2

=(

∂f1

∂v1

∂f2

∂v2− ∂f1

∂v2

∂f2

∂v1

)∂r∂u1

× ∂r∂u2

= det(J(f))∂r∂u1

× ∂r∂u2

Since r(u1, u2) is a simple surface, we have that∂r∂u1

× ∂r∂u2

�= 0. Since f is a coordinate transformation, det(J(f)) �= 0

by Theorem 4.1. Hence∂r∂v1

× ∂r∂v2

�= 0 and so r(v1, v2) is a regular surface. �

IMPORTANT NOTATION : If r(u1, u2) is a simple surface, we put r1(a, b) =∂r∂u1

(a, b) and r2(a, b) =∂r∂u2

(a, b).

Similarly, if r(u, v) is a simple surface, we put ru(a, b) =∂r∂u

(a, b) and rv(a, b) =∂r∂v

(a, b).

Definitions : Let r : U �→ R3 be a simple surface. Put P = r(a,b).

(a) The tangent plane to r at P is the plane through P and perpendicular to r1(a, b) × r2(a, b).

(b) The unit normal to r at P is n(a, b) =r1 × r2

‖r1 × r2‖ (evaluated at (a, b)).

18

Remark : It follows from the proof of Lemma 4.3 that the tangent plane is an intrinsic invariant : Let r : U �→ R3 be

a simple surface, f : V �→ U a coordinate transformation with f(c, d) = (a, b). Put P = r(a, b) = (r ◦ f)(c, d). Thenthe tangent plane to r(u1, u2) at P is the same as the tangent plane to (r ◦ f)(v1, v2) at P . Similarly, the unit normalto r(u1, u2) at P is equal to the unit normal to r(v1, v2) at P up to sign.

Definition : Let r : U �→ R3 be a simple surface and P = r(a, b). A vector X ∈ R

3 is a tangent vector to r at P if Xis the velocity vector at P of some curve in r(U). This means, there exist ε > 0 and a curve α : (−ε, ε) �→ r(U) such

that α(0) = P anddα

dt(0) = X.

Remark : Another way of describing a curve α on the surface r(U) is by giving u1 and u2 as functions of a newvariable t. So we get two real-valued functions α1(t) and α2(t) and we put α(t) = r(α1(t), α2(t)).

Lemma 4.4 Let r : U �→ R3 be a simple surface and P = r(a, b). Then the set of all tangent vectors to r at P is a

vector space.

Proof : Let X and Y be tangent vectors to r at P and λ, µ ∈ R. We have to show that λX + µY is also a tangentvector to r at P . Since X and Y are tangent vectors to r at P , there exist curves α(t) and β(t) in r(U) such that

α(0) = β(0) = P ,dα

dt(0) = X and

dβ

dt(0) = Y. Hence there exist functions α1(t), α2(t), β1(t) and β2(t) such that

α(t) = r(α1(t), α2(t)) and β(t) = r(β1(t), β2(t)). Using the chain rule, we get that

X =dα

dt

∣∣∣∣t=0

=[

∂r∂u1

dα1

dt+

∂r∂u2

dα2

dt

]t=0

and Y =dβ

dt

∣∣∣∣t=0

=[

∂r∂u1

dβ1

dt+

∂r∂u2

dβ2

dt

]t=0

Putγ1(t) = λ(α1(t) − a) + µ(β1(t) − a) + a and γ2(t) = λ(α2(t) − b) + µ(β2(t) − b) + b

Then γ1(0) = a and γ2(0) = b. Put γ(t) = r(γ1(t), γ2(t)). Then for small values of t, γ is a curve in r(U) andγ(0) = r(a, b) = P . Using the chain rule, we get that

dγ

dt

∣∣∣∣t=0

=[

∂r∂u1

dγ1

dt+

∂r∂u2

dγ2

dt

]t=0

=[

∂r∂u1

(λ

dα1

dt+ µ

dβ1

dt

)+

∂r∂u2

(λ

dα2

dt+ µ

dβ2

dt

)]t=0

= λ

[∂r∂u1

dα1

dt+

∂r∂u2

dα2

dt

]t=0

+ µ

[∂r∂u1

dβ1

dt+

∂r∂u2

dβ2

dt

]t=0

= λX + µY

Hence λX + µY is a tangent vector to r at P . �

Definitions : Let r : U �→ R3 be a simple surface and P = r(a, b).

(a) The u1-curve at P is the curve α(u1) = r(u1, b). The u2-curve at P is the curve β(u2) = r(a, u2).

(b) A parametric curve on r is a u1-curve or u2-curve at some point P on r.

Theorem 4.5 Let r : U �→ R3 be a simple surface and P = r(a, b). The set of all tangent vectors to r at P is a vector

space of dimension two with basis {r1(a, b), r2(a, b)}.

19

Proof : We first show that r1(a, b) and r2(a, b) are tangent vectors to r at P . Put α1(t) = t + a, α2(t) = b andα(t) = r(α1(t), α2(t)). Then α(t) is a curve in r(U) and α(0) = r(a, b) = P . Using the chain rule, we get that

dα

dt

∣∣∣∣t=0

=[

∂r∂u1

dα1

dt+

∂r∂u2

dα2

dt

]t=0

=∂r∂u1

(α1(0), α2(0)) · 1 +∂r∂u2

(α1(0), α2(0)) · 0 =∂r∂u1

(a, b) = r1(a, b)

So r1(a, b) is a tangent vector to r at P . Similarly, r2(a, b) is a tangent vector to r at P .By Lemma 4.4, the set of all tangent vectors to r at P is a vector space. Since r is a regular surface, r1(a, b)×r2(a, b) �= 0and so r1(a, b) and r2(a, b) are linearly independent. So it is enough to show that every tangent vector to r at P is alinear combination of r1(a, b) and r2(a, b).

Let X be a tangent vector to r at P . Then there exists a curve γ(t) in r(U) such that γ(0) = P anddγ

dt

∣∣∣∣t=0

= X.

Hence there exists functions γ1(t) and γ2(t) such that γ(t) = r(γ1(t), γ2(t)). Put λ =dγ1

dt(0) and µ =

dγ2

dt(0). Then

λ and µ are constants. Using the chain rule, we get that

dγ

dt

∣∣∣∣t=0

=[

∂r∂u1

dγ1

dt+

∂r∂u2

dγ2

dt

]t=0

= λ∂r∂u1

(γ1(0), γ2(0)) + µ∂r∂u2

(γ1(0), γ2(0)) = λ r1(a, b) + µ r2(a, b)

Hence X is a linear combination of r1(a, b) and r2(a, b). �

Lemma 4.6 Let r : U �→ R3 be a simple surface, f : V �→ U a coordinate transformation and f(c, d) = (a, b). Put

P = r(a, b). If X = λ∂r∂v1

+ µ∂r∂v2

then X = λ∂r∂u1

+ µ∂r∂u2

where

[λµ

]=

∂f1

∂v1

∂f1

∂v2

∂f2

∂v1

∂f2

∂v2

[ λµ

]= J(f)

[λµ

]

Proof : Note that r(v1, v2) = r(f1(v1, v2), f2(v1, v2)). Using the chain rule, we get that

X = λ∂r∂v1

+µ∂r∂v2

= λ

(∂r∂u1

∂f1

∂v1+

∂r∂u2

∂f2

∂v1

)+µ

(∂r∂u1

∂f1

∂v2+

∂r∂u2

∂f2

∂v2

)=(

∂f1

∂v1λ +

∂f1

∂v2µ

)∂r∂u1

+(

∂f2

∂v1λ +

∂f2

∂v2µ

)∂r∂u2

This is exactly what we get when we work out the matrix multiplication. �

4.2 Surfaces

Definitions :

(a) Let M be a subset of R3, P ∈ M and ε > 0. The ε-neighborhood of P is the set {Q ∈ M | d(P, Q) < ε} where d

stands for the distance function.

(b) Let M be a subset of R3, P ∈ M and g : M �→ R

2 a function. Then g is continuous at P if for every open subsetU of R

2 with g(P ) ∈ U there exists an ε-neighborhood M of P with g(M) ⊂ U .

(c) Let r : U �→ R3 be a coordinate patch. Then r is proper if the inverse function r−1 : r(U) �→ U is continuous at

each point of r(U).

(d) Let M be a subset of R3. Then M is a Ck surface if the following holds :

(1) For each point P ∈ M there is a proper Ck coordinate patch whose image is in M and which contains anε-neighborhood of P for some ε > 0.

(2) Suppose that r : U �→ R3 and s : V �→ R

3 are two such coordinate patches. Put U ′ = r(U) and V ′ = s(V).Then s−1 ◦ r : r−1(U ′ ∩ V ′) �→ s−1(U ′ ∩ V ′) is a Ck coordinate transformation .

20

Theorem 4.7 Let f : R3 �→ R

3 be a differentiable function. Put M = {(x, y, z) ∈ R3 | f(x, y, z) = 0} . Suppose that(

∂f

∂x,∂f

∂y,∂f

∂z

)�= 0 for all (x, y, z) ∈ M . Then M is a surface.

Proof : This proof uses the Implicit Function Theorem from advanced calculus. �

4.3 The First Fundamental Form and Arc Length

Definitions :

(a) Let r : U �→ R3 be a proper coordinate patch. We define

gij(u1, u2) = 〈ri(u1, u2), rj(u1, u2)〉 or gij = 〈ri, rj〉 for i, j = 1, 2

g = det([

g11 g12

g21 g22

])= det(gij)

gkl is the (k, l)-th entry of the inverse of (gij) for k, l = 1, 2

(b) Let M be a surface and P a point on M . The tangent space of M at P is the set TP M of all tangent vectors toM at P .

(c) The rule which assigns to any two vectors X,Y ∈ TP M their inner product 〈X,Y〉 is called the first fundamentalform of the surface M . If X = X1r1 + X2r2 and Y = Y1r1 + Y2r2 then we have the following formulas for thefirst fundamental form :

I(X,Y) =2∑

i,j=1

gijXiYj = [ X1 X2 ][

g11 g12

g21 g22

] [Y1

Y2

]

Remark : Since (gij) is the matrix representing the inner product restricted to the tangent space, we get that (gij)is a non-singular positive definite matrix. So g11 > 0, g22 > 0, g12 = g21 and g > 0.

Lemma 4.8 Let r : U �→ R3 be a proper coordinate patch. Then the following holds :

(a) g = ‖r1 × r2‖2

(b) g11 =g22

g, g22 =

g11

gand g12 = g21 = −g12

g

(c)2∑

k=1

gikgkj = δij for i, j = 1, 2

Proof : (a) Let θ be the angle between r1 and r2. Then

‖r1 × r2‖2 = ‖r1‖2‖r2‖2 sin2(θ)

= ‖r1‖2‖r2‖2(1 − cos2(θ))

= ‖r1‖2‖r2‖2

(1 − 〈r1, r2〉2

‖r1‖2‖r2‖2

)

= ‖r1‖2‖r2‖2 − 〈r1, r2〉2

= g11g22 − g12g21

= g

21

(b) Recall that[

a bc d

]−1

=1

ad − bc

[d −b−c a

]if ad − bc �= 0. Since g12 = g21, we have that

[g11 g12

g21 g22

]=[

g11 g12

g21 g22

]−1

=1g

[g22 −g12

−g12 g11

]

(c) For i, j = 1, 2, we have that2∑

k=1

gikgkj is the (i, j)-th entry of the matrix[

g11 g12

g21 g22

] [g11 g12

g21 g22

]. Since (gkl)

is the inverse of (gij), we get that this matrix product is[

1 00 1

]. �

IMPORTANT NOTATIONS :

Let r : U �→ R3 be a proper coordinate patch and f : V �→ U a coordinate transfor-

mation. Then we use the following notations :

gij = 〈ri, rj〉, (gkl) = (gij)−1 and g = det(gij)

s = r ◦ f , J = J(f) =(

∂ui

∂vj

)and J−1 = (J(f))−1 = J(f−1) =

(∂vi

∂uj

)gαβ = 〈sα, sβ〉, (gγδ) = (gαβ)−1 and g = det(gαβ)

We want to derive a relation between (gij) and (gαβ).

We have that r(u1, u2) = s(v1(u1, u2), v2(u1, u2)). Using the chain rule, we get that ri =2∑

α=1

sα∂vα

∂uifor i = 1, 2. Hence

gij = 〈ri, rj〉 =

⟨2∑

α=1

sα∂vα

∂ui,

2∑β=1

sβ∂vβ

∂uj

⟩=

2∑α,β=1

〈sα, sβ〉 ∂vα

∂ui

∂vβ

∂uj=

2∑α,β=1

gαβ∂vα

∂ui

∂vβ

∂uj

We can rewrite this in matrix form : g11 g12

g21 g22

=

∂v1

∂u1

∂v2

∂u1

∂v1

∂u2

∂v2

∂u2

g11 g12

g21 g22

∂v1

∂u1

∂v1

∂u2

∂v2

∂u1

∂v2

∂u2

or (gij) = J−t(gαβ)J−1

Taking determinants, we find that

g = det(gij) = det(J−t) det(gαβ) det(J−1) = g det(J−1)2 =g

det(J)2

Taking inverses, we find that

(gkl) = (gij)−1 = J(gαβ)−1J t = J(gγδ)J t or gkl =2∑

γ,δ=1

gγδ ∂uk

∂vγ

∂ul

∂vδfor k, l = 1, 2

(gij) = J−t(gαβ)J−1 or gij =2∑

α,β=1

gαβ∂vα

∂ui

∂vβ

∂ujfor i, j = 1, 2

(gkl) = J(gγδ)J t or gkl =2∑

γ,δ=1

gγδ ∂uk

∂vγ

∂ul

∂vδfor k, l = 1, 2

g =g

det(J)2

22

Reversing the role of (u1, u2) and (v1, v2), we get that

(gαβ) = J t(gij)J or gαβ =2∑

i,j=1

gij∂ui

∂vα

∂uj

∂vβfor α, β = 1, 2

(gγδ) = J−1(gkl)J−t or gγδ =2∑

k,l=1

gkl ∂vγ

∂uk

∂vδ

∂ulfor γ, δ = 1, 2

g = g det(J)2

4.4 Normal Curvature, Geodesic Curvature and Gauss’s Formulas

Definitions : Let r : U �→ R3 be a coordinate patch.

(a) If P = r(a, b) and M = r(U) then the normal space to M at P is the vector space NP M spanned by n(a, b). SoNP M = {λn |λ ∈ R}.

(b) Let γ(s) = r(γ1(s), γ2(s)) be a unit speed curve in r(U).

(1) The intrinsic normal of γ is S(s) = n(γ1(s), γ2(s)) × T(s) or S = n × T.

(2) The geodesic curvature of γ is κg(s) = 〈γ′′(s),S(s)〉.(3) The normal curvature of γ is κn(s) = 〈γ′′(s),n(s)〉.

Remarks :

(a) We have that R3 = TP M ⊕ NP M . So every w ∈ R

3 can be written uniquely as w = u + v with u ∈ TP M and

v ∈ NP M . Since {r1, r2} is a basis for TP M and n =r1 × r2

‖r1 × r2‖ is a basis for NP M , we have that 〈u,v〉 = 0

for all u ∈ TP M and all v ∈ NP M . Also, every w ∈ R3 can be written uniquely as w = ar1 + br2 + cn with

a, b, c ∈ R.

(b) Note that S ∈ TP M . Since T ∈ TP M , we have that 〈n,T〉 = 0. Hence ‖S‖ = 1. So S is a unit vector that istangent to the surface M but normal to the curve γ.

(c) The geodesic curvature is independend of the coordinate patch we’re using : it only depends on the geometricshape of the curve.

Lemma 4.9 Let r : U �→ R3 be a coordinate patch and γ(s) = r(γ1(s), γ2(s)) a unit speed curve in r(U). Then the

following holds :

(a) γ′′(s) = κg(s)S(s) + κn(s)n(s)

(b) κ2(s) = κ2g(s) + κ2

n(s)

Proof : (a) We write γ ′′(s) = X(s) + λ(s)n(s) where X(s) ∈ TP M and λ(s) ∈ R for all s. By the Frenet-Serretequations, we get that γ′′(s) = T′(s) = κ(s)N(s). So

0 = 〈κ(s)N(s),T(s)〉 = 〈γ′′(s),T(s)〉 = 〈X(s) + λ(s)n(s),T(s)〉 = 〈X(s),T(s)〉 + λ(s) 〈n(s),T(s)〉 = 〈X(s),T(s)〉

since 〈n(s),T(s)〉 = 0. Since X(s) ∈ TP M , we have that 〈X(s),n(s)〉 = 0. So X is perpendicular to both n(s)and T(s). Hence X(s) is parallel to n(s) × T(s) = S(s). So X(s) = µ(s)S(s) where µ(s) ∈ R for all s. Henceγ′′(s) = µ(s)S(s) + λ(s)n(s). Note that 〈S(s),n(s)〉 = 0. So

κg(s) = 〈γ ′′(s),S(s)〉 = µ(s) and κn(s) = 〈γ ′′(s),n(s)〉 = λ(s)

23

Hence γ ′′(s) = κg(s)S(s) + κn(s)n(s).(b) Since γ′′(s) = κ(s)N(s), we get that κ2(s) = ‖γ′′(s)‖2. By (a), we get that

κ2 = ‖γ′′‖2 = 〈κgS + κnn, κgS + κnn〉 = κ2g 〈S,S〉 + 2κgκn 〈S,n〉 + κ2

n 〈n,n〉 = κ2g + κ2

n �

Definitions : Let r : U �→ R3 be a coordinate patch.

(a) We put rij =∂r

∂ui∂ujfor i, j = 1, 2. Note that rij = rji.

(b) The coefficients of the second fundamental form are the functions Lij = 〈rij ,n〉 for i, j = 1, 2. Note thatLij = Lji.

(c) The second fundamental form of the surface M is the function II : TP M×TP M �→ R : (X,Y) �→∑2i,j=1 LijXiYj

where X = X1r1 + X2r2 and Y = Y1r1 + Y2r2.

(d) The Christoffel symbols are the functions Γkij =

∑2l=1 〈rij , rl〉 glk for i, j, k = 1, 2. Note that Γk

ij = Γkji.

Proposition 4.10 Let r : U �→ R3 be a coordinate patch. Then the following holds :

(a) (Gauss’s formulas) rij = Lijn +2∑

k=1

Γkijrk for i, j = 1, 2

(b) If γ(s) = r(γ1(s), γ2(s)) is a unit speed curve, then

κn =2∑

i,j=1

Lijγ′iγ

′j and κgS =

2∑k=1

γ′′k +

2∑i,j=1

Γkijγ

′iγ

′j

rk

Proof : (a) Pick i, j, k ∈ {1, 2}. Then there exist a, b1, b2 ∈ R such that rij = an + b1r1 + b2r2 = an +2∑

m=1

bmrm

Note that 〈rm,n〉 = 0 for m = 1, 2. Hence Lij = 〈rij ,n〉 = 〈an,n〉 = a. Since 〈n, rl〉 = 0 for l = 1, 2, we get

that 〈rij , rl〉 =2∑

m=1

bm 〈rm, rl〉 =2∑

m=1

bmgml for l = 1, 2. Multiplying this by glk and summing over l, we get that

Γkij =

2∑l=1

〈rij , rl〉 glk =2∑

m,l=1

bmgmlglk =

2∑m=1

bm

(2∑

l=1

gmlglk

). Note that

2∑l=1

gmlglk is the (m, k)-th entry of the

matrix (gml)(glk). Since these are inverse matrices, the (m, k)-th entry is δmk. So Γkij =

2∑m=1

bmδmk = bk. Hence

rij = Lijn + Γ1ijr1 + Γ2

ijr2.

(b) Since γ(s) = r(γ1(s), γ2(s)), we get that γ′ =dγ

ds=

∂r∂u1

dγ1

ds+

∂r∂u2

dγ2

ds= r1γ

′1 + r2γ

′2 =

2∑i=1

riγ′i. Note that

ri(s) = ri(γ1(s), γ2(s)) for i = 1, 2. Using the chain rule, we get thatdri

ds=

∂ri

∂u1

dγ1

ds+

∂ri

∂u2

dγ2

ds= ri1γ

′1 + ri2γ

′2 =

2∑j=1

rijγ′j

for i = 1, 2. Hence

γ′′ =dγ ′

ds=

2∑i=1

(dri

dsγ′

i + ridγ′

i

ds

)=

2∑i,j=1

rijγ′iγ

′j +

2∑i=1

riγ′′i =

2∑i,j=1

rijγ′iγ

′j +

2∑k=1

rkγ′′k

Using Gauss’s formulas, we get that

γ′′ =2∑

i,j=1

(Lijn +

2∑k=1

Γkijrk

)γ′

iγ′j +

2∑k=1

rkγ′′k =

2∑i,j=1

Lijγ′iγ

′j

n +2∑

k=1

γ′′k +

2∑i,j=1

Γkijγ

′iγ

′j

rk

24

By Lemma 4.9(a), we get that

κnn + κgS = γ′′ =

2∑i,j=1

Lijγ′iγ

′j

n +2∑

k=1

γ′′k +

2∑i,j=1

Γkijγ

′iγ

′j

rk

Since {n, r1, r2} is a basis for R3 and S is a linear combination of r1 and r2 (S is a tangent vector), we get that

κn =2∑

i,j=1

Lijγ′iγ

′j and κgS =

2∑k=1

γ′′k +

2∑i,j=1

Γkijγ

′iγ

′j

rk �

Definitions : Let r : U �→ R3 be a coordinate patch.

(a) A property of r is intrinsic if it only depends on (gij).

(b) Let γ(s) = r(γ1(s), γ2(s)) be a unit speed curve in r(U). A property of γ is intrinsic if it only depends on (gij)and γ1 and γ2.

Proposition 4.11 Let r : U �→ R3be a coordinate patch. Then

Γkij =

12

2∑l=1

(∂gil

∂uj− ∂gij

∂ul+

∂glj

∂ui

)glk for i, j, k = 1, 2

So the Christoffel symbols are intrinsic.

Proof : We get that

∂gij

∂ul=

∂

∂ul〈ri, rj〉 =

⟨∂ri

∂ul, rj

⟩+⟨ri,

∂rj

∂ul

⟩= 〈ril, rj〉 + 〈ri, rjl〉 for i, j, l = 1, 2

Permuting the indices around, we get that

∂gij

∂ul= 〈ril, rj〉 + 〈ri, rjl〉 ,

∂gil

∂uj= 〈rij , rl〉 + 〈ri, rlj〉 and

∂gjl

∂ui= 〈rji, rl〉 + 〈rj , rli〉

Hence we find that Γkij =

2∑l=1

〈rij , rl〉 glk =2∑

l=1

12

(∂gil

∂uj− ∂gij

∂ul+

∂glj

∂ui

)glk. �

4.5 Geodesics

Definition : A geodesic on a surface M is a regular (unit speed) curve on M with geodesic curvature equal to zeroeverywhere (so κg ≡ 0).

Proposition 4.12 Let γ(t) be a regular curve on M . Then γ is a geodesic if and only if [n, γ, γ] ≡ 0.

Proof : Let s be the arc length on γ. Using the chain rule, we get that

γ =dγ

dt=

dγ

ds

ds

dt= γ ′s and γ =

d

dt(γ ′s) = γ ′s + γ ′′s2

Hence we get that γ × γ = (γ′s) × (γ ′s + γ ′′s2) = ss γ′ × γ′ + s3 γ′ × γ′′ = s3 γ ′ × γ ′′. So

[n, γ, γ] = 〈γ × γ,n〉 =⟨s3 γ ′ × γ ′′,n

⟩= s3[n, γ ′, γ′′] = s3[n,T, γ ′′] = s3 〈n × T, γ ′′〉 = s3 〈S, γ′′〉 = s3κg

Note that s = ‖γ‖ > 0. So κg ≡ 0 if and only if [n, γ, γ] ≡ 0. �

25

Proposition 4.13 Let r : U �→ R3be a a coordinate patch and γ(s) = r(γ1(s), γ2(s)) a unit speed curve on r(U). Then

γ is a geodesic if and only if

γ′′k +

2∑i,j=1

Γkijγ

′iγ

′j ≡ 0 for k = 1, 2

Proof : By Proposition 4.10(b), we have that κgS =2∑

k=1

γ′′k +

2∑i,j=1

Γkijγ

′iγ

′j

rk. Since S is a unit vector and {r1, r2}

are linearly independent, we get that κg ≡ 0 if and only if γ′′k +

2∑i,j=1

Γkijγ

′iγ

′j ≡ 0 for k = 1, 2. �

Proposition 4.14 Let γ(s) be a unit speed curve on a surface M . Then γ is a geodesic if and only if γ′′(s) is normalto the surface for all s (so γ′′ is a multiple of n).

Proof : By Lemma 4.9(a), we have that γ′′(s) = κg(s)S(s) + κn(s)n(s). Note that S(s) and n(s) are linearlyindependent. So γ′′(s) is a multiple of n(s) for all s if and only if κg(s) = 0 for all s. �

Theorem 4.15 Let P be a point on a surface M , X a unit tangent vector to M at P (so X ∈ TP M and ‖X‖ = 1)and s0 ∈ R. Then there exists a unique geodesic γ(s) such that γ(s0) = P , γ ′(s0) = X and γ(s) is a unit speed curve.

Proof : The proof uses Picard’s Theorem for solving differential equations. �

Theorem 4.16 Let M be a surface, P and Q two points on M and γ(s) a unit speed curve going through P and Q.If γ is the shortest curve on M between P and Q, then γ is a geodesic.

Proof : Assume that P = γ(a) and Q = γ(b) with a < b. Let κg(s) be the geodesic curvature of γ. Pick s0 ∈ (a, b).We will show that κg(s0) = 0.Suppose that κg(s0) �= 0. Then there exist real numbers c and d and a coordinate patch r : U �→ R

3such thata < c < s0 < d < b and κg(s) �= 0 for all s ∈ [c, d] and γ([c, d]) ⊂ r(U). Then γ(s) = r(γ1(s), γ2(s)) for some functionsγi(s) : [c, d] �→ R for i = 1, 2. Let λ(s) : [c, d] �→ R be a C2 function such that λ(c) = λ(d) = 0, λ(s0) �= 0 andλ(s)κg(s) > 0 for all s ∈ (c, d) (for example λ(s) = (s − c)(d − s)κg(s) will work if γ is of class C4). Then there

exist functions vi(s) : [c, d] �→ R for i = 1, 2 such that λ(s)S(s) =2∑

i=1

vi(s)ri(γ1(s), γ2(s)). Since {r1, r2} are linearly

independent and λ(c) = λ(d) = 0, we get that vi(c) = vi(d) = 0 for i = 1, 2. Put f(t, s) = r(γ1(s) + t v1(s), γ2(s) +t v2(s)) for all s ∈ [c, d] and |t| small enough, say t ∈ (−ε, ε). Note that f(t, c) = r(γ1(c), γ2(c)) = γ(c). Similarly,f(t, d) = γ(d). So for t ∈ (−ε, ε), f(t, s) is a curve on M joining γ(c) and γ(d). Note that f(0, s) = γ(s). Let L(t) be

the length of the curve f(t, s) between γ(c) and γ(d). Then L(t) =∫ d

c

∥∥∥∥∂f

∂s

∥∥∥∥ ds =∫ d

c

⟨∂f

∂s,∂f

∂s

⟩ 12

ds. Hence

L′(t) =d

dt

(∫ d

c

⟨∂f

∂s,∂f

∂s

⟩ 12

ds

)∂

∂t

(⟨∂f

∂s,∂f

∂s

⟩ 12)

ds =

⟨

∂2f

∂s ∂t,∂f

∂s

⟩⟨

∂f

∂s,∂f

∂s

⟩ 12

ds

Note that⟨

∂f

∂s,∂f

∂s

⟩ 12

∣∣∣∣∣t=0

=⟨

dγ

ds,dγ

ds

⟩ 12

= 1 since γ is a unit speed curve. So

L′(0) =∫ d

c

⟨∂2f

∂s ∂t,∂f

∂s

⟩∣∣∣∣t=0

ds =∫ d

c

[d

ds

(⟨∂f

∂t,∂f

∂s

⟩∣∣∣∣t=0

)−⟨

∂f

∂t,∂2f

∂s2

⟩∣∣∣∣t=0

]ds

=[⟨

∂f

∂t,∂f

∂s

⟩∣∣∣∣t=0

]s=d

s=c

−∫ d

c

⟨∂f

∂t,∂2f

∂s2

⟩∣∣∣∣t=0

ds

26

We get that∂f

∂t(0, s) =

2∑i=1

ri(γ1(s), γ2(s))vi(s) = λ(s)S(s). But λ(c) = λ(d) = 0. So[⟨

∂f

∂t,∂f

∂s

⟩∣∣∣∣t=0

]s=d

s=c

= 0. Also,

∂2f

∂s2

∣∣∣∣t=0

=d2f(0, s)

ds2=

d2γ(s)ds2

= γ′′(s) = κg(s)S(s) + κn(s)n(s) by Lemma 4.9(a). So

⟨∂f

∂t,∂2f

∂s2

⟩∣∣∣∣t=0

= 〈λ(s)S(s), κg(s)S(s) + κn(s)n(s)〉 = λ(s)κg(s) > 0 for all s ∈ (c, d)

Hence L′(0) < 0. Since γ is the shortest curve on M joining P and Q, γ is also the shortest curve on M joining γ(c)and γ(d). But f(0, s) = γ(s). So L(t) has a minimum at t = 0. Hence L′(0) = 0, a contradiction.So κg(s0) = 0. Since s0 was arbitrary, we get that κg(s) = 0 for all s ∈ (a, b). �

4.6 Parallel Vector Fields

Definitions : Let M be a surface, γ(t) : [a, b] �→ M a regular curve on M and X(t) : [a, b] �→ R3 a vector field.

(a) X(t) is a vector field along γ if X(t) ∈ Tγ(t)M for all t ∈ [a, b].

(b) X(t) is parallel along γ(t) if X(t) is a vector field along γ anddXdt

is perpendicular to M for all t ∈ [a, b] (sodXdt

is a multiple of n(t) for all t ∈ [a, b]).

Proposition 4.17 Let r : U �→ R3 be a coordinate patch, γ(t) = r(γ1(t), γ2(t)) a regular curve in r(U) and X(t) =

X1(t)r1(t) + X2(t)r2(t) a vector field along γ. Then X is parallel along γ if and only if

dXk

dt+

2∑i,j=1

ΓkijXi

dγj

dt≡ 0 for k = 1, 2

Proof : We have that

X(t) is parallel along γ ⇐⇒⟨

dXdt

, rl

⟩≡ 0 for l = 1, 2

⇐⇒⟨

d

dt

(2∑

i=1

Xiri

), rl

⟩≡ 0 for l = 1, 2

⇐⇒⟨

2∑i=1

(dXi

dtri + Xi

dri

dt

), rl

⟩≡ 0 for l = 1, 2

⇐⇒2∑

i=1

dXi

dt〈ri, rl〉 + Xi

⟨2∑

j=1

rijdγj

dt, rl

⟩ ≡ 0 for l = 1, 2

⇐⇒2∑

i=1

dXi

dtgil +

2∑i,j=1

〈rij , rl〉Xidγj

dt≡ 0 for l = 1, 2 (∗)

Suppose first that X(t) is parallel along γ. Then (*) holds. Multiplying (*) by glk and summing over l, we get that

0 ≡2∑

i,l=1

dXi

dtgilg

lk +2∑

i,j,l=1

〈rij , rl〉 glkXidγj

dt=

2∑i=1

dXi

dtδik +

2∑i,j=1

ΓkijXi

dγj

dt=

dXk

dt+

2∑i,j=1

ΓkijXi

dγj

dtfor k = 1, 2

27

Suppose next thatdXk

dt+

2∑i,j=1

ΓkijXi

dγj

dt≡ 0 for k = 1, 2. Multiplying this by gkl and summing over k we get that

0 ≡2∑

k=1

dXk

dtgkl +

2∑i,j,k=1

ΓkijgklXi

dγj

dtfor l = 1, 2

=2∑

i=1

dXi

dtgil +

2∑i,j,k,p=1

〈rij , rp〉 gpkgklXidγj

dtfor l = 1, 2

=2∑

i=1

dXi

dtgil +

2∑i,j,p=1

〈rij , rp〉 δplXidγj

dtfor l = 1, 2

=2∑

i=1

dXi

dtgil +

2∑i,j

〈rij , rl〉Xidγj

dtfor l = 1, 2

So (*) holds. Hence X(t) is parallel along γ. �

Theorem 4.18 Let M be a surface, γ(t) : [a, b] �→ R3 a regular curve on M , t0 ∈ [a, b] and X ∈ Tγ(t0)M . Then there

exists a unique vector field X(t) : [a, b] �→ R3 that is parallel along γ with X(t0) = X.

Proof : �

Definition : That unique vector field X(t) is called the parallel translate of X along γ.

Proposition 4.19 Let M be a surface and γ(t) : [a, b] �→ R3 a regular curve on M . Then the following holds ;

(a) If X(t) is a vector field parallel along γ, then ‖X(t)‖ is constant.

(b) If X(t) and Y(t) are vector fields parallel along γ, then the angle between X(t) and Y(t) is constant.

Proof : Let X(t) and Y(t) be vector fields parallel along γ. Since Y(t) ∈ Tγ(t)M anddXdt

is a multiple of n(t), we

have that⟨

dXdt

,Y(t)⟩

= 0 for all t ∈ [a, b]. Similarly,⟨

dYdt

,X(t)⟩

= 0 for all t ∈ [a, b]. Hence

0 ≡⟨

dXdt

,Y⟩

+⟨X,

dYdt

⟩=

d

dt(〈X,Y〉) (∗)

(a) Substituting Y = X into (*), we get that 0 ≡ d

dt(〈X,X〉) =

d

dt

(‖X(t)‖2). Hence ‖X(t)‖ is constant.

(b) Let θ(t) be the angle between X(t) and Y(t) for all t ∈ [a, b]. From (*), it follows that 〈X(t),Y(t)〉 is constant.But 〈X(t),Y(t)〉 = ‖X(t)‖‖Y(t)‖ cos(θ(t)) for all t ∈ [a, b]. From (a), we get that ‖X(t)‖ and ‖Y(t)‖ are constant.Hence cos(θ(t)) is constant and so also θ(t) is constant. �

Proposition 4.20 Let M be a surface and γ(s) : [a, b] �→ R3 a unit-speed curve on M . Then γ′(s) is parallel along

γ if and only if γ is a geodesic.

Proof : We have that γ ′(s) is parallel along γ if and only if γ′′(s) is a multiple of n(s) for all s ∈ [a, b]. By Proposition4.14, this happens if and only if γ is a geodesic. �

28

4.7 The Weingarten Map

Definition : Let M be a surface, P a point on M , f a differentiable function defined on a neighborhood of P andX ∈ TP M . We define the directional derivative of f in the direction of X (notation : DXf) as follows :

Let α(t) be a curve on M with α(0) = P anddα

dt(0) = X. Then DXf =

d

dt(f ◦ α)(0).

Remark : We will prove in the next proposition that DXf is well-defined, namely that DXf does not depend on thechoice of α.

Proposition 4.21 Let M be a surface, P ∈ M , f a differentiable function defined on a neighborhood of P , r : U �→ R3

a coordinate patch of M with r(a, b) = P and X ∈ TP M . If X =2∑

i=1

Xiri, then DXf =2∑

i=1

Xi∂

∂ui(f ◦ r)

∣∣∣∣(a,b)

.

Proof : Let α(t) be a curve on M with α(0) = P anddα

dt(0) = X. Put α(t) = r(α1(t), α2(t)). Then (f ◦ α)(t) =

(f ◦ r)(α1(t), α2(t)). Hence

d

dt(f ◦ α)(t) =

2∑i=1

∂

∂ui(f ◦ r)

∣∣∣∣(α1(t),α2(t))

dαi

dt(t) (∗)

Note that r(α1(0), α2(0)) = α(0) = P = r(a, b). So α1(0) = a and α2(0) = b. Since

2∑i=1

Xiri(a, b) = X =dα

dt(0) =

d

dt(r(α1(t), α2(t)))|t=0 =

2∑i=1

ri(α1(0), α2(0))dαi

dt(0) =

2∑i=1

dαi

dt(0)ri(a, b)

we have thatdαi

dt(0) = Xi for i = 1, 2. Substituting t = 0 into (*), we get

d

dt(f ◦ α)(0) =

2∑i=1

∂

∂ui(f ◦ r)

∣∣∣∣(a,b)

Xi. �

Definitions : Let M be a surface and P a point on M .

(a) Let f = (f1, f2, f3) be a differentiable, vector-valued function defined on a neighborhood of P and X ∈ TP M .Then the directional derivative of f in the direction of X is DXf = (DXf1, DXf2, DXf3).

(b) The Weingarten Map at P is the function L : TP M �→ R3 : X �→ −DXn

(c) Lij =2∑

l=1

gilLlj for i, j = 1, 2 (so[

L11 L12

L21 L22

]=[

g11 g12

g21 g22

] [L11 L12

L21 L22

]=[

g11 g12

g21 g22

]−1 [L11 L12

L21 L22

]).

Remark : Since n only depends (up to a sign) on the surface M and not on a particular coordinate patch, the sameis true for the Weingarten Map.

Theorem 4.22 Let r : U �→ R3 be a coordinate patch and P = r(a, b). Then the following holds :

(a) The Weingarten map L is a linear transformation from TP M to TP M .

(b) (Weingarten equations) L(rj) = − ∂n∂uj

= −nj =2∑

i=1

Lijri for j = 1, 2

(c) If X =2∑

j=1

Xjrj ∈ TP M then L(X) =2∑

i,j=1

LijXjri. So in terms of the basis {r1, r2} of TP M , we have that

L(X) =[

L11 L12

L21 L22

] [X1

X2

]

29

Proof : We will show that L(r1) = −n1. Put α1(t) = a + t and α2(t) = b). Put α(t) = r(α1(t), α2(t)) = r(a + t, b).

Then α(t) is a curve on r(U), α(0) = r(a, b) = P anddα

dt(0) =

2∑i=1

ri(a, b)dαi

dt(0) = r1(a, b). By definition of directional

derivative, we get that

L(r1) = −Dr1n = − d

dt(n ◦ α)(t))

∣∣∣∣t=0

= − d

dt(n(a + t, b))

∣∣∣∣t=0

= −2∑

i=1

∂n∂ui

∣∣∣∣(a,b)

dαi

dt(0) = −n1(a, b)

Similarly, we get that L(r2) = −n2.(a) We have that L is a linear map from TP M to R

3. We need to show that the image of L is contained in TP M .Since L is linear and {r1, r2} is a basis for TP M , we only need to show that L(ri) ∈ TP M for i = 1, 2. Pick i ∈ {1, 2}.Note that 〈n,n〉 ≡ 1. Deriving this with respect to ui, we get that 0 = 〈ni,n〉 + 〈n,ni〉. So 〈ni,n〉 = 0 and ni isperpendicular to n. Hence ni ∈ TP M . So L(ri) = −ni ∈ TP M , which proves (a).

(b) Pick j ∈ {1, 2}. Then L(rj) =∑2

i=1 airi for some a1, a2 ∈ R. For k = 1, 2, we have that 〈n, rk〉 = 0. Deriving thiswith respect to uj, we get that

0 = 〈nj , rk〉 + 〈n, rkj〉 = 〈−L(rj), rk〉 + Lkj = Lkj −⟨

2∑i=1

airi, rk

⟩= Lkj −

2∑i=1

aigik for k = 1, 2

So Lkj =2∑

i=1

aigik for k = 1, 2. Multiplying this by gkl and summing over k, we get that

Llj =2∑

k=1

glkLkj =2∑

k=1

gklLkj =2∑

i,k=1

aigikgkl =2∑

i=1

ai

(2∑

k=1

gikgkl

)=

2∑i=1

aiδil = al for l = 1, 2

Hence L(rj) =2∑

l=1

alrl =2∑

l=1

Lljrl.

(c) Suppose that X =2∑

j=1

Xjrj . Then L(X) =2∑

j=1

XjL(rj) =2∑

i,j=1

LijXjri. �

Remark : When we were working with regular curves, {T,N,B} was an orthonormal basis for R3 at every point onthe curve. The Frenet-Serret equations gave us {T′,N′,B′} in terms of {T,N,B}. With surfaces, we still have that{r1, r2,n} is a basis for R

3 at every point on the surface. Gauss’s formulas (Proposition 4.10(a)) and the Weigartenequations (Theorem 4.22(b)) are the equivalent of the Frenet-Serret equations : for j = 1, 2, they give us the partialderivatives of {r1, r2,n} in terms of {r1, r2,n} :

for j = 1, 2 we have that{

rij = Lijn + Γ1ijr1 + Γ2

ijr2 for i = 1, 2nj = −L1jr1 − L2jr2

4.8 Gaussian Curvature and Normal Curvatures

Definition : Let M be a surface and P ∈ M . The Gaussian curvature of M at P (notation : K) is the determinantof the Weingarten map L at P .

Remark : Since the Weingarten map is determined by M up to a sign and det(L) = det(−L) for any two-by-twomatrix A, we get that the Gaussian curvature K is uniquely determined by the surface M . What is truly remarkable,is that K is actually intrinsic : it can be calculated using (gij).

Theorem 4.23 (Gauss’s Theorema Egregium) The Gaussian curvature K of a surface is intrinsic.

30

Proof : �

Remark : To calculate the Gaussian curvature K, we don’t need to know the Weingarten map :

K = det(Lij) = det((gkl)(Lij)) = det((gij)−1)(Lij)) =det([

L11 L12

L21 L22

])det([

g11 g12

g21 g22

])

Another invariant of a linear transformation are it’s eigenvalues. The eigenvalues of the Weingarten map L at P aredetermined by M up to a sign. We will see that these values are bounds for the normal curvature at P of any curveon the surface M through the point P .

Let r : U �→ R3 be a coordinate patch and P = r(a, b). If X ∈ TP M , we write X = X1r1 +X2r2. We use the following

notations :

1 =[

1 00 1

], G =

[g11 g12

g21 g22

], L =

[L11 L12

L21 L22

]and W =

[L11 L12

L21 L22

]Recall the first and second fundamental form :

• I(X,Y) = 〈X,Y〉 =[

X1 X2

]G

[Y1

Y2

]for all X,Y ∈ TP M

• II(X,Y) =[

X1 X2

]L

[Y1

Y2

]= 〈L(X),Y〉 = 〈X, L(Y)〉 for all X,Y ∈ TP M .

Since the Weingarten map is self-adjoint, the matrix W will have two real eigenvalues, say κ1 and κ2 (it is possbilethat κ1 = κ2). If α is a curve in r(U) going through P , what are the possible values that the normal curvature of αat P can be? There will be a minimum and maximum value that this normal curvature can take on. If α is a unitspeed curve, then the normal curvature of α is II(α′, α′). So we want to answer the following question :

What are the minimum and maximum values of II(X,X) where X is a unit vector in TP M?

We’ll use the method of Lagrange multipliers to answer this question. Put f(X1, X2, λ) = II(X,X) − λ(〈X,X〉 − 1).

Then a minimum/maximum can only occur when

∂f

∂X1= 0

∂f

∂X2= 0

〈X,X〉 = 1

We have that f(X1, X2, λ) = II(X,X) − λ 〈X,X〉 + λ = 〈L(X),X〉 − 〈λX,X〉 + λ = 〈L(X) − λX,X〉 + λ. In matrixform, this becomes

f(X1, X2, λ) = λ +[

X1 X2

]G(W − λ1)

[X1

X2

]Note that G(W − λ1) = GW − λG = GG−1L − λG = L − λG =

[L11 − λg11 L12 − λg12

L21 − λg21 L22 − λg22

]is symmetric since

g12 = g21 and L12 = L21. Put G(W − λ1) =[

a bb d

]. We get that

∂f

∂X1=[

1 0] [ a b

b d

] [X1

X2

]+[

X1 X2

] [ a bb d

] [10

]= 2[

1 0] [ a b

b d

] [X1

X2

]

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Similarly, we find that∂f

∂X2= 2

[0 1

] [ a bb d

] [X1

X2

]. We want to solve

∂f

∂X1=

∂f

∂X2= 0. So suppose that

∂f

∂X1=

∂f

∂X2= 0. Then

[00

]=[

a bb d

] [X1

X2

]= G(W − λ1)

[X1

X2

]. Hence

[Y1 Y2

]G(W − λ1)

[X1

X2

]= 0 for all Y1, Y2 ∈ R

Note that (W − λ1)[

X1

X2

]= L(X) − λX. Hence we get that 〈L(X) − λX,Y〉 = 0 for all Y ∈ TP M . This is only

possible when L(X) − λX = 0. So L(X) = λX. So we have the following result :

If α(s) is a unit speed curve on the surface going through the point P , then the normal curvature of α atthe point P will be maximal/minimal (among all such possible curves α) if α′(s) is an eigenvector of theWeingarten map L at the point P .

What are the extreme values of the normal curvature? Let λ be an eigenvalue of the Weingarten map L and X a uniteigenvector for λ. If α(s) is a unit speed curve through P with α′(s) = X, then the normal curvature of α is given by

II(X,X) = 〈L(X,X〉 = 〈λX,X〉 = λ 〈X,X〉 = λ

So the minimum and maximum value of the normal curvature are the eigenvalues of the Weingarten map!

Theorem 4.24 Let r : U �→ R3 be a coordinate patch, P = r(a, b), α(t) a curve in r(U) with α(0) = P and κ1 ≤ κ2

the eigenvalues of the Weingarten map L. Then the following holds :

(a) κ1(a, b) ≤ κn(0) ≤ κ1(a, b)

(b) Let i ∈ {1, 2}. Then κi(a, b) = κn(0) if and only ifdα

dtis an eigenvector for κi

(c) If κ1(a, b) < κ2(a, b) and Xi is an eigenvector for κi(a, b) for i = 1, 2 then 〈X1,X2〉 = 0.

Proof : We only have to show (c). We get that

κ1 〈X1,X2〉 = 〈κ1X1,X2〉 = 〈L(X1),X2〉 = 〈X1, L(X2)〉 = 〈X1, κ2X2〉 = κ2 〈X1,X2〉

So (κ1 − κ2) 〈L(X1),X2〉 = 〈X1, L(X2)〉 = 〈X1, κ2X2〉 = κ2 〈X1,X2〉 = 0. Since κ1 �= κ2, we get that 〈X1,X2〉 = 0.�

So we have the following result :

At each point of a surface M there are two orthogonal directions such that the normal curvaturetakes its maximum value along one direction and its minimum value along the other direction.

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