di erential geometry of curves and surfaces, 1st …deigen/chern.pdf · thomas bancho , shiing-shen...

Thomas Banchoff, Shiing-Shen Chern, and

William Pohl

Differential Geometry of Curvesand Surfaces, 1st EditionSPIN C346 Differential Geometry Banchoff/Chern/Pohl

Monograph – Mathematics –

July 17, 2002

Springer

Berlin Heidelberg NewYorkBarcelona Hong KongLondon Milan ParisTokyo

Table of Contents

Part I Introduction

1 Review of Euclidean Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.1 Motions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.1 Arc Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.2 Curvature and Fenchel’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . 132.3 The Unit Normal Bundle and Total Twist . . . . . . . . . . . . . . . . . 162.4 Moving Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.5 Curves at a Non-inflexional Point and the Frenet Formulas . . 212.6 Local Equations of a Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.7 Plane Curves and a Theorem on Turning Tangents . . . . . . . . . 262.8 Plane Convex Curves and the Four Vertex Theorem . . . . . . . . 302.9 Isoperimetric Inequality in the Plane . . . . . . . . . . . . . . . . . . . . . . 31

3 Fundamental Forms of a Surface . . . . . . . . . . . . . . . . . . . . . . . . . . 353.1 The First Fundamental Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

3.1.1 Geometry in the Tangent Plane . . . . . . . . . . . . . . . . . . . . 393.2 The Second Fundamental Form . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3.2.1 The Shape of a Surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423.2.2 Characterization of the Sphere . . . . . . . . . . . . . . . . . . . . . 443.2.3 Principal Curvatures, Principal Directions, and Lines

of Curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 453.3 Gauss Mapping and the Third Fundamental Form . . . . . . . . . . 473.4 Triply Orthogonal System; Theorems of Dupin and Liouville . 47

4 Fundamental Equations in Surface Theory . . . . . . . . . . . . . . . . 534.1 Weingarten and Gauss Equations . . . . . . . . . . . . . . . . . . . . . . . . . 534.2 Levi-Civita Parallelism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 564.3 Integrability Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 634.4 The Congruence Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

VI Table of Contents

5 Global Surface Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 715.1 Moving Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

5.1.1 INSERT TITLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 715.1.2 INSERT TITLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 725.1.3 INSERT TITLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

5.2 The Gauss-Bonnet Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 775.2.1 INSERT TITLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 775.2.2 Theorems of Jacobi and Hadamard . . . . . . . . . . . . . . . . . 795.2.3 Vector fields on a surface . . . . . . . . . . . . . . . . . . . . . . . . . . 81

5.3 Rigidity Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 815.3.1 Elliptic W-surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 825.3.2 Integral formulas; second proof of Liebmann’s theorem. 855.3.3 Cohn-Vossen’s rigidity theorem. . . . . . . . . . . . . . . . . . . . . 86

6 INSERT TITLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 896.1 Geodesics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

6.1.1 INSERT TITLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 896.1.2 INSERT TITLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 906.1.3 INSERT TITLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

6.2 Normal Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 946.2.1 INSERT TITLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 946.2.2 INSERT TITLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

6.3 The second variation of arc length . . . . . . . . . . . . . . . . . . . . . . . . 1026.3.1 INSERT TITLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1026.3.2 INSERT TITLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1056.3.3 INSERT TITLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1066.3.4 INSERT TITLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1086.3.5 INSERT TITLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114

Part I

Introduction

1 Review of Euclidean Geometry

1.1 Motions

Three-dimensional Euclidean space consists of points which have as coordi-nates ordered triples of real numbers x1, x2, x3. In vector notation, we writex = (x1, x2, x3). The distance between two points is given by the formula

d(x,y) =

√√√√ 3∑i=1

(xi − yi)2 . (1.1)

Note that d(x,y) ≥ 0 and d(x,y) = 0 if and only if x = y.An affine transformation T from R3 to R3 is defined by T (x) = Ax + b,

where A is a 3× 3 matrix and b is a vector in R3. An affine transformationthat preserves distance between points, so that d(T (x), T (y)) = d(x,y) forall x and y is called a motion of R3.

Proposition 1. An affine transformation is a motion if and only if A is anorthogonal matrix, i.e. a matrix with columns that are mutually perpendicularunit vectors.

Proof. Let the points (x1, x2, x3) and (y1, y2, y3) be mapped by an affinetransformation T into the points (x′1, x

′2, x′3) and (y′1, y

′2, y′3) respectively, so

that

x′i =3∑j=1

aijxj + bi

y′i =3∑j=1

aijyj + bi ,

where aij denotes the entries of the matrix A and the bi denotes thecomponents of b. If we subtract these two equations, we get

x′i − y′i =3∑j=1

aij (xj − yj) .

4 1 Review of Euclidean Geometry

Taking the sum of the squares gives us∑i

(x′i − y′i)2 =

∑i,j,k

aijaik (xj − yj) (xk − yk) ,

where all the indices run from 1 to 3. This equality will only hold true if

3∑i=1

aijaik = δjk , (1.2)

where δjk = 1 if j = k and 0 otherwise.

Given an affine transformation x′i =∑3j=1 aijxj + bi, we can solve ex-

plicitly for the xi in terms of the x′i. We first set x′i − bi =∑3j=1 aijxj then

multiply by aji and sum over i and j to get

3∑i=1

aji(x′i − bi) = xj .

Remark 1. The quantities δjk defined in (1.2) are called Kronecker deltas.We have illustrated their usefulness in the above proof, and they will be usedconsistently.

It will be convenient to introduce the matrices

A =

a11 a12 a13

a21 a22 a23

a31 a32 a33

AT =

a11 a21 a31

a12 a22 a32

a13 a23 a33

.

The second is obtained from the first by interchanging the rows and columnsand is called the transpose of A. Using this notation, (1.2) can be re-writtenas

ATA = I (1.3)

where I denotes the unit matrix (δij). A matrix A with this property is calledorthogonal.

We may rewrite the definition of a motion in terms of matrices asx′ = Ax + b where x is the column matrix with entries xi. We may thensolve explicitly for x in terms of x′ by writing x′ − b = Ax , so

AT (x′ − b) = AT (Ax)

= (ATA)x= Ix = x.

1.1 Motions 5

A basic result in linear algebra states that, for square matrices,

(AC)T = CTAT ,

where the order of the multiplication is important. From this it follows thatif A and C are orthogonal matrices, then

(AC)TAC = CTATAC = CTC = I

so the product AC is also an orthogonal matrix. For an orthogonal matrix A,we have A−1 = AT . Moreover, AT (A−1)T = (A−1A)T = IT = I so (A−1)T =(AT )−1 . It follows that if A is orthogonal, then so is A−1. A collection ofmatrices that is closed under multiplication and such that the inverse ofevery element of the collection is also in the collection is called a group.

The succesive application of two motions, as in (1.3) above, is called theirproduct. This multiplication is in general not commutative. It is easily seenthat all of the motions in E form a group under this multiplication, calledthe group of motions. Euclidean geometry studies the properties of E thatare invariant under the group of motions.

From (1.3) we find (det A)2 = 1 so that det A = ±1. The motion is calledproper if the determinant is +1, and improper if it is −1. It is easily verifiedthat the product of two proper motions is a proper motion, and it is a simpleresult that all proper motions form a subgroup of the group of motions.

Example 1. The mirror reflection, (x1, x2, x3)→ (−x1, x2, x3), is an impropermotion.

A motion of the form x′i = xi + bi, for i = 1, 2, 3 is called a translation. Amotion of the form

x′i =3∑i=1

aijxj

where j = 1, 2, 3 is called an orthogonal transformation. In matrix form, atranslation can be written x′ = x + b and an orthogonal transformation canbe written x′ = Ax. An orthogonal tranformation is called proper or improperaccording to the sign of det A. A proper orthogonal transformation can alsobe called a reflection. All of the translations form a group, as do all of theothogonal transformations and all the rotations. The group of all rotationscan be characterized as the subgroup of all proper motions leaving the originfixed.

Exercise 1. Prove that the quadratic polynomial∑i,j

αijξiξj


where αij = αji is zero for all ξi if and only if αij = 0. Show that this is nottrue without the symmetry condition on the coefficients αij . This result isused in the proof of (1).

Exercise 2. Show that the inverse motion of x′ = Ax + b is x′ = A−1x −A−1b . Let Ti be the motions x′ = Aix + bi where i = 1, 2 and show thatT1T2 is the motion x′ = A2A1x + A2b1 + b2 .

Exercise 3. Let G be a group with the elements {e, a, b, . . . }, where e is theunit element. The left and right inverses of an element a are defined by

a−1l a = e, and aa−1

r = e,

respectively. Prove that a−1l = a−1

r . Observe that the equivalence of theconditions AAT = I and ATA = I means group-theoretically that the matrixA has the same right and left inverse, which is AT .

Exercise 4. Prove that the translations form a normal subgroup of the groupof motions, while the rotations do not.

Exercise 5. Show that the helicoidal motions

x′1 = cos(t)x1 + sin(t)x2

x′2 = − sin(t)x1 + cos(t)x2

x′3 = x3 + bt,

where b is a constant and t is a parameter, form a group. Draw the orbit ofthe point (a, 0, 0), and distinguish the cases when b < 0 and b > 0.

Exercise 6. Prove that the rotation

x′i =∑j

aijxj

where i, j = 1, 2, 3 and det(aij) = 1 has a line of fixed points through the ori-gin, the axis of rotation. Hence prove that the group of rotations is connected.Prove also that the group of orthogonal transformations is not connected.(Note: A subgroup of motions is connected if any two points can be joinedby a continuous arc.)

1.2 Vectors

Two ordered pairs of points, p(x1, x2, x3), q(y1, y2, y3) and p′(x′1, x′2, x′3),

q′(y′1, y′2, y′3) are called equivalent if there is a translation T which maps p

to p′ and q to q′. The last property can be expressed by the conditionsx′i = xi+ bi and y′i = yi+ bi, where i = 1, 2, 3. It follows that a necessary and

1.2 Vectors 7

sufficient condition for the equivalence of the two ordered pairs of points isy′i− x′i = yi− xi . Such an equivalence class is called a vector. We denote thevector by V = −→pq and call

vi = yi − xi, (1.4)

where i = 1, 2, 3 denote its components. A vector is therefore completelydetermined by its components. Geometrically

−→pq =

−→p′q′ if and only if the

segments pq and p′q′ are of the same length and parallel in the same sense.Using the origin O of our coordinate system, we can set up a one-to-one

correspondence between the points p of R3 and the vectors −→Op. The latterwill be called the position vector of p. Notice that it is defined with referenceto the origin O.

Given two vectors v = (v1, v2, v3) and w = (w1, w2, w3) their sum isv+w = (v1+w1, v2+w2, v3+w3), and multiplication of a vector by a numberis defined by λv = (λv1, λv2, λv3). Throughout the book real numbers aresometimes called scalars, in order to emphasize their difference from vectors.

By (1.4) we see that under a motion the vectors are transformed accordingto the equations

vi =∑j

Aijvj (1.5)

where i, j = 1, 2, 3, and v′ = (v′1, v′2, v′3) is the image vector. Using (1.5) and (1)

we get

v′21 + v

′22 v

′23 = v2

1 + v22 + v2

3 .

This leads us to define

v2 = v21 + v2

2 + v23 . (1.6)

We call +√

v2 the length of v. Thus, the length of a vector is invariant undermotions.

More generally, we find

12{(v + w)2 − v2 −w2} =

∑i

viwi ,

where i, j = 1, 2, 3. Since the left-hand side involves only lengths of vectors,which are invariant under motions, the same property holds for the expres-sions at the right-hand side. Generalizing (1.6), we introduce the notation

v ·w = w · v =∑i

viwi ,

which is to be called the scalar or dot product of v and w.


Relative to addition and scalar multiplication of vectors, the scalar prod-uct has the following properties: (v1 + v2) · w = v1 · w + v2 · w and(λv) ·w = v · (λw) = λ(v ·w) where λ = scalar. The relation

(v + λw)2 = v2 + 2λv ·w + λ2w2 ≥ 0

is true for all λ. So by elementary algebra we get

v2w2 − (v ·w)2 ≥ 0, (1.7)

which is called the Cauchy-Schwartz inequality. The angle θ between v and wis defined by

cos θ =v ·w√v2w2

.

This is meaningful because by the Cauchy-Schwartz inequality the right-hand side has absolute value ≤ 1. The vectors v and w are perpendicular ororthogonal if v ·w = 0.

The determinant of three vectors u, v, w with components ui, vi, wi fori = 1, 2, 3 respectively, is defined by

det(u,v,w) =

∣∣∣∣∣∣u1 u2 u3

v1 v2 v3w1 w2 w3

∣∣∣∣∣∣ .Under the transformation in (1.5) the determinant (u,v,w) is multiplied bydet A = det (aij). Hence it is invariant under proper motions and changes itssign under improper motions. The following properties follow immediatelyfrom the definition:

(u + u1,v,w) = (u,v,w) + (u1,v,w),(λu,v,w) = λ(u,v,w) where λ = scalar,(u,v,w) = −(v,u,w) etc.

The vector product of two vectors v,w is the vector z such that therelation

(v,w,x) = z · x (1.8)

holds for all vectors x. It follows that z has the components

z1 = v2w3 − v3w2, z2 = v3w1 − v1w3, z3 = v1w2 − v2w1 . (1.9)

We write

1.2 Vectors 9

z = v ×w .

The vector product has the following properties:

v ×w + w × v = 0,(v1 + v2)×w = v1 ×w + v2 ×w,

(λv)×w = λ(v ×w) where λ = scalar.

The vector z can be described geometrically as follows: We see from (1.9)that v × w = 0 if and only if one of the vectors v,w is a multiple of theother. In this case we say that v and w are linearly dependent. Suppose nextthat v × w 6= 0, i.e., v and w are linearly independent. Putting x = v,wsuccessfully in (1.8), we get z · v = z ·w = 0, so that z is orthogonal to bothv and w. We write z = λu, where u is a unit vector orthogonal to v and w.Thus from (1.8) we get

(v,w,u) = z · u = λu2 = λ .

Hence we have

z = v ×w = (v,w,u) u 6= 0 .

The unit vector u is defined up to its sign and we can choose u such that(v,w,u) > 0. z is therefore a multiple of u and of length (v,w,u). Thiscompletely determines z.

Three vectors u,v,w are called linearly dependent if (u,v,w) = 0; other-wise they are linearly independent. An ordered set of three linearly indepen-dent vectors is called a right-handed or left-handed frame according to the signof its determinant. The property of right-handedness or left-handedness of aframe remains unchanged under proper motion, while they interchange un-der an improper motion. Also a right-handed (or left-handed) frame becomesleft-handed (or right-handed) when any two of its vectors are interchanged.

Remark 2. The importance of vectors in analytic geometry is due to the alge-braic structure. Two vectors can be added and a vector can be multiplied bya scalar. It is important to observe that corresponding operations are mean-ingless on the points of R3, because they are not invariant under motions.

Exercise 7. a) The vector equation of a line is x(t) = at+ b, where a,b =constant and a 6= 0. Find its angles with the coordinate axes.

b) The vector equation of a plane is a · x = b where a 6= 0. Give thegeometrical meaning of b when a is a unit vector.

c) The vector equation of a sphere is (x− a)2 = r2. What are a and r?In each case draw the relevant figure.


Exercise 8. Prove that x(t) = A cos t + b sin t for A,b 6= 0, represents anellipse.

Exercise 9. Let yi, for i = 1, 2, 3, be three linearly independent vectors.Prove that any vector x can be written

x =∑i

λiyi .

Hence prove that x = 0 if and only if x · yi = 0.

Exercise 10. Let (u,v,w) = 0, u × v 6= 0. Prove that w is a linear com-bination of u and v, i.e., w can be written w = λu + µv, where λ, µ arescalars.

Exercise 11. Prove Lagrange’s identity:

(v ×w) · (x× y) = (v · x)(w · y)− (v · y)(w · x) .

Hence prove that

(v ×w)× x = (vx)w − (wx)v .

Hint. To prove the first equation, write out both sides in components.

2 Curves

2.1 Arc Length

A parametrized curve in Euclidean three-space e3 is given by a vector function

x(t) = (x1(t), x2(t), x3(t))

that assigns a vector to every value of a parameter t in a domain interval[a, b]. The coordinate functions of the curve are the functions xi(t). In orderto apply the methods of calculus, we suppose the functions xi(t) to have asmany continuous derivatives as needed in the following treatment.

For a curve x(t), we define the first derivative x′(t) to be the limit of thesecant vector from x(t) to x(t+h) divided by h as h approaches 0, assumingthat this limit exists. Thus,

x′(t) = limh→0

(x(t+ h)− x(t)

h

).

The first derivative vector x′(t) is tangent to the curve at x(t). If we thinkof the parameter t as representing time and we think of x(t) as representingthe position of a moving particle at time t, then x′(t) represents the velocityof the particle at time t. It is straightforward to show that the coordinates ofthe first derivative vector are the derivatives of the coordinate functions, i.e.

x′(t) = (x1′(t), x2

′(t), x3′(t)) .

For most of the curves we will be concerning ourselves with, we will makethe “genericity assumption” that x′(t) is non-zero for all t. (MISSING SEC-TIONS) lengths of polygons inscribed in x as the lengths of the sides of thesepolygons tend to zero. By a theorem of calculus, this limit can be expressedas the integral of the speed s′(t) = |x′(t)| between the parameters of theend-points of the curve, a and b. That is,

s(b)− s(a) =∫ b

a

|x′(t)| dt =∫ b

a

√√√√ 3∑i=1

x′i(t)2 dt .

For an arbitrary value t ∈ (a, b), we may define the distance function

12 2 Curves

s(t)− s(a) =∫ t

a

|x′(u)| du ,

which gives us the distance from a to t along the curve.Notice that this definition of arc length is independent of the parametriz-

tion of the curve. If we define a function v(t) from the interval [a, b] to itselfsuch that v(a) = a, v(b) = b and v′(t) > 0, then we may use the change ofvariables formula to express the arc length in terms of the new parameter v:∫ b

a

|x′(t)| dt =∫ b

a

|x′(v(t))| v′(t) dt =∫ b

a

|x′(v)| dv .

We can also write this expression in the form of differentials:

ds = |x′(t)| dt = |x′(v)| dv.

This differential formalism becomes very significant, especially when we useit to study surfaces and higher dimensional objects, so we will reinterpretresults that use integration or differentiation in differential notation as we go

along. For example, the statement s′(t) =√∑3

i=1 x′i(t)2 can be rewritten as(

ds

dt

)2

=3∑i=1

(dxidt

)2

,

and this may be expressed in the form

ds2 =3∑i=1

dx2i ,

which has the advantage that it is independent of the parameter used todescribe the curve. ds is called the element of arc. It can be visualized as thedistance between two neighboring points.

One of the most useful ways to parametrize a curve is by the arc lengths itself. If we let s = s(t), then we have

s′(t) = |x′(t)| = |x′(s)| s′(t) ,

from which it follows that |x′(s)| = 1 for all s. So the derivative of x withrespect to arc length is always a unit vector.

This parameter s is defined up to the transformation s→ ±s+ c, wherec is a constant. Geometrically, this means the freedom in the choice of initialpoint and direction in which to traverse the curve in measuring the arc length.

Exercise 12. One of the most important space curves is the circular helix

x(t) = (a cos t, a sin t, bt) ,

where a 6= 0 and b are constants. Find the length of this curve over theinterval [0, 2π].

2.2 Curvature and Fenchel’s Theorem 13

Exercise 13. Find a constant c such that the helix

x(t) = (a cos(ct), a sin(ct), bt)

is parametrized by arclength, so that |x′(t)| = 1 for all t.

Exercise 14. The astroid is the curve defined by

x(t) =(a cos3 t, a sin3 t, 0

),

on the domain [0, 2π]. Find the points at which x(t) does not define animmersion, i.e., the points for which x′(t) = 0.

Exercise 15. The trefoil curve is defined by

x(t) = ((a+ b cos(3t)) cos(2t), (a+ b cos(3t)) sin(2t), b sin(3t)) ,

where a and b are constants with a > b > 0 and 0 ≤ t ≤ 2π. Sketch thiscurve, and give an argument to show why it is knotted, i.e. why it cannot bedeformed into a circle without intersecting itself in the process.

Exercise 16. (For the serious mathematician) Two parametrized curves x(t)and y(u) are said to be equivalent if there is a function u(t) such that u′(t) > 0for all a < t < b and such that y(u(t)) = x(t). Show that relation satisfiesthe following three properties:

1. Every curve x is equivalent to itself2. If x is equivalent to y, then y is equivalent to x3. If x is equivalent to y and if y is equivalent to z, then x is equivalent to

z

A relation that satisfies these properties is called an equivalence rela-tion. Precisely speaking, a curve is considered be an equivalence class ofparametrized curves.

2.2 Curvature and Fenchel’s Theorem

If x is an immersed curve, with x′(t) 6= 0 for all t in the domain, then we maydefine the unit tangent vector T(t) to be x′(t)

|x′(t)| . If the parameter is arclength,then the unit tangent vector T(s) is given simply by x′(s). The line throughx(t0) in the direction of T(t0) is called the tangent line at x(t0). We can writethis line as y(u) = x(t0) + uT(t0), where u is a parameter that can take onall real values.

Since T(t) · T(t) = 1 for all t, we can differentiate both sides of thisexpression, and we obtain 2T′(t) · T(t) = 0. Therefore T′(t) is orthogonalto T(t). The curvature of the space curve x(t) is defined by the conditionκ(t) = |T′(t)|

|x′(t)| , so = κ(t)s′(t) = |T′(t)|. If the parameter is arclength, thenx′(s) = T(s) and κ(s) = |T′(s)| = |x”(s)|.

14 2 Curves

Proposition 2. If κ(t) = 0 for all t, then the curve lies along a straight line.

Proof. Since κ(t) = 0, we have T′(t) = 0 and T(t) = a, a constant unitvector. Then x′(t) = s′(t)T(t) = s′(t)a, so by integrating both sides of theequation, we obtain x(t) = s(t)a + b for some constant b. Thus x(t) lies onthe line through b in the direction of a.

Curvature is one of the simplest and at the same time one of the mostimportant properties of a curve. We may obtain insight into curvature byconsidering the second derivative vector x”(t), often called the accelerationvector when we think of x(t) as representing the path of a particle at time t. Ifthe curve is parametrized by arclength, then x′(s)·x′(s) = 1 so x”(s)·x′(t) = 0and κ(s) = |x”(s)|. For a general parameter t, we have x′(t) = s′(t)T(t) sox”(t) = s”(t)T(t)+s′(t)T′(t). If we take the cross product of both sides withx′(t) then the first term on the right is zero since x′(t) is parallel to T(t).Moreover x′(t) is perpendicular to T′(t) so

|T′(t)× x′(t)| = |T′(t)||x′(t)| = s′(t)2κ(t) .

Thus

x”(t)× x′(t) = s′(t)T′(t)× x′(t)

and

|x”(t)× x′(t)| = s′(t)3κ(t) .

This gives a convenient way of finding the curvature when the curve is definedwith respect to an arbitrary parameter. We can write this simply as

κ(t) =|x”(t)× x′(t)||x′(t)x′(t)|3/2

.

Note that the curvature κ(t) of a space curve is non-negative for all t. Thecurvature can be zero, for example at every point of a curve lying along astraight line, or at an isolated point like t = 0 for the curve x(t) = (t, t3, 0).A curve for which κ(t) > 0 for all t is called non-inflectional.

The unit tangent vectors emanating from the origin form a curve T(t) onthe unit sphere called the tangential indicatrix of the curve x. To calculatethe length of the tangent indicatrix, we form the integral of |T′(t)| = κ(t)s′(t)with respect to t, so the length is κ(t)s′(t)dt = κ(s)ds. This significant integralis called the total curvature of the curve x.

Up to this time, we have concentrated primarily on local properties ofcurves, determined at each point by the nature of the curve in an arbitrarilysmall neighborhood of the point. We are now in a position to prove our firstresult in global differential geometry or differential geometry in the large.

By a closed curve x(t), a ≤ t ≤ b, we mean a curve such that x(b) = x(a).We will assume moreover that the derivative vectors match at the endpointsof the interval, so x′(b) = x′(a).

2.2 Curvature and Fenchel’s Theorem 15

Theorem 1 (Fenchel’s Theorem). The total curvature of a closed spacecurve x is greater than or equal to 2π.

κ(s)ds ≥ 2π

The first proof of this result was found independently by B. Segre in 1934and later independently by H. Rutishauser and H. Samelson in 1948. Thefollowing proof depends on a lemma by R. Horn in 1971:

Lemma 1. Let g be a closed curve on the unit sphere with length L < 2.Then there is a point m on the sphere that is the north pole of a hemispherecontaining g.

To see this, consider two points p and q on the curve that break g upinto two pieces g1 and g2 of equal length, therefore both less than π. Thenthe distance from p to q along the sphere is less than π so there is a uniqueminor arc from p to q. Let m be the midpoint of this arc. We wish to showthat no point of g hits the equatorial great circle with m as north pole. Ifa point on one of the curves, say g1, hits the equator at a point r, then wemay construct another curve g′1 by rotating g1 one-half turn about the axisthrough m, so that p goes to q and q to p while r goes to the antipodalpoint r′. The curve formed by g1 and g′1 has the same length as the originalcurve g, but it contains a pair of antipodal points so it must have length atleast 2π, contradicting the hypothesis that the length of g was less than 2π.

From this lemma, it follows that any curve on the sphere with length lessthan 2π is contained in a hemisphere centered at a point m. However if x(t)is a closed curve, we may consider the differentiable function f(t) = x(t) ·m.At the maximum and minimum values of f on the closed curve x, we have

0 = f ′(t) = x′(t) ·m = s′(t)T(t) ·m

so there are at least two points on the curve such that the tangential imageis perpendicular to m. Therefore the tangential indicatrix of the closed curvex is not contained in a hemisphere, so by the lemma, the length of any suchindicatrix is greater than 2π. Therefore the total curvature of the closed curvex is also greater than 2π.

Corollary 1. If, for a closed curve x, we have κ(t) ≤ 1R for all t, then the

curve has length L ≥ 2πR.

Proof.

L =∫ds ≥

∫Rκ(s)ds = R

∫κ(s)ds ≥ 2πR

16 2 Curves

Fenchel also proved the stronger result that the total curvature of a closedcurve equals 2π if and only if the curve is a convex plane curve.

I. Fary and J. Milnor proved independently that the total curvature mustbe greater than 4π for any non-self-intersecting space curve that is knotted(not deformable to a circle without self-intersecting during the process.)

Exercise 17. Let x be a curve with x′(t0) 6= 0. Show that the tangent lineat x(t0) can be written as y(u) = x(t0)+ux′(t0) where u is a parameter thatcan take on all real values.

Exercise 18. The plane through a point x(t0) perpendicular to the tangentline is called the normal plane at the point. Show that a point y is on thenormal plane at x(t0) if and only if

x′(t0) · y = x′(t0) · x(t0)

Exercise 19. Show that the curvature κ of a circular helix

x(t) = (r cos(t), r sin(t), pt)

is equal to the constant value κ = |r|r2+p2 . Are there any other curves with

constant curvature? Give a plausible argument for your answer.

Exercise 20. Assuming that the level surfaces of two functions f(x1, x2, x3) =0 and g(x1, x2, x3) = 0 meet in a curve, find an expression for the tangentvector to the curve at a point in terms of the gradient vectors of f and g(where we assume that these two gradient vectors are linearly independentat any intersection point.) Show that the two level surfaces x2 − x2

1 = 0 andx3x1 − x2

2 = 0 consists of a line and a “twisted cubic” x1(t) = t, x2(t) = t2,x3(t) = t3. What is the line?

Exercise 21. What is the geometric meaning of the function f(t) = x(t) ·mused in the proof of Fenchel’s theorem?

Exercise 22. Let m be a unit vector and let x be a space curve. Show thatthe projection of this curve into the plane perpendicular to m is given by

y(t) = x(t)− (x(t) ·m)m .

Under what conditions will there be a t0 with y′(t0) = 0?

2.3 The Unit Normal Bundle and Total Twist

Consider a curve x(t) with x′(t) 6= 0 for all t. A vector z perpendicular tothe tangent vector x′(t0) at x(t0) is called a normal vector at x(t0). Such avector is characterized by the condition z ·x(t0) = 0, and if |z| = 1, then z is

2.3 The Unit Normal Bundle and Total Twist 17

said to be a unit normal vector at x(t0). The set of unit normal vectors at apoint x(t0) forms a great circle on the unit sphere. The unit normal bundleis the collection of all unit normal vectors at x(t) for all the points on a curvex.

At every point of a parametrized curve x(t) at which x′(t) 6= 0, we mayconsider a frame E2(t), E3(t), where E2(t) and E3(t) are mutually orthogonalunit normal vectors at x(t). If E2(t), E3(t) is another such frame, then thereis an angular function φ(t) such that

E2(t) = cos(φ(t))E2(t)− sin(φ(t))E3(t)

E3(t) = sin(φ(t))E2(t) + cos(φ(t))E3(t)

or, equivalently,

E2(t) = cos(φ(t))E2(t) + sin(φ(t))E3(t)

E3(t) = sin(φ(t))E2(t) + cos(φ(t))E3(t) .

From these two representations, we may derive an important formula:

E′2(t) · E3(t) = E′2(t) · E3(t)− φ′(t)

Expressed in the form of differentials, without specifying parameters, thisformula becomes:

dE2E3 = dE2E3 − dφ .

Since E3(t) = T(t)× E2(t), we have:

E′2(t) · E3(t) = −[E′2(t), E2(t),T(t)]

or, in differentials:

dE2E3 = −[dE2, E2,T] .

More generally, if z(t) is a unit vector in the normal space at x(t), then wemay define a function w(t) = −[z′(t), z(t),T(t)]. This is called the connectionfunction of the unit normal bundle. The corresponding differential form w =−[dz, z,T] is called the connection form of the unit normal bundle.

A vector function z(t) such that |z(t)| = 1 for all t and z(t) · x′(t) =0 for all t is called a unit normal vector field along the curve x. Such avector field is said to be parallel along x if the connection function w(t) =−[z′(t), z(t),T(t)] = 0 for all t. In the next section, we will encounter severalunit normal vector fields naturally associated with a given space curve. Fornow, we prove some general theorems about such objects.

18 2 Curves

Proposition 3. If E2(t) and E2(t) are two unit normal vector fields that areboth parallel along the curve x, then the angle between E2(t) and E2(t) isconstant.

Proof. From the computation above, then:

E′2(t) · (−E2(t)×T(t)) = E′2(t) · (−E2(t)×T(t))− φ′(t) .

But, by hypothesis,

E′2(t) · (−E2(t)xT(t)) = 0 = E′2(t)(−E2(t)×T(t))

so it follows that φ′(t) = 0 for all t, i.e., the angle φ(t) between E2(t) andE2(t) is constant.

Given a closed curve x and a unit normal vector field z with z(b) = z(a),we define

µ(x, z) = − 12π

∫[z′(t), z(t),T(t)]dt = − 1

2π[dz, z,T] .

If z is another such field, then

µ(x, z)− µ(x, z) = − 12π

∫[z′(t), z(t),T(t)]− [z′(t), z(t),T(t)]dt

= − 12π

∫φ′(t)dt = − 1

2π[φ(b)− φ(a)] .

Since the angle φ(b) at the end of the closed curve must coincide withthe angle φ(a) at the beginning, up to an integer multiple of 2π, it followsthat the real numbers µ(x, z) and µ(x, z) differ by an integer. Therefore thefractional part of µ(x, z) depends only on the curve x and not on the unitnormal vector field used to define it. This common value µ(x) is called thetotal twist of the curve x. It is a global invariant of the curve.

Proposition 4. If a closed curve lies on a sphere, then its total twist is zero.

Proof. If x lies on the surface of a sphere of radius r centered at the origin,then |x(t)|2 = x(t) · x(t) = r2 for all t. Thus x′(t) · x(t) = 0 for all t, so x(t)is a normal vector at x(t). Therefore z(t) = x(t)

r is a unit normal vector fielddefined along x, and we may compute the total twist by evaluating

µ(x, z) = − 12π

∫[z′(t), z(t),T(t)]dt .

But

[z′(t), z(t),T(t)] = [x′(t)r

,x(t)r,T(t)] = 0

for all t since x′(t) is a multiple of T(t). In differential form notation, we getthe same result: [dz, z,T] = 1

r2 [x′(t),x(t),T(t)]dt = 0. Therefore µ(x, z) = 0,so the total twist of the curve x is zero.

2.4 Moving Frames 19

Remark 3. W. Scherrer proved that this property characterized a sphere, i.e.if the total twist of every curve on a closed surface is zero, then the surfaceis a sphere.

Remark 4. T. Banchoff and J. White proved that the total twist of a closedcurve is invariant under inversion with respect to a sphere with center notlying on the curve.

Remark 5. The total twist plays an important role in modern molecular bi-ology, especially with respect to the structure of DNA.

Exercise 23. Let x be the circle x(t) = (r cos(t), r sin(t), 0), where r is aconstant > 1. Describe the collection of points x(t)+z(t) where z(t) is a unitnormal vector at x(t).

Exercise 24. Let Σ be the sphere of radius r > 0 about the origin. Theinversion through the sphere S maps a point x to the point x = r2 x

|x|2 . Notethat this mapping is not defined if x = 0, the center of the sphere. Provethat the coordinates of the inversion of x = (x1, x2, x3) through S are givenby xi = r2xi

x21+x

22+x

23. Prove also that inversion preserves point that lie on the

sphere S itself, and that the image of a plane is a sphere through the origin,except for the origin itself.

Exercise 25. Prove that the total twist of a closed curve not passing throughthe origin is the same as the total twist of its image by inversion through thesphere S of radius r centered at the origin.

2.4 Moving Frames

In the previous section, we introduced the notion of a frame in the unit normalbundle of a space curve. We now consider a slightly more general notion. Bya frame, or more precisely a right-handed rectangular frame with origin, wemean a point x and a triple of mutually orthogonal unit vectors E1, E2, E3

forming a right-handed system. The point x is called the origin of the frame.Note that Ei · Ej = 1 if i = j and 0 if i 6= j.

Moreover,

E1 × E2 = E3, E2 = E3 × E1, andE3 = E1 × E2.

In the remainder of this section, we will always assume that small Latinletters run from 1 to 3.

Note that given two different frames, x, E1, E2, E3 and x, E1, E2, E3, thereis exactly one affine motion of Euclidean space taking x to x and taking Eito Ei. When x(t), E1(t), E2(t), E3(t) is a family of frames depending on aparameter t, we say we have a moving frame along the curve.

20 2 Curves

Proposition 5. A family of frames x(t), E1(t), E2(t), E3(t) satisfies a systemof differential equations:

x′(t) = Σpi(t)Ei(t)E′i(t) = Σqij(t)Ej(t)

where pi(t) = x′(t) · Ei(t) and qij(t) = E′i(t) · Ej(t).Since Ei(t) · Ej(t) = 0 for i 6= j, it follows that

qij(t) + qji(t) = E′i(t) · Ej(t) + Ei(t) · E′j(t) = 0

i.e. the coefficients qij(t) are anti-symmetric in i and j. This can be expressedby saying that the matrix ((qij(t))) is an anti-symmetric matrix, with 0 onthe diagonal.

In a very real sense, the function pi(t) and qij(t) completely determinethe family of moving frames.

Specifically we have:

Proposition 6. If x(t), E1(t), E2(t), E3(t) and x(t), E1(t), E2(t), E3(t) aretwo families of moving frames such that pi(t) = p

i(t) and qij(t) = q

ij(t) for

all t, then there is a single affine motion that takes x(t), E1(t), E2(t), E3(t)to x(t), E1(t), E2(t), E3(t)) for all t.

Proof. Recall that for a specific value t0, there is an affine motion tak-ing x(t0), E1(t0), E2(t0), E3(t0) to x(t0), E1(t0), E2(t0), E3(t0). We will showthat this same motion takes x(t), E1(t), E2(t), E3(t) to x(t), E1(t), E2(t), E3(t)for all t. Assume that the motion has been carried out so that the framesx(t0), E1(t0), E2(t0), E3(t0) and x(t0), E1(t0), E2(t0), E3(t0) coincide.

Now consider

(ΣEi(t) · Ei(t))′ = ΣE′i(t) · Ei(t) +ΣEi(t) · E′i(t)= ΣΣqij(t)Ej(t) · Ei(t) +ΣEi(t) ·Σqij(t)Ej(t)= ΣΣqij(t)Ej(t) · Ei(t) +ΣΣqij(t)Ei(t) · Ej(t)= ΣΣqij(t)Ej(t) · Ei(t) +ΣΣqji(t)Ej(t) · Ei(t)= 0 .

It follows that

ΣEi(t) · Ei(t) = ΣEi(t0) · Ei(t0) = ΣEi(t0) · Ei(t0) = 3

for all t. But since |Ei(t) · Ei(t)| ≤ 1 for any pair of unit vectors, we musthave Ei(t) · Ei(t) = 1 for all t. Therefore Ei(t) = Ei(t) for all t.

Next consider

(x(t)− x(t))′ = Σpi(t)Ei(t)−Σpi(t)Ei(t) = Σpi(t)Ei(t)−Σpi(t)Ei(t) = 0 .

2.5 Curves at a Non-inflexional Point and the Frenet Formulas 21

Since the origins of the two frames coincide at the value t0, we have

x(t)− x(t) = x(t0)− x(t0) = 0

for all t.This completes the proof that two families of frames satisfying the same

set of differential equations differ at most by a single affine motion.

Exercise 26. Prove that the equations E′i(t) = Σqij(t)Ej(t) can be writtenE′i(t) = d(t) × Ei(t), where d(t) = q23(t)E1(t) + q31(t)E2(t) + q12(t)E3(t).This vector is called the instantaneous axis of rotation.

Exercise 27. Under a rotation about the x3-axis, a point describes a circlex(t) = (a cos(t), a sin(t), b). Show that its velocity vector satisfies x′(t) =d× x(t) where d = (0, 0, 1). (Compare with the previous exercise.).

Exercise 28. Prove that (v ·v)(w ·w)−(v ·w)2 = 0 if and only if the vectorsv and w are linearly dependent.

2.5 Curves at a Non-inflexional Point and the FrenetFormulas

A curve x is called non-inflectional if the curvature k(t) is never zero. By ourearlier calculations, this condition is equivalent to the requirement that x′(t)and x′′(t) are linearly independent at every point x(t), i.e. x′(t)× x′′(t) 6= 0for all t. For such a non-inflectional curve x, we may define a pair of naturalunit normal vector fields along x.

Let b(t) = x′(t)×x′′(t)|x′(t)×x′′(t)| , called the binormal vector to the curve x(t).

Since b(t) is always perpendicular to T(t), this gives a unit normal vectorfield along x.

We may then take the cross product of the vector fields b(t) and T(t) toobtain another unit normal vector field N(t) = b(t)×T(t), called the principalnormal vector. The vector N(t) is a unit vector perpendicular to T(t) andlying in the plane determined by x′(t) and x′′(t). Moreover, x′′(t) ·N(t) =k(t)s′(t)2, a positive quantity.

Note that if the parameter is arclength, then x′(s) = T(s) and x′′(s)is already perpendicular to T(s). It follows that x′′(s) = k(s)N(s) so wemay define N(s) = x′′(s)

k(s) and then define b(s) = T(s) ×N(s). This is thestandard procedure when it happens that the parametrization is by arclength.The method above works for an arbitrary parametrization.

We then have defined an orthonormal frame x(t)T(t)N(t)b(t) called theFrenet frame of the non-inflectional curve x.

By the previous section, the derivatives of the vectors in the frame canbe expressed in terms of the frame itself, with coefficients that form an anti-symmetric matrix. We already have x′(t) = s′(t)T(t), so

22 2 Curves

p1(t) = s′(t) , p2(t) = 0 = p3(t) .

Also T′(t) = k(t)s′(t)N(t), so

q12(t) = k(t)s′(t) and q13(t) = 0 .

We know that

b′(t) = q31(t)T(t) + q32(t)N(t) , and q31(t) = −q13(t) = 0 .

Thus b′(t) is a multiple of N(t), and we define the torsion w(t) of the curveby the condition

b′(t) = −w(t)s′(t)N(t) ,

so q32(t) = −w(t)s′(t) for the Frenet frame. From the general computationsabout moving frames, it then follows that

N′(t) = q21(t)T(t) + q23(t)b(t) = −k(t)s′(t)T(t) + w(t)s′(t)b(t) .

The formulas for T′(t), N′(t), and b′(t) are called the Frenet formulas forthe curve x.

If the curve x is parametrized with respect to arclength, then the Frenetformulas take on a particularly simple form:

x′(s) = T(s)T′(s) = k(s)N(s)N′(s) = −k(s)T(s) + w(s)b(s)b′(s) = −w(s)b(s) .

The torsion function w(t) that appears in the derivative of the binormalvector determines important properties of the curve. Just as the curvaturemeasures deviation of the curve from lying along a straight line, the torsionmeasures deviation of the curve from lying in a plane. Analogous to the resultfor curvature, we have:

Proposition 7. If w(t) = 0 for all points of a non-inflectional curve x, thenthe curve is contained in a plane.

Proof. We have b′(t) = −w(t)s′(t)N(t) = 0 for all t so b(t) = a, a constantunit vector. Then T(t)a = 0 for all t so (x(t) · a)′ = x′(t) · a = 0 andx(t) · a = x(a) · a, a constant. Therefore (x(t) − x(a)) · a = 0 and x lies inthe plane through x(a) perpendicular to a.

If x is a non-inflectional curve parametrized by arclength, then

w(s) = b(s) ·N′(s) = [T(s),N(s),N′(s)] .

2.5 Curves at a Non-inflexional Point and the Frenet Formulas 23

Since N(s) = x′′(s)k(s) , we have

N′(s) =x′′′(s)k(s)

+ x′′(s)−k′(s)k(s)2

,

so

w(s) =[x′(s),

x′′(s)k(s)

,x′′′(s)k(s)

+ x′′(s)−k′(s)k(s)2

]=

[x′(s),x′′(s),x′′′(s)]k(s)2

.

We can obtain a very similar formula for the torsion in terms of an arbi-trary parametrization of the curve x. Recall that

x′′(t) = s”(t)T(t) + k(t)s′(t)T′(t) = s′′(t)T(t) + k(t)s′(t)2N(t) ,

so

x′′′(t) = s′′′(t)T(t) + s′′(t)s′(t)k(t)N(t) +[k(t)s′(t)2

]′N(t) + k(t)s′(t)2N′(t) .

Therefore

x′′′(t)b(t) = k(t)s′(t)2N′(t)b(t) = k(t)s′(t)2w(t)s′(t) ,

and

x′′′(t) · x′(t)× x′′(t) = k2(t)s′(t)6w(t) .

Thus we obtain the formula

w(t) =x′′′(t) · x′(t)xx′′(t)|x′(t)× x′′(t)|2

,

valid for any parametrization of x.Notice that although the curvature k(t) is never negative, the torsion w(t)

can have either algebraic sign. For the circular helix x(t) = (r cos(t), r sin(t), pt)for example, we find w(t) = p

r2+p2 , so the torsion has the same algebraic signas p. In this way, the torsion can distinguish between a right-handed and aleft-handed screw.

Changing the orientation of the curve from s to −s changes T to −T, andchoosing the opposite sign for k(s) changes N to −N. With different choices,then, we can obtain four different right-handed orthonormal frames, xTNb,x(−T)N(−b), xT(−N)(−b), and x(−T)(−N)b. Under all these changes ofthe Frenet frame, the value of the torsion w(t) remains unchanged.

A circular helix has the property that its curvature and its torsion areboth constant. Furthermore the unit tangent vector T(t) makes a constantangle with the vertical axis. Although the circular helices are the only curveswith constant curvature and torsion, there are other curves that have thesecond property. We characterize such curves, as an application of the Frenetframe.

24 2 Curves

Proposition 8. The unit tangent vector T(t) of a non-inflectional spacecurve x makes a constant angle with a fixed unit vector a if and only ifthe ratio w(t)

k(t) is constant.

Proof. If T(t) · a = constant for all t, then differentiating both sides, weobtain

T′(t) · a = 0 = k(t)s′(t)N(t) · a ,

so a lies in the plane of T(t) and b(t). Thus we may write a = cos(φ)T(t) +sin(φ)b(t) for some angle φ. Differentiating this equation, we obtain

0 = cos(φ)T′(t) + sin(φ)b′(t) = cos(φ)k(t)s′(t)N(t)− sin(φ)w(t)s′(t)N(t) ,

so w(t)k(t) = sin(φ)

cos(φ) = tan(φ). This proves the first part of the proposition andidentifies the constant ratio of the torsion and the curvature.

Conversely, if w(t)k(t) = constant = tan(φ) for some φ, then, by the same

calculations, the expression cos(φ)T(t) + sin(φ)b(t) has derivative 0 so itequals a constant unit vector. The angle between T(t) and this unit vectoris the constant angle φ.

Curves with the property that the unit tangent vector makes a fixedangle with a particular unit vector are called generalized helices. Just as acircular helix lies on a circular cylinder, a generalized helix will lie on ageneral cylinder, consisting of a collection of lines through the curve parallelto a fixed unit vector. On this generalized cylinder, the unit tangent vectorsmake a fixed angle with these lines, and if we roll the cylinder out onto aplane, then the generalized helix is rolled out into a straight line on the plane.

We have shown in the previous section that a moving frame is completelydetermined up to an affine motion by the functions pi(t) and qij(t). In thecase of the Frenet frame, this means that if two curves x and x have the samearclength s(t), the same curvature k(t), and the same torsion w(t), then thecurves are congruent, i.e. there is an affine motion of Euclidean three-spacetaking x(t) to x(t) for all t. Another way of stating this result is:

Theorem 2. The Fundamental Theorem of Space Curves. Two curves parametrizedby arclength having the same curvature and torsion at corresponding pointsare congruent by an affine motion.

Exercise 29. Compute the torsion of the circular helix. Show directly thatthe principal normals of the helix are perpendicular to the vertical axis, andshow that the binormal vectors make a constant angle with this axis.

Exercise 30. Prove that if the curvature and torsion of a curve are bothconstant functions, then the curve is a circular helix (i.e. a helix on a circularcylinder).

2.6 Local Equations of a Curve 25

Exercise 31. Prove that a necessary and sufficient condition for a curve xto be a generalized helix is that

x′′(t)× x′′′(t) · xiv(t) = 0 .

Exercise 32. Let y(t) be a curve on the unit sphere, so that |y(t)| = 1 andy(t) ·y′(t)×y′′(t) 6= 0 for all t. Show that the curve x(t) = c

∫y(u)×y′′(u)du

with c 6= 0 has constant torsion 1c .

Exercise 33. (For students familiar with complex variables) If the coordi-nate functions of the vectors in the Frenet frame are given by

T = (e11, e12, e13) ,N = (e21, e22, e23) ,b = (e31, e32, e33) ,

then we may form the three complex numbers

zj =e1j + ie2j

1− e3j=

1 + e3je1j − ie2j

.

Then the functions zj satisfy the Riccati equation

z′j = −ik(s)zj +i

2w(s)(−1 + z2

j ) .

This result is due to S. Lie and G. Darboux.

2.6 Local Equations of a Curve

We can “see” the shape of a curve more clearly in the neighborhood of apoint x(t0) when we consider its parametric equations with respect to theFrenet frame at the point. For simplicity, we will assume that t0 = 0, and wemay then write the curve as

x(t) = x(0) + x1(t)T(0) + x2(t)N(0) + x3(t)b(0) .

On the other hand, using the Taylor series expansion of x(t) about the pointt = 0, we obtain

x(t) = x(0) + tx′(0) +t2

2x′′(0) +

t3

6x′′′(0) + higher order terms .

From our earlier formulas, we have

x′(0) = s′(0)T(0) ,

x′′(0) = s′′(0)T(0) + k(0)s′(0)2N(0) ,

x′′′(0) = s′′′(0)T(0) + s′′(0)s′(0)k(0)N(0) + (k(0)s′(0)2)′N(0)

+ k(0)s′(0)2(−k(0)s′(0)T(0) + w(0)s′(0)b(0)) .

26 2 Curves

Substituting these equations in the Taylor series expression, we find:

x(t) = x(0) +(ts′(0) +

t2

2s′′(0) +

t3

6[s′′′(0)− k(0)2s′(0)3

]+ ...

)T(0)

+(t2

2k(0)s′(0)2 +

t3

6[s′′(0)s′(0)k(0) + (k(0)s′(0)2)′

]+ ...

)N(0)

+(t3

6k(0)w(0)s′(0)3 + ...

)b(0) .

If the curve is parametrized by arclength, this representation is muchsimpler:

x(s) = x(0) +(s− k(0)2

6s3 + ...

)T(0)

+(k(0)

2s2 +

k′(0)6

s3 + ...

)N(0)

+(k(0)w(0)

6s3 + ...

)b(0) .

Relative to the Frenet frame, the plane with equation x1 = 0 is the normalplane; the plane with x2 = 0 is the rectifying plane, and the plane with x3 = 0is the osculating plane. These planes are orthogonal respectively to the unittangent vector, the principal normal vector, and the binormal vector of thecurve.

2.7 Plane Curves and a Theorem on Turning Tangents

The general theory of curves developed above applies to plane curves. In thelatter case there are, however, special features which will be important tobring out. We suppose our plane to be oriented. In the plane a vector hastwo components and a frame consists of an origin and an ordered set of twomutually perpendicular unit vectors forming a right-handed system. To anoriented curve C defined by x(s) the Frenet frame at s consists of the originx(s), the unit tangent vector T(s) and the unit normal vector N(s). Unlikethe case of space curves this Frenet frame is uniquely determined, under theassumption that both the plane and the curve are oriented.

The Frenet formulas are

x′ = T ,

T′ = kN ,

N′ = −kT .

(2.1)

The curvature k(s) is defined with sign. It changes its sign when theorientation of the plane or the curve is reversed.

2.7 Plane Curves and a Theorem on Turning Tangents 27

The Frenet formulas in (2.1) can be written more explicitly. Let

x(s) = (x1(s) , x2(s)) (2.2)

Then

T(s) = ( x′1(s) , x′2(s)) ,N(s) = (−x′2(s) , x′1(s)) .

(2.3)

Expressing the last two equations of (2.1) in components, we have

x′′1 = −kx′2 (2.4)x′′2 = kx′1 . (2.5)

These equations are equivalent to (2.1).Since T is a unit vector, we can put

T(s) = (cos τ(s) , sin τ(s)) , (2.6)

so that τ(s) is the angle of inclination of T with the x1-axis. Then

N(s) = (− sin τ(s) , cos τ(s)) , (2.7)

and (2.1) gives

dτds

= k(s) (2.8)

This gives a geometrical interpretation of k(s).A curve C is called simple if it does not intersect itself. One of the most

important theorems in global differential geometry is the theorem on turningtangents:

Theorem 3. For a simple closed plane curve we have

12π

∮k ds = ±1 .

To prove this theorem we give a geometrical interpretation of the integralat the left-hand side of (3). By (2.8)

12π

∮k ds =

12π

∮dτ .

But τ , as the angle of inclination of τ(s), is only defined up to an integralmultiple of 2π, and this integral has to be studied with care.

Let O be a fixed point in the plane. Denote by Γ the unit circle aboutO; it is oriented by the orientation of the plane. The tangential mapping orGauss mapping

28 2 Curves

g : C 7→ Γ (2.9)

is defined by sending the point x(s) of C to the point T(s) of Γ. In otherwords, g(P ), P ∈ C, is the end-point of the unit vector through O parallelto the unit tangent vector to C at P . Clearly g is a continuous mapping. IfC is closed, it is intuitively clear that when a point goes along C once itsimage point under g goes along Γ a number of times. This integer is calledthe rotation index of C. It is to be defined rigorously as follows:

We consider O to be the origin of our coordinate system. As above wedenote by τ(s) the angle of inclination of T(s) with the x1-axis. In order tomake the angle uniquely determined we suppose O ≤ τ(s) < 2π. But τ(s) isnot necessarily continuous. For in every neighborhood of s0 at which τ(sc) = 0there may be values of τ(s) differing from 2π by arbitrarily small quantities.We have, however, the following lemma:

Lemma 2. There exists a continuous function τ(s) such that τ(s) ≡ τ(s)mod 2π .

Proof. We suppose C to be a closed curve of total length L. The continuousmapping g is uniformly continuous. There exists therefore a number δ > 0such that for |s1 − s2| < δ, T(s1) and T(s2) lie in the same open half-plane.Let s0(= O) < s1 < · · · < si(= L) satisfy |si− si−1| < δ for i = 1, . . . ,m. Weput τ(s0) = τ(s0). For s0 ≤ s ≤ s1, we define τ(s) to be τ(s0) plus the angleof rotation from g(s0) to g(s) remaining in the same half-plane. Carryingout this process in successive intervals, we define a continuous function τ(s)satisfying the condition in the lemma. The difference τ(L)−τ(O) is an integralmultiple of 2π. Thus, τ(L) − τ(O) = γ2π. We assert that the integer γ isindependent of the choice of the function τ. In fact let τ′(s) be a functionsatisfying the same conditions. Then we have τ′(s) − τ(s) = n(s)2π wheren(s) is an integer. Since n(s) is continuous in s, it must be constant. Itfollows that τ′(L) − τ′(O) = τ(L) − τ(O), which proves the independence ofγ from the choice of τ. We call γ the rotation index of C. In performingintegration over C we should replace τ(s) by τ in (2.8). Then we have

12π

∮k ds =

12π

∮d τ = γ . (2.10)

We consider the mapping h which sends an ordered pair of points x(s1), x(s2),O ≤ s1 ≤ s2 ≤ L, of C into the end-point of the unit vector throughO parallelto the secant joining x(s1) to x(s2). These ordered pairs of points can berepresented as a triangle 4 in the (s1, s2)-plane defined by O ≤ s1 ≤ s2 ≤ L.The mapping h of 4 into Γ is continuous. Moreover, its restriction to theside s1 = s2 is the tangential mapping g in (2.9).

To a point p ∈ 4 let τ(p) be the angle of inclination of Oh(p) to the x1-axis, satisfying O ≤ τ(p) < 2π. Again this function need not be continuous.We shall, however, prove that there exists a continuous function ˜τ(p), p ∈ 4,

2.7 Plane Curves and a Theorem on Turning Tangents 29

such that τ(p) ≡ τ(p) mod 2π. In fact, let m be an interior point of 4.We cover 4 by the radii through m. By the argument used in the proof ofthe above lemma we can define a function τ(p), p ∈ 4, such that τ(p) ≡ τ(p)mod 2π, and such that it is continuous on every radius through m. It remainsto prove that it is continuous in 4.

For this purpose let p0 ∈ 4. Since h is continuous, it follows from the com-pactness of the segment mp0 that there exists a number η = η(p0) > 0, suchthat for q0 ∈ mp0 and for any point q ∈ 4 for which the distance d(q, q0) < ηthe points h(q) and h(q0) are never antipodal. The latter condition can beanalytically expressed by

τ(q) 6≡ τ(q0) mod π . (2.11)

Now let ε > 0, ε < π2 be given. We choose a neighborhood U of p0 such

that U is contained in the η-neighborhood of p0 and such that, for p ∈ U ,the angle between Oh(p0) and Oh(p) is < ε. This is possible, because themapping h is continuous. The last condition can be expressed in the form

τ(p)− τ(p0) = ε′ + 2k(p)π , (2.12)

where k(p) is an integer. Let q0 be any point on the segment mp0. Drawthe segment qq0 parallel to pp0, with q on mp. The function τ(q) − τ(q0)is continuous in q along mp and equals O when q coincides with m. Sinced(q, q0) < η, it follows from (2.11) that |τ(q) − τ(q0)| < π. In particular, forqo = p0 this gives |τ(p) − τ(p0)| < π. Combining this with (2.12), we getk(p) = 0. Thus we have proved that τ(p) is continuous in 4, as assertedabove. Since τ(p) ≡ τ(p) mod 2π, it is clear that τ(p) is differentiable.

Now let A(O,O), B(O,L), D(L,L) be the vertices of 4. The rotationindex γ of C is, by (2.10), defined by the line integral

2πγ =∮AD

dτ .

Since τ(p) is defined in 4, we have∮AD

dτ =∮AB

dτ +∮BD

dτ .

To evaluate the line integrals at the right-hand side, we suppose the originO to be the point x(O) and C to lie in the upper half-plane and to be tangentto the x1-axis at O. This is always possible for we only have to take x(O)to be the point on C at which the x2-coordinate is a minimum. Then thex1-axis is either in the direction of the tangent vector to C at O or oppositeto it. We can assume the former case, by reversing the orientation of C ifnecessary. The line integral along AB is then equal to the angle rotated byOP as P goes once along C. Since C lies in the upper half-plane, the vectorOP never points downward. It follows that the integral along AB is equal

30 2 Curves

to π. On the other hand, the line integral along BD is the angle rotated byPO as P goes once along C. Since the vector PO never points upward, thisintegral is also equal to π. Hence their sum is 2π and the rotation index γ is+1. Since we may have reversed the orientation of C, the rotation index is±1 in general.

Exercise 34. Consider the plane curve x(t) = (t, f(t)). Use the Frenet for-mulas in (2.1) to prove that its curvature is given by

k(t) =f

(1 + f2)3/2. (2.13)

Exercise 35. Draw closed plane curves with rotation indices 0,−2, +3 re-spectively.

Exercise 36. The theorem on turning tangents is also valid when the simpleclosed curve C has ”corners.” Give the theorem when C is a triangle con-sisting of three arcs. Observe that the theorem contains as a special case thetheorem on the sum of angles of a rectilinear triangle.

Exercise 37. Give in detail the proof of the existence of η = η(p0) used inthe proof of the theorem on turning tangents. η = η(p0) .

2.8 Plane Convex Curves and the Four Vertex Theorem

A closed curve in the plane is called convex, if it lies at one side of everytangent line.

Proposition 9. A simple closed curve is convex, if and only if it can be sooriented that its curvature k is ≥ 0.

The definition of a convex curve makes use of the whole curve, while thecurvature is a local property. The proposition therefore gives a relationshipbetween a local property and a global property. The theorem is not true ifthe closed curve is not simple. Counter examples can be easily constructed.

Let ˜τ(s) be the function constructed above, so that we have k = dτds .

The condition k ≥ O is therefore equivalent to the assertion that τ(s)) is amonotone non-decreasing function. We can assume that ˜τ(O) = O. By thetheorem on turning tangents, we can suppose C so oriented that ˜τ(L) = 2π.

Suppose τ(s) , O ≤ s ≤ L, be monotone non-decreasing and that C isnot convex. There is a point A = x(s0) on C such that there are points ofC at both sides of the tangent λ to C at A. Choose a positive side of k andconsider the oriented perpendicular distance from a point x(s) of C to λ. Thisis a continuous function in s and attains a maximum and a minimum at thepoints M and N respectively. Clearly M and N are not on λ and the tangents

2.9 Isoperimetric Inequality in the Plane 31

to C at M and N are parallel to x. Among these two tangents and k itselfthere are two tangents parallel in the same sense. Call s1 < s2 the values ofthe parameters at the corresponding points of contact. Since τ(s) is monotonenon-decreasing and O ≤ τ(s) ≤ 2π, this happems only when τ(s) = ˜τ(s1) forall s satisfying s1 ≤ s ≤ s2. It follows that the arc s1 ≤ s ≤ s2 is a linesegment parallel to λ. But this is obviously impossible.

Next let C be convex. To prove that τ(s) is monotone non-decreasing,suppose ˜τ(s1) = ˜τ(s2), s1 < s2. Then the tangents at x(s1) and x(s2) areparallel in the same sense. But there exists a tangent parallel to them in theopposite sense. From the convexity of C it follows that two of them coincide.

We are thus in the situation of a line λ tangent to C at two distinctpoints A and B. We claim that the segment AB must be a part of C. In fact,suppose this is not the case and let D be a point on AB not on C. Drawthrough D a perpendicular µ to λ in the half-plane which contains C. Thenµ intersects C in at least two points. Among these points of intersection letF be the farthest from λ and G the nearest one, so that F 6= G. Then Gis an interior point of the triangle ABF . The tangent to C at G must havepoints of C in both sides which contradicts the convexity of C.

It follows that under our assumption, the segment AB is a part of C , sothat the tangents at A and B are parallel in the same sense. This proves thatthe segment joining x(s1) to x(s2) belongs to C. Hence τ(s) remains constantin the interval s1 ≤ s ≤ s2. We have therefore proved that τ(s) is monotoneand K ≥ O.

A point on C at which k′ = 0 is called a vertex. A closed curve has atleast two vertices, e.g., the maximum and the minimum of k. Clearly a circleconsists entirely of vertices. An ellipse with unequal axes has four vertices,which are its intersection with the axes.

Theorem 4 (Four-vertex Theorem.). A simple closed convex cuxve hasat least four vertices.

Remark 6. This theorem was first given by Mukhopadhyaya (1909). The fol-lowing proof was due to G.Herglotz. It is also true for non-convex curves, butthe proof will be more difficult.

2.9 Isoperimetric Inequality in the Plane

Among all simple closed curves having a given length the circle bounds thelargest area, and is the only curve with this property. We shall state thetheorem as follows:

Theorem 5. Let L be the length of a simple closed curve C and A be thearea it bounds. Then

L2 − 4πA ≥ 0 . (2.14)

32 2 Curves

Moreover, the equality sign holds only when C is a circle.

The proof given below is due to E. Schmidt (1939).We enclose C between two parallel lines g, g′, such that C lies between

g, g′ and is tangent to them at the points P , Q respectively. Let s = 0, s0be the parameters of P , Q. Construct a circle C tangent to g, g′ at P , Qrespectively. Denote its radius by r and take its center to be the origin of acoordinate system. Let x(s) = (x1(s), x2(s)) be the position vector of C, sothat

(x1(0), x2(0)) = (x1(L), x2(L)) .

As the position vector of C we take (x1(s), x2), such that

x1(s) = x1(s) ,

x2(s) = −√r2 − x2

1(s) , 0 ≤ s ≤ s0

= +√r2 − x2

1(s) , s0 ≤ s ≤ L .

Denote by A the area bounded by C. Now the area bounded by a closedcurve can be expressed by the line integral

A =∫ L

0

x1x′2ds = −

∫ L

0

x2x′1ds =

12

∫ L

0

(x1x′2 − x2x

′1)ds .

Applying this to our two curves C and C, we get

A =∫ L

0

x1x′2ds ,

A = πr2 = −∫ L

0

x2x′1ds = −

∫ L

0

x2x′1ds .

Adding these two equations, we have

A+ πr2 =∫ L

0

(x1x′2 − x2x

′1)ds ≤

∫ L

0

√(x1x′2 − x2x′1)2ds

≤∫ L

0

√(x2

1 + x22)(x′2

1 + x′22 )ds

=∫ L

0

√x2

1 + x22ds = Lr .

(2.15)

Since the geometric mean of two numbers is ≤ their arithmetic mean, itfollows that

√A√πr2 ≤ 1

2(A+ πr2) ≤ 1

2Lr .

2.9 Isoperimetric Inequality in the Plane 33

This gives, after squaring and cancellation of r, the inequality (2.14).Suppose now that the equality sign in (2.14) holds. A and πr2 have then

the same geometric and arithmetic mean, so that A = πr2 and L = 2πr.The direction of the lines g, g′ being arbitrary, this means that C has thesame ”width” in all directions. Moreover, we must have the equality signeverywhere in (2.15). It follows in particular that

(x1x′2 − x2x

′1)2 = (x2

1 + x22)(x

′21 + x

′22 ) ,

which gives

x1

x′2=−x2

x′1=

√x2

1 + x22√

x′21 + x

′22

= ±r .

From the first equality in (2.15) the factor of proportionality is seen to be r,i.e.,

x1 = rx′2 , x2 = −rx′1 .

This remains true when we interchange x1 and x2, so that

x2 = rx′1 .

Therefore we have

x21 + x2

2 = r2 ,

which means that C is a circle.

3 Fundamental Forms of a Surface

3.1 The First Fundamental Form

Let D be an open domain in the (u1, u2) plane. A parametrized surface is asmooth mapping

x : D 7→ E3 .

of D into three-dimensional Euclidean space E3 We will denote the value ofx at (u1, u2) by x(u1, u2), the position vector from the origin considered asa function of u1 and u2.

For the rest of our study of surfaces, we will be considering sums overthe indices i = 1, 2. We will therefore make adopt a convention, standardin tensor analysis, that all small Latin indices have the range 1 − −2, andthat they will be summed whenever repeated. We will also make the conven-tion that subscripts to a vector mean pertial derivatives with respect to thecorresponding parameter. Thus,

xi =∂x∂ui

and xik =∂2x

∂ui∂uketc.

The parametrized surface defined by x is said to be regular or immersedif x1(u1, u2) × x2(u1, u2) 6= 0 for every (u1, u2) in the domain D. A pointwhere x1(u1, u2)×x2(u1, u2) = 0 is said to be a singular point. In most caseswe will consider regular surfaces without singular points.

A point x of E3 can be given in terms of its coordinates (x1, x2, x3) andthe vector function x(u1, u2) can be expressed in terms of its component func-tions x(u1, u2) = (x1(u1, u2), x2(u1, u2), x3(u1, u2)). In terms of componentfunctions, the regularity condition from above means that the matrix

∂x1(u1, u2)∂u1

∂x2(u1, u2)∂u1

∂x3(u1, u2)∂u1

∂x1(u1, u2)∂u2

∂x2(u1, u2)∂u2

∂x3(u1, u2)∂u2

(3.1)

has rank two everywhere.

36 3 Fundamental Forms of a Surface

Example 2. The graph of the function f, x3 = f(x1, x2) can be expressed as aparametrized surface, x(u1, u2, f(u1, u2)). Since x1(u1, u2) = (1, 0, ∂f(u1,u2)

∂u1 )

and x2(u1, u2) = (0, 1, ∂f(u1,u2)

∂u2 ), the rank of the matrix in (3.1) is alwaystwo, so the regularity condition is satisfied at all points of the domain of f .

Example 3. The sphere of radius b is defined by the equation

x21 + x2

2 + x23 = b2

may be parametrized as

x(u1, u2) = (b sin(u1) cos(u2), b sin(u1) sin(u2), b cos(u1)) .

Then,

x1(u1, u2) = (b cos(u1) cos(u2), b cos(u1) sin(u2),−b sin(u1))

x2(u1, u2) = (−b sin(u1) sin(u2), b sin(u1) cos(u2), 0) .

So,

x1(u1, u2)× x2(u1, u2) = b sin(u1)x(u1, u2) .

At the points where sin(u1) = 0, i.e. at the north pole (0, 0, 1) and thesouth pole (0, 0,−1), the parametrized surface is not regular, although froma geometric point of view, all points of the sphere are like all others. Byrotating the sphere one quarter turn around the x1-axis, we may obtaina parametrization which is regular at the north and south poles, althoughsingular at two other points. This example shows the importance of defining asurface by using enough parametrizations so that each point is a regular pointof at least one of them. The process of defining a surface as a combination ofparametrized surfaces is at the foundation of differential geometry, and willbe discussed in detail in later chapters.

Example 4. The torus of revolution with the parametrization

x(u1, u2) = ((a+ b sin(u1)) cos(u2), ((a+ b sin(u1)) sin(u2), b cos(u1))

for constants a > b > 0 is a regular surface for all (u1, u2).

Let P = x(u10, u

20) be the point having the parameters (u1

0, u20). The equa-

tions ui = ui(t) for i = 1, 2 satisfying the conditions ui0 = ui(t0) for i = 1, 2define a curve through P on the parametrized surface. By the chain rule, thetangent vector to the curve at P is

dxdt

=2∑i=1

∂x∂ui

dui

dt=

2∑i=1

xidui

dt. (3.2)

3.1 The First Fundamental Form 37

When two indices appear in a formula, it will be assumed that they will besummed unless otherwise mentioned. Thus we may rewrite (3.2) as

dxdt

= xidui

dt.

Thus, the tangent vector to the curve is a linear combination of x1(t0) andx2(t0). All tangent vectors at P to curves passing through P lie in the planespanned by the linearly independent vectors x1(u1

0, u20) and x2(u1

0, u20). This is

called the tangent plane to the surface at P . The line through P perpendicularto this plane is called the normal line at P . It consists of all vectors of theform x(u1

0, u20) + r(x1(u1

0, u20)× x2(u1

0, u20)).

The unit normal for the parametrized surface x is defined by the condition

N =(x1 × x2)√

(x1 × x2) · (x1 × x2).

It is the unique unit vector perpendicular to the tangent plane such that thetriple product (vx1,x2,N) > 0.

By the chain rule, we observed above that a tangent vector v at P canbe written as a linear combination of the partial derivatives of x,

v = aixi = a1x1 + a2x2 ,

where a1 and a2 are the components of v (relative to the basis, x1,x2).The dot product of v with itself is then

v · v = a1a1x1 · x1 + a1a2x1 · x2 + a2a1x2 · x1 + a2a2x2 · x2 ,

which we abbreviate, using the summation convention, as

v · v = aiajxi · xj .

We introduce the metric coefficients

xi · xj = gij

so we have a final abbreviation

v · v = gijaiaj .

More generally, if w is another tangent vector at P with components bi wemany then write

w = aixi = a1x1 + a2x2 ,

and the dot product of the two vectors is


v ·w = aibjxi · xj = gijaibj .

We will write these expression using the notation

I(v,w) = v ·w and I(v) = I(v,v) = v · v.

This expression is a symmetric bilinear form on the tangent space at P , whichwe call the first fundamental form. Note that this form has the property thatI(v) ≥ 0, with I(v) = 0 if and only if v is the zero vector. A form with thisproperty is called positive definite.

The first fundamental form then gives a bilinear form on each of thetangent planes of a parametrized surface, and we can use this form to computethe length of a curve x(t) = x(u1(t), u2(t)) between the limits t1 and t2:

s =∫ t2

t1

√I(x′(t)) dt =

∫ t2

t1

√gijdui

dt

duj

dtdt (3.3)

or (ds

dt

)2

= gijdui

dt

duj

dt. (3.4)

In differential form notation, this formula for length can be written

ds2 = gijduiduj .

Exercise 38. Find the metric coefficients, gij , for the ellipsoid with equation

x(u1, u2) = (a sin(u1) cos(u2), b sin(u1) sin(u2), c cos(u1)) ,

where a, b, and c are constants. At which points does the surface fail to beregular? Find the unit normal vector at regular points.

Exercise 39. Find the metric coefficients, gij , for the surface of revolutionwith equation

x(u1, u2) = (u1 cos(u2), u1 sin(u2), f(u1)) .

At which points does the surface fail to be regular? What is the length of thecurve u1(t) = c, u2(t) = t? Find an expression for the unit normal vector.

Exercise 40. Find the metric coefficients, gij , for the right helicoid

x(u1, u2) = (u1 cos(u2), u1 sin(u2), g(u2)) .

What is the length of the curve u1(t) = t, u2(t) = c? Find an expression forthe unit normal vector.

3.1 The First Fundamental Form 39

Exercise 41. Find the metric coefficients, gij , for the catenoidwith equation

x(u1, u2) = (cosh(u1) cos(u2), cosh(u1) sin(u2), u1) .

What is the length of the curve u1(t) = t, u2(t) = c? What is the length ofthe curve u1(t) = c, u2(t) = t? Find an expression for the unit normal vector.

Exercise 42. Find the metric coefficients, gij , for the helicoid with equation

x(u1, u2) = (sinh(u1) sin(u2),− sinh(u1) cos(u2), u2) .

What is the length of the curve u1(t) = t, u2(t) = c? What is the length ofthe curve u1(t) = c, u2(t) = t? Find an expression for the unit normal vector.

Exercise 43. Find the metric coefficients, gij , and the unit normal vectorfor the function graph

x(u1, u2) = (u1, u2, f(u1, u2)) .

Exercise 44. Find the metric coefficients, gij , and the unit normal vector ofthe torus

x(u1, u2) = ((a+ b sin(u1)) cos(u2), ((a+ b sin(u1)) sin(u2), b cos(u1)) .

3.1.1 Geometry in the Tangent Plane

In order to study the geometry of the tangent plane at a point P on a surface,we need a notion of the angle φ between two non-zero vectors v and w. Thisis obtained by using the first fundamental form on the surface to define

cos(φ) =I(v,w)√I(v)I(w)

.

In particular, the angle φ1 between a vector v and the vector xi is given by

cos(φi) =v · xi√I(v)I(xi)

=ajgijgiiI(v)

with no summation on the index i.Two vectors v and w in the tangent plane at P are orthogonal if I(v,w) =

0. If v and w are orthogonal, so are the scalar multiples, av and bw. Thetotality of all non-zero scalar multiples of a vector v is called the directiondetermined by v. Thus, a direction is determined by the ratio a1 : a2 of itscomponents. Orthogonality is therefore a property of a pair of directions.

Suppose that two directions are determined by the quadratic conditioncija

iaj = 0 where cij = cji. Then these directions are orthogonal if and onlyif g22c11−2g12c12 +g11c22 = 0. To see this, let p and q denote the solutions of


the quadratic condition, c11 + 2c12x+ c22x2 = 0, so p and q satisfy pq = c11

c22

and p+ q = −2c12c22

. Let v = aixi and w = bjxj be two non-zero vectors with

directions p and q respectively. Set p = a2

a1 and q = b2

b1 . Then

I(v,w)a1b1

= g11 +b2

b1g12 +

a2

a1g21 +

a2

a1

b2

b1g22

= g11 + (p+ q)g12 + (pq)g22

= g11 − 2c12c22

g12 +c11c22

g22

=1c22

(g22c11 − 2g12c12 + g11c22) .

So, I(v,w) = 0 if and only if

g22c11 − 2g12c12 + g11c22 = 0 .

Exercise 45. The curves defined by u1 = constant for a fixed index i arecalled parametric curves of the surface. There are therefore two families ofparametric curves on a surface, and through each point of the surface therepasses exactly one parametric curve from each family. Find the conditionthat the parametric curves are orthogonal. Under which conditions will theparametric curves meet at a 45 ◦ angle?

Exercise 46. For each of the surfaces in the problem set for (3.1), find theangle between parametric curves.

Exercise 47. Consider a family of curves given by the conditions ciai = 0,for a fixed set of constants ci. Show that the orthogonal trajectories of thisfamily satisfy the condition

(g11c1 − g12c2)a1 + (g12c1 − g22c2)a2 = 0 .

Use this formula to find the orthogonal trajectories of the family of circlesgiven in polar coordinates by r = c cos(φ). Sketch the curves in this familyand sketch their orthogonal trajectories.

3.2 The Second Fundamental Form

The first fundamental form arises from studying the length of curves on asurface and the angle between vectors in the tangent planes. The study ofcurvature for curves on a surface leads to another bilinear form, the secondfundamental form. Let C be a curve on a parametrized surface x(u1, u2)determined by a curve (u1(t), u2(t)) in the parameter domain D. We writex(t) = x(u1(t), u2(t)), and we let T(t) and N(t) denote the unit tangent and

3.2 The Second Fundamental Form 41

(unit) principal normal vectors at x(t). As a curve in space, the curvaturek(t) of x(t) is given by the Frenet formula

k(t)s′(t)N(t) = T′(t) . (3.5)

We now wish to express this formula in terms of quantities associated withthe parametrized surface.

If φ(t) denotes the angle between the principal normal N(t) and the unitnormal ν(t) to the surface at x(t), then, taking the dot product of both sidesof (3.5) with ν(t) and using the fact that T(t) · ν(t) = 0, we obtain

k(t) cos(φ(t))s′(t) = T′(t) · ν(t) = −T(t) · ν′(t) . (3.6)

This can also be written

k(t) cos(φ(t))(s′(t))2 = −T(t)s′(t)ν′(t) = −x′(t) · n′(t) . (3.7)

By the chain rule, x′(t) = xi(t)dui

dt and ν′(t) = nj(t)duj

dt . Therefore we maywrite

−x′(t) · ν′(t) = −xi(t) · νj(t)dui

dt

duj

dt= hij(t)

dui

dt

duj

dt(3.8)

where hij = −xi ·νj are the coefficients of a differential form called the secondfundamental form. Notice that the condition xi · ν = 0 leads to the relation

xij · ν = −xi · νj = hij , (3.9)

from which it follows that hij ≤ hji, so the second fundamental form issymmetric. Recall that

(x1 × x2) · (x1 × x2) = (x1 · x1)(x2 · x2)− (x1 · x2)(x2 · x1) = g11g22 − g12g21

by Lagrange’s identity (1.15). We set g = g11g22 − g12g21, so we may writeI(x1 × x2) = g and ν = (x1×x2)√

g . Therefore (3.9) can be rewritten

hij =(x1,x2, ν)√g

. (3.10)

For two tangent vectors v = aixi and w = bjxj , we define the bilinear formII(v,w) = hija

ibj to be the second fundamental form on each tangent spaceof the parametrized surface. For a single tangent vector v, we set II(v,v) =II(v) = hija

iaj. Note that II(v,w) = II(w,v). We can now rewrite (3.5)as

k(t) cos(φ(t)) =II(x′(t))I(x′(t))

. (3.11)


The right hand side of this expression depends only on the direction of x′(t).This indicates that k(t) cosφ(t) will be the same for any two curves througha point that have the same tangent direction there. Given a vector v = aixiin the tangent plane at a point, there are many curves C = x(t) through thepoint on the surface so that the tangent vector x′(t) has the same directionas v. One such curve is obtained by cutting the surface by the plane deter-mined by v and the surface normal n at P , the so-called normal section atP in the direction of v. The curvature of the normal section in the directionof v is called the normal curvature and is denoted kN(v). Hence we mayrewrite (3.11) as

k(t) cos(φ(t)) =II(x′(t))I(x′(t))

= kN(x′(t)) . (3.12)

This relationship is known as Meusnier’s Theorem. It determines the curva-ture k(t) of a curve in terms of the normal curvature and the angle betweenthe principal normal and the surface normal, provided that the cosφ(t) isnot zero, i.e. provided that the principal normal does not lie in the tangentplane at the point. If N(t) does lie in the tangent plane at x(t), then theosculating plane of the curve coincides with the tangent plane of the surfaceat that point. Otherwise the tangent line of the curve is the intersection ofthe osculating plane of the curve and the tangent plane of the surface, andit follows that two curves through P on the surface with the same osculatingplane at P must have the same curvature there.

Exercise 48. Find the normal curvature at a point of the circle of latitude

x(u1, u2) = (cos(u1) cos(u2), cos(u1) sin(u2), sin(u1))

where u1 = c, a constant.

Exercise 49. Calculate the normal curvature at the point (0, 0, 0) for thecurve determined by u1(t) = t cos(b), u2(t) = tsin(b) on the surfacex(u1, u2) = (u1, u2, p(u1)2 + q(u2)2) for constants p and q.

Exercise 50. Find the normal curvature of the parametric curves in theexercises 1-6 in section 1.1. Find the coefficients of the second fundamentalform for each of these surfaces.

3.2.1 The Shape of a Surface

Consider a parametrized surface x(u1, u2) such that x(0, 0) = P. By Taylor’sTheorem, we have

x(u1, u2) = x(0, 0) + xi(0, 0)ai + 12xij(0, 0)aiaj +O(r3)

where r2 = (u1)2 + (u2)2. Then the distance from x(u1, u2) to the tangentplane at x(0, 0) is given by


(x(u1, u2)− x(0, 0)) · ν(0, 0) = 12hij(0, 0)aiaj +O(r3) ,

where we set xij(0, 0) · ν(0, 0) = hij(0, 0). This expression is dominated bythe quadratic term, so the shape of the surface will be similar to the shapeof the graph of the quadratic function f(x1, x2) = h11x

21 + 2h12x1x2 +h22x

22.

This shape is determined by the expression h = h11h22−(h12)2. When h > 0,the graph of the quadratic function intersects the horizontal plane only atthe origin, while if h < 0, the intersection is a pair of intersecting lines.

The point P of a parametrized surface is called elliptic, hyperbolic, orparabolic according to if h is positive, negative, or zero at P. At an ellipticpoint P , the tangent plane meets the surface only at P locally, i.e. there isa neighborhood of P that meets the tangent plane only at P. On the otherhand, if P is hyperbolic, the intersection of the tangent plane and a smallneighborhood of P will consist of two curves with distinct tangent lines, soarbitrarily close to P , there are points of the surface on both sides of thetangent plane. At a parabolic point, the behavior of the surface can be quitea bit more complicated (see the exercises at the end of this section for someexamples.)

The behavior of a parametrized surface in a neighborhood of P can alsobe described in terms of the normal sections at P , i.e. the intersections ofthe surface with planes through the normal line at P . If the curvature κN ofsuch a normal section is non-zero, then the center of the osculating circle tothis curve is P + 1

κNν. At an elliptic point, all normal curvatures will have

the same sign and the centers of osculating circles for all normal sectionslie on the same side of the tangent plane. At a hyperbolic point, the normalcurvature takes on both positive and negative values, and the centers of someosculating circles lie on one side of the tangent plane and some will lie on theother. At a parabolic point, at least one normal section has normal curvature0.

Exercise 51. For all surfaces in the exercises 1-6 of section 1.1, which pointsare elliptic, which are hyperbolic and which are parabolic?

Exercise 52. Let x(u1, u2) = (u1, u2, (u1)3−3u1(u2)2). Show that the originis the only parabolic point and that all of the metric coefficients hij are zeroat this point. (Such a point is called planar.) What is the intersection of thesurface and the tangent plane at this point? (This point is called a monkeysaddle. A saddle for a person riding a bicycle has two dips, but a saddle fora monkey requires a third depression for the tail.)

Exercise 53. Let x(u1, u2) = (u1, u2, (u1)4 + (u2)4). Show that the originis the only parabolic point which is planar (i.e. all of the metric coefficientshij are zero at this point.) What is the intersection of the surface and thetangent plane at this point?

Exercise 54. Let x(u1, u2) = (u1, u2, (u1)4 + c(u2)2) for a constant c. Showthat the origin is the only parabolic point. What is the intersection of thesurface and the tangent plane at this point?


Exercise 55. Let x(u1, u2) = (u1, u2, ((u1)2−u2)((u1)2−cu2)) for a constantc. Show that the origin is a parabolic point. What is the intersection of thesurface and the tangent plane at this point? Which points are parabolic,which are ellipic, and which are hyperbolic?

Exercise 56. Show that all points on a hyperboloid of one sheet are hyper-bolic:

x21

a21

+x2

2

a22

+x2

3

a23

= 1 .

Exercise 57. For the torus of revolution of problem 7 in section 1.1, whichpoints are parabolic, which are elliptic, and which are hyperbolic?

3.2.2 Characterization of the Sphere

A point P is called an umbilic if the curvatures of all normal sections areequal, so the normal curvature is independent of the direction in the tangentspace. Since κN = II(v)

I(v) , the condition for P to be an umbilic is equivalentto hij = cgij for all indices i, j. The constant c = c(u1, u2) depends on thepoint P .

For a plane, all points are umbilic with normal curvature 0. For a sphereof radius r, all points are umbilic with normal curvature 1

r . We shall nowprove the converse.

Proposition 10. A surface consisting entirely of umbilics is either a planeor a sphere.

We have c(u1, u2)xi · xj = −xij so (cxj + νj) · xi = 0. Furthermoreν · ν = 1 implies that ν · νj = 0, so (cxj + νj) · ν = 0 as well. Thereforecxj + νj , being perpendicular to three independent vectors, must be the zerovector, so cxj + xj = 0. By differentiating this expression, we have

cxij + cixj + νij = 0 (3.13)

and similarly,

cxji + cjxj + νij = 0 . (3.14)

Since we are assuming that the second partial derivatives of x and ν arecontinuous, we have xij = xji and νij = νji. Subtracting (3.13) from (3.14),we obtain cixj − cjxi = 0. Since x1 and x2 are linearly independent byhypothesis, it follows that ci = 0 = cj identically, so c is constant.

If c = 0, then νi = 0 = νj so ν is constant, equal to a fixed unit vector a.Then (x · a)i = xi · a = 0 for i = 1, 2 so x · a = d, a constant, and x thereforelies in the plane perpendicular to a at distance d from the origin.


If c 6= 0, then (x − ( 1c )n)i = 0fori = 1, 2, so x − (1/c) = c, a con-

stant vector, and x(u1, u2) − C = 1c (u1, u2) for all u1, u2. Thus the length

of x(u1, u2) − C is a constant, 1c , and the parametrized surface lies on the

sphere of radius 1c about the center C.

Corollary 2. A surface consisting entirely of planar points is a plane.

3.2.3 Principal Curvatures, Principal Directions, and Lines ofCurvature

We now wish to study the normal curvature κN as a function of the tangentvector v = aixi 6= 0. As remarked earlier, κN depends only on the directionof the vector v and not on its length, so we consider only unit vectors v, forwhich gijaiaj = 1. Then κN is a continuous function of the variables (a1, a2)on the unit circle in the tangent plane so it must have a maximum and aminimum. We set I = I(a1, a2) and II = II(a1, a2). Then ∂I

ai = 2gijaj and∂IIai = 2hijaj so

∂κN

ai=

[(2gijaj)II − (2hijaj)I]I2

=2(κNgija

j − hijaj)I

.

At a maximum or minimum of κN, we therefore have

hijaj − κNgija

j = 0 .

Thus,

(hi1a1 + hi2a2) = κN(gi1a1 + gi2a

2)

so, by eliminating κN, we obtain

(h11a1 + h12a

2)(g21a1 + g22a2) = (h21a

1 + h22a2)(g11a1 + g12a

2) ,

Expanding this out, we obtain the condition

(g11h21 − g21h11)(a1)2 + (g11h22 − g22h11)(a1a2) + (g12h22 − g22h12)(a2)2 = 0 .

At an umbilic this expression is satisfied identically, and at a non-umbilicpoint, there must be two solutions since the function must have at least onemaximum and one minimum. The solutions of this quadratic equation givethe directions for which κN attains its maximum and minimum, and theseare called principal directions. By our previous computation in section 1.2,,it follows that the principal directions at a non-umbilic point are orthogonal.

The maximum and minimum values of κN are called principal curvatures.To determine them, we rewrite

(hi1a1 + hi2a2) = κN(gi1a1 + gi2a

2)


as

(hi1 − κNgi1)a1 + (hi2 − κNgi2)a2 = 0 .

Since this system of equations has non-trivial solutions, we must have

(h11 − κNg11)(h22 − κNg22)− (h21 − κNg21)(h12 − κNg12)

so κN satisfies the quadratic equation

g(κN)2 −mκN + h = 0

where m = g11h22−2g12h12 + g22h11. The roots are the principal curvatures,denoted κ1 and κ2.

The average of the roots of this equation is called the mean curvature ofthe surface at the point, denoted by H, so

H =κ1 + κ2

2=m

2g=

(g11h22 − 2g12h12 + g22h11)2g

.

The product of the roots of this equation is called the total curvature orGaussian curvature of the surface at the point, so

K = κ1κ2 =h

g.

The principal curvatures, mean curvature, and Gaussian curvatures arescalar functions defined on the parametrized surface. They play an importantrole in the description of the geometric properties of the surface. For examplea point is elliptic, parabolic, or hyperbolic according to if K is positive, zero,or negative. Furthermore

H2 −K =(κ1 − κ2)2

4≥ 0 ,

with equality only at an umbilic.A curve for which all unit tangent vectors are principal directions is called

a line of curvature on the surface. For such a curve, x′(t) = xi dui

dt and thecondition that each such vector determine a principal direction is

g11h21 − g21h11)(du1

dt

)2

+ (g11h22 − g22h11)du1

dt

dtwo

dt+ (g12h22 − g2h12)

(du1

dt

)2

= 0 .

In the language of differential forms, this can be written

(g11h21 − g21h11)(du1)2 + (g11h22 − g22h11)du1du2 + (g12h22g22h12)(du2)2 = 0 .

This is a quadratic equation in the first derivatives of the componentfunctions u1(t) and u2(t), and by a standard existence theorem in the theoryof ordinary differential equations, it follows that in any portion of the domainwhere there are no umbilics, it is possible to find two families of solutioncurves, the lines of curvature. These form an orthogonal net, so that at eachpoint there are two lines of curvature with perpendicular tangent vectors.

3.4 Triply Orthogonal System; Theorems of Dupin and Liouville 47

Exercise 58. Find conditions in terms of gij and hij for the parametriccurves to be lines of curvature.

Exercise 59. For which of the six surfaces in the exercises 1-6 of section 1.1are the parametric curves lines of curvature?

Exercise 60. Find the lines of curvature for the right helicoid.

3.3 Gauss Mapping and the Third Fundamental Form

The unit normal vector ν = (u1, u2) can be considered as a point on the unitsphere S0 centered at the origin of our space E. The resulting mapping

ν : x 7→ S0

which assigns to the point x(u1, u2) the point ν(u1, u2), is called the Gaussmap. Using the first fundamental form on S, we define the third fundamentalform

III(v,w) = mijaibj

where v = aixi and w = bjxj and mij = νi · νj .

3.4 Triply Orthogonal System; Theorems of Dupin andLiouville

The notion of a parametrized surface can be generalized to that of a triplesystem, as follows: Let ∆ be an open domain in the (ul, u2, u3)-space. Considera mapping

x : ∆ 7→ E (3.15)

of ∆ into our Euclidean space E such that x(u1, u2, u3) has continuous firstand second partial derivatives and that the Jacobian determinant

∂(x1, x2, x2)∂(u1, u2, u3)

6= 0 . (3.16)

The last condition can also be written

det(x1,x2,x3) 6= 0 (3.17)

where

xi =∂xui

(3.18)


for i = 1, 2, 3.The name triple system derives from the fact that each of the equations

defines a family of parametric surfaces in E. The system is called a triplyorthogonal system, if

xi · xj = 0 (3.19)

for i, j = 1, 2, 3, i 6= j, i.e., two surfaces of different families are everywhereorthogonal.

Example 5. Consider the spherical coordinates defined by the equations

x = r cosu1 cosu2

x = r cosu1 sinu2

x = r sinu1 .

(3.20)

The three families of surfaces are:

1. The spheres r = constant.2. The cones of revolution u = constant.3. The half-planes u2 = constant.

We leave it as an exercise to verify that (3.20) defines a triply orthogonalsystem.

Theorem 6 (Dupin’s Theorem). The surfaces of a triply orthogonal sys-tem intersect each other in lines of curvature.

Proof. We take a surface u3 = constant. It suffices to show that on it theparametric curves u1 = constant and u2 =constant are lines of curvature. Wedifferentiate (3.19) with respect to uk, for k 6= i, j, getting xik ·xj+xjk ·xi = 0.These equations are, when written in detail,

x13 · x2 + x23 · x1 = 0x21 · x3 + x31 · x2 = 0x32 · x1 + x12 · x3 = 0 .

It follows that

x12 · x3 = x23 · x1 = x31 · x2 = 0 .

The first equation shows that x12 is orthogonal to x3. By (3.19), so are x1

and x2. Hence we have

det(x1,x2,x3) = 0 .

On the surface u3 = constant we have therefore g12 = h12 = 0.


A non-trivial example of a triply orthogonal system is given by the “con-focal quadrics.” For central quadrics this is the family defined by the equation

x21

t− a1+

x22

t− a2+

x23

t− a3= 1 (3.21)

where t is the parameter and where we assume 0 < a3 < a2 < a1. If (xl, x2, x3)are the coordinates of a given point distinct from the origin, then (3.21) is acubic equation in t, which is easily seen to have three real roots t1, t2, and t3satisfying the inequalities

t1 > a1 > t2 > a2 > t3 > a3 . (3.22)

Through this point there pass therefore three quadrics of the family: an el-lipsoid t1, a hyperboloid of one sheet t2, and a hyperboloid of two sheets t3.It can be verified that they are mutually orthogonal. (t1, t2, t3) can be takenas coordinates in a domain not containing the origin. They are called ellip-tic coordinates. These coordinates are useful in studying the properties ofellipsoids, which are among the most important surfaces.

Since every central quadric can be considered a member of a confocalfamily, the following property of its lines of curvature follows from Dupin’stheorem:

Proposition 11. The lines of curvature of a central quadric are its curvesof intersection by the confocal quadrics and are generally quartic curves.

An application of Dupin’s theorem gives Liouville’s theorem on conformalmappings in the three-dimensional Euclidean space E. A mapping of a do-main of E into another domain is called conformal, if it preserves the anglebetween two curves. A motion in E is conformal; so is an inversion (INSERTREFERENCE HERE). We say that a surface is totally umbilical if it consistsentirely of umbilics. By (10), a totally umbilical surface is either a plane ora sphere. In fact, we proved that such a surface satisfies the condition

νi + ρxi = 0 (3.23)

for i = 1, 2 where ρ is a constant.

Theorem 7 (Liouville’s Theorem). A conformal mapping E maps a to-tally umbilical surface into a totally umbilical surface.

Proof. Let S be a totally umbilical surface, given analytically by x(u1, , u2).We suppose that the parametric curves are orthogonal, so that

x1 · x2 = 0 . (3.24)

Let

x(u1, u2, u3) = x(u1, u2) + u3ν(u1, u2) . (3.25)


Then

xi = xi + u3νi = (1− ρu3)xi (3.26)

for i = 1, 2. So,

x3 =∂x∂u3

= ν . (3.27)

It follows that x defines a triply orthogonal system where 1− ρ u3 6= 0.Let S′ be the image of S under the conformal mapping T . Under a triply

orthogonal system goes into a triply orthogonal system. It follows by Dupin’stheorem that the parametric curves on S′ are lines of curvature. Let C ′ beany curve on S′. We can choose parameters u1 and u2 on S′ so that theparametric curves are orthogonal and C ′ is a parametric curve. Under themapping T−1 these can be used as parameters on S. T being conformal, Shas orthogonal parametric curves. By what was proved above, C ′ is a lineof curvature on S′. Since any curve on S′ is a line of curvature, S′ is totallyumbilic.

Remark 7. Liouville’s theorem is a local theorem. It can be proved that thelocally conformal transformations in E depends on ten parameters and aregenerated by the similarity transformations and inversions. In contrast thelocally conformal transformations in the plane depend on arbitrary functions.

Exercise 61. The cylindrical coordinates r, θ, z in E are defined by

x = r cos θx = r sin θx = z .

Show that they form a triply orthogonal system and describe the three fam-ilies of surfaces.

Exercise 62. Prove the following properties on confocal quadrics statedabove:

a) Through a point distinct from the origin there pass three confocal quadrics:an ellipsoid, a hyperboid of one sheet, and a hyperboloid of two sheets.

b) Two confocal quadrics of different families meet orthogonally.

Exercise 63. Let x∗i = fi(x1, x2, x3) for i = 1, 2, 3 define a mapping in Ewith

∂(x∗1,x∗2x∗3)

∂(x1,x2,x3)6= 0 .

Prove that it is conformal if

dx∗21 + dx∗22 + dx∗23 = λ(dx21 + dx2

2 + dx23)

for λ > 0. Hence prove that an inversion is conformal.


Exercise 64. Let z = x + ıy and w = u + ıv be complex variables andw = f(z) be a holomorphic function. The function defines a mapping(x, y) 7→ (u, v). Show that it is conformal at a point z0. where f ′(z0) 6= 0.Hence show that a conformal mapping in the plane generally does not mapa circle into a circle or a line.

4 Fundamental Equations in Surface Theory

In curve theory, the fundamental invariants are the curvature and torsion,and the fundamental equations are the Frenet formulas. These express thederivatives of the vectors T, N, and b of the Frenet frames of a curve in termsof these vectors T, N, and b themselves, with the coefficients essentially beingthe curvature and torsion.

The situation in surface theory is more complicated, but conceptually nodifferent. The fundamental invariants are now the coefficients of the first andsecond fundamental forms, namely the six functions gij and hij of the pa-rameters (u1, u2). These functions are not independent, and the relationshipbetween them is contained in the theorems of Gauss and Codazzi, to be es-tablished in (INSERT REF) of this chapter. Taking the place of the Frenetframe is the frame x1, x2, ν, where x1 and x2 are not orthogonal unit vectorsin general. The counterpart of the Frenet equations are the Weingarten andGauss equations, expressing the partial derivatives of x1, x2, ν in terms ofthe vectors x1, x2, ν themselves.

4.1 Weingarten and Gauss Equations

Since any vector in E can be written as a linear combination of the threelinearly independent vectors x1, x2, ν, we may write

νi = Ajixj +Biν .

To determine the coefficients Aji and Bi, we take the dot product with ν, andthe fact that ν · ν = 1 immediately leads to ν · νi = 0 so Bi = 0. Taking thedot product of νi with xj leads to the expression (MISSING EXPRESSION)and we may then solve for the coefficients Aji . To do so, we introduce theinverse matrix (gij) to the matrix (gij), so

gijgjk = dki .

We then get

−hikgkm = Ajigjkgkm = Ajid

mj = Ami .

54 4 Fundamental Equations in Surface Theory

Thus in the original expression, we have Aji = −hikgkj , which we now denoteby −hji . The Weingarten equations for the surface are then

νi = −hjixj .

From the Weingarten equations, we immediately derive the theorem,proved earlier:

Theorem 8. A surface consisting entirely of planar points is a plane.

This follows since the hypothesis implies that hij = 0 for all i, j, so νi = 0 =νj and is constant. This gives xi ·ν = (x ·ν)i = 0 and (x ·ν)j = 0 so x ·ν = c,a constant, the condition for the surface x to lie in a plane.

More subtle is another consequence of the Weingarten equations:

Proposition 12. A surface with Gaussian curvature K = 0 is, in the neigh-borhood of a non-planar point, generated by a family of lines such that thetangent plane remains the same along any generator.

Proof. By hypothesis, we have h11h22 − (h12)2 = 0 where h11 and h22 arenot both zero. Suppose h 6= 0. Then the equation of the asymptotic curves is

h11(du1)2 + 2h12du1du2 + h22(du2)2 = 0 =

(√h11du1 +

√h22du2

)2

du1 =√h22

h11du2 .

We can now make a change of parameters from (u1, u2) to (u1, u2) so that theparametric curve u1 = constant are asymptotic. Then h22 = 0 and h12 = 0as well so ν2 = 0 and ν = ν(u1) is a function of u1 alone. This implies thatthe normal vectors remain parallel to one another along an asymptotic curve.

Also, we have

(x · ν)2 = x2 · ν + x · ν2 = 0 ,

so x · ν = p(u1) is a function depending only on u1. Differentiating withrespect to u1 gives x · ν1 = p′(u1). Note that

ν1 = −hj1xj = −h1kgkjxj 6= 0

since h11 6= 0 6= g11. Therefore the asymptotic curve lies in this plane x ·ν1 =p′(u1). But it already lies in the plane x · ν = c and the vectors ν and ν1 aredistinct, so it follows that the asymptotic curve lies in the intersection of twodistinct plane, and therefore that it is a straight line. As remarked earlier,the normal vector is constant along this line, so the theorem is proved.

A surface with K = 0 and no planar points is called developable. The rea-son for this is that any such surface is isometric to a portion of the plane. We

4.1 Weingarten and Gauss Equations 55

may approach this fact in a different way by concentrating on ruled surfacesand singling out those for which the Gaussian curvature is everywhere 0. Aruled surface is of the form x(u1, u2) = y(u1) + u2z(u1) for a regular curvey(u1) and a unit vector field z(u1). Then x1(u1, u2) = y′(u1) + u2Z ′(u1)while X2(u1, u2) = z(u1). Thus

x1(u1, u2)× x2(u1, u2) = y′(u1)× z(u1) + u2Z ′(u1)× z(u1) .

When we normalize this vector, the result will be independent of u2 onlywhen y′(u1)× z(u1) = 0 identically, i.e. the vector z(u1) is a multiple of thetangent vector y′(u1) so we may assume z(u1) = T(u1), the unit tangent.Thus the ruled surface is the tangential surface of the curve y(u1) and wehave already seen that the metric coefficients of that surface depend only onthe curvature of the curve, not on its torsion. There is therefore a curve in theplane with the same curvature and torsion zero, and the tangential surfaceof this curve, a portion of the plane, will be isometric to the original surface.

The Gauss equations express the second partial derivatives Xik in termsof the vectors x1, x2, and ν. Let

xik = Γ jikxj + cikν

where Γ jik = Γ jki, cik = cki. Taking the dot product with ν gives cik =xik · ν = hik by the definition of the second fundamental form. To determinethe coefficients Γ jik, we take the dot product with xm to obtain

xik · xm = Γ jikxj · xm = Γ jikgjm .

We set Γimk = Γ jikgjm (being careful about the order of the subscripts). Notethat Γimk = Γkmi but in general there is no symmetry in the other subscripts.

Using the inverse of the first fundamental form, we may write Γ jik =Γimkg

jm. This process is called “raising or lowering the indices” and the setsof coefficients Γ jik and Γimk (or hji and hij) are called “associated”; one setcompletely determines the other.

We then have xik · xm = Γimk. It follows that

∂gim∂uk

=∂

∂ukxi · xm = xik · xm + xi · xmk = Γimk + Γmik ,

and similarly

∂gmk∂ui

= Γmki + Γkmi ,

∂gki∂um

= Γkim + Γikm .

Adding these last two equations and subtracting the first, we obtain:

Γikm =12

(∂gmk∂ui

+∂gki∂um

− ∂gim∂uk

).


It follows that the coefficients Γikm, and hence also the Γ jik, can be expressedcompletely in terms of the coefficients gik of the first fundamental form andtheir partial derivatives. The coefficients Γikm and Γ jik are called the Christof-fel symbols of the first and second type. It is remarkable that they are intrinsic,depending only on the first fundamental form of the surface.

In summary, the Gauss equations of the surface are

xik = Γ jikxj + hikν .

Exercise 65. When Gauss originally introduced the first and second funda-mental forms, he did not use multi-indexed quantities, rather writing

I = Edu2 + 2Fdudv +Gdv2

II = Ldu2 + 2Mdudv +Ndv2 .

Exercise 66. Show that H = 12EN−2FM+GL

EG−F 2 and K = LN−M2

EG−F 2 .

Exercise 67. Suppose that the lines of curvature are parametric curves, soF = 0 = M . Show that the Weingarten equations become νu = −LXuE andνv = −NXvG , and use these to prove the identity KI−2HII+III = 0 relatingthe three fundamental forms.

Exercise 68. Use the result of the previous problem to prove that for anasymptotic curve, we have w2 = −K, where w denotes the torsion of thecurve (a theorem of Beltrami and Enneper).

Exercise 69. If the parametric curves are orthogonal, so F = 0, show that

Γ 111 =

12EuE

, Γ 211 = −1

2EvG

,

Γ 112 =

12EvE

, Γ 212 =

12GuG

,

Γ 122 = −1

2GuE

, Γ 222 =

12GvG

.

Exercise 70. Determine the Christoffel symbols corresponding to polar co-ordinates in the plane.

Exercise 71. Prove that Γ iik = ∂∂uk

log g where g = g11g22 − (g12)2.

4.2 Levi-Civita Parallelism

Levi-Civita parallelism follows as an application of the Gauss equations.Consider a curve C on a surace x defined by ui = ui(t). At each point of

C, let y(t) = yi(t)xi(t) be a tangent vector to x, but not necessarily to C.Then

4.2 Levi-Civita Parallelism 57

y′(t) =dyi

dtxi(t) + yi(t)x′i(t)

=dyi

dtxi(t) + yj(t)

[xjk(t)

duk

dt

]=dyi

dtxi(t) + yj(t)

duk

dt

[Γ ijkxi(t) + hjkν

]=[dyi

dt+ Γ ijk

duk

dtyj(t)

]xi(t) + yj(t)

duk

dthjkν .

The vector

Dydt

=[dyi

dt+ Γ ijk

duk

dtyj(t)

]xi(t)

is the projection of the derivative of y(t) into the tangent space at y(t). Fromthe vector field y(t), we thus obtain another tangent vector field Dy

dt calledthe absolute or covariant derivative of yY.

In differential formalism, we may write the absolute or covariant differ-ential as

Dy =[dyi + Γ ijkdu

kyj(t)]xi(t) .

This differential form depends only on the first fundamental form of x, onthe equations ui(t) determining the curve C, and the vector field y(t).

Let z(t) be a second tangential vector field along C. Then

d(y · z) = dy · z + y · dz = Dy · z + y ·Dz ,

since the differences Dy−dy and Dz−dz are multiples of the normal vectorso their dot product with the tangent vectors z and y will be zero.

The vector field y(t) is said to be parallel along C if Dy = 0 for all t.It follows that if two vector fields y and z are parallel along C, then theirdot product is constant. In particular, if y is a parallel vector field, then thelength of y(t) is constant.

The condition that y(t) be parallel along C is then

dyi

dt+ Γ ijk

duk

dtyj(t) = 0 .

This is a system of linear homogeneous ordinary differential equations offirst order, with the coefficients yi(t) as dependent variables and t as theindependent variable. Note that the coefficients Γ ijk and duk

dt are all functionsof t. By a basic existence theorem in ordinary differential equations, thereexists a uniquely determined solution yi(t), which takes given initial valuesyi(t0) at t = t0. Geometrically this can be described by saying that a vectory(t0) at t0 can be displaced parallelly along C.


As an example, consider the two-dimensional Euclidean plane with firstfundamental form ds2 = (du1)2 + (du2)2. Then Γ ijk = 0 and the above equa-

tions show that a vector field y(t) is parallel if and only if dyi

dt = 0 for all t, sothe coefficients yi(t) are constant. Thus in this case, Levi-Civita parallelismcoincides with ordinary parallelism.

Consider two surfaces x and x intersecting along a curve C. The surfacesare said to be tangent along C if they have the same normal vectors at eachpoint of C. Since the absolute differential is defined by the use of the normalvector, it follows that when two suraces are tangent to each other along C,the absolute differential is the same whether C is considered as a curve oneither of the surfaces.

This property leads to the following geometrical construction of Levi-Civita parallelism: The tangent planes to x along C envelop a developablesurface S.

We can make this more explicit by observing that the tangent planes atx(t0) and x(t0 + h) will intersect in a line perpendicular to both ν(t0) andν(t0 + h). This line has direction

ν(t0)× ν(t0 + h) = ν(t0)× [ν(t0 + h)− (t0)] ,

and the limit of this as h tends to zero will have the same direction as ν(t0)×ν′(t0) = v(t0). Then the surface z(t, v) = x(t) + vv(t) will have

z1 = x′(t) + vv′(t) , z2 = v(t) ,

so

z1 × z2 = x′(t)× v(t) + vv′(t)× v(t) .

But v(t) = ν(t) × ν′(t) so v′(t) = ν(t) × ν′′(t) so v′(t) lies in the tangentplane. It follows that the normal to z is the same as the normal to x. Wewould like to show that the surface z has Gaussian curvature everywhere 0.But z22 = 0 so −h22 = 0 and z12 = v(t) so z12 · ν(t) = −h12 = 0 also.Therefore h11h22 − h2

12 = 0 and K = 0.To determine parallelism of a vector field y along C it suffices to consider

parallelism along C considered as a curve on S. But S is isometric to a regionin the plane, where the parallelism coincides with ordinary parallelism. Wemay then construct parallelism on the original surface x by reversing theprocess.

The curve C defined by the coordinate functions ui(t) is called a geodesicif it is auto-parallel, i.e. if the tangent vector field is parallel to itself alongC. If x(t) gives the equation for the curve C, then the tangent vector isx′(t) = dui

dt xi(t). Thus in the above calculations, we use yi(t) = dui

dt and thecondition for Dx′(t)

dt = 0 is

d2(ui)dt2

+ Γ ijkduk

dt

duj

dt= 0 .


This is the second-order system of ordinary differential equations satisfiedby the geodesics, with dependent variables ui(t) and independent variablet. By a basic existence theorem in ordinary differential equations, there is auniquely determined solution ui(t) that takes on given initial values ui(t0)and dui

dt at t = t0. Geometrically, this means that for each point and eachtangent vector at that point, there is a uniquely determined geodesic throughthat point with the given tangent vector as its derivative at the point.

Since the tangent vectors to the geodesic are parallel, in particular theyhave the same length. There is no loss of generality in assuming that thesetangent vectors are all of unit length, so t becomes the arc-length parameter.

We now show that the geodesics on a surface can be characterized interms of the previously defined geodesic curvature.

Proposition 13. The geodesics on a surface are the curves of zero geodesiccurvature.

Proof. The geodesic curvature of a curve x(t) is given by kg(t) = x′′(t)·u(t)x′(t)·x′(t) ,

where u(t) = ν(t)×T(t). It can also be determined by the equation

kg(t)s′(t) =dTdt· u(t) =

DTdt· u .

Thus the condition that DTdt = 0 implies that the geodesic curvature is zero.

Moreover, since T ·T = 1, we have

dTdt·T = 0 =

DTdt·T

so DTdt is a multiple of u, specifically

DTdt

= kg(t)s′(t)u(t) .

Thus if the geodesic curvature is zero along the curve, it follows that DTdt = 0

so the curve is a geodesic.

Corollary 3. A curve C with curvature k 6= 0 on a surface is a geodesic ifand only if the principal normal is a multiple of the surface normal at everypoint.

Proof. By a previous exercise, kg(t) = k(t) sin(φ(t)) where φ(t) is the anglebetween the principal normal to the curve and the normal vector to thesurface.

Exercise 72. On the sphere

x(u1, u2) = (a sin(u1) cos(u2), a sin(u1) sin(u2), a cos(u1))

consider the latitude circle u1 = a, constant. Show that by displacing a vectorparallelly once along the circle, the vector turns an angle of 2π(1− cos a).


Exercise 73. Two directions v = aixi and w = bjxj are said to be conjugateif hijaibj = 0. Consider the tangent planes to a surface x along C. Show thatthe limit of the intersection of two neighboring tangent planes is conjugateto the tangent vector to C.

Exercise 74. In the x − y-plane, consider the conic ax2 + 2bxy + cy2 = 1,ac−b2 6= 0, with center at the origin (0, 0). Two lines through the origin withslopes m and m are called conjugate if a+ b(m+m) + cmm = 0. Show thatthe midpoints of a family of parallel chords lie on a line through the originwhich is conjugate to the chord of the family through the origin.

Exercise 75. Show that the geodesic curvature of the latitude circle in (72)is cot a

a . Hence a latitude circle is a geodesic if and only if it is the equator.

The Levi-Civita parallelism follows as an application of the Gauss equa-tions (??).

Consider on the surface S a curve C defined by

ui = ui(t) , i = 1, 2 . (4.1)

At each point of C let

y(t) = yi(t)xi (4.2)

be a tangent vector to S, but not necessarily to C. By (??) its derivative is

dydt

=(dyi

dt+ Γijk

duk

dtyj)

xi + (...)ν .

The vector

Dydt

=(dyi

dt+ Γijk

duk

dtyj)

xi (4.3)

is its orthogonal projection into the tangent plane. y(t) defines a tangentvector field to S along C and DY

dt is called its absolute or covariant derivative.It will be convenient to omit the factor dt; then

Dy = (dyi + Γijkdukyj)xi

is called the absolute or covariant differential. The covariant differential de-pends only on the first fundamental form of S, and on C and y(t).

Let z(t) be a second tangent vector field to S along C. Then

d(yz) = dyz + yDz . (4.4)

This follows because Dy and Dz differ respectively from dy and dz by termsinvolving only the normal vector v, whose dot product with a tangent vectoris zero.


The vector field y(t) is said to be parallel along C in the sense of Levi-Civita if Dy = 0. It follows from (4.4) that if two vector fields are bothparallel along C, their scalar product is constant. In particular, the length ofa parallel vector field is constant.

It follows from (4.3) that the condition for the vector field (4.2) to beparallel along C is

dyi

dt+ Γijk

duk

dtyj = 0 . (4.5)

This is a system of linear homogeneous differential equations of the first orderwith yi as dependent variables and t as the independent variable. In fact,both Γijk and duk

dt are functions of t. By an existence theorem on differentialequations there exists a uniquely determined solution yi(t), which takes giveninitial values at t = t0. Geometrically this can be described by saying that avector y(t0) at t0 can be displaced parallelly along C.

Example 6. S is the plane with the first fundamental form

ds2 = (du1)2 + (du2)2 .

Then Γjik = 0 and equations (4.5) show that y(t) is parallel if yi = const.Thus Levi-Civita parallelism coincides with ordinary parallelism.

Consider two surfaces S and S∗ intersecting in a curve C. They are saidto be tangent to each other along C if they have the same normal along C.Since the absolute differential is defined by the use of the normal vector, wehave immediately: When two surfaces S and S∗ are tangent to each otheralong C, the absolute differential remains the same whether C is consideredas a curve on S or on S∗.

This property gives the following geometrical construction of the Levi-Civita parallelism: The tangent planes to S along C envelop a developableΣ. The parallelism is the same by considering C as a curve on Σ. Roll Σonto a plane and the parallelism becomes the ordinary one. The parallelismon S can be constructed by reversing this process. (INSERT REFERENCEHERE).

The curve C defined by (4.1) is called a geodesic if it is auto-parallel, i.e.,ifits tangent vector is parallel to itself along C. By putting yi = du

dt in (4.5) weget the differential system satisfied by the geodesics:

d2ui

dt2+ Γijk

duj

dt

duk

dt= 0 .

It is a second-order system with the dependent variables ui and the inde-pendent variable t. By a known existence theorem on ordinary differentialequations there is a uniquely determined solution ui(t) which takes given ini-tial values ui(t0), ui(t0) at t = t0. Geometrically this means that there is a


uniquely determined geodesic through a given point and tangent to a givenvector. All the tangent vectors to the geodesic, being parallel, are of the samelength. There is thus no loss of generality in supposing them to be of unitlength, which means that t is the arc length s.

Proposition 14. The geodesics are the curves of zero geodesic curvature.

Proof. By (INSERT REFERENCE HERE), the geodesic curvature of a curveis given by

c = T′Ni = (T,T′, ν) ,

where Ni = ν × T. Since DTds and dT

ds differ by a multiple of ν this can bewritten

c =DTds

Ni .

From T2 = 1 we have

DTds

T =dTds

T = 0 .

Being a tangent vector, DTds is also orthogonal to ν. Hence

DTds

= cNi . (4.6)

The vanishing of the left-hand side of (4.6) defines the geodesics, and theProposition follows.

Corollary 4. A curve C on S with curvature k 6= 0 is a geodesic if and onlyif it has the surface normal as the principal normal at every point.

Proof. By the same Exercise referred to above we have

c = k sinΘ , (4.7)

where Θ is the angle between the principal normal of C and the surfacenormal. The corollary is an immediate consequence of (4.7).

Exercise 76. On the sphere

x(ul, u2) = (a sinu1 cosu2, a sinu1 sinu2, a cosu1)

consider the latitude circle u1 = α = const. Show that by displacing a vectorparallelly once along the circle, the vector turns an angle 2π(1− cosα).

4.3 Integrability Conditions 63

Exercise 77. Two directions defined by the vectors

v = ξixi , w = ηjxj

are called conjugate if

hijξiηj = 0 .

Consider the tangent planes to a surface S along a curve C on S. Show thatthe limit of the intersection of two neighboring tangent planes is a line in theconjugate direction of the tangent direction of C.

Exercise 78. In the xy-plane consider the conic

ax2 + 2bxy + cy2 = 1 , ac− b2 6= 0

with center at 0 = (0, 0). Two lines through 0 with the slopes m, m′ arecalled conjugate if

a+ b(m+m′) + cmm′ = 0 .

Show that the mid-points of a family of parallel chords lie on a line through0, which is conjugate to the chord of the family through 0.

Exercise 79. Show that the geodesic curvature of the latitude circle in (IN-SERT REFERENCE HERE) is cot αa . Hence the latitude circle is a geodesicif and only if it is an equator.

4.3 Integrability Conditions

By successive applications of the Weingarten equations (??) and the Gaussequations (??) it is possible to express all partial derivatives of x and ν aslinear combinations of x1, x2, ν. The integrability conditions express thefact that they are independent of the order of differentiation. The resultingrelations are among the most fundamental in surface theory.

By differentiating (??) and using (??) we find

νik = −

(∂hji∂uk

+ hliΓjlk

)xj − hjihjkν .

Since the left-hand side is symmetric in i, k, the same is true of the right-handside, and of the coefficients of xj , ν at the right-hand side. The coefficient ofν is

hjihjk = gjlhijhlk ,


which is symmetric in i, k. Expressing that the coefficient of xj is symmetricin i, k, we get

∂hji∂uk−∂hjk∂ui

+ hliΓjlk − h

lkΓjli = 0 . (4.8)

These are called the Codazzi equations.These equations have an equivalent form expressed in terms of hij . For

we have, using (??),

∂hij∂uk

=∂(hpi gpj)∂uk

=∂hpi∂uk

gpj + hpi (Γpjk + Γjpk)

=(∂hpi∂uk

+ hqiΓpqk

)gpj + hpiΓjpk

which can be written

∂hij∂uk

− hipΓpjk =(∂hpi∂uk

+ hqiΓpqk

)gpj .

Hence we have

∂hij∂uk

− ∂hkj∂ui

− hilΓljk + hklΓlji = 0 , (4.9)

which are equivalent to (4.8).By differentiating (??) and using (??) and (??), we have

xipj =(∂Γlik∂uj

+ ΓpikΓlpj − hikhlj)

xl +(∂hik∂uj

+ Γlikhlj

)ν .

We introduce the Riemann-Christoffel tensor

−Rlikj = +Rlijk =∂Γlik∂uj

−∂Γlij∂uk

+ ΓpikΓlpj − ΓpijΓlpk . (4.10)

The symmetry in k, j of the coefficient of ν in the above equation for xikjgives the Codazzi equation (4.9), while the same symmetry of the coefficientof xl gives

−Rlikj − hikhlj − hijhlk = 0 . (4.11)

These are called the Gauss equations.Let

Rilkj = Rpkjgpl ,

4.3 Integrability Conditions 65

or, equivalently,

Rpikj = Rilkjgpl .

Then the Gauss equations (4.11) can be written

Rilkj + hikhjl − hijhkl = 0 .(CHECK THIS) (4.12)

From (4.12) we derive the following symmetry properties of Rilkj ;

Rilkj = −Rlikj = −Riljk ,Rilkj = Rkjil .

In particular, (4.12) gives

R1212 + h11h22 − h212 = 0 .(CHECK THIS)

By (INSERT REFERENCE HERE) we have

K =R1212

g,

where the right-hand side depends only on the first fundamental form. Thisfact is called:

Theorem 9. The Remarkable Theorem of Gauss. The Gaussian curvatureof a surface depends only on the first fundamental form.

Exercise 80. Show that in the notation of the Exercises in (INSERT REF-ERENCE HERE) the Codazzi equations can be written

∂L

∂v− ∂M

∂u= LΓ1

12 +M(Γ212 − Γ1

11)−NΓ211 ,

∂M

∂v− ∂N

∂u= LΓ2

22 +M(Γ222 − Γ1

12)−NΓ212 .

Exercise 81. With the same notation show that the Gauss equation can bewritten

EGK = − 12 (Guu + Evv) + 1

4

(EuGuE + EvGv

G

),

provided F = 0.There are therefore two Codazzi equations and one Gauss equation be-

tween the six functions which enter into the first and second fundamentalforms. This is in agreement with the fact that a parametric surface dependson three arbitrary functions in two variables: x(u, v).


4.4 The Congruence Theorem

The congruence theorem addresses to the question as to when two parametricsurfaces are congruent. More precisely, let ∆ be a connected domain in the(u1, u2)-plane and let

INSERT GRAPHIC HERE

be two parametric surfaces, E being the euclidean space. We wish to knowthe conditions that there exists a motion T so that the above diagram iscommutative.

Let I, II (resp I∗, II∗) be the.first and second fundamental forms of thesurface x (resp. x∗). If x and x∗ are congruent, we have I∗ = I. The form IIdepends on the orientation of the space E, because it depends on the sign ofthe unit normal vector ν(u1, u2). As usual suppose E be oriented. Then thesign of ν is defined by the condition

(x1,x2, ν) > 0 .

Under an improper motion this determinant changes sign. As a result IIchanges sign under an improper motion.

Theorem 10. Let

x,x∗ : ∆→ E

be two surfaces whose fundamental forms are to be denoted by I, II and I∗,II∗ respectively. If

I∗ = I , ±II∗ = II , (4.13)

then x∗ differs from x by a motion.

In terms of the coefficients of the fundamental forms conditions (4.13) canbe written

g∗ij = gij , h∗ij = ±hij , i, j = 1, 2 .

To prove the theorem we consider frames xx1x2ν(u1, u2), which satisfythe conditions

xixk = gik , xiν = 0 , ν2 = 1 , i, k = 1, 2 , (4.14)

and the partial differential equations

∂x∂ui

= xi ,

∂xk∂ui

= xki = Γjkixj + hkiν , i, j, k = 1, 2 ,

∂ν

∂ui= −hjixj .

4.4 The Congruence Theorem 67

The first equation is the definition of xi , the second is identical to the Gaussequations, and the third to the Weingarten equations. Their coefficients arecompletely determined by I and II.

We reduce the congruence theorem to a theorem on a one-parameterfamily of frames, whose special case on rectangular frames was proved earlier(INSERT REFERENCE HERE):

Proposition 15. Let xE1E2E3(t) and x∗E∗1E∗2E∗3 (t) be two smooth families

of frames in the parameter t which satisfy the conditions

EαEβ = E∗αE∗β = gαβ , α, β = 1, 2, 3,det(gαβ) 6= 0 , (4.15)

and the differential equations

x = pαEα , Eα = qβαEβ , (4.16)

x∗ = pαE∗α , E∗α = qβαE

∗β , (4.17)

with the functions pα, qβα, with dot denoting differentiation with respect to t.If the frames are identical at one value t0:

x∗(t0) = x(t0) , E∗α(t0) = Eα(t0) , (4.18)

they are identical for all values t:

x∗(t) = x(t) , E∗α(t0) = Eα(t) . (4.19)

Proof. This theorem is a special case of a uniqueness theorem on the solutionof a system of differential equations of the first order. The following directproof involves techniques which are useful in differential geometry.

We will write the equations in matrix notation. Let

Eα = (e1α, e2α, e

3α)

and introduce the matrix

M =

e11 e21 e31e12 e22 e

32

e13 e23 e

33

,

whose rows are therefore the components of Eα. Similary we define M∗. Thenthe second equations in (4.16) and (4.17) can be written

M = QM , M∗ = QM∗ , (4.20)

where


Q =

q11 q21 q31q12 q22 q

32

q13 q23 q

33

.

On the other hand, putting

G =

g11 g

21 g

31

g12 g

22 g

32

g13 g

23 g

33

,

equation (4.15) can be written

M tM = G . (4.21)

G being non-singular, we have, on taking the inverse of the above equation,

G−1 = tM−1M−1 ,

or

tMG−1M = I = identity matrix . (4.22)

Differentiating (4.21), we get, on using (4.20),

G = M tM +M tM = QG+GtQ .

Since GG−1 = I, we have

GG−1 +GG−1 = 0 ,

so that

G−1 = −G−1GG−1 = −G−1Q− tQG−1 . (4.23)

By using (4.20) and (4.23), we get

d(tM∗G−1M)dt

= tM∗tQG−1M + tM∗(−G−1Q− tQG−1)M + tM∗G−1QM = 0 .

Hence the matrix in the parentheses at the left-hand side is constant; it isequal to the matrix at t0. which, by (4.18), is equal to tMG−1M . The latteris I by (4.22). It follows that

tM∗G−1M = I .

Comparing this equation with (4.22), we get

tM∗ = tM , orM∗ = M .

This gives the second equation of (4.19).

4.4 The Congruence Theorem 69

From the first equations of (4.16) and (4.17) we get

d(x∗ − x)dt

= 0 .

Hence the vector in the parentheses is constant and is, by (4.18), zero.Proof of congruence theorem. By an improper motion if necessary we can

suppose II∗ = II. Let 0 be a fixed point of ∆. The frames xx1x2ν andx∗x∗1x

∗2ν∗ at 0, satisfying the conditions (4.14), differ by a motion, proper or

improper. By applying this motion, we can suppose them to be identical. Itremains to prove that xx1x2ν and x∗x∗1x

∗2ν∗ are identical at any other point

P of ∆.We join 0 to P by an arc, whose equations we suppose to be

ui = ui(t) , i = 1, 2 .

Along the arc, xx1x2ν satisfies the differential system

x = uixi ,

xk = Γjkiuixj + hkiu

iν ,

ν = −hji uixj ,

and x∗x∗1x∗2ν∗ satisfies the same differential system. Since they are identical

at 0, it follows from the above Proposition that they are identical at P . Thisproves the congruence theorem.

5 Global Surface Theory

We shall treat some selected problems in the theory of surfaces in the large,i.e., results which are valid only when the whole surface is taken into ac-count. For such global problems it will be most convenient to use the methodof “moving frames,” which will be developed in the first section. This ap-proach is consistent with our study of curves, where the Frenet formulaswere the culmination. In surface theory it is necessary to consider families offrames depending on two variables. The effectiveness of the moving framescan be theoretically justified by the observation that euclidean geometry isdominated by the group of its motions and the space of all frames can beidentified in an obvious way with the space of all motions.

5.1 Moving Frames

5.1.1 INSERT TITLE

We will consider differential forms in two variables, such as

ω = P (u, v)du+Q(u, v)dv ,

where the coefficients P , Q are smooth functions. When

ω′ = P ′du+Q′dv

is another form, we introduce a multiplication, called the exterior multipli-cation,which is anti-symmetric in the differentials, thus:

ω ∧ ω′ = −ω′ ∧ ω = (PQ′ −QP ′)du ∧ dv .

It follows that ω and ω′ are linearly dependent if and only if

ω ∧ ω′ = 0 .

The exterior derivative of ω is defined to be the form of degree two:

dω = (Qu − Pv)du ∧ dv .

72 5 Global Surface Theory

When it is applied to the differential df of a smooth function f , we haveclearly

d(df) = 0 . (5.1)

Also we have

d(fω) = df ∧ ω + f dω .

The exterior derivative allows the Green-Stokes formula to be written inthe form ∫∫

D

dω =∫∂D

ω ,

where D is a domain in the (u, v)-plane and ∂D is its boundary. The sameformula is true for a domain on a surface. This is proved by the “additivity”of the formula: Cut D into subdomains so small that each lies in a coordi-nate neighborhood; apply the formula to each subdomain; and then add theresults.

5.1.2 INSERT TITLE

As in Chapter 1 by a frame we mean the vectors xe1e2e3, satisfying therelations

eα · eβ = δαβ , α, β = 1, 2, 3 , (5.2)

(e1, e2, e3) = 1 . (5.3)

Condition (5.3) means that we restrict our frames to be right-handed. Thepoint with the position vector x is called the origin of the frame.

The local theory of surfaces can be developed as follows: Let, a surface Sbe described by x(u, v). We suppose S to be oriented and e3(u, v) to be theunit normal vector at (u, v). Choose a smooth field of unit tangent vectorse1(u, v), and let e2 = e3× e1. Since eα are linearly independent, we can write

dx =∑i

ωiei , i = 1, 2 , (5.4)

deα =∑β

ωαβeβ , α, β = 1, 2, 3 . (5.5)

The sum in (5.4) does not involve e3, because dx is a tangent vector.From (5.4) and (5.5) we find


ωi = dx · ei , ωαβ = deα · eβ . (5.6)

By differentiating (5.2) and using the second equation of (5.6), we get

ωαβ + ωβα = 0 .

Thus the matrix (ωαβ) is anti-symmetric; there are only three essential en-tries. observe that the equation de3 in (5.5) is the same as the Weingartenformulas and the equations for de1, de2 are equivalent to the Gauss equations.

The first fundamental form is

I = ds2 = dx · dx = ω21 + ω2

2 ,

and the element of area is

dA = ω1 ∧ ω2 6= 0 .

Thelatter implies that ω1 and ω2 are linearly independent. As a consequenceωαβ are their linear combinations. These coefficients, or their combinations,are to have geometrical meaning.

The equation (5.1) is true for each of the components of the vectors x, eαso that we have

d(dx) = d(deα) = 0 . (5.7)

These equations contain all the important information on local surface theoryand are to be studied in detail.

First we have

d(dx) =∑i

dωiei −∑i,β

ωi ∧ ωαβeβ , i = 1, 2 ; β = 1, 2, 3 .

Its vanishing implies the vanishing of the coefficients of eβ , which gives

dωi =∑j

ωj ∧ ωji , i, j = 1, 2 , (5.8)

ω1 ∧ ω13 + ω2 ∧ ω23 = 0 . (5.9)

If we write

ω13 = aω1 + bω2 , ω23 = b′ω1 + cω2 ,

equation (5.9) gives b′ = b.The second fundamental form is

II = −dx · de3 = ω1ω13 + ω2ω23 = aω21 + 2bω1ω2 + cω2

2 .


The two principal curvatures k1 and k2 are the roots of the equation

(a− k)(c− k)− b2 = 0 .

Their two elementary symmetric functions are respectively the mean curva-ture and the gaussian curvature and are given by

H =12

(k1 + k2) =12

(a+ c) , K = k1k2 = ac− b2 .

The equations (5.8) completely determine ω12. For, if

dωi =∑j

ωj ∧ ω′ji , i, j = 1, 2 ,

we get by subtraction ∑j

ωj ∧ (ω′ji − ωji) = 0 ,

or explicitly,

ω1 ∧ (ω′12 − ω12) = ω2 ∧ (ω′12 − ω12 = 0 ,

which implies ω′12 = ω12.Next we have

d(deα) =∑β

(dωαβ −

∑γ

ωαγ ∧ ωγβ

)eβ , α, β, γ = 1, 2, 3 .

The vanishing of this expression implies that the coefficient of eβ at theright-hand side is zero, i.e.,

dωαβ =∑γ

ωαγ ∧ ωγβ , α, β, γ = 1, 2, 3 . (5.10)

Because of the anti-symmetry of ωαβ this equation is identically satisfiedif α = β and remains the same equation when α and β are interchanged.Hence (5.10) involves only three independent equations. Written explicitly,the equation

dω12 = −ω13 ∧ ω23 = −Kω1 ∧ ω2 (5.11)

is commonly known as the Gauss equation and the equations

dω13 = ω12 ∧ ω23 , dω23 = ω21 ∧ ω13 (5.12)

are the Codazzi equations.


The absolute differential of a vector field is by definition the orthogonalprojection of the ordinary differential into the tangent plane. Hence from (5.5)we have

Dei =∑j

ωijej , i, j = 1, 2 .

It follows that

ω12 = e2 ·De1 = −e1 ·De2 ,

which says that ω12 depends only on the first fundamental form I (and onthe vector field e1). Since ω1∧ω2 is the element of area and depends also onlyon I, equation (5.11) contains the Gauss remarkable theorem: The Gaussiancurvature K depends only on the first fundamental form.

The above discussion contains everything that is needed to know on theformal aspect of local surface theory. Since it depends on the choice of avector field, we will have a complete control if we know the behavior of thedifferential forms under a change of the vector field. Such a change is givenby

e∗1 = cosΘe1 + sinΘe2 ,e = sinΘe1 + cosΘe2 ,

where Θ(u, v) is a function of u, v. Then

ω∗1 = dxe∗1 = cosΘω1 + sinΘω2 ,

ω∗2 = dxe∗2 = − sinΘω1 + cosΘω2 .

It follows that

dω∗1 = (dΘ + ω12) ∧ ω∗2 , dω∗2 = −(dΘ + ω12) ∧ ω∗1 .

But ω∗12 is completely determined by the equations (5.8) when the ω’s areprovided with asterisks. Comparing with the above equations, we have

ω∗12 = dΘ + ω12 .

Remark 8. When applied to the six-dimensional space of all frames in Eu-clidean space, the equations (5.7) lead to the structure equations of the groupof motions in the sense of the theory of Lie groups. Herein lies the basic reasonfor the importance of the Gauss and Codazzi equations.

5.1.3 INSERT TITLE

In this section we will relate the moving frames with our earlier treatment.Suppose the parametric curves be orthogonal, so that


ds2 = Edu2 +Gdv2 .

We choose

e1 =xu√E, e2 =

xv√G,

which implies

ω1 = E12 du , ω2 = G

12 dv . (5.13)

Then

dω1 =1

2√EEvdv ∧ du =

12√EG

Evω2 ∧ du ,

dω2 =1

2√GGudu ∧ dv =

12√EG

Guω1 ∧ dv .

From the fact that ω12 is completely determined by (5.8) we find

ω12 =1

2√EG

(−Evdu+Gudv) . (5.14)

Therefore

dω12 =12

[(Gu√EG

)u

+(

Ev√EG

)v

]du ∧ dv ,

and we get the classical Gauss equation

K = − 12√EG

[(Gu√EG

)u

+(

Ev√EG

)v

].

To compare the Codazzi equations observe that the second fundamentalform is

II = Ldu2 + 2Mdudv +Ndv2 = aω21 + 2bω1ω2 + cω2

2 .

Comparing with (5.13), we have

aE = L , b√EG = M , cG = N .

Hence

ω13 = aω1 + bω2 =1√E

(Ldu+Mdv) , ω23 = bω1 + cω2 =1√G

(Mdu+Ndv) .

Substituting these and (5.14) into (5.12), we get

2EG(Lv −Mu)− (EN +GL)Ev −M(EGu −GEu) = 0 ,2EG(Mv −Nu) + (EN +GL)Gu −M(EGv −GEv) = 0 ,

which are the classical Codazzi equations.

5.2 The Gauss-Bonnet Theorem 77

5.2 The Gauss-Bonnet Theorem

The Gauss-Bonnet theorem has its analytical basis in the Gauss equa-tion (5.11). It is nothing more than a geometrical interpretation of thatequation. But it has profound consequences and is perhaps one of the mostimportant theorems in mathematics.

5.2.1 INSERT TITLE

Let our surface be described by the position vector x(u, v) as a function ofthe parameters u, v. As in (INSERT REFERENCE HERE) we let e3(u, v)be the unit normal vector. We choose e1 to be the unit vector tangent to theu-curve, i.e., e1 = xu√

E; then e2 = e3 × e1. This defines a field of frames and

the formulas in (INSERT REFERENCE HERE) are valid.Consider an oriented curve C on the surface S. Let s be its arc length

and T = dxds be its unit tangent vector. We can write

T = cosΘe1 + sinΘe2 , (5.15)

so that Θ(s) is the angle that T makes with e1. Then

DT = (dΘ + ω12)N ,

where

N = − sinΘe1 + cosΘe2 (5.16)

is the unit normal vector to C. The geodesic curvature kg of C is defined bythe equation

dΘ + ω12 = kgds .

Consider now a domain D whose boundary is a sectionally smooth simpleclosed curve C. Then we have∫

C

kgds =∫C

dΘ +∫C

ω12 .

By the theorem of turning tangents, we have∫C

dΘ = 2π −∑i

(π − αi) ,

where π − αi are the exterior angles at the corners of C. On the other hand,by Stokes theorem and (5.11), we have∫

C

ω12 = −∫∫

D

KdA .

We have thus proved a special case of the Gauss-Bonnet theorem.


Theorem 11. Let D be a compact oriented domain on S bounded by a sec-tionally smooth curve C. Then∫

C

kgds+∫∫

D

KdA+∑i

(π − αi) = 2πχ , (5.17)

where kg is the geodesic curvature of C, π−αi are the exterior angles at thecorners of C, and χ is the Euler characteristic of D.

Proof. We have proved the theorem for the case that D lies in a neighborhoodwhere the same set of parameters u, v is valid, with χ = 1. In the generalcase suppose D be subdivided into a union of polygons Dλ, λ = 1, ..., f , suchthat: 1) each Dλ lies in one coordinate neighborhood; 2) two Dλ’s have eitherno point, or one vertex, or a whole side in common. Such a subdivision isclearly possible. We orient Dλ coherently with D and apply (5.17) to eachDλ, for which the theorem has been proved. Adding the resulting equations,we have, because the integrals of geodesic curvature along the interior sidescancel, ∫

C

kgds+∫∫

D

KdA+∑λ

∑i

(π − αλ,i) = 2πf . (5.18)

We will evaluate the double sum in the formula, by summing it withrespect to the vertices of the subdivision. Let vI , vE be respectively thenumbers of interior and exterior vertices and let s1 (respectively s2 be thenumber of sides which join two interior vertices (respectively one interiorvertex and one exterior vertex). Then at the interior vertices the contributionof∑λ

∑i π is 2πs1 +πs2 while the contribution of −

∑λ

∑i αλ,i is −πvI . At

the exterior vertices the contribution of∑λ

∑i(π−αλ,i) is πs2 +

∑i(π−αi),

where the last sum is the sum of the exterior angles of D. Since C has thesame number of sides as vertices, the total number of sides of the subdivisionis s = s1 + s2 + vE . Let v = vI + vE be the total number of vertices. Thenumber

χ = v − s+ f = vI − s1 − s2 + f

is called the Euler characteristic of the subdivision. Substituting the abovevalues into (5.18), we have proved the Gauss-Bonnet theorem.

It follows from (5.17) that the Euler characteristic, which is defined interms of a subdivision, depends only on D.

Remark 9. The theorem is true on an abstract riemannian manifold of twodimensions, and essentially the same proof applies.

Special cases: 1) S is a surface of constant curvature K and D is boundedby a triangle formed by three geodesic arcs. Then (5.17) becomes

KA = α1 + α2 + α3 − π , (5.19)

5.2 The Gauss-Bonnet Theorem 79

where A is the area of D and αi, i = 1, 2, 3, are the angles of the triangle. Itfollows that α1 + α2 + α3 is greater than, equal to, or less than π, accordingas K > 0, = 0, or < 0. Moreover, in the case K 6= 0, A is proportional to theexpression at the right-hand side of (5.19).

2) S is a compact orientable surface without boundary. Then (5.17) gives∫∫D

KdA = 2πχ .

It follows that if K = 0, the Euler characteristic of S is zero and S is home-omorphic to a torus. Also if K > 0, then χ > 0 and S is homeomorphic tothe sphere.

5.2.2 Theorems of Jacobi and Hadamard

Consider a closed curve x(s) in space, referred to the arc length s as param-eter. We suppose that the curvature k(s) never vanishes, and there is no lossof generality in supposing k(s) > 0. Then the unit principal normal vectorN(s) = x′′(s)

k and the unit binormal vector b(s) = x′′(s)×N(s) are both welldefined over the whole curve. Their end-points describe on the unit sphere theprincipal normal indicatrix and the binormal indicatrix respectively. Jacobi’stheorem is the following:

Theorem 12. Let x′(s) be a closed space curve whose curvature is nowherezero. If its principal normal indicatrix does not intersect itself, it divides theunit sphere in two domains with the same area.

Proof. To prove the theorem we define τ by the equations

k = (k2 + w2)12 cos τ , w = (k2 + w2)

12 sin τ ,

where w is the torsion of the curve. Then we have

d(− cos τT + sin τb) = (sin τT + cos τb)dτ − (k2 + w2)12 Nds ,

where T = x′ is the unit tangent vector. By Frenet’s formula the vectorbetween parentheses at the left-hand side is the unit tangent vector to theprincipal normal indicatrix. It follows that, if σ denotes the arc length ofN(s), dτ

dσ is the geodesic curvature of N(s) on the unit sphere. Let D beone of the domains bounded by N(s), and A its area. By the Gauss-Bonnetformula we have, since K = 1,∫

N(s)

dτ +∫∫

D

dA = 2π .

The first integral at the left-hand side is zero because it is integrated over aclosed curve. Hence we have A = 2π and the theorem is proved.


Another application of the Gauss-Bonnet theorem gives Hadamard’s the-orem on convex surfaces:

Theorem 13. If the Gaussian curvature of a closed orientable surface inspace is everywhere positive, the surface is convex. (That is, it lies at oneside of every tangent plane.)

Observe that for curves in the plane a similar theorem is true only un-der the additional assumption that the curve is simple. (INSERT REFER-ENCE HERE) For a surface it is not necessary to suppose that it has noself-intersection.

Let S denote the surface. It follows from the Gauss-Bonnet formula thatits Euler characteristic χ(S) is positive, so that χ(S) = 2 and we have∫∫

S

KdA = 4π .

Suppose S be oriented. If SO is the unit sphere about a fixed point O, theGauss mapping

g : S → SO

is defined by assigning to every point p ∈ S the end-point of the unit vectorthrough O parallel to the unit normal vector to S at p. The condition K > 0implies that g has everywhere a nonzero functional determinant and is locallyone-to-one. It follows that g(S) is an open subset of SO. Since S is compact,g(S) is a compact subset of SO, and hence is also closed. Therefore g mapsonto SO.

Suppose that g is not one-to-one, that is, there exist points p and q ofS, p 6= q, such that g(p) = g(q). Then there is a neighborhood U of q suchthat g(S−U) = SO. Since

∫∫S−U KdA is the area of g(S−U), counted with

multiplicities, we will have ∫∫S−U

KdA ≥ 4π .

But ∫∫U

KdA > 0 ,

so that ∫∫S

KdA =∫∫

U

KdA+∫∫

S−UKdA > 4π ,

which is a contradiction. Hence g is one-to-one and we see easily that S isconvex.

5.3 Rigidity Theorems 81

5.2.3 Vector fields on a surface

The tools developed above also provide a convenient means to study thebehavior at the singularities of a vector field on the surface. We use thenotations of (INSERT REFERENCE), where Θ(u, v) is now a function of u,v, defines a vector field on S; then equation (INSERT REFERENCE) definesthe vector field perpendicular to it. We suppose that T has a singularity atthe point p0 with the parameters u = v = 0. Let U be a neighborhood of p0.In U − p0 we have

DT ·N = dΘ + ω12 .

Let γε be a circle of radius ε about p0. The limit

12π

limε→0

∫γε

dΘ + ω12

is an integer, called the index of the vector field at p0. Clearly the index iszero if p0 is no singularity.

We will prove the theorem:

Theorem 14. Let S be a closed orientable surface and ξ be a smooth vectorfield with a finite number of singularities. Then the sum of the indices at thesingularities is equal to the Euler characteristic χ(S).

It follows that no continuous vector field ξ exists on a closed orientablesurface other than the torus, such that ξ 6= 0. For if such a ξ exists, then

ξ

(ξ·ξ)12

defines a unit vector field without singularity on S. By our theorem

this is only possible when χ(s) = O.

Proof. To prove the theorem let pi, 1 ≤ i ≤ n, be the singularities of ξ. Letγi(ε) be a circle of radius ε about pi and ∆i(ε) the disk bounded by γi(ε).By (INSERT REFERENCE) we have

d(dΘ + ω12) = −KdA .

Applying Stokes Theorem to the domain S −⋃i∆i(ε), we get∫∫

S−Si∆i(ε)

KdA =∑i

∫γi(ε)

dΘ + ω12 ,

where γi(ε) is so oriented that it is the boundary of ∆i(ε). The theoremfollows by letting ε→ 0.

5.3 Rigidity Theorems

We shall establish some global rigidity (or uniqueness) theorems on closedsurfaces. They characterize surfaces by the validity of certain local propertiesthroughout. The two most elementary methods are maximum principle andintegral formulas. We shall show how they are applied.


5.3.1 Elliptic W-surfaces

For definiteness and for historical reason we consider the problem of deter-mining the closed surfaces of constant Gaussian curvature K. We shall beginby some general analytical considerations.

We take a frame field over a neighborhood of S and use the notations andresults of (INSERT REFERENCE). Let

yα = x · eα , α = 1, 2, 3 .

Geometrically y3 is the oriented distance from the origin to the tangent planeat (u, v). By (INSERT REFERENCE) and (INSERT REFERENCE) we have

dy1 = ω1 + ω12y2 + ω13y3 , (5.20)dy2 = ω2 − ω12y1 + ω23y3 , (5.21)dy3 = −ω13y1 − ω23y2 . (5.22)

The following lemma results from the elementary study of the behaviorof a geometrically defined function at a local maximum:

Lemma 3. Let pO ∈ S be a point where the function w = x2 attains a localmaximum. Then the Gaussian curvature of S at pO is positive.

The function w on S is the square of distance from O to a point of S. Wehave

12dw = x · dx = y1ω1 + y2ω2 .

If the subscript O denotes the value of a function at pO, we have

(y1)O = (y2)O = O .

By (5.20) the second differential of w at pO is given by

12

(d2w)O = IO + (y3)OIIO ,

which is negative semi-definite. Using the expressions (INSERT REF) and(INSERT REF), this implies

1 + (y3)OaO ≤ O , 1 + (y3)OcO ≤ O ,

(1 + (y3)OaO)(1 + (y3)OcO)− (y3)2Ob2O ≥ O .

From these we derive KO = aOcO − b2O > O by a simple manipulation.It follows that a closed surface S must have a point with positive Gaussian

curvature, i.e., the point at a maximum distance from a fixed point in space.Hence if its Gaussian curvature is constant, the constant is positive and, byHadamard’s theorem, the surface is convex.

The answer to our problem is given by Liebmann’s theorem:


Theorem 15. A closed surface of constant Gaussian curvature K is a sphereof radius 1√

K.

The proof that we will give was due to Hilbert. It relies on a lemmawhich is similar to the above but is more sophisticated. Hilbert’s lemma isthe following:

Lemma 4. Let pO ∈ S be a non-umbilical point and let k1(p) > k2(p) bethe two principal curvatures, where p is in a neighborhood of pO. Supposek1 has a local maximum and k2 a local minimum at pO. Then the Gaussiancurvature of S at pO is ≤ 0.

Proof. By hypothesis the principal directions are well defined in a neighbor-hood of pO. We choose our frame field such that e1, e2 are in these directions.Observe that the principal directions are given by the equation∣∣∣∣ω1 aω1 + bω2

ω2 bω1 + cω2

∣∣∣∣ = 0

or

(a− c)ω1ω2 − b(ω21 − ω2

2) = 0 .

Our choice implies b = 0. Then a and c are clearly the normal curvatures inthe principal directions and we can write

ω13 = k1ω1 , ω23 = k2ω2 .

Taking the exterior derivatives of these equations and using (INSERT REF)and (INSERT REF) (Codazzi equations), we get

dk1 ∧ ω1 + (k1 − k2)ω12 ∧ ω2 = 0 , (5.23)dk2 ∧ ω2 + (k1 − k2)ω12 ∧ ω1 = 0 . (5.24)

We will use ω1, ω2 (instead of du, dv) to express the differentials of functionsand we put

dki =∑j

kijωj ,

dkij =∑k

kijkωk , i, j, k = 1, 2 .

Substituting into (5.23), we get

(k1 − k2)ω12 = k12ω1 + k21ω2 . (5.25)

By the extremal properties of k1, k2 at pO we have, at pO,


(kij)O = 0 .

The second differential of ki at pO is given by

(d2ki)O =∑j,k

(kijk)Oωjωk ,

which is a negative (respectively positive) semi-definite form when i = l(respectively i = 2). Hence

(k1jj)O ≤ 0 , (k2jj)O ≥ 0 , j = 1, 2 .

Taking the exterior derivative of (5.25) and using the Gauss equation(INSERT REF), we get

−(k1 − k2)Kω1 ∧ ω2 = (k211 − k211)ω1 ∧ ω2 + ... ,

where the dots denote terms which vanish at pO. Equating the coefficients ofω1 ∧ ω2 at both sides and taking their values at pO, we get

KO =(k122 − k211

k1 − k2

)O

≤ 0 .

This proves the lemma.

From the lemma follows a theorem more general than Liebmann’s. Asurface S is called a Weingarten surface or a W-surface if there is a functionalrelation between the principal curvatures:

W (k1, k2) = 0 .

If, moreover, the relation

∂W

∂k1

∂W

∂k2> 0 (5.26)

holds, S is called an elliptic W-surface. Examples of elliptic W-surfaces in-clude the surfaces of constant mean curvature and surfaces of constant posi-tive Gaussian curvature. The condition (5.26) means that k2 is a monotonedecreasing function of k1.

We will prove the theorem:

Theorem 16. A closed convex (i.e. K > 0) elliptic W-surface is a sphere.

We choose e3 to be the unit inward normal vector, so that the secondfundamental form is positive definite everywhere. Let k1(p) ≥ k2(p) > 0,p ∈ S, be the two principal curvatures. Then k1(p) attains a maximum ata point pO, say. Since k2 is a decreasing function of k1, pO is at the sametime a minimum of k2. If pO is non-umbilical, Hilbert’s lemma asserts that


the Gaussian curvature K(pO) ≤ 0. This contradicts our hypothesis K > 0.Hence pO is an umbilic, i.e., k1(pO) = k2(pO). But, we have, for any pointp ∈ S,

k1(pO) ≥ k1(p) ≥ k2(p) ≥ k2(pO) .

The equality of the two extreme sides implies that p is an umbilic. ThereforeS consists entirely of umbilics and is a sphere.

Clearly our theorem implies Liebmann’s theorem. We will state also thefollowing corollary:

Corollary 5. A closed convex surface of constant mean curvature is a sphere.

5.3.2 Integral formulas; second proof of Liebmann’s theorem.

By (5.20) and the equations (INSERT REF), (INSERT REF) we get

d(y2ω1 − y1ω2) = −2(1 + y3H)ω1 ∧ ω2 ,

d(y2ω13 − y1ω23) = −2(H + y3K)ω1 ∧ ω2 .

The expressions under d at the left-hand sides are independent of the choiceof the frame field, for we have

(x, e3, dx) = y2ω1 − y1ω2 ,

−(x, e3, de3) = y2ω13 − y1ω23 .

These are therefore linear differential forms defined over the whole surface S.Applying Stokes theorem, we have the integral formulas:

Let S be a closed orientable surface without boundary. Then∫∫S

(1 + y3H)dA = 0 , (5.27)∫∫S

(H + y3K)dA = 0 . (5.28)

Let A be the total area of S and let

M =∫∫

HdA

(the integral of mean curvature). Then (5.27) can be written

A = −∫∫

y3HdA , (5.29)

M = −∫∫

y3KdA . (5.30)


To illustrate the way that integral formulas can be used to derive globalgeometrical theorems we shall use (5.29) to give a second proof of Liebmann’stheorem. In fact, let

K = κ2 > 0 , κ > 0 .

By the Gauss-Bonnet theorem we have∫∫KdA = κ

2A = 4π ,

from which follows that∫∫(k1 − κ)(k2 − κ)dA = 8π − 2κM .

On the other hand, using (5.29), we get∫∫(k1 − κ)(k2 − κ)y3dA = 2

∫∫y3KdA− 2κ

∫∫y3HdA = −2M + 2κA = −2M +

8πκ

.

If the integrals at the left-hand sides of the last two equations are denotedby J1, J2 respectively, we have

J1 = κJ2 .

As above, we choose e3 to be the unit inward normal vector, so that thesecond fundamental form is positive definite, i.e., k1 > 0, k2 > 0, H > 0, andy3 = x · e3 < 0. Then we have

(k1 − κ)(k2 − κ) = 2(K − κH) = 2κ(κ −H) ≤ 0 ,

since κ is the geometric mean of k1, k2 and H is their arithmetic mean. Itfollows that

J1 ≤ 0 , J2 ≥ 0 .

This is possible only if J1 = J2 = 0. Since the integrands keep constant signs,this implies κ −H = 0. Hence every point is an umbilic and S is a sphere.

5.3.3 Cohn-Vossen’s rigidity theorem

Cohn-Vossen’s theorem is the following:

Theorem 17. An isometry between two closed convex surfaces is establishedeither by a motion or by a motion and a reflection.


In other words, such an isometry is always trivial. The theorem is clearlynot true locally. For example, the plane and the circular cylinder are isometricbut are not congruent. Also the convexity assumption cannot be removed(See figure - INSERT FIG). The question answered by the theorem was long-outstanding, till it was solved by Cohn-Vossen in 1927. The following proof,based on an integral formula, was due to G. Herglotz.

We shall begin by an algebraic lemma:

Lemma 5. Let ax2+2bxy+cy2 and a′x2+2b′xy+c′y2 be two positive definitequadratic forms, with

ac− b2 = a′c′ − b′2 .

Then ∣∣∣∣a′ − a b′ − bb′ − b c′ − c

∣∣∣∣ ≤ 0 , (5.31)

and the equality sign holds in (5.31) only when the two forms are identical.

To prove the lemma we observe that the statement of the lemma remainsunchanged under a linear transformation of the variables. Applying such alinear transformation when necessary, we can assume b′ = b. Then the left-hand side of (5.31) becomes

(a′ − a)(c′ − c) = − c

a′(a′ − a)2 ≤ 0 ,

as to be proved. Moreover, the quantity equals zero only when we also havea′ = a and c′ = c.

Remark 10. Geometrically the lemma means that two concentric ellipses withthe same area meet in four distinct points or are congruent.

To prove the theorem let S and S∗ be the two surfaces which are relatedby an isometry T : S → S∗. Suppose both be oriented by inward normals.The isometry T either preserves or reverses the orientation. In the latter casewe replace S∗ by its mirror image so that we can assume without loss ofgenerality that T preserves orientation.

We now study local properties of the isometry. Let u, v be parameters onboth surfaces such that corresponding points have the same parameters. Tmaps tangent vectors of S to tangent vectors of S∗ and preserves their scalarproducts. Since it also preserves orientation, it maps a (local) frame field on Sinto a frame field on S∗. We will use the results developed in (INSERT REF)on the differential geometry of S relative to a frame field. This also appliesto S∗ relative to the image frame field and we will denote the correspondingquantities by the same notations with asterisks. By (INSERT REF) we have

ωi = dxei = (xuei)du+ (xvei)dv , i = 1, 2 .


A similar equation holds for ω∗i . Since T preserves scalar products of tangentvectors, we have

ω∗i = ωi , i = 1, 2 .

By exterior differentiation and use of (INSERT REF) we find that the con-nection forms are equal, i.e.,

ω∗12 = ω12 .

Exterior differentiation of this equation and use of the Gauss equation (IN-SERT REF) give the equality K∗ = K of Gaussian curvatures at correspond-ing points. This is to be expected, because we know that the Gaussian curva-ture depends only on the first fundamental form and thus remains invariantunder an isometry.

To prove the theorem it suffices to show that the second fundamentalforms are equal. By (5.20) and the Codazzi equations (INSERT REF) appliedto S∗, we find

d(y1ω∗23 − y2ω∗13) = 2H∗ω1 ∧ ω2 + (ac∗ + a∗c− 2bb∗)y3ω1 ∧ ω2 .

The expression under d at the left-hand side can be written

y1ω∗23 − y2ω∗13 = −(x, e3, ω∗13e1 + ω∗23e2) .

It remains invariant under a rotation of the frame field on S; it is therefore alinear differential form defined everywhere on S. By applying Stokes theoremwe get the integral formula∫∫

y3JdA = −2∫∫

H∗dA ,

where we define

J = ac∗ + a∗c− 2bb∗ .

Combining with the second equation of (5.29), we get∫∫S

∣∣∣∣a∗ − a b∗ − bb∗ − b c∗ − c

∣∣∣∣ y3dA = 2(M∗ −M) ,

where M and M∗ are the integrals of mean curvature of S and S∗ respec-tively. The integrand at the left-hand side is ≥ 0. Hence the same is true ofthe integral and we have M∗ ≥ M . But the relation of the two surfaces issymmetric, so that the above inequality remains true when the two sides areinterchanged. This gives M∗ = M . Hence the integrand at the left-hand sidemust be zero. By our algebraic lemma this is possible only when

a∗ = a , b∗ = b , c∗ = c .

This completes our proof of Cohn-Vossen’s theorem.

6 INSERT TITLE

6.1 Geodesics

6.1.1 INSERT TITLE

Consider a curve C on a Riemannian surface. In terms of the local coordinatesu, v it is defined by the equations

u = u(t) , v = v(t) . (6.1)

The vector with components dudt , dvdt (relative to the coordinates u,v) is called

its tangent vector. From (INSERT REF) and (INSERT REF) it follows thatit is a unit vector when t is the arc length of C. When we use arc length asthe parameter we will write s for t.

We use the notation of the last section and, at every point of C, we lete1 be the unit tangent vector. Then u, v and τ are all functions of s and wecan write ω12 = kgds on C. The number kg clearly generalizes the curvatureof a curve in the Euclidean plane. We call it the geodesic curvature of C (atthe point with parameter s). It is to be observed that kg is a well-definedquantity only when both M and C are oriented (the latter means that wehave made a choice of the sign of the arc length). If M is non-orientable orunoriented, then only the absolute value of kg is defined.

We will give a geometrical interpretation of the geodesic curvature, whichillustrates even more clearly its analogy with the case of the Euclidean plane.In fact. let v(s) be a family of unit vectors parallel along C. Denote by τ ′(s)the angle from ∂

∂u to v(s), while, as in section 7.1 (CHANGE THIS?), τ(s)is the angle from ∂

∂u to the tangent vector e1(s) of C. Then we have

dτ(s) + αθ1 + βθ2 = kgds

dτ ′(s) + αθ1 + βθ2 = 0 .

where in α, β, θ1, θ2 the coordinates u, v are to be considered as functionsof s, as given by (6.1). From these two equations we get

d(τ − τ ′) = kgds .

Since τ − τ ′ is the angle from v(s) to e1(s), the geodesic curvature is the rateof change of this angle with respect to arc length. Clearly it is independentof the choice of the family of parallel vectors v(s) along C.

90 6 INSERT TITLE

6.1.2 INSERT TITLE

A geodesic is a curve whose geodesic curvature vanishes at every point. Analternative way of saying this is that the family of unit tangent vectors isparallel along the curve. This property is preserved if we reverse the orienta-tion of the curve or the orientation of the surface, so that being a geodesic isa property of an unoriented curve on an unoriented Riemannian surface.

We next derive the “differential equation” of geodesics. For simplicity weconsider only the geodesics which can be defined by an equation of the formv = v(u). This restriction is not essential, for, by interchanging u and v,we can get hold of the geodesics which escape our consideration. Where ageodesic is defined by such an equation, its tangent vector has components(l, v′), where v′ = dv

du . We assume now that the curves u = const., v = const.form an orthogonal net, and use the equations of (INSERT REF). We observethat θ1, θ2 has as dual basis

a1 =1p

∂

∂u, a2 =

1q

∂

∂v,

(which is orthonormal). Hence the tangent to the curve can be written as

pa1 + v′qa2 .

From this we find that

τ = tan−1 q

pv′ .

By (INSERT REF) the geodesics are characterized by the condition

dτ

du+ αp+ βqv′ = 0 .

Substituting into this the expression for τ we get a differential equation ofthe form

v′′ = A(u, v) +B(u, v)v′ + C(u, v)v′2 +D(u, v)v′3 , (6.2)

where A, B, C, D are functions of u, v. If (u0, v0) is any point of U , v′0 anarbitrary constant, the differential equation (6.2) has a uniquely determinedsolution v(u) satisfying the initial conditions

v(u0) = v0 , v′(u0) = v′0 .

Geometrically this means that through any point of M and tangent to anyvector at the point, there passes exactly one geodesic.

We do not care about the explicit expressions for the coefficients A, B, C,D in (6.2). We only remark that in order that (6.2) defines the geodesics thesecoefficients are not arbitrary. The question of determining what conditionsthey must satisfy is unsolved. It involves very complicated computations.

6.1 Geodesics 91

Exercise 82. Determine the geodesics on a plane, cylinder, and sphere, bothanalytically and by direct geometric arguments. Note: for a curve on a surfacein ordinary space, if we take frames along the curve such that e1 is tangentto the curve and e2 is normal to the curve but tangent to the surface, thenω12 = de1 · e2.

6.1.3 INSERT TITLE

One of the standard methods in geometry is to establish. for each problem, asuitable coordinate system. In what follows we are going to give some differentcoordinate systems, each with certain advantages. Let C0 be a curve. It isgiven locally by (u(t), v(t)), where u′2 + v′2 6= 0. By interchanging u andv, if necessary, we can arrange that v′ 6= 0. Then by the Inverse FunctionTheorem t is locally a differentiable function of v, so that the curve is locallyrepresented by u = u(t(v)) = f(v). Let

u = u− f(v) + u0

u = v0 ,

where u0 is some number. The Jacobian is

∂(u, v)∂(u, v)

=∣∣∣∣1 −f ′(v)0 1

∣∣∣∣ = 1 ,

so that u, v are valid local coordinates. Writing u, v in place of u, v, we havecoordinates such that C0 is locally defined by u = u0. Construct throughevery point (u0, v) of C0 the geodesic perpendicular to C0. Such a geodesiccan be defined by an equation of the form v = v(u, v0). By the theorem onthe dependence of the solution of (6.2) on the initial conditions, the functionv(u, v0) is differentiable with respect to both u and v0. Since v(u, v0) = v0,we have

∂v

∂v0

∣∣∣∣u=u0

= 1 .

Therefore at (u0, v0) the Jacobian of u, v(u, v0)

∂(u, v)∂(u, v0)

=∣∣∣∣ 1 0∂v∂u 1

∣∣∣∣ = 1 .

Hence we can take u and v0 as local coordinates in a sufficiently small neigh-borhood of (u0, v0). We now write v for v0, so that the geodesics perpendicularto C0 are u-curves, that is curves with v = const.

We wish to have coordinates such that the u-curves are orthogonal to thev-curves. To do this, we observe that the condition that a curve defined byu = u(v), whose tangent vector has components (u′, 1), be orthogonal to theu-curves, whose tangent vectors have components (1, 0), is that

92 6 INSERT TITLE

E(u, v)du

dv+ F (u, v) = 0 . (6.3)

by (INSERT REF) using the notation of (INSERT REF). Considered asa differential equation for u, this has a unique solution u(v, u) such thatu(v0, u0) = u0, provided that (v0, u0) lies in the domain of validity of u, v. Itis readily checked that the Jacobian of u, v with respect to u, v is 1 at (u, v0),so that u and v can be taken as local coordinates in a neighborhood of (u0, v0).The v-curves are orthogonal to the u-curves, since (6.3) is satisfied. Writingu for u, we can say, in summary, that the u-curves are geodesics orthogonalto the curve u = u0, and the v-curves are their orthogonal trajectories. Aneighborhood in which such a system of local coordinates u, v is valid iscalled a geodesic field. The v-curves are called geodesic parallels.

Suppose that u, v form such a local coordinate system. Then since theparameter curves form an orthogonal net we may apply the formulas of (IN-SERT REF). We take a field of frames such that e1 is in the u direction whichmakes τ = 0. Since the u-curves are geodesics, we have on each u-curve by(INSERT REF)

0 = ω12 = dτ − pvqdu+

qupdv = −pv

qdu .

Hence pv = 0, which says that p depends on u only, p = p(u). We changecoordinate u to u defined by

u =∫p(u)du .

Writing again u for u we have

ω1 = θ1 = du , ω2 = θ2 = qdv ,

which implies by (INSERT REF) that

ds2 = du2 +Gdv2 , G = q2 .

Thus ds2 assumes a quite simple form in our local coordinate system u, v.The arc length on a u-curve is given by

s =∫ u2

u1

du = u2 − u1 , if , u2 ≥ u1 .

Hence we have the theorem:

Theorem 18. The geodesics of a field are cut by any two geodesic parallelsin arcs of the same length.

Conversely, suppose that we have a coordinate system u, v such that theparameter curves form an orthogonal net. Suppose that the v-curves cut equallengths on the u-curves. Then

6.1 Geodesics 93∫ u2

u1

pdu

is independent of v, so that

0 =d

dv

∫ u2

u1

pdu =∫ u2

u1

pvdu .

Since this is true for any values of u1, u2, v, we must have pv = 0. Takingframes such that e1 is in the u direction so that τ = 0 we have on any u-curve,where v = const.,

ω12 = dτ − pvqdu+

qupdv = 0 .

Hence the u-curves are geodesics. We have proven the converse of the lasttheorem: if the v-curves cut equal lengths on their orthogonal trajectories,the latter are geodesics.

Using coordinates in a geodesics field as above we can see that thegeodesics are the curves of shortest length in a certain sense. Let (u1, v1)and (u2, v1) two points, u2 > u1 and v = v(u) a curve joining them whichlies in the geodesic field. Its length between these two points is∫ u2

u1

√1 +Gv′2du ≥ u2 − u1 .CHECK THIS

However u2 − u1 is the length of the geodesic v = v1 joining them. Moreoverthe inequality is an equality only if v′ = 0 which says that the v = const. andthe curve is a u-curve. This gives the theorem:

Theorem 19. A geodesic of a field is the unique shortest path in two of itspoints and lying in the geodesic field.

Example 7. Let M be the unit sphere with its first fundamental form asRiemannian metric. Let C0 be the equator. The orthogonal geodesics are thelines of longitude and their orthogonal trajectories are the usual parallels, orlines of latitude. The coordinates in this geodesic field are u = ϕ and v = θ,the usual angular spherical coordinates.

Exercise 83. Show that the meridians of a surface of revolution in ordinaryspace are geodesics.

Exercise 84. a) Show that a Riemannian surface M is locally isometric toa surface of revolution in ordinary space if and only if there exist local coor-dinates u, v on M such that

ds2 = du2 + p(u)dv2

with |p′(u)| ≤ 1.

94 6 INSERT TITLE

b) Show that a Riemannian manifold M is locally isometric to a surfaceof revolution in ordinary space if and only if the orthogonal trajectories ofthe curves K = const. are geodesics and at least one of the curves K = const.has constant geodesic curvature.

c) Which Riemannian surfaces are locally isometric to a surface of revo-lution in more than one way?

Exercise 85. Let P ∈ M and C an arc through P . Show that there existlocal coordinates u, v in a neighborhood of P such that on the arc C we havep = q = 1, pv = qu = 0. Such coordinates are called Fermi coordinates relativeto C. Their significance might be described as follows. A two-dimensionalcreature living on a Riemannian surface cannot determine experimentallyby measurements along an arc whether the Gauss curvature of his worldvanishes. To do so he must make measurements around a closed path.

Hint: Let a1a2 be a family of frames on C which is parallel on C and suchthat a2 is tangent to C at P . Show that coordinates in a geodesic field canbe chosen in such a way that the geodesics of the field are tangent to a1. Inthese coordinates p = 1; show that qu = 0 on C. Now find a function v(v)such that dv = q(u, v)dv on C.

6.2 Normal Coordinates

6.2.1 INSERT TITLE

Similar to the above coordinate system is the system of so-called geodesic po-lar coordinates. We take a point O of M and construct the geodesics throughO. If ϕ is the angle which such a geodesic makes with some fixed direction atO and we go along the geodesic a distance r, then r, ϕ are the geodesic polarcoordinates with center O of the resulting point. We will show that r, ϕ arevalid local coordinates in a neighborhood of each point of some neighborhoodof O, excluding O itself. In order to show this we first construct the so callednormal coordinates centered at O.

To construct these coordinates we start with some system of coordinatesin a geodesic field u, v. By a linear transformation, if necessary, we arrangethat at O,

u = v = O , p = q = 1 .

We now consider the geodesics (u(t), v(t)) with (u(0), v(0)) = (0, 0), where tis a constant non-zero multiple c of arc length from 0. By (INSERT REF)the components of the tangent vector of (u(t), v(t)) with respect to a1, a2

are (u′, qv′), where the prime denotes differentiation with respect to t. Nowt = cs. Hence

1c

=ds

dt=√u′2 + q2v′2 . (6.4)

6.2 Normal Coordinates 95

By (INSERT REF), (INSERT REF) and taking into account that p = 1 wemust have

0 = ω12 = dτ + qudv , τ = tan−1 qv′

u′.

Combining the last two equations, and differentiating (6.4) we obtain

−qv′u′′ + u′(qv′)′ = −quv′(u′2 + q2v′2) (6.5)u′u′′ + qv′(qv′)′ = 0 . (6.6)

Solving this system for u′′ and (qv′)′, using Cramer’s rule, the denominatorbeing −(u′2 + q2v′2), we obtain

u′′ = qquv′2 , (qv)′ = −quu′v′

or equivalently,

u′′ = qquv′2 (6.7)

v′′ = −1q

(2quu′v′ + q′v2) . (6.8)

Thus our geodesics through (0, 0) parametrized by a constant non-zero mul-tiple of arc length are solutions of (6.7). Conversely any solution of (6.7) withu′2 + q2v′2 6= 0 satisfies (6.5) and is therefore a geodesic parametrized by aconstant multiple of arc length. On the other hand if u′2 + q2v′2 = 0 thenu′ = v′ = 0 so u(t), v(t) are constant.

By the Fundamental Theorem of Differential Equations the system (6.7)has a unique solution u(t), v(t) characterized by the properties that u(0) =v(0) = 0 and u′(0) = u′0, v′(0) = v′0, where u′0, v′0 are arbitrary constants.This solution is defined in an interval containing 0, and we take this intervalto be as large as possible. Furthermore, if we express u and v as functions ofthe initial conditions u′0, v′0 as well as of t:

u(t, u′0, v′0), v(t, u′0, v

′0) ;

then these functions are differentiable and defined at least provided

|t| ≤ ε ,√u′o

2 + v′02 ≤ η ,

for some positive numbers ε and ηLet us now suppose that t, u′0, v′0 are numbers satisfying

|t|√u′o

2 + v′02 ≤ εη . (6.9)

Define functions u, v by

u(t, u′0, v′0) = u(Ct,C−1u′0, C

−1v′0)

v(t, u′0, v′0) = v(Ct,C−1u′0, C

−1v′0) ,

96 6 INSERT TITLE

where

C = η−1(u′02 + v′0

2)12 .

The right-hand sides are defined and differentiable in all three variables since|Ct| ≤ |t|(u′02 + v′0

2)12 η−1 ≤ ε, and

((C−1u′0)2 + (C−1v′0)2)12 = η .

Considered as functions of t, u, and v give a parametrization of a geodesicthrough 0 by a constant multiple C−1c of arc length where c = ((C−1u′0)2 +(C−1v′0)2)−

12 by (6.4). It follows that u, v satisfy the system (6.7). They

satisfy the initial conditions

u(0, u′0, v′0) = v(0, u′0, v

′0) = 0 ,

and

d

dt

∣∣∣∣t=0

u(t, u′0, v′0) = Cu1(0, C−1u′0, C

−1v′0) = u′0

d

dt

∣∣∣∣t=0

v(t, u′0, v′0) = Cv1(0, C−1u′0, C

−1v′0) = v′0 .

It follows from the uniqueness part of the Fundamental Theorem of OrdinaryDifferential Equations that

u(t, u′0, v′0) = u(t, u′0, v

′0)

v(t, u′0, v′0) = v(t, u′0, v

′0) ,

which tells us in effect that u, v are defined and differentiable in t, u′0, v′0provided (6.9) is satisfied.

Suppose that t, u′0, v′0 satisfy (6.9) and let C be any number differentfrom zero. Then

|Ct|((C−1u′0)2 + (C−1v′0)2)12 ≤ εη ,

so that u(Ct,C−1u′0, C−1v′0), v(Ct,C−1u′0, C

−1v′0) are defined and differen-tiable. As before, u(0, C−1u′0, C

−1v′0) = v(0, C−1u′0, C−1v′0) = 0,

d

dt

∣∣∣∣t=0

u(Ct,C−1u′0, C−1v′0) = u′0 ,

d

dt

∣∣∣∣t=0

v(Ct,C−1u′0, C−1v′0) = v′0 ,

and these functions as functions of t represent a parametrization of a geodesicby a constant multiple of arc length and therefore satisfy (6.7). It follows asabove that

u(Ct,C−1u′0, C−1v′0) = u(t, u′0, v

′0)

v(Ct,C−1u′0, C−1v′0) = v(t, u′0, v

′0) .


Letting C = 1t gives

u(1, tu′0, tv′0) = u(t, u′0, v

′0) , (6.10)

v(1, tu′0, tv′0) = v(t, u′0, v

′0) . (6.11)

The functions

u(x, y) = u(1, x, y)v(x, y) = v(1, x, y)

are defined and differentiable provided that√x2 + y2 < εη, as we have seen.

Using (6.10) we find that

∂u

∂x

∣∣∣∣(0,0)

=d

dx

∣∣∣∣0

u(x, 1, 0) = 1 .

The other derivatives of u, v with respect to x, y at (0, 0) may be found in asimilar fashion, with the result that the Jacobian matrix of u, v with respectto x, y at (0, 0) is the identity matrix. Hence x, y can be taken as localcoordinates on M in some neighborhood of P called a normal neighborhood ;x, y are called normal coordinates. Since by choice of u, v, ds2 = du2 + dv2

at P , and the Jacobian matrix is the identity matrix, we have

ds2 = dx2 + dy2 at P . (6.12)

We now introduce polar coordinates

x = r cosϕ , y = r sinϕ (6.13)

in the x, y plane in the usual way. These form a local coordinate systemin a neighborhood of any point in the plane except (0, 0), and consequentlythey can be taken as local coordinates in some neighborhood of any point ofour normal neighborhood except for P itself. They are called geodesic polarcoordinates. If we hold ϕ fixed we note that u(x, y) = u(l, r cosϕ, r sinϕ) =u(r, cosϕ, sinϕ) and similarly for v. Since cosϕa1 + sinϕa2 is a unit vectorthe resulting curve u(r), v(r) is a geodesic through P parametrized by arclength from P . We wish next to express our metric in these coordinates. LetU ⊂ R2 be the domain of validity of the normal coordinates centered at Pand consider the map (6.13), which we call Π, from the r, ϕ-plane to thex, y-plane. Here it will be convenient to allow r to be negative. Ordinarily inusing geodesic polar coordinates we take r ≥ 0 just as we do in the familiarcase of the plane. Let Π−1(U) = V . Then we have a map T from V to Mdefined by (6.13), taking x, y as normal coordinates. Let e1 = T∗( ∂∂r ). Thene1 is a unit tangent vector, since for ϕ fixed, as we just saw, u(r cosϕ, r sinϕ),v(r cosϕ, r sinϕ) represents a geodesic through P with arc length as param-eter. Let e2 be the unit normal in M to e1, chosen so that e1e2 agrees with

98 6 INSERT TITLE

the orientation of M . Thus we obtain a map to =, the bundle of frames ofM . Pull back ω1, ω2, ω12. We write these forms in terms of dr and dϕ. Onthe geodesic dϕ = 0, we have ω1 = dr, ω2 = 0, and ω12 = 0. Consequentlythese forms can be written as

ω1 = dr +mdϕ

ω2 = qdϕ

ω12 = Qdϕ ,

for some functions m, q, Q. The structure equation gives

dω1 = mrdr ∧ dϕ= ω12 ∧ ω2 = 0 .

Hencemr = 0. But for r = 0 the map T is constant, that is to say T (0, ϕ) = P ,so that ω1 = ω2 = 0 on the line r = 0. Consequently m(0, ϕ) = 0. Sincemr = 0 everywhere we have that m vanishes identically. Our metric thereforetakes the form ds2 = ω2

1 + ω22 = dr2 +Gdϕ2, where G = q2.

We note that C is well-defined and differentiable at every point wherethe geodesic polar coordinates are valid, that is to say at every point inthe normal neighborhood except at P . We now show that G extends to afunction differentiable at P and we prove certain expansion formulas. Invertformulas (6.13) in the usual way:

r2 = x2 + y2 , ϕ = tan−1 y

x.

Differentiating these and substituting in our expression for ds2 we obtain

ds2 =x2dx2 + 2xydxdy + y2dy2

r2+G

y2dx2 − 2xydxdy + x2dy2

r4(6.14)

=r2x2 +Gy2

r4dx2 +

(1r2− G

r4

)2xydxdy +

r2y2 +Gx2

r4dy2 . (6.15)

Now the coefficients of dx2, dxdy, and dy2 must extend differentiably to P ,since x, y are valid local coordinates. By (6.12) their values must be 1, 0, 1respectively, at P . Adding the coefficients of dx2 and dy2 and simplifying, weobtain

1 +G

r2,

which shows that Gr2 extends differentiably to P with limiting value 1. (It

follows that C itself extends differentiably to P .) Consequently we may write

G

r2= H , where H(0, 0) = 1 ,


with H everywhere differentiable. The coefficient of dxdy in (6.14) may nowbe written as

J(x, y) =1−Hr2

2xy , (6.16)

which, as we said, must extend differentiably to P and have 0 as value at P .Now J(0, y) = 0. Hence

J(x, y) = J(x, y)− J(0, y) =∫ 1

0

d

dtJ(tx, y)dt

= x

∫ 1

0

J1(tx, y)dt .

The last integral is differentiable in x and y since it can be differentiated underthe integral sign. We write J(x, y) = L(x, y)x. Now J(x, 0) = 0, by (6.16)and continuity of J at P . Hence L(x, 0) = 0 for all x. By an argument similarto what we have just given, we find that

L(x, y) = yM(x, y) ,

for some differentiable function M(x, y). This gives

xyM(x, y) =1−H(x, y)

r22xy ,

and cancelling xy, which is permitted by continuity, we have

1−H(x, y)r2

=12M(x, y) ,

so that

G(x, y) = r2H(x, y) = r2(

1− 12r2M(x, y)

),

which we write as

G(x, y) = r2(1 + r2Q(x, y)

), (6.17)

where Q = −12M is differentiable in x and y for all x, y in the normal

neighborhood, P included. Equation (6.17) can also be written as

G = r2 + αr4 + r5(β cosϕ+ γ sinϕ) +O(r6) , (6.18)

where α, β, γ are constants and O(r6) denotes a function of x, y which is amultiple of r6. Substituting into (6.14), we get

ds2 = dx2 + dy2 + α(xdy − ydx)2 + ... (6.19)

100 6 INSERT TITLE

where the dots represent a quadratic differential form in x, y whose coeffi-cients are O(r3). This equation expresses the ds2 in terms of normal coordi-nates. Now the normal coordinates are determined up to orthogonal transfor-mation about (0, 0) depending on the choice of the vector a1 from which theangle ϕ is measured and the sense of the angle; it follows that the coefficientsof (6.19), or their combinations, which are invariant under an orthogonaltransformation on x, y, are invariants of the Riemannian metric at P . Thisis in particular the case with α. Its relation with known invariants is easilyfound. In fact, from (INSERT REF) we have

K = − 1√G

∂2√G

∂r2.

By (6.18) we find

K(P ) = −3α ,

where K(P ) is the Gaussian curvature at P .Using geodesic polar coordinates, we can obtain some simple geometric

interpretations of the Gaussian curvature. Call the curves r = const. geodesiccircles, with center P and radius r. The perimeter of a geodesic circle ofradius r is

L =∫ π

−π

√Gdϕ =

∫ π

−π

(r − K(P )

6r3 + ...

)dϕ

= 2πr − π

3K(P )r3 + ... .

Hence we get

K(P ) = limr→0

3π

2πr − Lr3

.

Similarly, the area of a geodesic circle of radius r is

A =∫∫

θ1 ∧ θ2 =∫ r

0

∫ π

−π

√Gdϕdr =

∫ r

0

Ldr = πr2 − π

12K(P )r4 + ... .

From this it follows that

K(P ) = limr→0

12π

πr2 −Ar4

.

6.2.2 INSERT TITLE

From (6.2.1) we can derive the following useful result.

Theorem 20. Every point P of N has a neighborhood U such that thegeodesic lying in U and joining P to any point Q ∈ U is the only curveof shortest length in M joining these two points.


Proof. To prove this it suffices to take the geodesic polar coordinates r, ϕabout P and to define U by r < c, where c is a positive number so small thatthe geodesic polar coordinates are valid for all r, ϕ such that r < c. If C is acurve joining P to Q and lying in U then its length is∫

C

ds =∫C

√dr2 +Gdϕ2 ≥

∫C

dr = r(Q) .

The inequality is an equality if and only if dϕ = 0, which is to say if and onlyif C is the unique geodesic lying in U joining P to Q. On the other hand ifC goes outside of U then it must meet the geodesic circle r = c and so itslength is greater than c and so is longer than the unique geodesic joining Pand Q and lying in U .

The notions we have introduced above, the parallelism of Levi-Civita, thegeodesic curvature of a curve, the Gaussian curvature, etc., all pertain to aneighborhood. Some of the most interesting results in Riemannian geometryare concerned with the relations of such properties with the properties of themanifold as a whole, or, more specifically, with its topological properties. Tostudy these an effective tool is the distance function in M . Such a functionassigns to any two points P , Q of M a number d(P,Q) > 0, such that thefollowing conditions are satisfied:

1. d(P,Q) = 0 if and only if P = Q.2. d(P,Q) = d(Q,P ).3. d(P,Q) + d(Q,R) ≥ d(P,R).

In topology we say that the distance function defines a metric in M and thatM is a metric space. Notice that such a metric is entirely different from whatwe call a Riemannian metric.

We wish to define a distance function in M which bears a close relationwith its Riemannian metric. For any two points P,Q ∈M we define d(P,Q)to be inf PQ, where PQ is the arc length of a sectionally smooth curve, calleda path, joining P , Q. To justify this definition we have to show that such acurve exists. In fact, from the connectedness of M , which we assume here, itfollows that there is a sequence of coordinate neighborhoods U1, ..., Um, suchthat P ∈ U1, Q ∈ Um, and Ui ∩Ui+1 6= 0, i = 1, ...,m− 1. Let Pi ∈ Ui ∩Ui+l.Then any two consecutive points of the sequence P, P1, ..., Pm−l, Q lie in thesame coordinate neighborhood and can be joined by a smooth arc. These arcsmake up a sectionally smooth arc joining P , Q.

This distance function clearly has the properties:

d(P, P ) = 0 , d(P,Q) = d(Q,P ) ,d(P,Q) + d(Q,R) ≥ d(P,R) .

Moreover, if P 6= Q, it follows from the statement in the beginning of thissection that d(P,Q) > 0. Hence our distance function has the properties 1)-3)and makes M into a metric space.

It is important to observe that the topology defined in (MISSING TEXT)

102 6 INSERT TITLE

6.3 The second variation of arc length

Earlier in this chapter we have seen that geodesics have the following twominimizing properties: firstly, if C is a geodesic of a field then it is the shortestpath joining two of its points and lying in the geodesic field; secondly, everypoint O has a “normal neighborhood” Dr such that for any P ∈ Dr thereis a unique geodesic lying in Dr and joining O to P , and this is the uniqueshortest path in M joining O and P . Our approach to these results has beenlocal in nature; we now want to consider how they can be extended globally.In particular, if g is a geodesic which starts from a point O of M , and ifP is a variable point on C, then C furnishes the unique shortest path inM joining O and P at least as long as P lies in a normal neighborhoodcentered at O. Beyond that, if G between O and P belongs to a geodesicfield, then G furnishes the unique shortest path joining O and P lying in thegeodesic field. It is natural to ask when, if ever, G fails to provide the shortestpath between O and P as P moves farther and farther away from O. Thestudy of this question leads to relations between the topology of M and itsdifferential geometry, and to information on the topology of M , particularlyin higher dimensions. We are not in a position here to say much more aboutthis development, but we trust that the results we present in two dimensionswill already be found sufficiently attractive.

6.3.1 INSERT TITLE

We begin by examining how the arc length of a curve is changed if the curveis smoothly deformed. Let M be a Riemannian manifold of dimension 2 andR the rectangle in the (x, t)-plane

R = {(x, t) | 0 ≤ x ≤ V , −ζ < t < ζ} ,

where V , ζ are two positive numbers. By a 1-parameter family of curves inM we mean a differentiable map F from R to M which has rank 1 if t isfixed, i.e. in local coordinates u, v on M .(

∂u

∂x,∂v

∂x

)6= (0, 0) .

To each (x, t) we assign the frame Pe1e2 of M such that

P = F (x, t) , e1 along(∂u

∂x

∂

∂u+∂v

∂x

∂

∂v

);

we assume that M is oriented, so that e2 is then uniquely determined. Underthis map the various differential forms on the frame bundle of M are pulledback to R.

The exterior derivative on R may be decomposed into the sum of differ-entiation with respect to x and differentiation with respect to t:

6.3 The second variation of arc length 103

d = dC + dt , dC = dx ∧ ∂

∂x, dt = dt ∧ ∂

∂t.

Likewise, the forms ω1, ω12, pulled back from the frame bundle of M , aredecomposed as

ωi = θi + aidt , ω12 = θ12 + a12dt .

where θ1, θ12 are characterized by

θi ∧ dx = θ12 ∧ dx = 0 .

For t fixed, let Ct denote the curve F (t, x) in M . (It is parametrized byx.) The length of Ct, L(t), is given by

L(t) =∫t×[0,V ]

θ1

and its derivative by

dL

dt(t) ≡ d

dt

∫t×[0,V ]

∂θ1∂t

. (6.20)

To find the differential form in the second integral we calculate dω, in twodifferent ways (differentiate and then decompose, decompose and then differ-entiate):

dω1 = ω12 ∧ ω2 = (a2θ12 − a12θ2) ∧ dt

dω1 = d(θ1 + a1dt) =(−∂θ1∂t

+ dCa1

)∧ dt .

Cancelling dt gives

dθ1dt

= dCa1 + a12θ2 − a2θ12 . (6.21)

Now θ2 = 0, since e1 is tangent to Ct; θ12 is the connection form, whichvanishes on Ct if and only if Ct is a geodesic. By combining (6.20) and (6.21)we obtain

dL

dt(t) =

∫Ct

(−a2θ12 + dCa1) = −∫Ct

a2θ12 + a1(V, t)− a1(0, t) , (6.22)

where the second equality is gotten by using the Fundamental Theorem ofCalculus, and where we have indicated the domain of integration t × [O, V ]by the somewhat imprecise but simpler expression Ct. Any family Ct is calleda variation of C0 and a1(0, t), a2(0, t) are called, respectively, the tangentialand normal components of the variation vector. On Ct, θ12 = kgds, where kgis the geodesic curvature and ds is the element of arc.

104 6 INSERT TITLE

We next show that given any curve C(x), 0 ≤ x ≤ V , in a Riemanniansurface and any differentiable real-valued function f(x), 0 ≤ x ≤ V , wecan find a 1-parameter family of curves Ct in M such that C = C0 anda2(x, 0) = f(x). In fact, taking frames e1e2 along C0, with e1 tangent to Cwe let F (x, t) be the point obtained by going along the geodesic perpendicularto C at C(x) a distance f(x)t from C(x) in the e2-direction. We must finda positive ζ such that F (x, t) is defined for all (x, t) with 0 ≤ x ≤ V and−ζ < t < ζ,

(∂u∂x ,

∂v∂x

)6= 0. Note that the last condition is assumed to hold

when t = 0. The Fundamental Theorem of Ordinary Differential Equationsguarantees that for each y, 0 ≤ y ≤ V , there are positive numbers εy, ζy suchthat F (x, t) is defined for all (x, t) with |x−y| < εy, |t| < ζy, and the requiredζ for all x is then gotten by a compactness argument. It remains to show thata2(x, 0) = f(x). But along the geodesic through C(x) perpendicular to C theelement of arc is ds = f(x)dt, while ω2 = a2dt is the element of arc at t = 0,i.e. ω2 = ds at t = 0. Hence a2(x, 0) = f(x), as was to be shown. This one-parameter family of curves is called the one-parameter family correspondingto a2. Note that if f(O) = f(V ) = 0 then the endpoints of the correspondingfamily are fixed at C0(0) and C0(V ), respectively. If f is continuous andpiecewise differentiable this construction can still be carried out; the resultingcurves are continuous but may have finitely many corners.

Formula (6.22) has various geometric consequences. First of all, if C is ageodesic and we extend it to a 1-parameter family of curves, as above, wehave θ12 = 0 on C = C0. If the curves of the family have the same fixedendpoints P , Q, then a1(V, t) = a1(0, t) = 0, and hence

δL =dL

dt(0) = 0 .

Conversely if C is a given curve while is not a geodesic, we can find a 1-parameter family of curves Ct with fixed endpoints such that C0 = C, andsuch that a2(x, 0) is non-zero only on an interval where the geodesic curvatureof C is bounded away from zero. Taking a2 to have the same sign as thegeodesic curvature, we find that δL < 0. (This implies that there are curvesof the family Ct of strictly shorter length than C0 = C.) What we have showncan be summed up in the following way: for curves joining fixed points P ,Q of M , the geodesic or geodesics, if any, joining P and Q are precisely thecurves for which the first variation of arc length, δL, vanishes.

We mention another consequence of formula (6.22). Suppose that D isa curve in M and consider the family of geodesics perpendicular to D; al-ternatively, let D be a point and consider a 1-parameter family of geodesicsthrough D. Suppose that the geodesics in question envelop a curve E (thatis to say are tangent to a curve E). We parametrize the geodesics so that x0

lies on D and x = V is the point of tangency with E. Then in formula (6.22)θ12 = 0 because the curves are geodesics, and a1(0, t) = 0 because the end-points at 0 are fixed or move perpendicularly to the geodesics as t varies. Onthe curve endpoints at V , i.e. E, the geodesics, and hence the vectors e1, are


tangent to E, and therefore ω1 = a1(V, t)dt = ds is the element of arc on E.Hence formula (6.22) reduces to

dL = ds .

Integrating this gives the following result: the difference in length of twogeodesics of the family perpendicular to (or passing through) D and envelopingE is equal to the length of the arc of E lying between the points of tangencyof the two geodesics.

(INSERT FIGURE)

Exercise 86. Let D be a plane curve and let E be the locus centers ofcurvature of D. Let P,Q ∈ D and let k denote the curvature of D. Find thelength of the arc of E between the centers of curvature of D at P and at Qby a) direct computation using the Frenet formulas, and b) as a special caseof the result just proved.

6.3.2 INSERT TITLE

We next calculate d2Ldt2 (0) for a 1-parameter family of curves Ct on M . For

simplicity, and because it suffices for our needs, we will assume that C0 is ageodesic. This condition can be expressed analytically as

θ12 = 0 when t = 0 . (6.23)

The first task will be to find ∂θ12∂t . To do this we calculate dω12 in two different

ways:

dω12 = −Kω1 ∧ ω2 = K(a1θ2 − a2θ1) ∧ dt

dω12 = (dC + dt ∧ ∂

∂t)(θ12 + a12dt)

=(−∂θ12

∂t+ dCa12

)∧ dt .

Remembering that θ2 = 0, and cancelling dt, we obtain

∂θ12∂t

= dCa12 +Ka2θ1 . (6.24)

In order to determine dC(a12) along t = 0, we differentiate ω2 in two differentways:

dω2 = ω1 ∧ ω12 = (θ1 + a1dt) ∧ (θ12 + a12dt)= (a12θ1 − a1θ12) ∧ dt

dω2 = d(a2dt) = dCa2 ∧ dt .

At t = 0, where θ12 = 0, we obtain

106 6 INSERT TITLE

dCa2 = a12θ1 . (6.25)

Now θ1 is the element of arc along the curve C0, and we will henceforth writeit as dr. Formula (6.25) then becomes

a12 =da2

dr

and therefore

dCa12 =d2a2

dr2dr . (6.26)

From (6.22), taking into account restriction (6.23) and formulas (6.24)and (6.26), we obtain the formula for the second variation

δ2L =d2L

dt2(0) = −

∫C

a2∂θ12∂t

+∂a1

∂t(V, 0)− ∂a1

∂t(0, 0)

= −∫C

a2

(d2a2

dr2+Ka2

)dr +

∂a1

∂t(V, 0)− ∂a1

∂t(0, 0) .

where we write C for C0 and recall that this is proved under the assumptionthat C is a geodesic. Integration by parts yields another formula for thesecond variation:

δ2L =∫C

((da2

dr

)2

−Ka22

)dr + a2(0)

da2

dr(0)− a2(L)

da2

dr(L)

+∂a1

∂t(V, 0)− ∂a1

∂t(0, 0) .

where, to simplify the notation, we have written

a2(0) = a2(0, 0) , a2(L) = a2(V, 0) .

where L is the length of C. We now regard a2 as a function of r, 0 < r < L,with r = 0 corresponding to x = 0.

6.3.3 INSERT TITLE

In order to draw out the geometric consequences of the second variationformula, we must first re-examine geodesic polar coordinates, which we con-structed in (INSERT REF), from a global point of view. Let O ∈ M be apoint of a Riemannian surface and let v be a tangent vector of M at O. ThenexpO(v) is defined to be the point of M obtained by going along the geodesicthrough O tangent to v a distance |v| in the direction of v, provided that thispoint exists. The mapping expO goes by the name of the exponential mappingat O. We note that expO is defined on some neighborhood of O in the tangent


space V (O) of M at O, where it is essentially nothing more than the normal,or geodesic polar coordinates, which we have previously introduced, but itmay not be defined on all of V (O). (If the metric on M defined in (INSERTREF) has the so-called completeness property, then it can be proved thatexpO is defined on all of V (O).)

We claim that expO, which is defined on an open subset of V (O), is dif-ferentiable. To prove the claim we let vO ∈ V (O) be such that expO(vO)is defined. Note that if |vO| is sufficiently small then the claimed differen-tiability is part of the conclusion of the Fundamental Theorem of OrdinaryDifferential Equations. In general, assuming that vO is not the 0-vector, welet vO = |vO|eO, where eO is a vector of unit length. Let C(r), 0 ≤ r ≤ |vO|,be the geodesic expO(reO), which is parametrized by arc length. Using theFundamental Theorem and an argument from elementary topology, we findthat there is a sequence of numbers rk, 0 = r0 < r1 < ... < rn = |vO|, suchthat if ek denotes the tangent vector of C(r) at rk (which is a unit vectorsince C(r) is parametrized by arc length), then there is a neighborhood Ukof ek in the unit tangent bundle F of M for each k, 0 ≤ k ≤ n − 1, hav-ing the property that if e ∈ Uk then expO(λe) is defined and differentiablein λe provided |λ| < 2(rk+1 − rk). If e ∈ Uk ∩ V (P ) let e′ be the tangentvector of D(λ) = expP (λe) at λ = rk+1 − r. Note that e′k = ek+l and thatthe correspondence e → e′ is differentiable. We call e′ the k-translate of e.Now by continuity there is a neighborhood U ′n−2 ⊂ Un−2 of en−2 such thatif e ∈ U ′n−2 then the (n − 2)-translate of e lies in Un−1; likewise there ex-ists a neighborhood U ′n−3 ⊂ Un−3 of en−3 such that if e ∈ U ′n−3 then the(n − 3)-translate of e lies in U ′n−2. Continuing in this fashion we arrive at aneighborhood U ′O of eO such that if e ∈ U ′O ∩ V (P ) then the composition ofthe k-translates of eO ≤ k ≤ n − 2, that is to say the tangent vector e′′ ofD(r) = expP (re) at r = rn−1, lies in Un−1 and depends differentiably on re.From this it follows that D(r) is defined for all r such that 0 < r < 2rn−rn−1

and depends differentiably on re. The same is true if e is restriced to lie inV (O), which is a submanifold of the unit tangent bundle. Hence expO isdifferentiable at VO, which proves the claim.

For each v in the tangent space atO for which expO is defined, we associatethe right-handed frame e1e2 of M at expO(v) such that e1 is tangent toC(r) = expO(rv) in the direction of increasing v. Pulling back ω1, ω2 fromthe frame bundle of M , and taking polar coordinates r, ϕ in the tangentspace to M at O with respect to some right-handed orthonormal frame, wehave, by the argument given in (INSERT REF) in connection with geodesicpolar coordinates,

θ1 = dr (6.27)θ2 = qdϕ , (6.28)

for some function q(r, ϕ). Now expO may fail to be one-to-one, and its rankmay fall below 2. By (6.27) and (6.28), and the fact that expO is non-singular

108 6 INSERT TITLE

at O, we find that the rank falls exactly where r > O and q = O. (A pointwhere the rank of a mapping falls is called a singular point.) At such a pointthe rank is 1. Note that the metric pulled back to V (O) is given by

ds2 = θ21 + θ22 = dr2 +Gdθ2 , G = q2 ,

and, by (INSERT REF), q satisfies the differential equation

∂2q

∂r2+Kq = 0 , (6.29)

where K is the Gaussian curvature.

Exercise 87. On the sphere of radius ρ, using spherical coordinates, and onthe circular cylinder of radius ρ, using cylindrical coordinates, determine theexponential map expO, where O is the north pole, in the case of the sphere,and O is any convenient point in the case of the cylinder. Determine thedomain of definition and the singular points of expO in each case.

6.3.4 INSERT TITLE

Our preparations are now completed and we have in hand the tools for car-rying out a rather powerful and global analysis of the minimizing propertiesof geodesics. Let C(r), A < r < B be a geodesic in M parametrized by arclength r. By a Jacobi field along C we mean a differentiable function a(r),A < r < B, which satisfies the differential equation

d2a

dr2+K(C(r))a = 0 , (6.30)

where K is the Gaussian curvature. If r1 6= r2, and there is a non-zero Jacobifield along C which vanishes at r1 and r2, we say that r1 and r2 are conjugatealong C. We specify this relation in terms of the parameter values, insteadof the image points P = C(r1), Q = C(r2), since C may fail to be one-to-one. However, if r1 and r2 are conjugate along C and when there is nodanger of confusion, we say, simply that P and Q are conjugate along C. Thedifferential equation (6.30) is linear in a. Consequently, if a is a Jacobi fieldthen λa is also a Jacobi field, for any constant λ. Moreover, by the uniquenesspart of the Fundamental Theorem of Ordinary Differential Equations, if a, aare two non-zero Jacobi fields which vanish at r1, then

a =a′(r1)a′(r1)

a ,

since both sides are Jacobi fields and have the same first derivative at r1.Let rO be a point in the domain of C, O = C(rO). Let us change the

parameter so that arc length is measured from rO and O = C(O). Let e be a


unit tangent vector to C at O in the direction of increasing arc length. ThenC(r) = expO(re). Let us introduce polar coordinates r, ϕ in V (O) so thate is in the direction ϕ = ϕO. From equation (6.29) we see that q =

√G is

a Jacobi field on C. Moreover, q vanishes at r = 0, as we saw in (INSERTREF). Let r1 be such that C(r1) is defined. Reversing the parametrizationof C if necessary, we can assume that r1 > 0. As we showed in (INSERTREF), q(r1, ϕO) = 0 if and only if expO is singular at r1e. Now any Jacobifield which vanishes at 0 is a constant multiple of q, as we just saw. Hencer1 is conjugate to 0 along C if and only if q(r, ϕfO) = 0. It follows that r1is conjugate to O along C(r) if and only if expO is singular at r1e1. Nowlet e be a unit vector tangent to C(r) at r1 and directed in the directionof decreasing r. Let P = C(r1). Then since r1 and O are conjugate expP issingular at r1e1. This remarkable fact can be expressed simply but crudely asfollows: expO is singular at P if and only if expP is singular at O. We notealso that the conjugate points of O along C are isolated. For otherwise thezeros of q would have an accumulation point on C and therefore, by the MeanValue Theorem, ∂q

∂r would vanish at the accumulation point. But q and ∂q∂r

cannot vanish simultaneously because q satisfies (6.30) and the FundamentalTheorem of Ordinary Differential Equations tells us that a solution of (6.30)which vanishes simultaneously with its derivative must be identically zero,whereas q 6= 0 in the domain of validity of the geodesic polar coordinates.

Let O ∈ M and let C be a geodesic parametrized by arc length throughO with unit tangent vector e. Suppose that expO(re) = P is defined and thatthere are no conjugate points to O along D between O and r, r included,which is to say that expO(re) is non-singular for all r such that 0 ≤ r ≤ r.By continuity of the derivatives the set of points of V (O) where expO isnon-singular forms an open neighborhood N of re|0 ≤ r ≤ r. Now let C(t),0 ≤ t ≤ V , be a path in M joining O to P . We say that C(t) is nearby toC(r) if it can be “lifted” under the exponential map to N , i.e. there is a curveC(t) in N such that C(0) = 0, C(V ) = re, and expO(C(t)) = C(t). If C(t)is nearby to C(s), then since its “lift” C(t) lies in N where the pull-back ofds2 is given by

ds2 = dr2 +Gdϕ2 , G > 0 if r > 0 ,

its length is given by

L(C) =∫C(t)

√dr2 +Gdϕ2 ≥

∫C(t)

dr = r ,

with equality holding if and only if ϕ = const., that is to say if and only ifC(t) lies on the line re. Thus we have shown the following

Theorem 21. If, along the geodesic C(r), 0 ≤ r < r, there are no conjugate-points of O, then C(r) is shorter than any other path nearby to C(r) andjoining C(O) and C(r).

110 6 INSERT TITLE

On the other hand we will next show the following

Theorem 22. If O and r, r > 0, are conjugate along the geodesic C(r), andr > r, then C(r) is not the shortest path joining C(O) to C(r). There exists,in fact, a one-parameter family of paths Ct with CO = C such that for allt 6= 0 Ct is strictly shorter than C.

We may express this informally by saying that a geodesic starting from 0fails to be minimizing beyond the first conjugate point.

Remark 11. The curves to be constructed are not smooth but have each onea corner. It can be shown that the corner of any one of them can be roundedoff to give a smooth curve of length shorter than that of D(r).

Proof. Recall the second variation formula

δ2L = −∫C

a2

(d2a2

dr2+Ka2

)dr +

∂a1

∂t(V, 0)− ∂a1

∂t(0, 0) , (6.31)

valid where C is a geodesic.Suppose that O and r, r > O, are conjugate along the geodesic C(r) and

that r > r with C defined at r. Take a Jacobi field which vanishes at O andr, we can suppose that it is q, referring to the exponential map at C(O) andlet f(r) be the function defined by

f(r) ={q(r) , O ≤ r ≤ rO , r ≤ r ≤ r .

Then f(r) is continuous, but is not differentiable at r, although right and leftderivatives exist at r, with

f ′(r−) 6= 0 , f ′(r+) = 0 .

Let

a2(r) = f(r) + εη(r) ,

where η is a function and ε a number yet to be determined. Then∫ r

0

a2

(d2a2

dr2+Ka2

)dr =

∫ r

0

f

(d2f

dr2+Kf

)dr

+ ε

(∫ r

0

f

(d2η

dr2+Kη

)dr +

∫ r

0

η

(d2f

dr2+Kf

)dr

)

+ ε2∫ r

0

η

(d2η

dr2+Kη

)dr .


Now the first integral on the right, and the second integral in the ε term,vanish because the integrand vanishes. Integrating by parts twice on [O, r]and [r, r] gives∫ r

0

f

(d2η

dr2+Kη

)dr =

∫ r

0

fd2η

dr2dr +

∫ r

r

fd2η

dr2dr +

∫ r

0

fKηdr

= f(r)dη

dr(r)− f(0)

dη

dr(0) + f(r)

dη

dr(r)− f(r)

dη

dr(r)

−∫ r

0

df

dr

dη

drdr −

∫ r

r

df

dr

dη

drdr +

∫ r

0

fKηdr

= −dfdr

(r−)η(r) +df

dr(0)η(0)− df

dr(r)η(r) +

df

dr(r+)η(r)

+∫ r

0

η

(d2f

dr2+Kf

)dr

= −dfdr

(r−)η(r) +df

dr(0)η(0) .

Hence

−∫ r

0

(d2a2

dr2+Ka2

)dr = ε

(df

dr(r−)η(r)− df

dr(0)η(0)− ε

∫ r

0

η

(d2η

dr2+Kη

)dr

).

We now let η be a continuous function on [0, r] such that η(0) = η(r) = 0,and η(r) is non-zero and has the opposite sign to

df

dr(r−) =

dq

dr(r) .

(Note that this quantity is non-zero, since q and q′ cannot vanish simultane-ously. For example, let η(r) = ± sin(πr−1r).

Then

−∫ r

0

a2

(d2a2

dr2+Ka2

)dr = ε

dq

dr(r)η(r)− ε2

∫ r

0

η

(d2η

dr2+Kη

)dr .

(6.32)

Our aim is to calculate δ2L for the variation corresponding to a2. We havejust calculated the integral in the formula; we next calculate the other terms.Since a2(0) = a2(r) = 0, the endpoints are fixed and therefore

a1(0, t) = a1(r, t) = 0 .

The only remaining non-zero terms are

∂a1

∂t(r−, 0)− ∂a1

∂t(r+, 0) .

112 6 INSERT TITLE

We take coordinates u, v in a geodesic field in a neighborhood of C(r) so thatC is defined by u = 0 and the v-curves are the geodesics perpendicular to C(see (INSERT REF)). v can be taken to be arc length on C, in fact, so thatv = r on C. The Riemannian metric can be written as

ds2 = ρ21 + ρ2

2

where

ρ1 = du

ρ2 = Q(u, v)dv ,

for some function Q. Change the sign of ρ2 if necessary to make Q positive.Let g(r) be a function and consider the variation of C corresponding to g(r).For each curve t = const. we take frames e1e2 such that e1 is tangent toCt, as before. Let ω1ω2 be the dual basis. Then, after changing the sign ofρ1 = du, if necessary, we can write

ω1 = cosψρ1 + sinψρ2

ω2 = − sinψρ1 + cosψρ2 ,

for some function ψ, since ρ1, ρ2 is also dual to an orthonormal basis. Onany curve t = const., ω2 = 0, so that

0 = ω2 = − sinψρ1 + cosψρ2 .

On u = 0, that is to say, on the geodesic C, we have ρ1 = 0, and sincedr2 = ds2 = ρ2

2, it follows that Q = 1. Hence, on C we have

dr = ω1 = sinψρ2

0 = ω2 = cosψρ2 .

It follows that on u = 0 sinψ = 1, cosψ = 0, so that ψ = π2 . At any point of

C we therefore have ω2 = −ρ1 = −du, so that u decreases in the a1-direction.Using this we find that the variation corresponding to g is given by

u = −g(r)tv = r .

Hence

ρ1 = −g′tdr − gdtρ2 = Qdr

ω1 = (− cosψg′t+ sinψQ)dr − cosψgdtω2 = (sinψg′t+ cosψQ)dr + sinψgdt .


Since ω2 = 0 on t = const., we have

sinψg′t+ cosψQ = 0 . (6.33)

From the expression for ω1 we see that

a1 = − cosψg .

Hence ∂a1∂t (r, 0) = ψ′g. Differentiating (6.33) with respect to t gives, at t = 0,

g′ − ψ′ = 0 .

Hence

∂a1

∂t= gg′ .

We apply this last formula to calculate δ2L. If

g(r) = f(r) + εη(r) ,

then

∂a1

∂t(r−, 0)− ∂a1

∂t(r+, 0) = εη(r)

[∂q

∂r(r) + ε

∂η

∂r(r)− ε∂η

∂r(r)]

= εη(r)∂q

∂r(r) .

From (6.31), (6.32), and the last formula, we obtain

δ2L = ε

(2∂q

∂r(r)η(r)− ε

∫ r

0

η

(d2η

dr2+Kη

)dr

).

Now η has been so chosen that

∂q

∂r(r)η(r) < 0 .

If we choose ε so that

0 < ε <

∣∣∣∣∣∣dqdr (r)η(r)∫ r

0η(d2ηdr2 +Kη

)dr

∣∣∣∣∣∣ .we have δ2L < 0. Hence the length of Ct, t 6= 0, is smaller than that of C0

for all t sufficiently small, that is to say that C0 is a strict relative maximumin this family. This proves the theorem.

114 6 INSERT TITLE

6.3.5 INSERT TITLE

We next ask whether conjugate points exist.

Theorem 23. a) If the Gaussian curvature K ≤ 0 on a surface M , then Mhas no pairs of conjugate points.

b) If K > b > 0, for some constant b, then for every point 0 and everygeodesic C starting at 0, there is a conjugate point r on C to 0 at distance

r <π√b,

provided that C extends to distance π√b

from 0.

Proof. Part a) is a calculus exercise. Suppose that K ≤ 0 and that thereis a geodesic C(r) having two conjugate points along it, 0 and r. Since theconjugate points of 0 along C(r) are isolated, proceeding from 0 along Ctoward r there is a first conjugate point of 0 which we take to be r. Let a(r)be a non-zero Jacobi field which vanishes at 0 and r. Then a′(0) 6= 0, sinceotherwise a is identically zero. If a′(0) < 0, we replace a by −a, so we mayassume that a′(0) > 0. Then, we have a(r) > 0 for r sufficiently close to 0and therefore a(r) > 0 on (0, r). Since a(r) = 0 this implies that a′(r) ≤ 0and therefore that a′(r) < 0. But

a′′ = −Ka ≥ 0 on (0, r) ,

which says that a′ is non-decreasing on [O, r], contradicting the conditiona′(O) > 0, a′(r) < 0. This proves part a).

Part b) follows from the following, which is quite remarkable in itself.

Theorem 24. The Sturm Comparison Theorem. If K(r), L(r) are continu-ous functions on [r1, r2] with

K(r) > L(r)

for all r ∈ [r1, r2], and if a, a are functions on [r1, r2] satisfying

a′′ +Ka = 0 ,a′′ + La = 0 ,

respectively, and if a(r1) = a(r2) = 0, a(r) 6= 0 for r ∈ (r1, r2) then a(r) = 0for some r ∈ (r1, r2).

Proof. Assume that a(r) 6= 0 for r ∈ (r1, r2). Replacing a by −a, a by −a,as needed, we may assume a, a > 0 on (r1, r2), and the differential equations,being linear, are still satisfied. Multiplying the left-hand side of the firstdifferential equation by a, multiplying the left-hand side of the second by a,subtracting and integrating gives


0 =∫ r2

r1

(a(a′′ +Ka)− a(a′′ + La)) dr .

Integrating by parts then gives

0 = a(r2)a′(r2)− a(r1)a′(r1)− a(r2)a′(r2)

+ a(r1)a′(r1) +∫ r2

r1

aa(K − L)dr ,

or ∫ r2

r1

aa(K − L)dr = a(r2)a′(r2)− a(r1)a′(r1) .

Since a(r) > 0 for r ∈ (r1, r2), a(r1) = a(r2) = 0, we have a′(r1) ≥ 0,a′(r2) ≤ 0, and hence a′(r1) > 0, a′(r2) < 0, since otherwise a(r) is identicallyzero, by the Fundamental Theorem, since it satisfies the linear differentialequation a′′ + La = 0. Also, a(r1), a(r2) ≥ 0. Hence

0 <∫ r2

r1

aa(K − L)dr = a(r2)a′(r2)− a(r1)a′(r1) ≤ 0 ,

which is a contradiction. Hence a(r) = 0 for some r ∈ (r1, r2), which provesthe Sturm Comparison Theorem.

Returning now to the proof of the previous theorem, Part b), we supposethat M is a surface whose Gaussian curvature satisfies K > b > 0, for b aconstant. Let C(r) be a geodesic parametrized by arc length which starts atC(0) = 0 and extends at least for r ≤ π√

b. Let q =

√G, where the pull-back

of the Riemannian metric on M under expO is given by

ds2 = dr2 +Gdϕ2 .

Then q is a Jacobi field, which is to say that q satisfies

q′′ +Kq = 0

along C(r). Now let

q = sin√br .

Then q satisfies

q′′ + bq = 0 .

But K > b and q(0) = q(π√b

)= 0, with q(r) 6= 0 for r ∈

(0, π√

b

). The Sturm

Comparison Theorem now implies that q(r) = 0, for some r ∈(

0, π√b

). Since

q(0) = 0, by (INSERT REF), we have r conjugate to 0, and the proof of Partb) is complete.

116 6 INSERT TITLE

The theorem we have just proved has interesting topological consequences,one of which we now wish to describe. It can be shown that if the metric onM defined in (INSERT REF) has the so-called “completeness” property, thenany geodesic C(r) parametrized by arc length r is defined for all values ofr, which is to say that for any O ∈ M expO(v) is defined for all v ∈ V (O);secondly, any two points of M are joined by a minimizing geodesic, that isa geodesic arc whose length is less than or equal to the length of any pathjoining the two points. Let us suppose that M is a complete surface whoseGaussian curvature satisfies K(s) > b > 0, b constant. Then any minimizinggeodesic has length < π√

b, since otherwise, by previous theorems, it would

contain a conjugate point of one of the endpoints and would therefore notminimize the distance between its endpoints. Thus the distance of any pointP to a fixed point O is less than π√

b, and from this it can be shown that M

is compact, that is to say that M is a closed surface without boundary.As always in mathematics, one must consider the theory we have pre-

sented as it is realized in some examples. First of all, consider a unit sphereand O a point on it. If C(r) is a geodesic through O parametrized by arclength from O then π is conjugate to O along C(r); in fact C(π) = O′ isthe antipodal point, and if we express the exponential map expO in terms ofpolar coordinates (r, ϕ) in V (O), we have expO(π, ϕ) = O′, for all ϕ. HenceexpO fails to be locally one-to-one at (π, ϕ), so that (π, ϕ) is a singular pointof expO and π is conjugate to O along C(r). By the same reasoning 2π isconjugate to O along C(r). But C(2π) = 0, so that one might say that O isconjugate to itself. For rO < π, C is the unique minimizing geodesic joiningO to C(rO) since M − O′ is a normal neighborhood of O. If π < rO ≤ 2π,then C(r), 0 ≤ r ≤ r0, is no longer a minimizing geodesic; the minimizinggeodesic joining O to C(rO) is now C(−(2π − rO)).

A right circular cylinder of radius 1 provides a second interesting example.In this case K = 0, so there are no conjugate points. A base circle C(r) is ageodesic since its geodesic curvature vanishes, as may be seen by drawing theright picture. If we start at r = 0, then C(r) is a unique minimizing geodesicuntil r = π; for r > π it fails to be minimizing, even though there are noconjugate points. It remains, however, the shortest path among all nearbypaths joining its endpoints, even for r = 2π, when it joins C(O) to itself.

In this section we have covered only some of the most basic aspects of thegeometry of geodesics on a surface. More on the subject can be found in W.Blaschke, Vorlesungen uber Differentialgeometrie, vol. I, Chapters 6 and 7.New York: Chelsea, 1967. Blaschke’s approach, which is in certain respectsvery attractive, sets up the variation in a way different from ours.

di erential geometry of curves and surfaces, 1st …deigen/chern.pdf · thomas bancho , shiing-shen...

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