Math 162A - Introduction to Differential Geometry

Neil Donaldson

Winter 2024

Introduction

Classical Differential Geometry is the study of curves and surfaces in the plane and three-dimensional

space using multi-variable calculus, linear algebra & differential equations. At a more advanced level,

topology, analysis and abstract algebra become more important, but none of this is required for our

treatment.

Of particular interest is the notion of curvature: a measure of the ‘bendiness’ of a curve or surface.

Intuitively, a straight line should have zero curvature, while the curvature of a circle should vary

inversely as the radius: a very large circle should have very small curvature.

Zero curvature Small curvature Larger curvature Variable curvature

Understanding and quantifying this concept for more complicated curves is our ﬁrst important goal.

The rough idea is to imagine a curve as a roller-coaster along which you travel at a constant speed;

the curvature is then the force necessary to keep you travelling along the curve.

Curvature is a more difﬁcult concept for surfaces. In particular, we will hunt for quantities which

measure how much a surface appears to be dome- or saddle-shaped.

Dome-shaped Saddle-shaped More complicated

The third surface is saddle-shaped near the narrow neck and dome-shaped away from it.

1 Curves in Euclidean Space

1.1 Euclidean Space, Tangent Vectors & Regular Curves

We begin by refreshing and developing a little notation.

Deﬁnition 1.1. The set of n-tuples of real numbers is denoted R

An element can be thought of either as a point P or as its position vector

p =

−→

OP connecting the origin O = (0, . . . , 0) to P.

In co-ordinates, points are typically written as row vectors

P = (p

, . . . , p

) where each p

∈ R

P has position vector p

For vectors, either row or column vector notation is acceptable.

For each i, the co-ordinate function x

: R

→ R returns the i

co-ordinate of a point: x

(P) = p

Since the focus of the course is curves and surfaces in 2- and 3-dimensions, we’ll mostly restrict to

n ≤ 3 and quote theorems in this context.

We typically use x, y, z for the standard (rectangular)

co-ordinate functions

x(P) = p

, y(P) = p

, z(P) = p

You should be comfortable with this notation from previous classes and, in particular, with partial

derivatives of functions deﬁned in terms of the co-ordinate functions x, y, z.

Examples 1.2. 1. If P = (3, 1, 5) ∈ R

, then y(P) = 1.

2. The function f : R

→ R deﬁned by f = x

sin(yz) has partial derivatives

∂ f

∂x

= 3x

sin(yz)

∂ f

∂y

= x

z cos(yz)

∂ f

∂z

= x

y cos(yz)

A vector is a directed line segment joining two points. We’ve already seen the position vector of a point

P, namely

−→

OP. In differential geometry it is crucial to distinguish the vectors based at a given point.

Deﬁnition 1.3. A tangent vector v

is a pair of elements of R

: a base point p and

a direction v. It is the directed line segment from the point with position vector

p to the point with position vector p + v.

The tangent space at p is the set T

of all tangent vectors based at p. At each

point, R

has a different tangent space!

p + v

Be aware that v

= w

⇐⇒ p = q and v = w: the same direction at different base points means a

different tangent vector!

For simplicity’s sake, we’ll almost always state theorems in R

. The majority are valid in R

with a simple notational

modiﬁcation {x, y, z} ⇝ {x

, . . . , x

}. For R

just delete z = x

; many results even make sense in R = R

The tangent space at p is suitably named, for it is indeed a

vector space: to add tangent vectors v

, w

∈ T

, simply

sum the direction vectors

+ w

:= (v + w)

(∗)

Scalar multiplication is similar: λ v

:= (λv)

+ w

In chapter 2 we will see a more abstract discussion of tangent vectors, vector ﬁelds, and their appli-

cation.

Euclidean Space: E

versus R

To describe curves and surfaces in differential geometry, we parametrize using functions.

Example 1.4. There are multiple ways to do this for a given curve: for instance

x : (−π, π] → R

: t 7→



cos t, sin t



and y : R → R

: s 7→



1 − s

1 + s



both parametrize (most of) the unit circle in the plane (y ignores the point (−1, 0)).

Plainly the codomain R

is where the geometric action is: in the above we have the same circle, and

concepts such as length and angle can be measured. This extra structure motivates us to distinguish

the codomain with new notation.

Deﬁnition 1.5. Euclidean space E

is R

equipped with the usual dot product. Speciﬁcally in E

The dot product of p and q is p ·q = p

q = (p

)





= p

+ p

The length of p is

√

p ·p =

+ p

The angle θ between p and q satisﬁes cos θ =

p ·q

||||

Vectors are orthogonal/perpendicular if p ·q = 0, equivalently θ =

; we write p ⊥ q.

Curves in E

and E

This course is primarily concerned with functions x : U ⊆ R

→ E

. In particular:

Plane curves: m = 1 and n = 2; for example the above circle.

Spacecurves: m = 1 and n = 3; we’ll see several momentarily.

Surfaces: m = 2 and n = 3. For instance, the parametrization x : R

7→ E

: (u, v) 7→ (u, v, u

+ v

) of

a paraboloid should be familiar.

Surfaces are the focus of Chapter 3. It is now time for the formal deﬁnition of a curve.

Deﬁnition 1.6. A (smooth parametrized) curve is a function, x : I → E

, x( t) =



x(t), y(t), z(t)



deﬁned on an interval I and whose components x, y, z are inﬁnitely differentiable

everywhere on I.

Its derivative (also velocity or tangent vector) is denoted

′

( t) =



′

( t), y

′

( t), z

′

( t)



The curve’s speed is the continuous scalar function

v(t) =



′

( t)



′

( t)

+ y

′

( t)

+ z

′

( t)

A curve is regular if its tangent vector x

′

( t) is everywhere non-zero.

x(t)

′

(t)

In the context of Deﬁnitions 1.1 and 1.3, note that for each t ∈ I:

x(t) is a position vector whose nose describes the location of a point on the curve.

′

( t) ∈ T

is a tangent vector based at the point p with position vector x(t).

A parametrized curve has an orientation (indicated by the blue arrow): as t increases along the interval

I, the point x(t) moves in a particular direction along the curve.

Examples 1.7. Straight line: The line through points with position

vectors a, b may be parametrized by

x(t) = a + t( b −a) = (1 − t)a + tb

The tangent vector at x(t) is the constant x

′

( t) = b −a and the

parametrization has constant speed

b −a

. For instance,

x(t) =



2 + t, 3 −2t



has constant velocity x

′

( t) = (1, −2) and speed v(t) =

√

Circle (Example 1.4) The parametrization x(t) =



cos t, sin t



has ve-

locity x

′

( t) =



−sin t, cos t



and constant speed v(t) = 1.

By contrast, y(s) =

1+s



1 − s

, 2s



has non-constant speed

′

( s) =

(1 + s

)



−2s, 1 −s



v(s) =

1 + s

A particle moves quickest at s = 0 when v(0) = 2 and the speed

tends to zero as s → ±∞ (see the linked animation).

x(0) = a

x(1) = b

y(−2)

′

(− 2)

y(0)

′

(0)

y(0.4)

′

(0.4)

y(3)

′

(3)

The meaning of smooth depends on the author: at a minimum it means that x, y, z must be differentiable with continuous

derivative. We take the maximal approach for simplicity.

Helix x(t) =



cos t, sin t, t



parametrizes a helix (ascending spiral).

To help visualize this, imagine sitting on top of the z-axis and look-

ing down; you’d see its horizontal projection t 7→



cos t, sin t



counter-clockwise circle). Since z(t) = t, the curve moves upwards

at constant speed. One can similarly project onto the xz- and yz-

planes.

−1

−1 1

−π

2π

3π

−1 1

−π

2π

3π

The tangent vector at x(t) is x

′

( t) =



−sin t, cos t, 1



and the speed is constant v(t) =

√

Tangent Line Let x : I → E

be regular and t

∈ I be ﬁxed. The tangent line

at x(t

) is simply the straight line through the point with position

vector x(t

) oriented in the direction of the tangent vector x

′

( t

). It

is itself a parametrized curve, y : R → E

y(s) = x(t

) + sx

′

( t

)

For example, the tangent line to the above helix at t

7π

y(s) =

√

7π

−

√

, 1

The tangent line has the same speed as the helix

√

Self-intersections These are no problem for our formulation! The curve

x(t) =



sin

, cos t



, t ∈ [0, 6π)

passes through the origin at both t

3π

and t

9π

, with

corresponding tangent vectors

′

(

3π

) =



−

, 1



, x

′

(

9π

) =



−

, −1



In this example, we shouldn’t talk about the tangent vector to the

curve at the origin, since it is non-unique. Rather we should refer

to the co-ordinates

3π

9π

−1

−1 1

The linked animation shows the variable speed v(t) =

cos

+ sin

t of this curve.

Corners and Cusps To ensure that a tangent direction exists, a regular curve has everywhere non-

zero derivative. Here are a couple of examples of curves with non-regular points.

Examples 1.8. Corner A curve might enter and leave a point in dif-

ferent directions. For example, x(t) =



t, 1 −



has deriva-

tive

′

( t) =

(

(1, 1) if t < 0

(1, −1) if t > 0

At x(0) = (0, 1) the curve is non-differentiable and thus non-

smooth and non-regular.

Cusp The curve x(t) =



, t



has derivative

′

( t) =



, 2t



The origin is a cusp, a special type of corner where the curve

leaves the point in the opposite direction to how it entered.

In this case the curve is differentiable at the origin, but is non-

regular since its speed v(0) is zero.

−1 0 1

Exercises 1.1. 1. A twice-differentiable curve x(t) has the property that its second derivative x

′′

( t)

is identically zero. What can be said about x?

2. Find the unique curve such that x(0) = ( 1, 0, 5) and x

′

( t) = (t

, t, e

3. An ellipse in the plane has equation

= 1. By modifying the standard parametrization

of the circle, ﬁnd a regular parametrization of this ellipse. What is its speed?

4. Show that x(t) =



−t

−e

−t

) parametrizes half of the hyperbola x

− y

= 1. How would

you parametrize the other half?

5. (a) Find the speed of the re-parametrized standard helix y(s) = x(s

) =



cos s

, sin s

, s



(b) More generally, if x(t) is a regular curve, show that y(s) := x(s

) is non-regular.

6. Verify that our cusp example (above) may instead be parametrized y(u) =



u, u

2/3



. Is the new

parametrization still non-regular at the origin? Explain.

7. Show that the tangent vectors to the regular curve x( t) =



3t, 3t

, 2t



make a constant angle

with the vector ( 1, 0, 1).

8. Consider the plane curve x(t) =



t −1 + e

−t

, e

−t



. Find the equation of its tangent line at t = t

and ﬁnd where the tangent line intersects the x-axis.

9. Let f : R → R be a smooth function. Find a parametrization for the graph of y = f (x) and ﬁnd

its tangent line when x = x

10. Find a parametrization of the straight line through the points (1, −3, −1) and (6, 2, 1). Does this

line meet the line through the points (−1, 1, 0) and (−5, −1, −1)?

1.2 The Arc-length Parametrization and Curvature

As we’ve seen, the same ‘curve’ (viewed as a subset of E

) may be parametrized in different ways.

For instance, in Exercise 1.1.5, the standard helix x(t) = (cos t, sin t, t) was re-parametrized to obtain

y(s) = (cos s

, sin s

, s

) (∗)

This new parametrization is non-regular at s = 0; it slows down and stops before resuming its journey

up the helix! Regularity is not, therefore, an intrinsic property of a curve viewed as a set (range(x)),

rather it is a property of the parametrization.

Thankfully it is easy to create new parametrizations that remain regular.

Lemma 1.9. If x : I → E

is regular and α : J → I is smooth with nowhere-zero derivative, then we

obtain a new regular parametrization

y : J → E

, y(s) := x



α(s)



Proof. By the chain rule,

= α

′

( s)

, which is non-zero by assumption.

By contrast, if x(t) is non-regular, no smooth reparametrization can possible regularize it.

Since α

′

( s) is continuous and non-zero, there are two distinct cases:

α(s) increasing We call this an orientation-preserving re-parametrization, since a ‘particle’ travels along

the curve in the same direction.

α(s) decreasing The re-parametrization is orientation-reversing.

In the language of the Lemma, (∗) turned a regular parametrization into a non-regular one because

α(s) = s

has α

′

( s) = 3s

which is zero at s = 0.

Our next goal is to develop a special parametrization for regular curves. First we recall a concept

from multi-variable calculus.

Deﬁnition 1.10. The (signed) arc-length of a curve x : I → E

measured from x(t

) to x(t) is the

integral of the speed

s(t) =



′

(T)



dT =

v(T) dT

The arc-length is signed because it is negative if t < t

: we are measuring length against the orienta-

tion of the curve. Of course if x : [a, b] → E

has domain a closed bounded interval, then it is most

sensible to measure arc-length from t

= a so that s(t) ≥ 0 everywhere on the curve.

Example 1.11. The standard helix x(t) =



cos t, sin t, t



has constant speed

√

2, whence the arc-

length measured from x(0) is simply s(t) =

√

2t.

Observe that α

′

(s) is always positive or always negative. In particular, α(s) is 1–1. If, in addition, α is onto, then x and y

parametrize precisely the same subset of E

Recall the Fundamental Theorem of Calculus: is s(t) is the arc-length of a regular curve, then



′

(T)



dT =



′

( t)



= v(t)

is the curve’s speed, which is positive and continuous. The same is therefore true for its inverse function

′

( t)

v(t)

> 0

Deﬁnition 1.12. An arc-length parameter for a regular curve x(t) is the inverse α( s) = t(s) of an

arc-length function s(t).

Lemma 1.9 tells us that y(s) = x



α(s)



is a regular re-parametrization of our original curve. Indeed

it is a re-parametrization with a very special property:



′

( s)



= α

′

( s)



′



α(s)





v(t)

v(t) = 1 (†)

The curve y(s) has unit-speed. We have therefore proved a key result.

Theorem 1.13. Every regular curve has a unit-speed parametrization, namely by an arc-length pa-

rameter (measured from wherever you like).

The usefulness of the Theorem is abstract; by assuming that we have a unit-speed parametriza-

tion, certain analyses become much simpler. As a practical matter, explicitly ﬁnding an arc-length

parametrization might be essentially impossible (evaluate an integral then invert a function. . . ).

Examples 1.14. 1. Since the standard helix has arc-length parameter s(t) =

√

2t, it is trivial to

observe that the re-parametrization

y(s) =



cos

√

, sin

√



has unit speed.

2. More generally, if x( t) has constant speed v, then s(t) = vt is an arc-length parameter and

y(s) = x(

) a unit-speed re-parametrization.

3. The graph of y =

3/2

(t ≥ 0) may be parametrized by x(t) = (t,

3/2

). The arc-length

measured from the origin is then

s(t) =

√

1 + T dT =

(1 + t)

3/2

−1

=⇒ α(s) = t(s) =



1 +



2/3

−1

We’ve obtained an explicit unit-speed parametrization

y(s) = x



α(s)







1 +



2/3

−1,



1 +



2/3

−1

3/2



though is it really something you ever want to compute with?!

Armed with unit-speed curves, we can now deﬁne our principal notion of bendiness.

Deﬁnition 1.15. The curvature of a unit-speed curve x : I → E

κ(s) =



′′

( s)



We modify this slightly for curves in the plane: κ(s) is positive/negative if the tangent vector rotates

counter-clockwise/clockwise as we traverse the curve. This corresponds to the usual right hand rule.

By Newton’s second law, a unit mass travelling along the curve at unit speed experiences a transverse

force of magnitude κ(s).

Examples 1.16. 1. A straight line has curvature zero. For example, the line joining (1, 4) and (−3, 1)

has unit-speed parametrization x(s) =



−3 +

s, 1 +



, whence x

′′

( s) = 0 =⇒ κ(s) = 0.

2. The circle of radius r has unit-speed parametrization x(s) = r



cos

, sin



, whence

′′

( s) = −



cos

sin



=⇒ κ(s) =

This is positive since the tangent vector rotates counter-clockwise. Observe that κ =

is in-

versely proportional to the radius: smaller circles have larger curvature.

3. The standard helix with unit-speed parametrization x(s) =



cos

√

, sin

√



has

′′

( s) = −



cos

√

, sin

√

, 0



=⇒ κ(s) =

Since ﬁnding a unit-speed parametrization is difﬁcult, there few curves for which this approach is

sensible. What we want is a method that works for arbitrary parametrization. This is indeed possible,

though for spacecurves it will take a while. For curves in the plane however, things are fairly easy.

Curvature of Plane Curves If y : I → E

has unit-speed, we can write

′

( s) =



cos θ(s)

sin θ(s)



where θ(s) is the angle between the tangent line and the positive x-axis.

Now observe that

′′

( s) = θ

′

( s)



−sin θ(s)

cos θ(s)



Since



−sin θ, cos θ



points to the left of y

′

( s), we conclude:

(

cos θ

sin θ

)

(

−sin θ

cos θ

)

θ(s)

κ(s) = θ

′

(s) > 0

Theorem 1.17. The curvature of a unit-speed plane curve is the rate of change κ(s) = θ

′

( s) of the

angle of its tangent line.

This should be intuitive for constant curvature examples such as the straight line and the circle.

Now suppose x(t) =



x(t), y(t)



is any regular parametrization of the same curve; its speed satisﬁes

v(t) =

′

( t)

+ y

′

( t)

= s

′

( t)

where s(t) is an arc-length function for x(t). Moreover, the angle θ(s) plainly satisﬁes

θ(s) = tan

−1

′

( t)

′

( t)

Now differentiate and applying the chain rule:

κ(s) =

tan

−1

′

( t)

′

( t)

tan

−1

′

( t)

′

( t)

= ···

The result is a formula for the curvature as a function of an arbitrary regular parametrization.

Corollary 1.18. A regular curve x(t) =



x(t), y(t)



has curvature

κ(t) =

′′

′

− x

′′

′

′2

+ y

′2

]

3/2

′′

′

− x

′′

′

′′

· Jx

′

where Jx

′



0 −1

1 0





′





−y

′



. In particular, the graph of a smooth function y = f (x) has

curvature

κ(x) =

′′

(x)

1 +



′

(x)



3/2

Examples 1.19. 1. The graph of y =

3/2

has curvature

κ(x) =

−1/2

(1 + x)

3/2

x(1 + x)

2. If f (x) = sin x, then κ(x) =

−sin x

(1 + cos

3/2

3. The spiral x(t) =



t cos t, t sin t



has

′

( t) =



cos t − t sin t

sin t + t cos t



, x

′′

( t) =



−2 sin t −t cos t

2 cos t −t sin t



=⇒ κ(t) =

(2 cos t −t sin t)(cos t −t sin t) − (−2 sin t − t cos t)(sin t + t cos t)

[

(cos t −t sin t)

+ (sin t + t cos t)

]

3/2

2 + t

[

1 + t

]

3/2

Exercises 1.2. 1. Compute the arc-length of the following curves by parametrizing and evaluating

an integral:

(a) The straight line between points (3, 1, 2) and (1, 1, 0).

(b) The circle centered at ( 1, −2) with radius 5 measured clockwise from (6, −2) to (1, 3).

3/2

−

1/2

for 1 ≤ x ≤ 9.

2. Find the curvature of the following plane curves (use Corollary 1.18).

(a) The graph of y = x

(b) The catenary: the graph of y =

( e

+ e

−x

) = cosh x



cos t, sin 2t



(d) The exponential spiral x(t) =



cos t, e

sin t



3. Find a unit-speed parametrization of the straight line between points with position vectors

a = b in E

and hence verify that its curvature is zero.

4. Suppose x : R → E

has unit speed. Verify that x is parametrized by an arc-length parameter.

5. Find the curvature of the spacecurve x(s) =



cos s, sin s,

cos s



. What is this curve?

6. (a) Find the arc-length of the standard helix x(t) =

(

cos t, sin t, t

)

between t = −π and t = 2π.

(b) Suppose a particle travels down the standard helix so that y(0) = (1, 0, 2π) and such that

its speed is v(t) = 2

√

2t. Find a parametrization which describes this motion.

x(t) =



r cos t, r sin t, ht



and interpret how it depends on r and h.

7. Check the evaluation of κ(t) and κ(x) in the proof of Corollary 1.18.

8. We ﬁnd the curvature of the exponential spiral x(t) =



cos t, e

sin t



the hard way.

(a) Calculate the arc-length s(t) measured from x(0).

(b) Find a unit-speed parametrization y(s) where y(0) = (1, 0).

9. A circle of radius 1 rolls at constant speed with-

out slipping along the x-axis so that the angle

indicated in the picture is t at time t.

The curve described by a point on the circum-

ference of the rolling circle is a cycloid.

π 2π

3π

(a) Find a parametrization x : [0, 2π] → E

(b) Find the curvature of the cycloid as a function of t.

1.3 Orthogonality, Moving Frames & The Structure Equations

Our plan is to analyze a curve with respect to a family of moving orthonormal bases. Before embark-

ing on this, we summarize the relevant ideas from linear algebra. Hopefully most of the concepts

are familiar. Most proofs are omitted, but will be met in a standard linear algebra class. As usual,

deﬁnitions and results are stated in 3-dimensions, but are valid in others, particularly 2-dimensions.

In E

, points are typically denoted with reference to the standard basis {i, j, k}. For instance,

v =









= 3i + 4j + 6k

The numbers 3, 4, 6 are the co-ordinates of v with respect to the standard basis. Of course other bases

are available. . .

Deﬁnition 1.20. A set β = {e

, e

} ⊆ E

is a basis if every vector v ∈ E

can be expressed

uniquely

as a linear combination of e

, e

: that is

v = c

+ c

(∗)

for unique c

, c

∈ R, the co-ordinates of v with respect to β.

A basis is orthonormal if e

·e

= δ

(

1 if j = k

0 if j = k

Consider the (invertible) matrix E = (e

) whose columns are the elements of β viewed as

column vectors (with respect to the standard basis). A basis is positively oriented if det E > 0.

Examples 1.21. 1.

√





√



−1

−2



√



−1

o

is a negatively oriented orthonormal basis of E

(det E = −1 < 0).

2. Every orthonormal basis of E

has the form



cos θ

sin θ





−sin θ

cos θ





cos θ

sin θ





sin θ

−cos θ



for some angle θ. The ﬁrst is positively oriented (det = 1 > 0)

and the second negatively (det = −1 < 0).

(

cos θ

sin θ

)

(

−sin θ

cos θ

)

(

sin θ

−cos θ

)

A positively oriented orthonormal basis in E

satisﬁes the right-hand rule: e

= e

×e

. In E

, positive

orientation means that e

is obtained by rotating e

counter-clockwise by 90°: we can write this as

= Je



0 −1

1 0



In a linear algebra class this is usually broken into two deﬁnitions which imply, respectively, the existence and unique-

ness of the linear combination (∗).

Spanning Set Every v ∈ E

can be expressed as a linear combination v = c

+ c

for some c

, c

∈ R.

Linear Independence The only linear combination summing to 0 is trivial: c

+ c

= 0 =⇒ c

= c

= 0.

Finding the co-ordinates of a vector with respect to a basis (∗) is really a matrix problem

v = E









=⇒









= E

−1

Inverting a 3 ×3 matrix is tedious. Thankfully the co-ordinates can be found more easily if the basis

is orthonormal just by taking dot products!

v ·e

= (c

+ c

) ·e

= c

Lemma 1.22. If β = {e

, e

} is an orthonormal basis, then for any vector v ∈ E

v = (v ·e

) e

+ ( v ·e

) e

+ ( v ·e

) e

Example 1.23. β = {e

, e

} =









−3



is a positively oriented orthonormal basis of E

With respect to β, the vector v =





can be written





= (v ·e

) e

+ ( v ·e

) e

Orthogonal Matrices Recall Deﬁnition 1.5. Given β = {e

, e

} and its associated matrix E =

( e

), observe that

E =









( e

) =







·e







When β is an orthonormal basis, this matrix is very simple.

Deﬁnition 1.24. A 3 ×3 matrix A is orthogonal if A

A = I (equivalently AA

= I). The set of all

such is denoted O

(R ). In addition, if det A = 1, we write A ∈ SO

(R ) (special orthogonal matrices).

Lemma 1.25. 1. If A ∈ O

(R ), then it is invertible with inverse A

(also orthogonal).

2. The product of two orthogonal matrices is orthogonal.

3. A is orthogonal if and only if (Ax) ·(Ay) = x ·y for all vectors x, y ∈ E

4. Let β = {e

, e

} and E = (e

) ∈ M

(R ):

(a) E ∈ O

(R ) ⇐⇒ β is an orthonormal basis.

(b) E ∈ SO

(R ) ⇐⇒ β is a positively oriented orthonormal basis.

Parts 1 and 2 together say that O

(R ) forms a group under matrix multiplication; it is therefore known

as the orthogonal group.

For obvious reasons, this is known as the change of co-ordinate matrix from β to the standard basis {i, j, k}.

Knowledge of group theory is not necessary for these notes. Lie Groups, of which O

(R) is an example, are critical to

more advanced differential geometry.

Examples (1.21 cont). 1. It is no fun to check E

E = I directly, but since we have an orthonormal

basis, the lemma tells us that

E =







√

−1

√

−1

√

−2

√







∈ O

(R )

2. Every 2 ×2 orthogonal matrix has one of two forms:

Rotations A



cos θ −sin θ

sin θ cos θ



∈ SO

(R )

The effect of the map x 7→ A

x is to rotate x

counter-clockwise by θ radians.

Reﬂections B



cos θ sin θ

sin θ −cos θ



(det B

= −1)

The effect of x 7→ B

x is to reﬂect x across the

line making angle

with the positive x-axis.

−1 1



cos θ

sin θ





−sin θ

cos θ



−1 1



cos θ

sin θ





sin θ

−cos θ



π + θ

Motivated by the 2 ×2 case, it is common to refer to every orthogonal matrix in O

(R ) as a rotation

(det = 1) or a reﬂection (det = −1).

Part 3 of Lemma 1.25 says that multiplication by an orthogonal matrix preserves the dot product and

thus (Deﬁnition 1.5) the lengths of vectors and the angles between them. We use this to deﬁne a useful

family of transformations of E

Deﬁnition 1.26. An isometry is a function S : E

→ E

acting on points/position vectors by

S(x) = Ax + b

where b is a constant vector and A ∈ O

(R ). We call S a direct isometry or rigid motion if det A = 1

(A ∈ SO

(R )), and an indirect isometry otherwise.

Isometry literally means equal length; it can be seen that every function S : E

→ E

which pre-

serves distances between all pairs of points is an isometry. Congruent geometric objects (in standard

Euclidean geometry) are precisely those which are related by an isometry.

Recall that the matrix of a linear map is found by evaluating the map on the standard basis: the 1

column of A

the column vector A

i =



cos θ

sin θ



. The pictures should verify the remaining columns; for B

you might ﬁnd it helpful to

consider how the required reﬂections of the standard basis vectors i, j may be computed using rotations.

A full analysis is more complicated. For instance, the map x 7→ Ex in the ﬁrst example is the composition of a reﬂection

across a plane in E

followed by a rotation in that plane.

Moving Frames

Thus far we have analyzed curves with reference to the standard orthonormal basis ϵ = {i, j, k}. We

now replace this static frame of reference with one that moves. The goal is eventually to describe a

special moving frame with respect to which the fundamental properties of the curve are clear.

Deﬁnition 1.27. Let x : I → E

be a smooth curve. Suppose that e

, e

are smooth functions on I

such that, for each t ∈ I,

( t), e

( t) } is a positively oriented orthonormal basis of the tangent space T

x(t)

We call this family of functions a moving frame along x.

Equivalently, E(t) =



( t) e

( t)



is a smooth function E : I → SO

(R ). We will often refer to

this matrix-valued function as a moving frame.

(t)

A moving frame in E

The smoothness criterion needs a little unpacking. At each point on the curve, the tangent space

x(t)

has a standard basis of tangent vectors {i

x(t)

, j

x(t)

, k

x(t)

}, with respect to which

( t) = a

( t) i

x(t)

+ b

( t) j

x(t)

+ c

( t) k

x(t)





( t)





We require that the functions a

, b

, c

: I → R be smooth. Strictly speaking, e

( t) is a smooth vector

ﬁeld along the curve.

Example 1.28. We deﬁne a moving frame along the unit

circle x(t) =



cos t, sin t



via

( t) =



cos 2t

sin 2t



( t) =



−sin 2t

cos 2t



Click on the picture to see how the frame rotates twice as one

travels once round the circle!

In accordance with the deﬁnition, for each t,

E(t) =



cos 2t −sin 2t

sin 2t cos 2t



∈ SO

(R )

−1

−1 1

(t)

The obvious disadvantage of a moving frame is that we have to understand how such a frame moves!

Theorem 1.29 (Structure equations). Suppose {e

( t), e

( t) } is a moving frame (orthonormal

positive orientation). For each j, express the derivative as a linear combination

′

= e

+ e

where each w

( t) = e

·e

′

is a scalar function. Then the matrix W = (w

) is skew-symmetric:



′











0 w

−w

0 w

−w





(matrix form E

′

= EW)

In E

, there is only a single function w

= e

· e

′

. As we’ve done already, we often drop the (t) to

make things more readable; just remember that everything is still a function!

Proof. Since e

·e

is constant (equals 0 or 1), the product rule says that

0 =



·e



= e

′

·e

+ e

·e

′

=⇒ w

+ w

= 0

Examples 1.30. 1. Example 1.28 described a moving frame in E

( t) = e

( t) ·e

′

( t) =



cos 2t

sin 2t





−2 cos 2t

−2 sin 2t



= −2

The structure equations are therefore

( e

′

) = (e

)



0 −2

2 0



2. A moving frame can be described without mentioning a speciﬁc curve x(t):

( t) =





cos

cos t sin t

sin t





( t) =





sin t

−cos t





( t) =





sin t cos t

sin

−cos t





The structure equations are easily computed

( t) = e

·e

′

= cos

t + cos t sin

t = cos t,

( t) = e

·e

′

= cos

t(cos

t −sin

t) + 2(cos t sin t)

+ sin

t = 1

( t) = e

·e

′

= sin t(cos

t −sin

t) −cos t(2 cos t sin t) = −sin t



′











0 cos t 1

−cos t 0 −sin t

−1 sin t 0





The set of skew-symmetric matrices is sometimes denoted so

(R). The structure equations are an example of the

relationship between a Lie group and its Lie algebra, a foundation on which much advanced differential geometry rests.

Exercises 1.3. 1. Express v =





= a

+ a

as a linear combination with respect to the or-

thonormal basis β = {e

, e

} =









−5



of E

2. (a) Show that β =















√





−1





√





−1











is an orthonormal basis of E

. Is it positively

oriented?

(b) Find the co-ordinates of





with respect to β.

3. (a) Explain why the product rule

( x ·y) = x

′

·y + x ·y

′

holds for differentiable curves x, y.

(b) Let x, y be differentiable on an interval and use the product rule to answer the following:

i. Suppose x(t

) and x

′

( t) are orthogonal to a ﬁxed vector v (the latter for all t). Show

that x(t) is always orthogonal to v.

ii. If y(t

) is a point on y which is closest to the origin, show that y(t

) ⊥ y

′

( t

4. Find the function w

for the moving frame {e

, e

} =



1 + t



1 − t



1 + t



−1



5. Find the structure equations for the moving frame {e

, e

} =

n

cos t

sin t









−sin t

cos t

o

6. (a) Explain why every moving frame in E

has the form {e

, e

} =

n

cos θ(t)

sin θ(t)





−sin θ(t)

cos θ(t)

o

for some function θ.

(b) Find the structure equations for this frame: how does w

relate to θ?

( t) = x

′

( t), what is w

( t)?

7. (a) Let E(t) be a square matrix-valued function. Show that

(E(t))

−1

= −E

−1

′

−1

(b) Suppose E : I → O

(R ) is differentiable and deﬁne W(t) := E

−1

( t)E

′

( t). Use part (a) to

prove that W(t) is skew-symmetric (W

= −W).

8. (a) Verify parts 2 and 3 of Lemma 1.25.

(b) Suppose f , g are rigid motions. Show that f ◦ g and f

−1

are also rigid motions.

9. Let i =





. Suppose p ∈ E

and a unit vector v are given. Prove that there is a unique rigid

motion S : x 7→ Ax + b such that

S(0) = p and S(i) = p + v

Write i



0, i



∈ T

and v



p, v



∈ T

as tangent vectors, explain why it is reasonable

to write v

= S(i

) = (Ai)

: i.e., only A affects the directional part of a tangent vector.

10. (Hard) Suppose that a moving frame has structure equations

′

= −

√

( e

+ e

), e

′

√

, e

′

√

(a) By considering e

′′

, show that the vector e

×e

′

is constant.

(b) Show that

′

is constant.

( t) = cos ta + sin tb and compute e

, e

in terms of this basis.

1.4 The Frenet Frame for a Spacecurve

In this section we analyze spacecurves with respect to a moving frame adapted to the curve. To do

this, we need to restrict our class of curves slightly. For this section, we work exclusively in E

Deﬁnition 1.31. A regular spacecurve x : I → E

is biregular if it has non-zero curvature κ.

Every biregular curve is necessarily regular, but the converse is false. For instance, a straight line is

regular but not biregular. Indeed for a biregular curve, the vectors x

′

( t) and x

′′

( t) must be linearly

independent.

Deﬁnition 1.32. Let x : I → E

be a biregular unit-speed curve. The Frenet frame E(t) = (T N B) is

the moving frame deﬁned as follows:

T := x

′

is the unit tangent vector ﬁeld

N :=

′

is the principal normal vector ﬁeld

B := T ×N is the binormal vector ﬁeld

We verify that the Frenet frame is indeed a moving frame:

1. Since x is unit-speed, T has unit length.

2. By the product rule, T ·T = 1 =⇒ 2T

′

· T = 0 =⇒ N · T = 0. Moreover, the deﬁnition of

curvature tells us tht



′



′′



= 1

so that N is a unit vector perpendicular to T.

3. Standard properties of the cross product ﬁnish things off:

• B has unit length since

||||

sin θ = 1 (θ = 90° is the angle between T, N).

•



T ×N) · z = det



T N z



with z = T or N says that B is perpendicular to T, N. Finally, let

z = B to see that the Frenet frame is positively oriented.

Theorem 1.33. The Frenet frame is a moving frame. Its structure equations are



′





T N B







0 −κ 0

κ 0 −τ

0 τ 0















′

= κN

′

= −κT + τB

′

= −τN

where κ > 0 is the curvature and τ = N

′

·B = −N ·B

′

is called the torsion.

The structure equations for the Frenet frame are also known as the Frenet–Serret equations. The mov-

ing planes spanned by pairs of these vectors have special names:

Span{T, N}, Span{T, B} and Span{N, B} are the osculating, rectifying and normal planes.

At any point, the tangent line lies in the osculating plane.

Examples 1.34. 1. We compute the Frenet frame and its structure equations for the standard helix

x(s) =



cos

√

, sin

√



parametrized by arc-length (3D pic)(animation)

T(s) = x

′

( s) =

√







−sin

√

cos

√







=⇒ T

′

( s) = −







cos

√

sin

√







=⇒ N(s) = −







cos

√

sin

√







, κ(s) =

=⇒ B(s) = T(s) ×N(s) =

√







sin

√

−cos

√







τ(s) = N

′

( s) ·B( s) =







sin

√

−cos

√













sin

√

−cos

√







The Frenet–Serret equations for the helix are therefore



′





T N B







0 −





2. Let x(s) =



(1 + s)

3/2

√

(1 −s)

3/2



for s ∈ (−1, 1). First we verify this is unit-speed

′

( s) =





√

1 + s

√

−

√

1 − s





=⇒ v(s) =



′

( s)



√

1 + s + 2 + 1 −s = 1

It follows that T = x

′

. Now compute the rest of the Frenet apparatus:

′





(1 + s)

−1/2

(1 −s)

−1/2





=⇒ κ =



′



1 + s

1 − s

√

1 − s

N =

′

√

1 − s





(1 + s)

−1/2

(1 −s)

−1/2





√





√

1 − s

√

1 + s





B = T ×N =





√

1 + s

−

√

−

√

1 − s





=⇒ τ = N

′

·B =

−1

√

1 − s



′





T N B



√

1 − s





0 −1 0

1 0 1

0 −1 0





The Frenet Frame in arbitrary parametrization

Since there are relatively few curves for which an explicit unit-speed parametrization can be found,

we want to be able to compute the Frenet frame for any biregular curve, regardless of parametriza-

tion. This requires nothing more than the careful application of the chain rule. . .

Example 1.35. We compute for the exponential spiral x(t) =



cos t, e

sin t, e



′

( t) = e





cos t −sin t

sin t + cos t





=⇒ v(t) =

√

=⇒ T(t) =

√





cos t −sin t

sin t + cos t





Since T(t) has unit length, T

′

⊥ T. But then

′

( t) =

√





−sin t −cos t

cos t −sin t





=⇒ N(t) =

√





−sin t −cos t

cos t −sin t





(unit length ∥ T

′

)

=⇒ B(t) = T ×N =

√





−cos t + sin t

−sin t −cos t





It is tempting to think that the curvature should be

′

( t)

, but this is not so. Since x is not

unit-speed, we need to use the chain rule:

κ =



T(t)



T(t)



v(t)



′

( t)



√

−t

The torsion may be computed similarly

τ =

·B =

v(t)

′

( t) ·B( t) =

−t

For the general result, simply(!) repeat the example in the abstract.

Corollary 1.36. Let x(t) be a biregular spacecurve with arbitrary parametrization. The speed, cur-

vature, torsion, Frenet frame, and structure equations are as follows.

v(t) =



′

( t)



κ(t) =

′

×x

′′

τ(t) =

( x

′

×x

′′

) ·x

′′′

T(t) =

′

N(t) =

′′

−v

′

B(t) =

′

×x

′′



′





T N B







0 −vκ 0

vκ 0 −vτ

0 vτ 0





The curvature formula also holds if x(t) is merely regular.

Exercises 1.4. 1. Compute the curvature and torsion of the spiral x(t) =



cos t, e

sin t, e



directly

using the expressions in Corollary 1.36.

2. A circular helix has the form x(t) =



r cos t, r sin t, ht



, where r > 0 and h are constants. Find

its Frenet frame and show that its curvature and torsion are given by

κ =

+ h

, τ =

+ h

3. Find the curvature and torsion of the curve x(t) =



t, t

, t



4. Given x(t) =

√



√

1 + t

, 2t, ln



t +

√

1 + t





, ﬁnd the Frenet frame, curvature and torsion.

5. Let f (t) =

√

1 − e

−2u

du, and deﬁne the curve x(t) =

√



−t

cos t, e

−t

sin t, f (t)



, t > 0.

(a) Verify that x(t) has unit speed.

(b) Calculate the curvature of x and show that lim

t→∞

κ(t) = 0.

6. Let a, b be positive constants and x(t) =



4a cos

t, 4a sin

t, 3b cos 2t



where 0 < t <

. Find the

Frenet frame, curvature and torsion of x.

7. Let x : I → E

be a twice-differentiable regular curve.

(a) Prove the formula for κ in Corollary 1.36:

κ(t) =

′

×x

′′

Hence conclude that κ(t

) = 0 ⇐⇒ x

′

( t

) and x

′′

( t

) are parallel.

(Hint: let x(t) = y



s(t)



where y(s) has unit speed)

(b) Prove as much else as you can tolerate of Corollary 1.36.

8. Suppose x : I → E

is a curve lying on the surface of the unit sphere (

= 1).

(a) If x has unit speed, show that x

′′

·x = −1.

(b) Hence or otherwise, prove that the curvature of x is at least 1 everywhere.

(Hint: x and x

′

are orthonormal. . . )

= r of radius r > 0?

(d) (Hard) If a unit-speed curve lies on a sphere of radius r, show that

−1) = (κ

′

)

(Hint: compute the coefﬁcients of x with respect to the Frenet frame)

9. (Hard) Let d(t) > 0. Suppose x(t) and y(t) = x(t) + dN(t) are unit-speed curves such that

the principal normal vector ﬁeld N of x is the translate

of the binormal vector ﬁeld

B of y.

Prove that the distance d between corresponding points of the curves is constant. Prove also

that the curvature and torsion of x satisfy 2κ = d(κ

+ τ

(Hint: Compute

T and take dot products with something useful. . . )

That is, the directional parts of N,

B are identical: of course these are members of different tangent spaces.

1.5 The Fundamental Theorem of Biregular Spacecurves

Our goal for this section is to see that curvature and torsion determine a spacecurve uniquely up to

rigid motions. We do this by recognizing the Frenet–Serret equations satisﬁed by the Frenet frame as

a system of ordinary differential equations; provided sufﬁcient initial conditions (starting point and

orientation), the usual existence and uniqueness theorem for initial value problems shows that there

is a unique curve with this data.

As a precursor, we consider how to interpret curvature and torsion, and how they change (or don’t!)

under rigid motions of a curve.

Theorem 1.37. 1. A regular spacecurve has κ ≡ 0 if and only if it is a straight line.

2. A biregular spacecurve has τ ≡ 0 if and only if it is contained in a ﬁxed plane (the unmoving

osculating plane of the curve).

Proof. In both cases, we assume, without loss of generality, that x(s) is a unit-speed parametrization

of our spacecurve.

1. κ(s) =

′′

( s)

= 0 ⇐⇒ x

′′

( s) = 0. Thus x is a straight line.

2. (⇐) Suppose the curve lies in a ﬁxed plane. Then x

′

and x

′′

are parallel to this plane, whence

T and N are also. But then B is a continuous unit vector orthogonal to the plane and is therefore

constant. From the Frenet equations, −τN = B

′

= 0 =⇒ τ ≡ 0.

(⇒) As above, if τ ≡ 0, then B is constant. But then

( x ·B)

′

= x

′

·B + x ·B

′

= T ·B = 0

from which x ·B is constant. The curve therefore lies in a ﬁxed plane perpendicular to B.

Curvature measures the deviation of a curve from a straight line; its bending. Torsion measures how

badly a curve fails to be planar; its twisting.

To visualize the difference, the pictures below show a segment of a standard helix. In the ﬁrst we

look down the binormal onto the osculating plane; the non-zero curvature is clearly visible. In the

second we look along the principal normal vector N and across the osculating plane; the positive

torsion (τ =

) indicates that the curve crosses the plane similarly to how the cubic function y = x

crosses the x-axis. The full 3D curve is linked via either picture.

Theorem 1.38. Under an isometry

x := Ax + b (recall Deﬁnition 1.26), the curvature and torsion of

a biregular spacecurve transform as follows:

Direct isometry/rigid motion:

κ = κ,

τ = τ.

Indirect isometry:

κ = κ,

τ = −τ.

Proof. Suppose x(s) has unit-speed. We relate the Frenet frame (

B) of

x to the original.

Since orthogonal matrices preserve the dot product (Lemma 1.25),

x has unit-speed also:

′

( s) = Ax

′

( s) =⇒

v(s) =



′

( s)



′

( s)



= 1 =⇒

T = AT

Moreover, since A is constant and both

N and N have unit length,

N =

′

= AT

′

AN =⇒

κ = κ and

N = AN

Curvature is therefore invariant under any isometry. Since A preserves angles, AB is perpendicular

to both

T and

N, and so AB = ±

B. Since the Frenet frame E = (T N B) is a special orthogonal

matrix, AE is also orthogonal, and moreover

det AE = det A det E = det A

We conclude that det A = 1 ⇐⇒ AE = (AT AN AB) = (

N AB) is positively oriented, whence

B = (det A)AB =

(

AB if the isometry is direct,

−AB if the isometry is indirect.

Finally, we compute the torsion:

τ =

′

B = AN

′



(det A)(AB)



= (det A)(AN

′

) ·(AB) = (det A)N

′

·B = (det A)τ

Existence and Uniqueness of Solutions to ODEs

Our classiﬁcation of spacecurves depends on the ‘usual’ existence and uniqueness result for ODEs.

Here is a version suitable for our needs.

Theorem 1.39 (Existence/Uniqueness for Linear ODE (Picard, Lindel¨of, etc.)).

Let t

∈ R and c ∈ R

be given, and let M(t) ∈ M

(R ) be a continuous matrix-valued function

deﬁned on an interval

t −t

≤ T. Then the initial value problem

= M(t)E, E(t

) = c

has a unique solution E : [t

− T, t

+ T] → R

Recall Exercise 1.3.9: when we write

T = AT we mean that the directional parts of the tangent vectors are thus related.

The rough idea of the proof is to deﬁne a sequence of functions

( t) := c, E

( t) := c +

M(u)E

( u) du, E

( t) := c +

M(u)E

( u) du, . . .

which are seen to converge to the required solution; this last step requires advanced ideas from

topology/analysis. A simple example should convince you of the approach.

Example 1.40. Given the initial value problem

= 2tE, E(0) = 1, we obtain

( t) = 1, E

( t) = 1 +

2u du = 1 + t

, E

( t) = 1 +

2u(1 + u

) du = 1 + t

, . . .

The Picard iteration builds up the correct solution as a power series

E(t) = e

∞

∑

n=0

= 1 + t

+ ···

Corollary 1.41. Let O be an orthogonal matrix, I = [t

− T, t

+ T] an interval, and W : I → M

(R )

a matrix-valued function such that each W(t) is skew-symmetric. Then:

1. There exists a unique solution E : I → O

(R ) to the initial value problem

= EW, E(t

) = O

2. If det O = 1, then E : I → SO

(R ).

Proof. 1. The initial value problem is a system of nine linear ﬁrst order ODEs in the entries of the

3 × 3 matrix E. We are therefore in the case of Picard’s theorem where E : I → R

. There

therefore exists a unique solution E : I → M

(R ). Now differentiate:

(EE

) = E

′

+ E(E

′

)

= EWE

+ E(EW)

= EWE

+ EW

= EWE

+ E(−W)E

= 0 (W

= −W!)

Thus EE

is constant. However E(t

)E(t

)

= I since E(t

) = O is orthogonal. We conclude

that E(t) is always orthogonal.

2. Determinant is continuous (it is a polynomial!); E is differentiable, and so det E : I → R is

continuous on an interval. But det E = ±1 since E is orthogonal. It follows that det E is the

constant 1.

For simple W, we might be able to state the solution using the matrix exponential; for instance

W constant =⇒ E(t) = Oe

This is of limited utility: the matrix exponential is rarely computable except as an inﬁnite series, and

the approach fails for general W(t).

Corollary 1.42 (Fundamental theorem of biregular spacecurves).

Suppose we are given the following data:

• Smooth functions κ > 0 and τ on an interval I = [t

− T, t

+ T].

• A position vector c ∈ E

and a positively oriented orthonormal basis (T

) of T

Then there exists a unique unit-speed biregular spacecurve x : I → E

with curvature κ, torsion τ,

initial position x(t

) = c and Frenet frame E(t

) = (T

) at x(t

Proof. The structure equations E

′

= EW put us in the situation of Corollary 1.41; there exists a unique

solution E = (T N B) : I → SO

(R ). Integrate the unit tangent vector ﬁeld to ﬁnish:

x(t) = c +

T(u) du

This is plainly the unique curve with the required initial conditions, curvature and torsion.

Alternatively, a biregular curve is determined up to rigid motions by its curvature and torsion.

Corollary 1.43. Given two biregular spacecurves with the same curvature and torsion functions,

there exists a unique direct isometry transforming one to the other.

Proof. Suppose x

: I → E

and x

: I → E

have Frenet frames E

, E

, and the same curvature and

torsion functions. Choose some (any!) t

∈ I. The required rigid motion S : x 7→ Ax + b must satisfy

the conditions at t

, whence



( t

)



= x

( t

) and AE

( t

) = E

( t

)

Plainly A = E

( t

)



( t

)



−1

and b = x

( t

) − Ax

( t

) provide the unique isometry S. Moreover

det A = 1 since both E

and E

do so also.

By Theorem 1.38, x

:= S(x

) is a spacecurve with the same initial conditions (at t

), curvature and

torsion as x

. The Fundamental Theorem says that x

= x

= S(x

Compare what we’ve done to the standard acceleration/position kinematics problem, where three

scalar functions x

′′

( t) =



′′

( t), y

′′

( t), z

′′

( t)



and six scalar initial conditions x( t

) and x

′

( t

) recover the

motion by twice integrating.

The Fundamental Theorem says that a spacecurve is determined uniquely by three scalar functions

v(t), κ(t), τ(t) and the initial conditions x(t

), T(t

), N(t

), which also amount to six scalar constants.

One beneﬁt of our result is that, by standardizing v( t) ≡ 1 and ignoring rigid motions, we see that

the physical shape of a curve depends only on two scalar functions κ(t) and τ(t).

As in Theorem 1.38, S acts on position vectors but Frenet frames consist of tangent vectors and thus only see A.

You don’t need explicitly to specify B(t

) = T(t

) × N(t

)! The position x(t

) requires three constants; T(t

) needs

two angles (spherical polar co-ordinates), and N(t

) a single angle in the plane (T(t

))

⊥

We ﬁnish this discussion with a quick application of the Fundamental Theorem.

Corollary 1.44. Every biregular curve with κ and τ constant is a circular helix (circle if τ ≡ 0).

Proof. By (Exercise 1.4.2), the circular helix x(t) =



r cos t, r sin t, ht



has constant curvature κ =

and torsion τ =

Given constant κ, τ, it is a simple exercise to ﬁnd suitable r, h. By Corollary 1.43, this is the only such

curve up to direct isometry (and constant speed re-parametrization).

What changes in other dimensions?

For plane curves things are a little simpler. Here is a summary.

Assume: x : I → E

is regular; we don’t need biregularity.

Frenet frame: Deﬁne T :=

′

and N := JT, where J =



0 −1

1 0



is rotation by 90° counter-clockwise; no differentiation is re-

quired to compute N!

Curvature: κ =

′

·N is signed as we saw in Section 1.2:

Frenet–Serret equations: In arbitrary parametrization



′





T N





0 −vκ

vκ 0



Isometries: Direct preserves κ, indirect changes its sign.

Fundamental Theorem: Given κ(s) , x(s

) ∈ E

and T

∈ T

x(s

)

there exists a unique unit-speed curve with curvature κ(s)

and these initial data. Exercise 7 gives an elementary proof.

T(t)

N(t)

κ > 0: x bends towards N

T(t)

N(t)

κ < 0

κ < 0: x bends away from N

Example 1.45. Constant κ curves are circles, as we can see explicitly in a couple of ways. The unit-

speed structure equations E

′

= E



0 −κ

κ 0



become T

′′

= −κ

T which may be explicitly integrated.

Alternatively, θ

′

( t) = κ =⇒ θ(t) = κt + c, yields an explicit circle of radius

T(t) =



cos(κt + c)

sin(κt + c)



=⇒ x(t) = x(t

) +

T(u) du = x(t

) +



−sin(κt + c)

cos(κt + c)



We can also play this game in higher dimensions. Given a unit-

speed curve x : I → E

whose ﬁrst n −1 derivatives at each

point are linearly independent, apply Gram-Schmidt orthog-

onalization to obtain a moving frame E = (e

··· e

) and

functions κ

, . . . , κ

n−1

(the generalized curvatures) satisfying

the structure equations shown.

′

= E







0 −κ

0 ··· 0 0

0 −κ

0 0

0 κ

0 0 0

0 0 0 0 −κ

n−1

0 0 0 ··· κ

n−1







Conversely, the n −1 generalized curvatures determine the curve up to rigid motions.

Exercises 1.5. 1. Find an explicitly parametrized curve with constant curvature κ and torsion τ.

2. Reﬂection in the xy-plane S(x) =



1 0 0

0 1 0

0 0 −1



x is an indirect isometry. Explicitly compare the

curvature and torsion of the standard helix x(t) =



cos t, sin t, t



with those of S(x).

3. In the manner of Example 1.40, compute the Picard iteration process up to E

( t) for the initial

value problem



0 −1

1 0



E, E(0) =





Verify that this comports with the correct solution E(t) =

(

cos t

sin t

)

to this system of ODEs.

4. Suppose f is a function such that x(t) =



cos t, sin t, f (t)



lies in a ﬁxed plane. Show that f

satisﬁes the 3

-order linear ODE f

′′′

( t) + f

′

( t) = 0 and thus ﬁnd all possible functions f .

(Hints: What is the torsion of a plane curve?)

5. Assume that all principal normals of a biregular curve in E

pass through a ﬁxed point: ∃α(t)

and a constant n such that x(t) + α(t)N(t) = n. Show that the curve is (part of) a circle.

6. Let x : I → E

be a regular curve and let y = S(x) = Ax + b be a new curve resulting from a

rigid motion. Prove that the curvatures of x and y are identical.

7. For regular curves in E

, the Fundamental Theorem is relatively simple to prove.

(a) Suppose you are given a smooth function κ : I → R on an interval I containing t

, an initial

position x(t

) = (a, b) and an initial direction θ(t

) = θ

(angle with positive x-axis).

Use the Fundamental Theorem of Calculus to describe the unique unit-speed curve x :

I → E

with curvature κ and given initial data.

(Hints: use θ(t) := θ

κ(u) du to deﬁne T(t) and integrate! Your answer will contain

deﬁnite integrals.)

(b) Suppose x : R → E

is unit-speed with κ( t) =

1+t

, x(0) = (0, 0), and x

′

(0) = (1, 0). Find

x(t).

8. (Hard) A cylindrical helix is a curve x(t) whose unit tangent ﬁeld T( t) makes a constant angle

θ ∈ (0,

) with a ﬁxed vector n.

(a) If x(t) =



cos t, sin t, t



is the standard circular helix, describe a suitable vector n.

(b) Use the Frenet–Serret formulas to prove that a (unit-speed) non-planar curve is a cylindri-

cal helix if and only if κ/τ is constant.

9. (Very hard) Suppose a moving frame e

, e

has structure equations where all three func-

tions w

, w

are constant. Find the moving frame f

, f

where f

= e

such that f

, f

is the Frenet frame of a unit-speed circular helix. Calculate the curvature κ and torsion τ of this

helix in terms of w

, w

. Can you ﬁnd an orthogonal matrix A such that

−1





0 w

−w

0 w

−w





A =





0 −κ 0

κ 0 −τ

0 τ 0





1.6 Radii of curvature

We have seen how curvature measures the deviation of a curve from a straight line and that the only

planar curves with constant curvature κ are circles of radius

. We could have started with this as our

deﬁnition; at a given point, a curve has curvature κ if the circle which best approximates the curve

has radius

. Of course, we have to deﬁne what is meant by best approximation.

Deﬁnition 1.46. Unit-speed curves x, y have n

order contact at an intersection point x(t

) = y(s

if their ﬁrst n derivatives agree there: x

(j)

( t

) = y

(j)

( s

) for all 1 ≤ j ≤ n.

Let x(t) be a unit-speed curve in E

, ﬁx r = 0 and consider the

unit-speed circle c(s) with (signed) radius r for which

c(0) = x(t

) and c

′

(0) = x

′

( t

)

We take r > 0 ⇐⇒ the circle lies on the same side of the curve

as the principal normal vector N.

The circle is straightforward to parametrize:

c(s) = x(t

) + rN(t

)

| {z }

center

+ r sin(s/r)T(t

) −r cos(s/r)N(t

)

| {z }

rotation

T( t

)

N( t

)

r > 0

r < 0

Certainly this circle has 1

-order contact with the curve: c(0) = x(t

) and

′

( s) = cos(s/r)T(t

) + sin(s/r)N(t

) =⇒ c

′

(0) = T(t

) = x

′

( t

)

Moreover,

′′

( s) = −

sin(s/r)T(t

) +

cos(s/r)N(t

) =⇒ c

′′

(0) =

N(t

)

The circle has second-order contact with the curve if and only if

′′

(0) = x

′′

( t

) ⇐⇒

= κ(t

)

There is nothing stopping us from ﬁnding this circle for an arbitrary speed regular curve, since all

we need is the curvature and the Frenet frame at the relevant point.

Deﬁnition 1.47. Let x(t) be a regular curve. At a point x(t

) with non-zero curvature:

• The radius of curvature is r =

κ(t

)

• The center of curvature is the point with position vector x(t

) + rN(t

• The osculating circle is the radius r circle centered at the center of curvature. It has unit-speed

parametrization

c(s) = x(t

) +

κ(t

)



sin(s/r)T(t

) + (1 −cos(s/r))N(t

)



Osculating means ‘kissing.’ If κ(t

) = 0, some consider the tangent line to be an osculating circle with

inﬁnite radius!

Example 1.48. We ﬁnd the osculating circles for the parabola y = x

parametrized in the obvious

manner x(t) =



t, t



. The relevant ingredients are

′

( t) =





=⇒ T(t) =

√

1 + 4t





N(t) =

√

1 + 4t



−2t



′′

( t) =





, κ(t) =

(1 + 4t

)

3/2

The center of curvature when t = t

has position vector

x(t

) +

κ(t

)

N(t

) =



−4t

+ 3t



Several osculating circles are drawn and their centers of

curvature indicated.

−2 −1 0 1 2

The centers of curvature describe a curve that is interesting in its own right.

Deﬁnition 1.49. Let x(t) be a regular plane curve with non-zero curvature. The curve e(t) deﬁned

by the centers of curvature is the evolute of x(t):

e(t) = x(t) +

κ(t)

N(t)

Example (1.48 cont). The evolute of the parabola

x(t) = (t, t

) was found above:

e(t) = x(t) +

κ(t)

N(t) =



−4t

+ 3t



Alternatively, this is the graph y =

+ 3





2/3

notice that this isn’t regular at x = 0.

The picture now animates to show the osculating

circles and the construction of the evolute.

−4 −3 − 2 −1 0 1 2 3 4

The gray lines are the normal lines to the parabola, and are also tangent to the evolute.

′

= x

′

−

′

N +

( −vκT) = −

′

This last means that the evolute is a focal curve for the family of normal lines. The same equation

shows that the evolute is regular precisely when κ

′

( t) = 0.

A related notion is the involute, which may be imagined by rolling a line along a curve and seeing

what curve the end of the line traces out.

Deﬁnition 1.50. Suppose x(t) has unit speed. Its involute is the curve

i(t) := x(t) − tx

′

( t) = x(t) − tT(t)

An involute depends crucially on its parametrization: it intersects its source curve when t = 0.

Examples 1.51. 1. The unit speed unit circle x(t) =



cos t, sin t



. Its involute is therefore

i(t) = x(t) − tT(t) =



cos t + t sin t

sin t − t cos t



2. The involute of the unit speed catenary x(t) =



sinh

−1

√

1 + t



is the tractrix:

i(t) =



sinh

−1

t −t(1 + t

)

−1/2

(1 + t

)

−1/2



This is the curve obtained when an object starting at the point (0, 1) is dragged (subjected to

traction) by attaching a rope of length 1 to a vehicle moving along the x-axis.

−6

−4

−2

−4 −2 2

−2 −1 0 1 2

Circle and involute (spiral) Catenary and involute (tractrix)

Another way to visualize the involute of the catenary is to imagine attaching a weight at (0, 1)

to a long string wrapped tightly along the catenary and then releasing the weight. Similarly,

imagine a string is wound tightly around the circle and then uncoiled; the result is the involute.

Theorem 1.52. The evolute of any involute is the original curve, except where t = 0 or κ = 0.

We leave the argument as an exercise. The reverse process fails, as an observation of the parabola

example should convince you: remember that an involute intersects its source curve at t = 0. . .

Exercises 1.6. 1. Find the center of curvature for the curve x(t) = (1 − t

−1

, 1 + t) at t = 1.

2. Consider the ellipse x(t) =



a cos t, b sin t



where a > b > 0.

(a) Compute the curvature of the ellipse.

(b) Show that its evolute is the astroid e(t) = (a

−b

)



−1

cos

−b

−1

sin



ture has at least four points where κ

′

= 0. Show that the ellipse has precisely four.

3. Describe the involutes of a straight line.

(Hint: this is a trick question!)

4. In Example 1.51.2 we constructed the tractrix as the involute of the catenary.

(a) Use sinh

−1

t = ln



t +

√

1 + t



to verify that x(t) has unit speed and thus conﬁrm the

derivation of i(t).

(b) Compute the tangent line to the tractrix when t > 0 and show that this line cuts the x-axis

a distance 1 from the curve, thus justifying the traction claim.

5. Suppose that the graph of a smooth function y = f (x) passes horizontally through the origin:

f (0) = 0 = f

′

(0). Show that its Maclaurin series is

f (x) ≈

κ(0)x

+ higher order terms

Use this to quickly state the curvature at x = 0 of the graph of y = x

(7x

−29).

6. Let x(t) be unit speed with non-zero curvature κ and Frenet frame {T, N}. Moreover, let

i(t) = x(t) − tT(t) be an involute and denote the speed and corresponding data for the in-

volute

κ,

N. For simplicity, suppose κ, t > 0.

(a) Compute the Frenet frame of i(t) in terms of T and N.

(b) Show that

κ(t) =

7. We see how an involute of the evolute fails to recover the original curve.

Let x(t) be regular with non-zero curvature, κ

′

( t) = 0, and evolute e(t) = x(t) +

κ(t)

N(t). Since

e(t) is regular, we may assume it is parametrized by arc-length.

(a) If κ

′

> 0, explain why κ

′

= κ

(b) Show that the natural involute of the evolute is

e(t) −te

′

( t) = x(t) +

κ(0)

N(t)

that is, the original curve shifted a constant distance

κ(0)

in its normal direction.

(Hint: the ODE in part (a) is separable)

2 Vector Fields & Differential Forms

In preparation for our study of surfaces, we further develop the notion of

a tangent vector. To permit easy differentiation, throughout this section all

functions are assumed to be smooth (inﬁnitely differentiable) and U ⊆ R

will denote a connected open set: (informally) a region consisting of a single

piece without edge points. As previously, n will always be 1, 2 or 3: when

n = 1, U = (a, b) is an open interval; the picture illustrates n = 2.

2.1 Directional Derivatives, Tangent Vectors & Vector Fields

First recall some basic objects and facts from elementary multivariable calculus.

Deﬁnition 2.1. The gradient of f : U ⊆ R

→ R is the function ∇f : U → R

deﬁned by

∇f (x

, . . . , x

) =



∂ f

∂x

, . . . ,

∂ f

∂x



Given a point p ∈ U, a vector v = (v

, . . . , v

) ∈ R

, and a function f : U → R, the directional

derivative of f at p in the direction v is the scalar

f (p) :=

∑

k=1

∂ f

∂x



= v ·



∇f (p)



Example 2.2. Suppose f (x, y, z) = x

−z cos y, p = (1, π, 0), and v = (3, 5, 1). Then

∇f =





z sin y

−cos z





=⇒ D

f (p) =

















= 7

The directional derivative describes the rate of change of the value of f in a given direction.

Lemma 2.3. 1. By the chain rule, if x(t) is a curve such that x(0) = p and x

′

(0) = v, then



t=0



x(t)



∑

k=1

∂ f

∂x



′

(0) = D

f (p)

is the rate of change of f at p as one travels along the curve.

2. If t is small, then f (p + tv) ≈ f (p) + D

f (p) t.

3. If v is a unit vector making angle θ with ∇f (p), then

f (p) = v ·∇f (p) =

∇f (p)

cos θ

0 f (p)

is maximal when v points in the same direction as ∇f (p). Otherwise said, ∇f (p) points in the

direction of greatest increase of f at p; its magnitude measures the rate of change.

By placing the function f at the end of the directional derivative, we are tempted to create an operator

∑

k=1

∂

∂x



which takes a function f : U → R and returns the scalar D

f (p). This operator is a map (function)

from the set of smooth functions f : U → R to the real numbers. It is even more tempting to drop

the point p and allow the components of v to be smooth functions. This yields a new deﬁnition of an

old concept.

Deﬁnition 2.4. The set of directional derivative operators D

is the tangent space T

at p ∈ R

A vector ﬁeld v on U ⊆ R

is a smooth choice for each p ∈ U of an element of T

: that is

v =

∑

k=1

∂

∂x

where each v

: U → R is smooth

Each operator

∂

∂x

is termed a co-ordinate vector ﬁeld.

If f : U → R is smooth, we write v[ f ] =

∑

∂ f

∂x

for the result of applying the vector ﬁeld v to f ; this

is itself a smooth function v[ f ] : U → R.

Each tangent space T

is a vector space, with natural basis

∂

∂x



, . . . ,

∂

∂x



. In this brave new

world, a tangent vector v

∑

∂

∂x



corresponds to our previous notion v

= (v

, . . . , v

). While

this might seem artiﬁcially complicated, the rational is simple: the purpose of tangent vectors is to

measure how functions change in given directions (Lemma 2.3!).

Examples 2.5. 1. The vector ﬁeld v = 3x

∂

∂x

+ 2xz

∂

∂y

− x

∂

∂z

on R

corresponds to the vector-valued

function v(x, y, z) = ( 3x, 2xz, −x). Given f (x, y, z) = xy

+ z, we have

v[ f ] = 3x

∂ f

∂x

+ 2xz

∂ f

∂y

− x

∂ f

∂z

= 3xy

+ 4x

yz − x

which, as expected, is a smooth function v[ f ] : R

→ R.

2. Suppose, in R

, that we are given a vector ﬁeld v = y

∂

∂x

− x

∂

∂y

, a function f (x, y) = x

y, and a

point p = (2, −1) . These may be combined in various ways, for instance:

Vector ﬁeld on R

f v = x



∂

∂x

− x

∂

∂y



= x

∂

∂x

− x

∂

∂y

Tangent vector ( f v)(p) = f (p)v

= −4

∂

∂x



+ 8

∂

∂y



∈ T

Function R

→ R v[ f ] = y

∂

∂x

y) − x

∂

∂y

y) = 2xy

− x

Number



v[ f ]



(p) = −4 −8 = −12

Note the use of different brackets! Note also that f v denotes the vector ﬁeld obtained by multi-

plying v by the value of f at each point. It does not mean apply the function f to the vector ﬁeld v,

which makes no sense!

Here are the basic rules of computation for vector ﬁelds. These are all essentially trivial if you take

v =

∑

∂

∂x

, etc., as in Deﬁnition 2.4. Just be careful with notation!

Lemma 2.6. Let v, w be vector ﬁelds on U, let f , g : U → R be smooth, and a, b ∈ R constant. Then,

1. f v + gw is a vector ﬁeld: at each p ∈ U, ( f v + gw)(p) := f (p)v

+ g(p)w

2. Vector ﬁelds act linearly on smooth functions: v[a f + bg] = av[ f ] + bv[g]

3. (Leibniz rule) Vector ﬁelds obey a product rule: v[ f g] = f v[g] + gv[ f ]

Examples 2.7. 1. We verify the Leibniz rule for the vector ﬁeld v =

∂

∂x

−xy

∂

∂y

and functions f (x, y) =

x and g(x, y) = ye

v[ f g] =



∂

∂x

− xy

∂

∂y



[xye

] = ye

+ xye

− x

f v[g] + gv[ f ] = x



∂

∂x

− xy

∂

∂y



[ye

] + ye



∂

∂x

− xy

∂

∂y



[x] = x(ye

− xye

) + ye

2. (Polar co-ordinates) Let U be the plane without the non-positive x-axis. On U, the standard

rectangular co-ordinates (x, y) are related to the polar co-ordinates (r, θ) via

(

x = r cos θ

y = r sin θ

↭

(

r =

+ y

θ = tan

−1

(or ±

if x = 0)

The chain rule tells us that the co-ordinate vector ﬁelds

∂

∂x

∂

∂y

∂

∂r

∂

∂θ

are related via

∂

∂r

∂x

∂r

∂

∂x

∂y

∂r

∂

∂y

= cos θ

∂

∂x

+ sin θ

∂

∂y

+ y



∂

∂x

+ y

∂

∂y



∂

∂θ

∂x

∂θ

∂

∂x

∂y

∂θ

∂

∂y

= −r sin θ

∂

∂x

+ r cos θ

∂

∂y

= −y

∂

∂x

+ x

∂

∂y

∂

∂x



∂

∂y



∂

∂r



∂

∂θ



At p, these point in the direction of maximal increase for the corresponding co-ordinate.

We could similarly compute

∂

∂x

and

∂

∂y

by differentiating. For variety, we instead use linear

algebra:



∂

∂r

∂

∂θ





cos θ sin θ

−r sin θ r cos θ



∂

∂x

∂

∂y

=⇒

∂

∂x

∂

∂y



r cos θ −sin θ

r sin θ cos θ



∂

∂r

∂

∂θ



=⇒











∂

∂x

= cos θ

∂

∂r

−

sin θ

∂

∂θ

∂

∂y

= sin θ

∂

∂r

cos θ

∂

∂θ

The ﬁrst matrix is the familiar Jacobian

∂(x,y)

∂(r,θ)

from multivariable calculus. Strictly, we are view-

ing U as subsets of two different versions of R

• In rectangular co-ordinates, U = R

\{(x, 0) : x ≤ 0} is a cut plane.

• In polar co-ordinates, U = (0, ∞) × (−π, π) is an inﬁnite open rectangle.

In practice, particularly since we are so familiar with polar co-ordinates, it is easier to stick to

the ﬁrst interpretation and draw all four co-ordinate tangent vectors on the same picture.

Exercises 2.1. 1. You are given the following vector ﬁelds and functions

u = 7

∂

∂x

−3

∂

∂y

v = x

∂

∂x

+ 2y

∂

∂y

w = sin x

∂

∂x

−2 cos x

∂

∂y

f (x, y) = xy

g(x, y) = −y

Compute the functions:

(a) u[ f ] (b) v[ f ] (c) w[ f ]

(d) v[ f g] (e) f u[g] (f) v



w[g]



2. Revisit Example 2.7.2 on polar co-ordinates.

(a) Use the chain rule to compute

∂

∂x

and

∂

∂y

directly in terms of r, θ,

∂

∂r

and

∂

∂θ

and verify that

you obtain the same expressions as the linear algebra approach.

(b) Suppose T

is equipped with the standard dot product so that

∂

∂x



and

∂

∂y



are con-

sidered orthonormal.

i. Show that

∂

∂r



and

∂

∂θ



are perpendicular.

ii. What are the lengths of

∂

∂r



and

∂

∂θ



3. Consider the spherical polar co-ordinate system











x = r cos θ cos ϕ

y = r sin θ cos ϕ

z = r sin ϕ

where r > 0, 0 < θ < 2π and −

< ϕ <

Show that

∂

∂r



∂

∂x

+ y

∂

∂y

+ z

∂

∂z



4. Prove the Leibniz rule (Lemma 2.6 part 3).

5. If f , g, h are smooth functions and v is a vector ﬁeld, expand v[ f gh] using the Leibniz rule.

6. Let s = x

−y

and t = 2xy. Compute

∂

∂s

∂

∂t

in terms of

∂

∂x

and

∂

∂y

(Hint: use the chain rule to ﬁnd

∂

∂x

and

∂

∂y

, then invert the Jacobian)

2.2 Differential 1-forms

Make sure you are comfortable with vector ﬁelds before you tackle this section and the next! There is

a lot of new notation to get used to here, but with a little practice it is very easy to use.

Deﬁnition 2.8. Let (x

, . . . , x

) be co-ordinates on U ⊆ R

and p ∈ U. The (co-ordinate) 1-form dx

p is the linear map

: T

→ R deﬁned by

∂

∂x



= δ

(

1 j = k

0 j = k

A 1-form α =

∑

k=1

on U is a smooth assignment (a

: U → R smooth) of 1-forms.

If v is a vector ﬁeld on U, we write α(v) for the function U → R obtained by mapping p 7→ α(v

Examples 2.9. 1. Consider the vector ﬁeld v = xy

∂

∂x

−2

∂

∂y

on R

. At each p ∈ R

, the components

xy and −2 are scalars and thus ignored by the linear map dx : T

→ R. We therefore obtain

a function dx(v) : R

→ R

dx(v) = dx



∂

∂x

−2

∂

∂y



= xy dx



∂

∂x



−2 dx



∂

∂y



= xy

2. Again on R

, let α = 2x dx + dy and v = x

∂

∂x

−e

∂

∂y

. Then

α(v) = (2x dx + dy)



∂

∂x

−e

∂

∂y



= 2x

y −e

Remember that a 1-form α is linear when restricted to each tangent space T

: if v

∈ T

and

f : U → R

, we obtain a real number



f (p)v



= f (p)α





∈ R

by pointwise multiplication by the value of f . Taken over all points p, this means that scalar functions

come straight through a 1-form: if v is a vector ﬁeld on U, then

α( f v) = f α(v)

Deﬁnition 2.10. Let f : U → R be smooth. The exterior derivative of f is the 1-form

d f =

∑

k=1

∂ f

∂x

∂ f

∂x

+ ···+

∂ f

∂x

If a 1-form is the exterior derivative of a function, we say that it is exact.

For those who’ve met dual vector spaces in linear algebra, the set of 1-forms at p is the cotangent space T

∗

, or the

space of covectors. At each p, {dx

, . . . , dx

} is the dual basis to



∂

∂x



, . . . ,

∂

∂x





Our approach essentially splits a derivative into two pieces: for each k, we have d f



∂

∂x



∂ f

∂x

Moreover, since a linear map (d f

: T

→ R) is determined by what it does to a basis, the exterior

derivative d f is the unique 1-form with the property that d f (v) = v[ f ] for all vector ﬁelds v on U.

This says that the deﬁnition is co-ordinate independent (does not depend on x

, . . . , x

Examples 2.11. 1. Let f (x, y) = x

y, then d f = α = 2xy dx + x

dy. As a sanity check, consider a

general vector ﬁeld v = a

∂

∂x

+ b

∂

∂y

(remember that a, b are smooth functions!) and compute

d f (v) = 2axy + bx

= v[x

2. If α = 4xy

dx + (4x

y + 1) dy = f

dx + f

dy is exact, then ‘partial integration’ forces

f (x, y) =

4xy

dx = 2x

+ g(y) =

y + 1 dy = 2x

+ y + h(x)

for some functions g, h. Plainly g, h must be constant and α = d(2x

+ y).

3. We could a similar game to see that α = 3x

y dx + 2 dy is not exact on R

. Alternatively, note

that if α = d f = f

dx + f

dy, we obtain a contradiction by observing that the mixed partial

derivative is simultaneously

∂ f

∂y

= f

∂ f

∂x

= 0

See Exercise 6 for the general result.

Lemma 2.12. If f , g are smooth functions, then

1. d( f + g) = d f + dg

2. d( f g) = f dg + g d f

3. d f = 0 ⇐⇒ f is a constant function

Proof. These follow straight from the deﬁnition of d f . For instance

d f = 0 ⇐⇒

∂ f

∂x

= d f



∂

∂x



= 0 for all j = 1, . . . , n ⇐⇒ f is constant

Example (2.7.2 cont). The exterior derivative and part 2 of the Lemma make it easy to compute the

relationship between the 1-forms dx, dy, dr, dθ:

(

x = r cos θ

y = r sin θ

=⇒

(

dx = cos θ dr −r sin θ dθ

dy = sin θ dr + r cos θ dθ

=⇒

(

dr =

(x dx + y dy)

dθ =

( −y dx + x dy)

We may also verify directly that the dual basis relations hold; for instance,



∂

∂r



(x dx + y dy)



cos θ

∂

∂x

+ sin θ

∂

∂y



(x cos θ + y sin θ)

= cos

θ + sin

θ = 1

Elementary Calculus & Line Integrals

It is worth reviewing some staples from basic calculus in our new language.

If f : R → R is differentiable, then its exterior derivative d f = f

′

(x) dx feels familiar.

To make

sense of this as a relation between 1-forms we need vector ﬁelds: the derivative of f isn’t the ratio of

two 1-forms, rather it is the application of the 1-form d f to the vector ﬁeld

d f

[ f ] = d f





Vector ﬁelds in R are written with a straight d rather than partial ∂ since there is only one direction

in which to differentiate!

You’ve seen 1-forms before when integrating: we integrate 1-forms over oriented curves.

Deﬁnition 2.13. Let α be a 1-form on U ⊆ R

and suppose x : [a, b] → U parametrizes a smooth

curve C. Our usual identiﬁcation (Deﬁnition 2.4) produces the tangent vector ﬁeld

′

( t) = x

′

( t)

∂

∂x

+ ···+ x

′

( t)

∂

∂x

along the curve. Now deﬁne the integral of α along C by

α :=



′

( t)



dt =



′

( t)

∂

∂x

+ ···+ x

′

( t)

∂

∂x



Examples 2.14. 1. We integrate α = x dy over the unit-circle x(t) = (cos t, sin t) counter-clockwise.

Differentiate to obtain the tangent vector ﬁeld x

′

( t) = −sin t

∂

∂x

+ cos t

∂

∂y

, then

α =

2π



′

( t)



2π

cos

t dt =

2π

1 + cos 2t dt = π

2. Integrate α = y

dx −x

dy over the curve x(t) = (t, t

) between ( 0, 0) and (1, 1):

α =



′

( t)



dt =



∂

∂x

+ 2t

∂

∂y



dt =





y(t)



−2t



x(t)





−2t

dt =

−

= −

Lemma 2.15. The integral of a 1-form along a curve is independent of the choice of (orientation-

preserving) parametrization.

Otherwise said, if x(t) = y



s(t)



parametrizes the same curve where s

′

( t) > 0, then



′

( t)



dt =

s(b)

s(a)



′

( s)



The proof is an easy exercise in interpreting old material (the chain rule/substitution).

Consider the equivalence of notations

d f

= f

′

(x), linear approximations (differentials) & integration by substitution.

Our ﬁnal result from elementary calculus shows that integrals of exact forms are independent of

path. This is essentially the fundamental theorem of calculus for curves.

Theorem 2.16 (Fundamental Theorem of Line Integrals). If f is a function on U ⊆ R

and C is a

curve in U, then the integral of d f depends only on the values of f at the endpoints of C:

d f = f



end of C



− f



start of C



The converse also holds: if

α is independent of path, then α is exact.

Proof. Suppose x : [a, b] → U parametrizes C, then

d f =

d f (x

′

) dt =

′

[ f ] dt =



′

( t)

∂ f

∂x

+ ···+ x

′

( t)

∂ f

∂x





f (x(t))



dt = f



x(b)



− f



x(a)



The converse is sketched in an exercise.

In elementary multivariable calculus this result was written

∇f · dx = f



x(b)



− f



x(b)



which

comports with our new notation when we view dx as a vector of 1-forms:

∇f ·dx =







∂ f

∂x

∂ f

∂x



















∂ f

∂x

+ ···+

∂ f

∂x

= d f

The exterior derivative d f is just the gradient in disguise!

Example 2.17. If α = cos(xy)(y dx + x dy), ﬁnd the integral of α over any curve C joining the points



π,



and



, π



. Since α = d sin(xy) is exact on R

, we see that

α = sin(xy)



(

,π

)

(

π,

)

= sin

−sin

= 1 −

√

Summary

• Tangent vectors & vector ﬁelds encode directional derivatives, measuring how functions change

in given directions.

• Vector ﬁelds and 1-forms break standard derivatives into two pieces: the result is a more ﬂexible

and extensible language for describing familiar results from multi-variable calculus.

The real pay-off comes once our new language is applied to surfaces and higher-dimensional objects.

Here is a pr

ecis. A parametrized surface is a function x : U ⊆ R

→ E

; its exterior derivative dx is a

vector-valued 1-form which, at each point p ∈ U, describes a linear map between tangent spaces

: T

→ T

x(p)

which maps the co-ordinate ﬁelds

∂

∂x

∂

∂y

on U to corresponding vector ﬁelds tangent to the surface.

Exercises 2.2. 1. In R

, let α = 2y dx −3 dy and v = 3x

∂

∂x

∂

∂y

. Compute α (v), and v



α(v)



2. On R

, suppose f (x, y, z) = x

cos(yz) and v = e

∂

∂x

+ 2y

∂

∂z

. Verify that d f (v) = v[ f ].

3. Find dr directly by taking the exterior derivative of the equation r

= x

+ y

4. Prove parts 1 and 2 of Lemma 2.12.

5. Continuing Example 2.7.2, verify that dθ



∂

∂θ



= 1, and dr



∂

∂θ



= 0 = dθ



∂

∂r



6. Suppose that α =

∑

is exact. Prove that

∂a

∂x

∂a

∂x

for all j, k.

7. Decide whether the 1-forms α are exact on R

. If yes, ﬁnd a function f such that α = d f .

(a) α = 2x dx + dy (b) α = dx + 2x dy

y)(2y dx + x dy) (d) α = x cos(x

y)(2y dx + x dy)

8. We consider a partial converse to Exercise 6.

(a) Suppose α = a dx + b dy is a 1-form on a rectangle [p, q] × [r, s], where

∂a

∂y

∂b

∂x

. Deﬁne

f (x, y) :=

a(s, y) ds +

b(p, t) dt

Prove that df = α is exact.

(b) Let α =

−y dx+x dy

= a dx + b dy be deﬁned on the punctured plane R

\{(0, 0)}.

Show that

∂a

∂y

∂b

∂x

but that α is not exact: the full converse to Exercise 6 is therefore false.

(Hint: α = dθ except on the non-positive real axis; why is this a problem?)

9. Evaluate the integral

α given C and α.

(a) α = dx −x

−1

dy, where C is parametrized by x(t) = (t

, t

), 0 ≤ t ≤ 1.

(b) α = 2x tan

−1

y dx +

1+y

dy, where C is parametrized by x(t) =



t+1

, 1



, 0 ≤ t ≤ 2.

10. Which of the integrals in the previous question are path-independent?

11. Prove Lemma 2.15. Moreover, show that if we reverse the orientation of the curve (s

′

( t) < 0)

then the order of the limits is reversed and

α becomes −

α.

12. Let p ∈ U ⊆ R

and let α = a dx + b dy be a 1-form on U. For each q deﬁne f (q) :=

α where

we additionally assume this value is independent of the path C joining p to q.

Let h be small and C

the straight line from q to q + hi. Integrate over C

to show that

∂ f

∂x



= lim

h→0

(

q + hi

)

− f (q)

= a(q)

Make a similar argument to conclude that α = d f is exact.

13. (If you’ve done complex analysis) Let f (x, y) = u(x, y) + iv(x, y) be a complex-valued func-

tion f : R

→ C where u, v are real-valued. Viewing z = x + iy and z = x −iy as co-ordinates

on R

, prove that df



∂



= 0 if and only if u, v satisfy the Cauchy-Riemann equations:

= v

, v

= −u

2.3 Higher-degree Forms

We introduce a new operation on forms which generalizes the cross product of vectors.

Deﬁnition 2.18. Given 1-forms α, β on U, their wedge product α ∧ β is the function which takes two

vector ﬁelds and returns the smooth function

α ∧ β



u, v



= det



α(u) α(v)

β(u) β(v)



: U → R

We call α ∧ β a 2-form.

Example 2.19. Let x, y be the usual co-ordinates on R

. The standard area form is the object dx ∧dy

which takes two vector ﬁelds u

∂

∂x

+ u

∂

∂y

and v = v

∂

∂x

+ v

∂

∂y

and returns the determinant

dx ∧dy



u, v





dx(u) dx(v)

dy(u) dy(v)



This gets its name since, at each point p, it returns the (signed) area of the

parallelogram spanned by the tangent vectors u

, v

For instance, if u = 3x

∂

∂x

+ 2y

∂

∂y

and v = y

∂

∂x

+ 4x

∂

∂y

, then

dx ∧dy



u, v





3x y

2y 4x



= 12x

−2y

0 x x+y x+3x

y+4x

y+2y

Recall that determinants change sign if you switch its rows or columns, and that they are linear

functions of both their rows and columns. This has two consequences for α ∧ β.

Lemma 2.20. 1. (Columns) At each p ∈ U, a wedge product of 1-forms is an alternating, bilinear

function α ∧ β : T

× T

→ R: given vector ﬁelds u, v, w and functions f , g : U → R,

α ∧ β



v, u



= −α ∧ β



u, v



(alternating)

α ∧ β



f u + gv, w



= f α ∧ β



u, w



+ g α ∧ β



v, w



(linear in 1

slot)

2. (Rows) Wedge products are alternating and addition distributes over ∧

β ∧α = −α ∧ β and α ∧α = 0 (alternating)

( α + γ) ∧ β = α ∧ β + γ ∧ β (distributivity in 1

slot)

Linearity/distributivity in the second slot is similar in both cases.

The linearity and alternating properties tell us that every wedge product of 1-forms on R

may be

written

α ∧ β =



dx + a



∧



dx + b



= (a

− a

) dx ∧dy

Notice the determinant again!

For higher order forms, we extend the same approach.

Deﬁnition 2.21. The wedge product of 1-forms α

, . . . , α

on U ⊆ R

takes k vector ﬁelds and returns a

smooth function:

∧···∧α



, . . . , v





) ··· α

)

) ··· α

)



: U → R

Let x

, . . . , x

be co-ordinates on U. A k-form on U (alternating form of degree k) is an expression

α =

∑

∧···∧dx

, a

: U → R smooth

where we sum over all increasing multi-indices I = {i

< i

< ··· < i

} ⊆ {1, 2, . . . , n} of length k.

The wedge product of a k-form α and an l-form β is the (k + l)-form

α ∧ β =

∑

I,J

∧···∧dx

∧dx

∧···∧dx

where the 1-forms dx may be rearranged/cancelled using the alternating property (Lemma 2.20.2).

By convention, a 0-form is a smooth function f : U → R, whose wedge product with anything is

pointwise multiplication. At each point p ∈ U, the k-forms comprise the vector space of alternating

multilinear maps with basis {dx

∧···∧dx

: i

< ··· < i

} and dimension

(

)

k!(n−k)!

In this course we’ll never have reason to work in more then three dimensions!

The table describes all k-forms in 2 and 3 di-

mensions written in standard co-ordinates.

Analogous to Example 2.19, dx ∧ dy ∧ dz is

the standard volume form on R

k R

0 function f f

1 f dx + g dy f dx + g dy + h dz

2 f dx ∧dy f dx ∧dy + g dx ∧dz + h dy ∧dz

3 None f dx ∧dy ∧dz

4+ None None

Examples 2.22. 1. Given 1-forms α = 2 dx −3x dy and β = y

dx + y dy on R

α ∧ β = (2 dx −3x dy) ∧



dx + y dy



= 2y

dx ∧dx + 2y dx ∧dy −3xy

dy ∧dx −3xy dy ∧dy



2y + 3xy



dx ∧dy

2. Given the 1-forms α = dx + 2 dy + x dz and 2-form β = 3z dx ∧dy −dy ∧dz on R

, the wedge

product α ∧ β is the 3-form

α ∧ β = dx ∧(−dy ∧dz) + 3xz dz ∧dx ∧dy

= (3xz −1) dx ∧dy ∧dz

Note how dz ∧ dx ∧ dy = −dx ∧ dz ∧ dy = dx ∧ dy ∧ dz requires two swaps, so the sign is

ultimately unchanged!

Lemma 2.23. For any forms α, β,

β ∧α = (−1)

deg α deg β

α ∧ β

where deg α = k means that α is a k-form.

This is true by deﬁnition when α , β are 1-forms, and trivially true when α is a 0-form. Check the

previous examples to make sure they agree.

Example 2.24 (Polar co-ordinates). Changing to polar co-ordinates, the standard area form on R

becomes

dx ∧dy = (cos θ dr −r sin θ dθ) ∧(sin θ dr + r cos θ dθ) = r dr ∧dθ

This should remind you of change of variables in integration: if f (x, y) = g(r, θ), then

f (x, y)dxdy =

g(r, θ)r drdθ

The example illustrates one of the advantages of forms: change of variables (Jacobians) are built in!

The Exterior Derivative Just as with functions, we can apply ‘d’ to forms.

Deﬁnition 2.25. The exterior derivative of a k-form α =

∑

∧···∧dx

is the (k + 1)-form

dα =

∑

∧dx

∧···∧dx

where da

∑

∂a

∂x

is the usual exterior derivative of a function (Deﬁnition 2.10).

Example 2.26. In R

, let α = xy

z dx − xz dz. Then

dα = d(xy

z) ∧dx −d(xz) ∧dz

= (y

z dx + 2xyz dy + xy

dz) ∧dx −(z dx + x dz) ∧dz

= −2xyz dx ∧dy −(xy

+ z) dx ∧dz

Since dx ∧dx = 0 = dz ∧dz, there was no need to write the blue terms.

Theorem 2.27. Let α, β be forms:

1. d(α + β) = dα + dβ (α, β must have the same degree)

2. d(α ∧ β) = dα ∧ β + (−1)

deg α

α ∧dβ

3. d(dα) = 0. This is often written

α = 0, or just d

= 0.

A k-form α is closed if dα = 0, and exact if ∃β such that α = dβ. The result says that every exact form is closed. Poincare’s

Lemma gives a partial converse: every closed form on an open ball/hypercube is exact. Exercise 2.2.8 is a simple version.

Example (2.26 cont). We verify that d

α = 0:

d( dα) = d(−2xyz) ∧dx ∧dy −d(xy

+ z) ∧dx ∧dz

= −2xy dz ∧dx ∧dy −2xy dy ∧dx ∧dz = 0

Proof. This is very easy to prove explicitly for the only forms we’ll ever see (up to 3-forms in R

Here are general arguments that work in any dimension.

For simplicity of notation, write dx

= dx

∧···∧dx

, whenever I = {i

< ··· < i

}. Then

d(α + β) =

∑

∧dx

+ db

∧dx

∑

(da

+ db

) ∧dx

= dα + dβ

Part 2 is an exercise. For part 3, we extend Exercise 2.2.6 which in fact shows that d

f = 0 for any

function (0-form)

d( dα) = d

∑

∧dx

= d

∑

j∈I

∂a

∂x

∧dx

∑

i,j∈I

∂

∂x

∧dx

∑

i<j∈I



∂

∂x

−

∂

∂x



∧dx

= 0

since mixed partial derivatives commute.

A New Take on Vector Calculus

The standard vector calculus operations of div, grad and curl in E

are closely related to the exterior

derivative. For instance, compare the curl of a vector ﬁeld v = a

i + a

j + a

k with the exterior

derivative of the 1-form α = a

dx + a

dy + a

dz:

∇×v =







∂

∂x

∂

∂y

∂

∂z

















∂a

∂y

−

∂a

∂z



i +



∂a

∂z

−

∂a

∂x



j +



∂a

∂x

−

∂a

∂y



dα =



∂a

∂y

−

∂a

∂z



dy ∧dz +



∂a

∂z

−

∂a

∂x



dz ∧dx +



∂a

∂x

−

∂a

∂y



dx ∧dy

Comparing coefﬁcients gives part of the dictionary for comparing forms and traditional vector ﬁelds.

function f



←→ function f

∇

grad



dx + a

dy + a



←→ a

i + a

j + a

∇×

curl



dy ∧dz + b

dz ∧dx + b

dx ∧dy



←→ b

i + b

j + b

∇·

div



c dx ∧dy ∧dz ←→ function c

The exterior derivative d is div, grad and curl all in one tidy package! Moreover:

• The identity d

= 0 translates to two familiar results from vector calculus:

∇×( ∇f ) = 0 and ∇ · (∇ × v) = 0

• Under the above identiﬁcation, the wedge product of 1-forms corresponds to the cross product,

and the wedge product of a 1-form and a 2-form to the dot product. Various identities may be

obtained this way: for instance, if α is a 1-form, then

d( f α) = d f ∧ α + f dα ←→ ∇× f v = ∇f × v + f ∇ × v

• Changes of co-ordinates are built into forms (e.g. Example 2.24).

• The exterior derivative and wedge product apply in any dimension, thus extending standard

vector calculus and the cross product to arbitrary dimensions.

None of what we’ve done in this chapter is strictly necessary for the analysis of surfaces in E

. How-

ever, forms are the language of modern differential geometry (and other things besides) and it is

easier to meet them ﬁrst in a familiar setting. And if you want to do higher-dimensional geometry

(e.g., general relativity), this new language becomes almost essential.

Exercises 2.3. 1. Compute α(u, v), given α = dx ∧dy + z dy ∧dz, u =

∂

∂x

−

∂

∂z

and v = y

∂

∂y

∂

∂z

2. Let α = y

dx ∧dz −dy ∧dz and u = x

∂

∂x

+ xy

∂

∂y

−

∂

∂z

and v = −y

∂

∂x

+ y

∂

∂y

(a) Compute α(u, v).

(b) Find the 3-form dα.

3. Given s = x

−y

and t = 2xy, compute ds ∧dt in terms of dx ∧dy

4. Revisit Lemma 2.20. State what it means for a wedge product of 1-forms α ∧ β to be linear in

the second slot.

5. Let f , g be functions and consider the 1-form α = g d f . Show that α ∧dα = 0. Can the 1-form

dx + y dz be written in the form g d f ?

6. (a) Check the claim that the wedge product of 1-forms on R

corresponds to the cross product.

(b) Suppose α is a 2-form on R

. To what vector calculus identity does d( f α) = d f ∧ α + f dα

correspond?

∇·( u ×v) = (∇ × u) · v −u ·(∇ × v)

7. Let r, θ, ϕ be the spherical polar co-ordinate system in Exercise 2.1.3. Show that

dx ∧dy ∧dz = r

cos ϕ dr ∧dθ ∧dϕ

8. A 2-form is decomposable if it can be written as a wedge product α ∧ β for some 1-forms α, β.

(a) Show that every 2-form on R

is decomposable.

(b) If w, x, y, z are co-ordinates on R

, show that the 2-form dw ∧dx + dy ∧dz is not decom-

posable.

(Hint: if a 2-form γ is decomposable, what is γ ∧ γ?)

9. (Hard) Suppose α, β are forms, sketch an argument for why

α ∧ β = (−1)

deg α deg β

β ∧α

Now prove that

d(α ∧ β) = dα ∧ β + (−1)

deg α

α ∧dβ

10. (Hard) Given vector ﬁelds u, v, their Lie bracket [u, v] is the vector ﬁeld such that

[u, v][ f ] := u



v[ f ]



−v



u[ f ]



for all functions f .

(a) Compute [u, v][ f ] where u = 3x

∂

∂x

∂

∂y

and v =

∂

∂x

− x

∂

∂y

and f (x, y) = x

(b) If u =

∑

∂

∂x

and v =

∑

∂

∂x

, show that [u, v] really is a vector ﬁeld by explicitly comput-

ing [u, v][ f ] in the form

∑

∂ f

∂x

: how do the coefﬁcients c

of the vector ﬁeld [u, v] depend

on those of u, v? Find the ﬁeld [u, v] when u, v are as in part (a).

dα



u, v



= u



α(v)



−v



α(u)



−α



[u, v]



This provides a co-ordinate-free deﬁnition of dα; similar expressions exist for k-forms

(Hint: Write everything out as sums over j, k so that all differentiations of scalars are with respect

to the single variable x

; now compare!)

3 Surfaces

3.1 Regular Parametrized Surfaces

We approach surfaces in E

similarly to how we considered curves; a parametrized surface is a

function x : U → E

where U is some open subset of the plane R

. Our main purpose is to develop

and measure the curvature of a surface in terms of the parametrizing function x.

Our primary deﬁnition should mostly be familiar from elementary multivariable calculus.

Deﬁnition 3.1. A (smooth local) surface is the range S = x(U) of a smooth function x : U → E

, where

U is a connected open subset of R

Given co-ordinates u, v on U, the co-ordinate tangent vector ﬁelds are the partial derivatives x

∂x

∂u

and x

∂x

∂v

The exterior derivative or differential of the surface is the vector-valued 1-form dx = x

du + x

dv.

A surface is regular at P = x(p) if the tangent vectors x

(p) and x

(p) are linearly independent: other-

wise said, at P, the surface has a well-deﬁned

Tangent plane T

S = Span



(p), x

(p)



(a 2-dim subspace of T

), and

Unit normal vector n(p) =

(p) ×x

(p)

(p) ×x

(p)

∈ T

S is regular if it is regular everywhere. An orientation is a smooth choice of unit normal vector ﬁeld n.

The M¨obius strip (Exercise 9) shows that not every surface is orientable!

For brevity, we will often refer to the parametrizing function x as the surface, though many different

parametrizations will exist! A general surface typically needs to be parametrized by several overlap-

ping functions x

, x

, . . .. Our deﬁnition is local since there is only one x.

∂

∂v



∂

∂u



The partial derivatives x

(p), x

(p) are tangent to the surface at x(p): if p = (u

, v

) then the curve

y(t) := x(t, v

) lies in the surface and passes through P = x(p); its tangent vector at P is then

′

( u

) = lim

h→0

y(u

+ h) −y(u

)

= lim

h→0

x(u

+ h, v

) −x(p)

= x

(p)

To help distinguish between domain and codomain, we standardize notation.

Domain U ⊆ R

: Points are written lower case or as row vectors: e.g., p = (u

, v

) ∈ U. Typically we’ll

use u, v as co-ordinates unless it is more natural to use angles such as ϕ, θ.

Tangent vectors/ﬁelds are written with an arrow in our new notation: e.g.,



∂

∂u



∈ T

Codomain E

: Points are written upper case or as row vectors, e.g., P = (3, 4, 8) ∈ E

. Co-ordinates on

will typically be x, y, z.

Vectors are written bold-face as either row or column vectors: e.g., x(u, v) =



u, v, u

+ v



Tangent vectors/ﬁelds use the old notation:

e.g., if P = x(p), then x

(p) =

∂x

∂u



∈ T

Example 3.2. Consider the sphere of radius a parametrized using spherical polar co-ordinates:

x(θ, ϕ) = a





cos θ cos ϕ

sin θ cos ϕ

sin ϕ





, dx = x

dθ + x

dϕ = a cos ϕ





−sin θ

cos θ





dθ + a





−cos θ sin ϕ

−sin θ sin ϕ

cos ϕ





dϕ

The unit normal ﬁeld is simply n =

x. The domain U = (0, 2π) × (−

) is an open rectangle

whose image S = x(U) is the sphere minus the (dashed) semicircle x(0, ϕ). While we could extend θ

to wrap round the equator, we cannot extend to the north or south poles without sacriﬁcing regularity:

= a cos ϕ





−sin θ

cos θ





= 0 when ϕ = ±

∂

∂ϕ



∂

∂θ



3π

2ππ

−

This illustrates the term local: indeed the famous hairy ball theorem from topology says that it is im-

possible to ﬁnd a regular parametrization of the entire sphere by a single function.

Also observe how the tangent vectors

∂

∂ϕ



∂

∂θ



∈ T

are mapped by dx to tangent vectors

dϕ



= dx

∂

∂ϕ



dθ



= dx

∂

∂θ



∈ T

x(p)

To use our new notation in E

would require a subtle redeﬁnition of dx: if



w is a vector ﬁeld on U, then dx(



w) is the

vector ﬁeld on S such that



dx(





[ f ] =





f ◦x



for all f : S → R. In co-ordinates this beneﬁts from tensor notation:

x(u

, u

) =



, u

), x

, u

), x

, u

)



=⇒ dx =

∑

i,j

∂x

∂u

∂

∂x

⊗du

In more general situations this approach is necessary, but it is overkill for our purposes!

Theorem 3.3. Let S = x(U) be a smooth surface containing the point P = x(p):

1. The differential at p is a linear map dx : T

→ T

mapping tangent vectors in R

to vectors

tangent to S.

2. S is regular at P if and only if dx is injective (1–1) at p. In such a case we can view it as an

invertible linear map dx : T

→ T

Proof. 1. The differential at p is linear since the co-ordinate 1-forms du, dv are linear: indeed

∂

∂u



+ b

∂

∂v



= x

(p) du

∂

∂u



+ b

∂

∂v



+ x

(p) dv

∂

∂u



+ b

∂

∂v



= ax

(p) + bx

(p) = a dx

∂

∂u



+ b dx

∂

∂v



This expression is moreover tangent to S at x(p): if this last assertion is unconvincing, see

Exercise 8.

2. The range of dx at p is plainly Span{x

(p), x

(p)}. This is 2-dimensional (and thus deﬁnes the

tangent plane) if and only if rank dx = 2 ⇐⇒ dx is 1–1.

It is worth reiterating two crucially important properties of dx:

• At a regular point, dx : T

→ T

S is an invertible linear map. We shall shortly use this to

pull-back calculations from S to U.

• The differential is co-ordinate independent and thus does not depend on the parametrization

of S. This follows since dx is the unique 1-form satisfying dx(



w) =



w[x] for all vector ﬁelds



on U; a description that does not depend on co-ordinates.

Aside: change of co-ordinates To more clearly spell this out, suppose we choose a new

parametrization y(s, t) = x



F(s, t)



where F(s, t) = (u, v) is a change of co-ordinates on U. By

the chain rule,







∂u

∂s

∂v

∂s

∂u

∂t

∂v

∂t





and (du dv) = (ds dt)



∂u

∂s

∂v

∂s

∂u

∂t

∂v

∂t



from which

dy = (ds dt)





= (du dv)



∂u

∂s

∂v

∂s

∂u

∂t

∂v

∂t



−1



∂u

∂s

∂v

∂s

∂u

∂t

∂v

∂t





= (du dv)





= dx

The matrix of partial derivatives is the Jacobian of the co-ordinate change.

To be completely strict, dx and dy are not identical since they feed on tangent vectors with respect to

different co-ordinates. Formally

y = x ◦ F =⇒ dy = dx ◦dF

where dF maps tangent vectors in Span{

∂

∂s

∂

∂t

} to those in Span{

∂

∂u

∂

∂v

}: in matrix language, dF is

precisely the above Jacobian!

Common Surfaces

You should have met many of these families/examples in multi-variable calculus.

Graphs If f (x, y) is a smooth function, its graph may be parametrized by x(u, v) =



u, v, f (u, v)



Its differential and unit normal ﬁeld are

dx =









du +









dv n =

1 + f

+ f





−f





This is regular at all points, regardless of f .

Examples 3.4. 1. The standard circular paraboloid may be parametrized x(u, v) =



u, v, u

+ v



2. The upper half of the unit sphere is the graph of z = f ( x, y) =

1 − x

−y

where x

+ y

< 1.

3. A plane has equation ax + by + cz = d where a, b, c, d are constant. Since at least one of a, b, c

must be non-zero, this may be written as a function and graphed. For instance, if b = 0 we

have y = f (x, z) =

( d − ax −cz) and n =

√



a, b, c



Surfaces of Revolution If a smooth positive function x = f (z) is rotated around the z-axis, we

obtain a parametrization

x(θ, v) =



f (v) cos θ, f (v) sin θ, v



, ( θ, v) ∈ (0, 2π) ×dom( f )

with differential and unit normal ﬁeld

dx =





−f (v) sin θ

f (v) cos θ





dθ +





′

( v) cos θ

′

( v) sin θ





dv n =

1 + f

′

( v)





cos θ

sin θ

−f

′

( v)





Examples 3.5. 1. The simplest example ( f (z) ≡ 1) is the right circular cylinder of radius 1.

2. We may rotate around any axis! For instance, if we

rotate the curve z = 2 + cos x around the x-axis, the

resulting surface may be parametrized

x(θ, v) = (2 + cos v)





cos θ

sin θ













This time v measures distance along the x-axis and θ

the angle of rotation around it.

The differential and unit normal ﬁeld are

dx = (2 + cos v)





−sin θ

cos θ





dθ +





−sin v cos θ

−sin v sin θ





dv n =

1 + sin





sin v

cos θ

sin θ





Note the orientation of the surface: the unit normal ﬁeld points outward, away from the x-axis.

Ruled Surfaces Given functions y(u) , z(u) , deﬁne

x(u, v) = y(u) + vz(u)

Through each point P = x(u

, v

) passes a line t 7→ x(u

, t) = y(u

) + tx(u

) lying in the surface. The

surface can be visualized as moving a ruler through space. Ruled surfaces are common in engineering

applications since they may be constructed using straight beams.

Deﬁnition 3.6. The tangent developable of a smooth curve y is the special case when z = y

′

Examples 3.7. 1. Every plane is a ruled surface! Let y be a line in the plane and z any other tan-

gent direction. For instance, the plane passing through (1, 0, 9) and spanned by (2, −3, −5) ad

(1, 2, 3) may be parametrized

x(u, v) =



1, 0, 9





2, −3, −5



| {z }

y(u)



1, 2, 3



| {z }

z(u)

2. A helicoid is built by joining each point of a helix to its axis of rotation. From the standard helix,

we obtain the helicoid x(u, v) =



v cos u, v sin u, u



for v > 0.

3. The hyperboloid of one sheet is a doubly ruled surface: through each point there are two lines lying

on the surface. It may be parametrized as a ruled surface by

x(u, v) =









+ u









+ v





−1

+ 1





though convincing yourself there are two lines through each point takes a little more work. . .

Helicoid Hyperboloid

Implicitly Deﬁned Surfaces

Deﬁnition 3.8. A regular implicitly deﬁned surface is the zero set of a smooth function f : E

→ R for

which d f = 0 (equivalently ∇f = 0).

Recall that the directional derivative of f in the direction v is D

f (P) = v · ∇f (P). This is zero if

and only if v is orthogonal to ∇f (P). In particular, this says that ∇f provides a normal ﬁeld to an

implicitly deﬁned surface.

Examples 3.9. 1. Let a, b, c, d be constants. The function f (x, y, z) = ax + by + cz −d has

d f = a dx + b dy + c dz

which is non-zero provided at least one of a, b, c are non-zero. This deﬁnes a plane with unit

normal ﬁeld n =

∇f

∇f =

√



a, b, c



2. The sphere of radius a is the zero set of f (x, y, z) = x

+ y

+ z

− a

. It has unit normal ﬁeld

n =

∇f

∇f =



x, y, z



The sphere is everywhere regular since at least one of x, y, z is non-zero at all points of the

sphere. Contrast this with our earlier example of the parametrized sphere which could not be

made regular at the north and south poles. The lack of regularity in this case is an aspect of the

parametrization, not the surface itself.

3. The function f (x, y, z) = x

+ y

−z

−c has

d f = 2(x dx + y dy −z dz)

which is non-zero away from (x, y, z) = ( 0, 0, 0). Depending on the sign of c, the zero set is a

hyperboloid or a cone; visualize the horizontal cross-sectional circles to determine which.

c > 0 Hyperboloid of 1-sheet: x

+ y

= z

+ c > 0 for all z

c = 0 Cone: x

+ y

= z

contains a non-regular point (0, 0, 0)

c < 0 Hyperboloid of 2-sheets: x

+ y

= z

−

≥ 0 only when

≥

Our next result, a corollary of the famous implicit function theorem, ties together the notions of regular-

ity. In particular, it says that we can always assume the existence of local co-ordinates.

Theorem 3.10. A regular implicitly deﬁned surface f (x, y, z) = 0 is (locally) the image of a regular

local surface.

Proof. Suppose P = (x

, y

, z

) lies on the surface and ∇f (P) = 0. At least one of the partial deriva-

tives of f is non-zero; suppose WLOG that f

(P) = 0. By the implicit function theorem, there exists

U ⊆ R

and a function g : U → R for which g(x

, y

) = z

and f



x, y, g(x, y)



= 0. The surface is

then (locally) the graph of z = g(x, y).

x : U → E

: (u, v) 7→



u, v, g(u, v)



Example 3.11. The zero set of f (x, y, z) = x

+ y

−z

−6 is a hyperboloid of one sheet. It has unit

normal vector ﬁeld

n(x, y, z) =

∇f

∇f =

+ y

+ z





−z





√

6 + 2z





−z





whenever (x, y, z) is a point on the hyperboloid. For instance, at P = (3, 1, 2) the unit normal is

n(P) =

√

(3, 1, 2) and the tangent plane has equation

3x + y −2z = 6

Alternatively, the hyperboloid can be parametrized in several ways.

(a) In the language of the proof, near P = ( 3, 1, 2) it is the graph of z = g(x, y) =

+ y

−6.

This results in a (local) regular parametrization

x(u, v) =



u, v,

+ v

−6



(b) The hyperboloid is a surface of revolution around the z-axis:

x(θ, v) =





√

6 + v

cos θ

√

6 + v

sin θ





For this parametrization, the differential and normal ﬁeld are

dx =

6 + v





−sin θ

cos θ





dθ +

√

6 + v





v cos θ

v sin θ

√

6 + v





n =

×x

√

6 + 2v





√

6 + v

cos θ

√

6 + v

sin θ

−v





which is precisely what we obtained above.

Yet another expression could be obtained using a parametrization as a ruled surface (e.g., page 51).

Exercises 3.1. 1. Show that parametrization x(r, θ) =



r cos θ, r sin θ,

√

1 −r



of the upper hemi-

sphere is non-regular at r = 0.

2. Explain why the parametrization in Example 3.11(a) is local: what is left out?

3. (a) Compute dx and n for the paraboloid x(u, v) =



u, v, u

+ v



(b) Repeat for the polar co-ordinate parametrization y(r, θ) =



r cos θ, r sin θ, r



. Is this

parametrization everywhere regular?

(d) By viewing the paraboloid as the zero set of f (x, y, z) = z − x

− y

, ﬁnd another expres-

sion for the unit normal ﬁeld.

4. (a) Find a parametrization for the tangent developable of the helix

y(u) =



cos u, sin u, u



. Compute dy and the unit normal ﬁeld n.

(The picture covers v ∈ (−3, 6) with the original curve y(u) in green)

(b) If y is a unit speed biregular curve, prove that its tangent devel-

opable x(u, v) = y( u) + vy

′

( u) is a regular surface except when

v = 0. Express the differential and unit normal ﬁeld in terms of the

Frenet frame of y.

5. Let f (x, y, z) = z

. Show that the zero set of f has a regular parametriza-

tion despite the gradient of f vanishing at z = 0.

6. Let a, b, c be positive constants and deﬁne x(θ, ϕ) =



a cos θ cos ϕ

b sin θ cos ϕ

c sin ϕ



, (θ, ϕ) ∈ (0, 2π) ×(−

)

(a) Show that x parametrizes the ellipsoid

= 1. What part(s) of the ellipsoid are

‘missing’ from the parametrization?

(b) Describe geometrically the curves θ = constant and ϕ = constant on the ellipsoid.

7. The tube of radius a > 0 centered on a curve y(t) may be parametrized

in terms of the Frenet frame of y:

x(ϕ, t) = y(t) + a cos ϕ N(t) + a sin ϕ B(t)

(a) Brieﬂy explain why the normal ﬁeld is n = cos ϕ N(t) + sin ϕ B(t).

(b) Suppose y is unit speed. Prove that x is everywhere regular if and

only if κ(t) <

at all points of the generating curve.

8. Let c(t) : (−ϵ, ϵ) → U be a curve and y(t) = x



c(t)



the corresponding

curve in the surface x : U → E

. Prove that dx



′

(0)



= y

′

(0).

(Hint: Recall how to write c

′

( t) as a vector ﬁeld)

9. (M

obius strip) Show that x(u, v) =

(2+v cos

) cos u

(2+v cos

) sin u

v sin

is regular and ori-

entable whenever 0 < u < 2π and −1 < v < 1. By computing n(0, 0)

and n(2π, 0), explain what happens if we try to extend u to [0, 2π].

3.2 The Fundamental Forms

Our immediate goal is to use differentials to describe the shape of a surface. Before making the main

deﬁnition, we need another product of 1-forms.

Deﬁnition 3.12. Given 1-forms α, β on U, deﬁne the symmetric 2-form αβ by

αβ(



w) =



α(



v)β(



w) + α(



w)β(





where



w are vector ﬁelds on U. Note that α

(



w) := αα(



w) = α(



v)α(



w).

Symmetric 2-forms behave the way you (hopefully!) think they should.

Lemma 3.13. On each tangent space, αβ : T

× T

→ R is a symmetric and bilinear.

Moreover αβ = βα, and the product is linear in each slot:

α(β + γ) = αβ + αγ and (α + β)γ = αγ + βγ (∗)

Take care when using co-ordinate 1-forms; convention dictates that dx

= (dx)

is a symmetric 2-

form, not the exterior derivative (1-form) d(x

) = 2x dx.

Example 3.14. Let



v = a

∂

∂x

+ b

∂

∂y

and



w = c

∂

∂x

+ d

∂

∂y

. Then

(



w) = ac, dy

(



w) = bd, dx dy(



w) =

(ad + bc)

In particular,



+ dy



(



w) = ac + bd is the dot product in disguise.

To evaluate symmetric 2-forms with respect to co-ordinates, linearity and distributivity (∗) are all

you need. For instance, if α = x dx −dy and β = xy dy, then αβ = x

y dxdy − xy dy

If α, β take values in E

, we use the dot product for multiplication of the resulting vectors α(



v), etc.

( α ·β)(



w) :=



α(



v) · β(



w) + α(



w) · β(





Deﬁnition 3.15. The ﬁrst and second fundamental forms of a regular local surface x : U → E

are

I = dx ·dx, I = −dx ·dn

where dn is the differential of the unit normal ﬁeld (I requires that the surface be oriented). The ﬁrst

fundamental form is also commonly denoted ds

(see Example 3.17 and Theorem 3.20 for why).

Example 3.16. If x(u, v) =



u, uv, 1 + u



, then

dx =









du +









dv, n =

√





−1





, dn = 0

from which I = (2 + v

) du

+ 2uv du dv + u

and I = 0.

Why should we care about I & I?

Basic interpretation I(



w) = dx(



v) ·dx(



w) pulls back the dot product from T

S to T

. The length

of and angle between tangent vectors to the surface S at P may now be computed in T



w) = −



dx(



v) · dn(



w) + dx(



w) · dn(





describes how the normal ﬁeld n changes over

the surface. In the example, I ≡ 0 encapsulates the constancy of the normal ﬁeld: the surface is

(part of) the plane x ·(1, 0, −1) = −1.

Co-ordinate invariance Since dx is independent of co-ordinates, so also is I. The unit normal ﬁeld is

independent of oriented co-ordinate changes. More formally, if y(s, t) = x(u, v) parametrize the

same surface, then

= I

and I

(

if the orientations are identical

−I

if the orientations are reversed

The upshot is that the fundamental forms provide a co-ordinate independent way to compute infor-

mation about a surface from within the parametrization space U.

Example 3.17. For the sphere of radius a in spherical polar co-ordinates, recall Example 3.2:

x(θ, ϕ) = a





cos θ cos ϕ

sin θ cos ϕ

sin ϕ





=⇒ dx = a cos ϕ





−sin θ

cos θ





dθ + a





−cos θ sin ϕ

−sin θ sin ϕ

cos ϕ





dϕ

=⇒ I = a



cos

ϕ dθ

+ dϕ



If you revisit the pictures in Example 3.2, the effect of I is easy to visualize:

• I(

∂

∂θ

∂

∂θ

) =

= a

cos

ϕ: the tangent vector x

is shorter near the poles, where cos ϕ → 0.

• I(

∂

∂ϕ

∂

∂ϕ

) =



= a

: the tangent vector x

always has the same length.

• I(

∂

∂θ

∂

∂ϕ

) = x

·x

= 0: the co-ordinate tangent vectors are always orthogonal.

At a point P = x(p) on the sphere, if we increase the co-ordinates by tiny quantities ∆p =



∆θ, ∆ϕ),

then the distance ∆s travelled along the surface approximately satisﬁes

( ∆s)

≈

x(p + ∆p) − x(p)

≈



∆θ + x

∆ϕ



= a

cos

ϕ (∆θ)

+ a

( ∆ϕ)

with equality in the limit ∆θ, ∆ϕ → 0. Near the poles, a change in longitude ∆θ corresponds to a

smaller distance on the sphere. This is analogous to how a standard map of the Earth works, with

distances appearing distorted near the poles. We’ll return to this idea shortly. . .

Computing I is very easy for the sphere, since n =

x is merely the scaled position vector:

I = −dx ·dn = −

dx ·dx = −

I = −a



cos

ϕ dθ

+ dϕ



As in the Aside on page 49, we strictly have I

= I

◦dF, etc., where y(s, t) = x



F( s, t)



= x( u, v). The ±-sign in the

expressions for I is that of the determinant of the Jacobian dF.

The fundamental forms I, I may be computed directly in terms of co-ordinates u, v.

Theorem 3.18. If x : U → E

is a regular (oriented) surface, then

I = E du

+ 2F du dv + G dv

and I = l du

+ 2m du dv + n dv

where the smooth functions E, F, G, l, m, n : U → R are deﬁned by

E = x

·x

F = x

·x

G = x

·x

l = x

·n = −x

·n

m = x

·n = −x

·n

= −x

·n

n = x

·n = −x

·n

The expressions for I come from differentiating x

· n = 0 = x

· n, and are particularly helpful

because they avoid computing derivatives of n (which likely contains a square-root).

Example 3.19. Parametrize the graph of z = f (x, y) by x(u, v) =



u, v, f (u, v)



to obtain,









, x









=⇒ E = 1 + f

, F = f

, G = 1 + f

=⇒ I = (1 + f

) du

+ 2 f

du dv + ( 1 + f

) dv









, x









, x









, n =

1 + f

+ f





−f





=⇒ l =

1 + f

+ f

, m =

1 + f

+ f

, n =

1 + f

+ f

=⇒ I =

1 + f

+ f



+ 2 f

du dv + f



As a particular example, the circular paraboloid z = x

+ y

has fundamental forms

I = (1 + 4u

) du

+ 8uv du dv + (1 + 4v

) dv

= du

+ dv

+ 4



u du + v dv



I =

√

1 + 4u

+ 4v



+ dv



As a sanity check, compare with the parametrization of the same paraboloid in polar co-ordinates

y(r, θ) =



r cos θ, r sin θ, r



(Exercise 3.1.3). By computing the partial derivatives y

, y

rθ

, y

θθ

directly, one easily veriﬁes that

I = (1 + 4r

) dr

+ r

dθ

, I =

√

1 + 4r



+ r

dθ



These expressions are identical to the originals (same orientation!) since

(

du = cos θ dr −r sin θ dθ

dv = sin θ dr + r cos θ dθ

=⇒

(

+ dv

= dr

+ r

dθ



u du + v dv



= r

Curves in Surfaces: interpreting I and I

Given a regular (oriented) surface x : U → E

and a curve c(t) in U, we may transfer this curve to

the surface y(t) = x



c(t)



. Its tangent vector (Exercise 3.1.8) and speed are then

′

( t) = dx



′

( t)



=⇒



′

( t)





′

( t)



·dx



′

( t)





′

( t), c

′

( t)



We can do something similar for the second fundamental form.

Theorem 3.20. Let y(t) = x



c(t)



parametrize a curve in a surface x with unit normal ﬁeld n.

1. If a < b, then the arc-length of y between y(a) and y(b) is



′

( t), c

′

( t)



dt.

2. The normal acceleration of the curve is y

′′

( t) ·n = I(c

′

, c

′

This puts some ﬂesh on our earlier observations (page 56). I measures inﬁnitesimal squared-distance

on the surface, while I measures how the surface bends away from the normal ﬁeld: recall how

force/acceleration motivated the curvature κ of a curve (Deﬁnition 1.15).

Proof. 1. Arc-length is the integral of the speed

′

( t)



′

( t), c

′

( t)



2. Since y

′

lies in the tangent plane, we have y

′

·n ≡ 0. Differentiate to obtain

0 =

( y

′

·n) = y

′′

·n + y

′



c(t)



= y

′′

·n + dx(c

′

) ·dn(c

′

) = y

′′

·n −I(c

′

, c

′

)

Example (3.17, cont). Consider the curve c(t) = (θ(t), ϕ(t)) =



2t, t



where 0 ≤ t ≤

. This has

tangent ﬁeld c

′

( t) = 2

∂

∂θ

∂

∂ϕ

. Translated to the unit sphere, the resulting curve has arc-length



′

, c

′

) dt =

4 cos

t + 1 dt ≈ 1.619

In the parametrization space U, c(t) is a straight line. The shortest path between the endpoints of the

curve on the sphere is the great circle arc with length

2π

≈ 1.571; its pre-image in U appears longer

but isn’t due to the cos

ϕ factor in the ﬁrst fundamental form. By spending more time at northerly

latitudes, I is smaller for more of the great circle arc and the resulting arc-length is shorter.

If a map of the Earth covers a small latitude range (almost constant ϕ ≈ ϕ

), the ﬁrst fundamental

form is almost similar to a standard dot product I ≈ (a cos ϕ

dθ)

+ (a dϕ)

. If not, say when we

travel by plane, the distortion becomes much more apparent.

The picture shows the shortest path from Irvine (California) to Irvine (Scotland), as ﬂown by an

aircraft in ideal conditions. The straight line on the map corresponds to a longer real-world path.

If we travel at constant speed, it can be checked that great circles are precisely those curves whose

acceleration is entirely normal to the surface. This observation, and its relation to geodesics (paths

minimizing distance), is a matter for another course.

Example 3.21. A skater descends into a paraboloidal bowl z =

following the path described

by c(t) = (r(t), θ(t)) = (1 − t, 4t

) in polar co-ordinates. If we parametrize the bowl in polar co-

ordinates x(r, θ) =



r cos θ, r sin θ,



, the fundamental forms are seen to be

I = (1 + r

) dr

+ r

dθ

I =

√

1 + r

(dr

+ r

dθ

)

For the skater’s path, c

′

( t) = −

∂

∂r

+ 8t

∂

∂θ

, whence

I(c

′

, c

′

) = (1 + (1 −t)

) + 64t

(1 −t)

The path therefore has arc-length

I(z

′

, z

′

) dt =

1 + (64t

+ 1)(1 −t)

dt ≈ 1.82

and normal acceleration

′′

·n = I(c

′

, c

′

) =

1 + (1 − t)



1 + 64t

(1 −t)



′′

· n

0 1

By Newton’s second law, this is proportional to the component of the force experienced by the skater

pushing perpendicularly out from the surface.

Exercises 3.2. 1. Verify the ﬁnal details of Example 3.19: that is, compute I, I directly using the polar

co-ordinate parametrization y(r, θ) =



r cos θ, r sin θ, r



2. Find the fundamental forms for the surface of revolution x(θ, v) =



f (v) cos θ, f (v) sin θ, v



3. Compute the ﬁrst fundamental forms of each parametrized surface wherever they are regular

(a, b, c are non-zero constants). Where does each parametrization fail to be regular?

(a) Ellipsoid x(θ, ϕ) = (a cos θ cos ϕ, b sin θ cos ϕ, c sin ϕ)

(b) Elliptic paraboloid x(r, θ) = (ar cos θ, br sin θ, r

)

4. Calculate the fundamental forms of Enneper’s surface

x(u, v) =



u −

+ uv

, v −

+ vu

, u

−v



5. Compute dy for the parametrization y(r, θ) =



r cos θ, r sin θ,

√

1 −r



of the upper unit hemi-

sphere. Verify that the ﬁrst fundamental form is the same as in Example 3.17.

6. Let x be the tangent developable of a unit speed biregular curve y (Exercise 3.1.4).

(a) Compute the fundamental forms of x in terms of the curvature and torsion of y.

(b) If y(u) =



cos

√

, sin

√



is the unit speed helix, show that

I =



1 +



+ 2 dudv + dv

, I = −

7. Prove that I ≡ 0 if and only if x is (part of) a plane.

8. Parametrize the great circle in Example 3.17 (cont) by z(t) =



cos t,

√

sin t,

√

sin t



, 0 ≤ t ≤

Verify that the arc has length

and that the acceleration of z is entirely normal; z

′′

= (z

′′

·n)n.

9. Equip the upper half plane y > 0 with the abstract ﬁrst fundamental form I =



+ dy



Compare the arc-length between the points (1, 1) and (−1, 1):

(a) Over the circular arc c(t) =

√



cos t, sin t



centered at the origin.

(b) Over the ‘straight’ line y = 1.

This is the Poincar´e half-plane model of hyperbolic space. There is neither a surface x : U → E

nor a

second fundamental form I!

10. (Hard) The torus obtained by rotating the unit circle in the x, z-plane centered at (2, 0, 0) around

the z-axis may be parametrized

x(u, v) =



(2 + cos ϕ) cos θ, (2 + cos ϕ) sin θ, sin ϕ



, ( θ, ϕ) ∈ R

Let k = 0 be constant and consider the curve y( t) = x(kt, t) on the torus.

(a) Prove that y(t) has a self-intersection (∃s = t such that y(t) = y(s)) if and only if k ∈ Q.

(b) If k ∈ Q, show that the curve is periodic in that there exists a minimum positive T for which

y(t + T) = y(t) for all t. Find T in terms of k and write down (don’t evaluate!) the integral

for the arc-length of the curve over one period.

3.3 Principal, Gauss & Mean Curvatures

Since I and I are symmetric bilinear forms on each tangent space T

, they may be expressed in

matrix form: their matrices with respect to linearly independent vector ﬁelds



t are

[I] :=





s) I(



t) I(





and [I] :=





s) I(



t) I(





Otherwise said





s + g



t, h



s + k







f g



[I]





and similarly for I. Matters are simplest when these matrices are diagonal. . .

Deﬁnition 3.22. Linearly independent vector ﬁelds



t are said to be orthogonal if I(



t) = 0. They

additionally describe curvature directions if I(



t) = 0.

Co-ordinates u, v are orthogonal/curvature-line if the above apply to the the co-ordinate ﬁelds

∂

∂u

∂

∂v

In the language of Theorem 3.18, the matrices of the fundamental forms with respect to

∂

∂u

∂

∂v

are

A :=



E F

F G



and B :=



l m

m n



(∗)

Co-ordinates are orthogonal iff F = x

· x

≡ 0 (I has no du dv term), and are curvature-line iff I is

also diagonal:

I = E du

+ G dv

and I = l du

+ n dv

While the meaning of orthogonal is clear, the reason for the term curvature-line will take a little work.

Examples 3.23. 1. Since the sphere of radius a has I = −

I, any orthogonal co-ordinates on the

sphere are curvature-line! E.g., spherical polar co-ordinates: I = a



cos

ϕ dθ

+ dϕ



2. (Example 3.2.3.19) Standard polar co-ordinates are curvature-line for the paraboloid z = r

I = (1 + 4r

) dr

+ r

dθ

, I =

√

1 + 4r



+ r

dθ



3. In curvature-line co-ordinates n

= −

and n

= −

(see Exercise 11).

A Little Linear Algebra The existence of curvature directions is equivalent to the simultaneous diag-

onalization of both matrices (∗). This requires an extension of the concepts of eigenvalues/vectors.

Deﬁnition 3.24. Let A, B be square matrices of the same dimension. A non-zero vector



v is an

eigenvector of B with respect to A with eigenvalue λ if

(B −λA)



v =



If A = I is the identity matrix, these are standard eigenvalues/vectors. We compute in the usual

manner: solve the characteristic polynomial and ﬁnd



v ∈ N(B −λA) in the nullspace. . .

Example 3.25. Let A =



2 3

3 5



and B =



0 1

1 3



det(B −λA) =



−2λ 1 −3λ

1 −3λ 3 − 5λ



= λ

−1 = 0 ⇐⇒ λ = ±1



∈ N(B − A) = N



−2 −2



= Span



−1





∈ N(B + A) = N =



2 4

4 8



= Span



−1



Note that {



} =



−1





−1



is a basis of R

consisting of eigenvectors of B with respect to A.

Theorem 3.26. Let A, B be symmetric matrices of the same dimension, with A positive-deﬁnite.

1. There exists a basis of eigenvectors of B with respect to A. Moreover, all eigenvalues are real.

2. If



t are eigenvectors corresponding to distinct eigenvalues, then



t = 0 =



Proof. 1. This follows from the famous spectral theorem in linear algebra.

2. Assume B



s = k



s and B



t = k



t where k

= k

, and apply the symmetry of A and B,



t =



( k



t) = k



∥



s =



( k



s) = k



s = k













=⇒ (k

−k

)



t = 0 =⇒



t = 0

Application to Regular Surfaces With respect to independent vector ﬁelds, the matrices A, B of I, I

are symmetric. Moreover, the regularity of x guarantees the positive-deﬁniteness of A:

∀



w =



0 =⇒ I(



w) = dx(



w) ·dx(



w) =

dx(



> 0

We may therefore apply Theorem 3.26 to the matrices of the fundamental forms.

Deﬁnition 3.27. The principal curvatures k

, k

: U → R of an oriented surface x : U → E

are the

eigenvalues of I with respect to I. Corresponding eigenvectors are curvature directions.

The Gauss and mean curvatures are, respectively, K := k

and H :=

( k

+ k

A point x(p) is umbilic if k

(p) = k

(p).

For all non-zero vectors,



v > 0. Equivalently, all eigenvalues of A are positive. This means that

⟨



⟩



deﬁnes an inner product on R

. In Example 3.25, A has eigenvalues

(7 ±

√

45) > 0.

In case you’re interested: A has an orthogonal eigenbasis {



, . . . ,



} by the spectral theorem. Since its eigenvalues

are positive, we may scale such that



. Let X = (



···



) so that X

AX = I is the identity matrix. But then,

det(B −λA) = det(X

)

−1

det



BX −λI



det(X

−1

) = 0 ⇐⇒ det(X

BX −λI) = 0

Since X

BX is symmetric (spectral theorem again), it has an orthogonal eigenbasis {



, . . . ,



} and real eigenvalues λ

Each



:= X



is an eigenvector of B with respect to A with eigenvalue λ

. Since X is invertible, {



, . . . ,



} is a basis.

The curvatures are independent of oriented co-ordinate changes. If we reverse orientation, then k

, k

and H change sign, while K = k

is unchanged.

At non-umbilic points, Theorem 3.26 says that curvature directions diagonalize both fundamental

forms, in line with Deﬁnition 3.22.

At umbilic points, I = kI and all directions are curvature directions; any orthogonal directions nec-

essarily diagonalize both fundamental forms.

Example 3.28. Here are two totally umbilic surfaces where the curvatures are constant.

1. A plane: I ≡ 0 =⇒ all curvatures are zero.

2. A sphere of radius a: I = −

I =⇒ k

= k

= −

, K =

and H = −

In fact these comprise all totally umbilic surfaces (see Exercise 12).

Theorem 3.29. 1. In co-ordinates, the Gauss and mean curvatures are given by

K =

ln − m

EG − F

det B

det A

= det(A

−1

B) and H =

lG + nE −2mF

2(EG − F

)

tr A

−1

2. At non-umbilic points, the curvatures k

, k

, K, H are smooth functions and the curvature direc-

tions may be described locally by (smooth) vector ﬁelds.

Proof. 1. The principal curvatures are the solutions to the quadratic equation

det



l m

m n



−λ



E F

F G



= (EG − F

) λ

−( lG + nE −2mF)λ + (ln −m

)

of which K and H are the product and half the sum of the roots.

2. The roots

−b±

√

−4ac

of a quadratic are smooth functions of the coefﬁcients unless b

−4ac = 0,

in which case we have a repeated root (k

= k

). At non-umbilic points, each eigenspace is

one-dimensional, so there is no obstruction to choosing smooth eigenvectors.

Examples 3.30. 1. (Example 3.19) For the paraboloid x(r, θ) =



r cos θ, r sin θ, r



, standard polar

co-ordinates are curvature-line:

A = [I] =



1 + 4r

0 r



B = [I] =

√

1+4r

√

1+4r

The curvatures are therefore

(1 + 4r

)

3/2

, k

√

1 + 4r

, K =

(1 + 4r

)

, H =

2 + 4r

(1 + 4r

)

3/2

The curvatures make sense at the single umbilic point (r = 0), but the co-ordinates are not

curvature-line there since the parametrization fails to be regular (x

(0, θ) = 0).

At an umbilic point x(p), the eigenspace is 2-dimensional so lim

q→p



v(q) need not exist and



v need not be continuous.

2. Parametrize a graph z = f (x, y) in the usual manner x( u, v) =



u, v, f (u, v)



. Then

A = [I] =



1 + f



B = [I] =

1 + f

+ f





Theorem 3.29 tells us that

K =

− f

(1 + f

+ f

)

H =

(1 + f

) + f

(1 + f

) −2 f

2( 1 + f

+ f

)

3/2

In the abstract, solving for the curvatures and directions is disgusting. As a sanity check, you

should verify that f (u, v) = u

+ v

recovers exactly the curvatures in the previous example!

3. (Exercise 3.2.6) The tangent developable of the unit-speed helix has

A = [I] =



1 +

1 1



B = [I] =



−

0 0



Now solve for the curvatures:



−

−λ



1 +



−λ

−λ −λ



λ = 0 =⇒ k

= 0, k

= −

, K = 0, H = −

In this case an explicit computation of the curvature directions is not difﬁcult:

= 0 =⇒ N(B −k

A) = N



−

0 0



= Span





⇝



s =

∂

∂v

= −

=⇒ N(B −k

A) = N





= Span



−1



⇝



t =

∂

∂u

−

∂

∂v

where we made the natural choice of vector ﬁelds



t. As a sanity check, here are the matrices

of the fundamental forms with respect to



s) =



0 1







= 1 . . . =⇒ [I] =





s) I(



t) I(







1 0



[I] =



0 0

0 −



in which the principal curvatures are clearly visible: 1k

= 0,

= −

Constant Gauss & Mean Curvature Surfaces

Minimal Surfaces H ≡ 0: Among all surfaces whose boundary is a given closed curve, a surface with

minimal surface area has H ≡ 0. This is the shape made by a soap ﬁlm whose boundary is the

curve: it minimizes the ‘total tension’ of the soap ﬁlm. More gen-

erally, constant mean curvature (CMC) surfaces model soap bubbles.

Constant Gauss Curvature Surfaces: We’ve see that planes, cones and

cylinders have K = 0, and that spheres have constant positive

Gauss curvature. A pseudosphere with constant K = −1 is shown

in the picture.

Existence of (Curvature-Line) Co-ordinates

At non-umbilic points, Theorems 3.26 and 3.29 tell us how to ﬁnd curvature directions as vector ﬁelds



t. Unfortunately, being able to compute explicit curvature co-ordinates is exceptionally unlikely.

Example (3.30.3 cont). Recall that we chose curvature direction ﬁelds



s =

∂

∂v

and



t =

∂

∂u

−

∂

∂v

. By

inspection, the functions s = u + v and t = u satisfy the required equations:



s[s] = 1 =



t[t],



s[t] = 0 =



t[s] (∗)

It follows that



s =

∂

∂s

and



t =

∂

∂t

for curvature-line co-ordinates s, t, as you can easily verify using the

chain rule. Indeed

I =

+ d(u + v)

= ds

, I = 0 ds

−

so that the co-ordinates really do diagonalize both fundamental forms.

The simple reason the example is so unlikely is that mixed partial derivatives must commute: if



s =

∂

∂s

and



t =

∂

∂t

are co-ordinate ﬁelds (∃s, t : U → R), then their Lie bracket (Exercise 2.3.10) vanishes:

[



t] =



∂

∂s

∂

∂t



∂

∂s

◦

∂

∂t

−

∂

∂t

◦

∂

∂s

= 0

The astonishing fact is that this simple condition is locally sufﬁcient.

Theorem 3.31 (Co-ordinate ﬁelds). Let



t be linearly independent vector ﬁelds on U ⊆ R

1. If there exist functions s, t : U → R such that



s =

∂

∂s



t =

∂

∂t

, then [



t] = 0.

2. Suppose [



t] = 0 and let p ∈ U. Then there exists a neighborhood V of p and co-ordinate

functions s, t : V → R for which



s =

∂

∂s



t =

∂

∂t

Examples 3.32. 1. The ﬁelds



s =

∂

∂x

+ y

∂

∂y

and



t =

∂

∂y

do not arise simultaneously from co-ordinates:

[



t] =

∂

∂x∂y

+ y

∂

∂y

−

∂

∂y∂x

−

∂

∂y

−y

∂

∂y

= −

∂

∂y

= 0

2. The cylindrical paraboloid x(u, v) = (u, v,

+ v) has curvatures and curvature directions

= 0,



s =

∂

∂v

, k

[2 + u

]

3/2



t = 2

∂

∂u

−u

∂

∂v

The Lie bracket condition [



t] = 0 is satisﬁed, so co-ordinates s, t corresponding to these ﬁelds

must exist. You can try to ﬁnd such by inspection, though simultaneously solving (∗) is messy.

Alternatively, following the proof of part 2 (Exercise 13), observe that the dual 1-forms are

α =

u du + dv, β =

du (α(



s) = β(



t) = 1, α(



t) = β(



s) = 0)

These forms are exact: α = d(

+ v) and β = d(

u). We therefore conclude that s =

+ v

and t =

u are suitable curvature-line co-ordinates.

The Lie bracket condition says that explicit co-ordinates corresponding to given vector ﬁelds are very

unlikely to exist. This is no matter: we typically only require co-ordinates s, t whose ﬁelds

∂

∂s

∂

∂t

are

parallel to



t: that is

∂

∂s

= f



s and

∂

∂t

= g



t for some functions f , g (equivalently



s[t] = 0 =



t[s])

Such co-ordinates indeed exist, though only locally, as shown by one of the most important founda-

tional results in differential geometry.

Theorem 3.33 (Frobenius). Let



t be linearly independent vector ﬁelds on a domain U. Then there

exist local co-ordinates s, t whose co-ordinate ﬁelds are parallel to



In particular, if x(p) is a non-umbilic point on a surface x : U → E

, then there exists a neighborhood

V of p and curvature-line co-ordinates s, t on V.

Frobenius’ theorem comes in many guises and generalizes to higher dimensions, taking the place

of Picard’s ODE existence/uniqueness theorem (1.39) for particular classes of PDE. Its proof is too

involved for us, though the informal idea is to search for functions f , g such that [ f



s, g



t] = 0, a

lengthy process that indeed depends on Picard’s theorem.

Exercises 3.3. 1. Find the eigenvalues of B =



−1 −1

−1 1



with respect to A =



1 1

1 2



. If



t are

corresponding eigenvectors, verify that



t = 0 =



2. Parametrize the graph of x = z

; compute I, I and the principal, Gauss and mean curvatures.

3. Use Theorem 3.29 to ﬁnd the Gauss and mean curvatures of the graph of y = x

−z

4. Show that Enneper’s surface (Exercise 3.2.4) is minimal.

5. Let x(u, v) = y(u) + vy

′

( u) be the tangent developable of a unit speed biregular curve y.

(a) Find the principal curvatures, Gauss and mean curvatures of x.

(b) Compute the curvature directions and ﬁnd curvature line co-ordinates.

(This is very similar to Example 3.30.3 - keep track of the changes!)

6. With respect to some co-ordinates u, v, suppose that a surface has fundamental forms

I = u

+ v

, I = u

+ 2uv du dv + v

(a) Show that the principal curvatures are constant: k

= 0 and k

= 2.

(b) Show that



s = v

∂

∂u

−u

∂

∂v

and



t = v

∂

∂u

+ u

∂

∂v

are curvature directions.



t] to show that these are not vector ﬁelds with respect to some

curvature-line co-ordinates s, t.

(d) Find explicit curvature-line co-ordinates for the surface; functions s, t such that

∂

∂s

∂

∂t

are

parallel to



t and express I, I with respect to s, t.

(Hint: try to guess solutions to



s[t] = 0 =



t[s])

7. Rotate y = f (x) around the x-axis and parametrize the surface via

x(ϕ, v) =



v, f (v) cos ϕ, f (v) sin ϕ



(a) Verify that the co-ordinates ϕ, v are curvature-line, compute the principal curvatures, and

show that the Gauss and mean curvatures are

K = −

′′

( v)

f (v)



1 + f

′

( v)



H =

f (v) f

′′

( v) −1 − f

′

( v)

2 f (v)



1 + f

′

( v)



3/2

(b) Demonstrate the following by choosing suitable f (v):

i. A cylinder has K = 0;

ii. A cone has K = 0;

iii. A sphere of radius a has K =

iv. A catenoid f (v) = a

−1

cosh(av −c) is a minimal surface.



′

( v)



, show that

1 + f

′2

= g = a

for some constant a

By substituting f (v) = a

−1

cosh



ah(v)



, show that the surface is a catenoid.

(d) Plainly K ≡ 0 if and only if f

′′

( v) ≡ 0. What are these surfaces? More generally, if the

surface has constant non-zero Gauss curvature K, show that f satisﬁes a non-linear ODE

K f

= (1 + f

′2

)

−1

+ c for some constant c

(Solving for f requires an elliptic integral when c = 0, so don’t try!)

8. The tractrix is parametrized by y(t) =



sinh

−1

t −t(1 + t

)

−1/2

, (1 + t

)

−1/2



. By revolving this

curve around the x-axis, show that the resulting surface is a pseudosphere with K ≡ −1.

9. We know that the Gauss and mean curvature are deﬁned in terms of the principal curvatures.

By writing down a suitable quadratic polynomial, prove that knowing of H, K is sufﬁcient to

recover the principal curvatures.

10. The graph of a function z = f (x, y) is parametrized by x(u, v) =



u, v, f (u, v)



. What can you

say about the surface if (u, v) are curvature-line co-ordinates?

(Hint: recall Example 3.19)

11. Suppose u, v are curvature-line co-ordinates for a surface x. Explain why n

= −k

and

= −k

12. Suppose that a surface x : U → E

is totally umbilic I = kI for some function k : U → R.

(a) Use Exercise 11 and n

= n

to prove that k is constant.

(b) Prove that x is (part of) a plane or a sphere (recall Example 3.28).

(Hint: If k = 0 consider c := x +

n. . . )

13. We prove part 2 of Theorem 3.31. Given the assumptions, deﬁne the dual 1-forms to



α(



s) = 1 = β(



t) and α(



t) = 0 = β(



Use Exercise 2.3.10 to prove that dα = 0 = dβ. Hence conclude (footnote, page 43) that (locally)

α = ds and β = dt for some functions s, t.

3.4 Power Series Expansions and Euler’s Theorem

In this section we intersect a surface with certain planes and consider the resulting curves. The

curvatures provide data about these curves and thus tell us something about the local shape of the

surface. The key is to see how curvatures describe a quadratic approximation to a surface.

At a regular point P on a surface S, choose axes such that P is

the origin and the (x, y)-plane is tangent

to S. By Theorem

3.10, S is locally the graph of a function z = f (x, y), which we

may parametrize in the usual manner

x(u, v) =



u, v, f (u, v)



The unit normal vector n

= k is therefore the standard vertical basis vector. Since the tangent plane

at P is the (x, y)-plane, we see that f

(0, 0) = 0 = f

(0, 0); substituting into Example 3.19 yields the

fundamental forms at P:

= du

+ dv

= f

+ 2 f

dudv + f

[I]



1 0

0 1



[I]

= Hess f =





The last matrix is the Hessian of f , and the Gauss and mean curvatures at P are

K(P) = det Hess f (0, 0) and H(P) =

tr Hess f (0, 0)

It bears repeating that these expressions are only valid at the origin O ∈ U (equivalently P ∈ S).

Although the co-ordinates u, v will extend nearby on the surface, the ﬁrst fundamental form need

not be diagonal anywhere except at the origin.

Now suppose we rotate the (x, y)-plane so that the axes point in the principal directions. Then the

Hessian is also diagonal ( f

(0, 0) = 0) and the principal curvatures at P are

= f

(0, 0) and k

= f

(0, 0)

Theorem 3.34. If the graph of z = f (x, y) is tangent to the (x, y)-plane at the origin O so that the

axes are the curvature directions, then the Maclaurin approximation of the function f (x, y) is

f (x, y) ≈ f (O) + (x y) ∇f

(x y) Hess f (O)





+ higher order terms

(O)x

(O) y

+ higher order terms

Example 3.35. Let f (x, y) = x

−y

(above picture). At the origin, x(u, v) =



u, v, u

−v



has

I = du

+ dv

, I = 2(du

−dv

), k

= 2, k

= −2, K = −4, H = 0

In this case the Maclaurin approximation is exact!

= x

−y

= f (x, y)

This amounts to applying a rigid motion (direct isometry) to the surface, which does nothing to the fundamental forms.

Level Curves: intersections with planes parallel to the tangent plane

If c is small, then the intersection of S with a plane cn

+ T

S parallel to the tangent plane is a level

curve; in our analysis, they correspond to level curves f (x, y) = constant. Theorem 3.34 tells us how

level curves depend on the curvatures. For instance, if k

, k

have opposite signs, then for small c,

+ k

≈ 2c

is approximately a hyperbola.

Deﬁnition 3.36. Suppose k

, k

, K, H are the curvatures of a surface S at a point P. We say that P is:

Elliptic ⇐⇒ K > 0 ⇐⇒ k

, k

= 0 and have the same sign.

Level curves near P are approximately ellipses.

Hyperbolic ⇐⇒ K < 0 ⇐⇒ k

, k

= 0 and have opposite signs.

Level curves near P are approximately hyperbolæ.

Parabolic ⇐⇒ K = 0 and H = 0 ⇐⇒ exactly one of k

, k

is zero.

Level curves near P are approximately a pair of parallel lines, e.g. x = ±c.

Planar ⇐⇒ K = H = 0 ⇐⇒ k

= k

= 0.

The curvatures provide no data as to the level curves near P.

Example (3.35, mk. II). For the graph of z = x

−y

, the level curve x

−y

= c = 0 is a hyperbola.

In fact this is true everywhere on this surface: under the usual

parametrization x(u, v) =



u, v, u

−v



, we have

K = −

(1 + 4u

+ 4v

)

and H =

4(v

−u

)

(1 + 4u

+ 4v

)

3/2

Since K < 0 everywhere, all points are hyperbolic.

In the picture, shifted tangent planes cn

+ T

S and their intersections with the surface are drawn

for two points. In both cases the level curves are genuine hyperbolæ.

Normal Curvature: intersections with planes containing the normal vector

Theorem 3.34 is the surface analogy of Exercise 1.6.5: a regular curve in E

passing through the origin

horizontally at t = 0 has its graph given locally by

y =

κ(0)x

+ higher order terms (∗)

We put this to work by considering the curvature of curves passing through a point on a surface.

Deﬁnition 3.37. Let S be a surface and v

∈ T

S a non-zero tangent vector.

The normal curvature ν(v

) is the curvature at P of the curve

deﬁned by the intersection of the

surface S and the normal plane Span{v

, n

We say that v

is asymptotic if ν(v

) = 0.

Strictly, the curve is the connected component of S ∩Span{v

, n

} containing P.

Example (3.35, mk. III). Consider the hyperbolic paraboloid z = x

− y

at the origin P = O.

Fix an angle ψ and let v

= (cos ψ, sin ψ). The intersection curve y ⊆ S ∩ Span{v

, n

} may be

parametrized using polar co-ordinates:

y(r) =



r cos ψ, r sin ψ, r

(cos

ψ −sin

ψ)



which amounts to the graph of the function g(r) = r

cos 2ψ.

The normal curvature is the curvature at r = 0 of this curve:

ν(v

) = κ(0) =

′′

(0)

[1 + g

′

(0)

]

3/2

= 2 cos 2ψ

Think about how the this corresponds to the picture and observe that

is asymptotic ⇐⇒ cos 2ψ = 0 ⇐⇒ ψ = ±

Our next result generalizes the method in the example.

Theorem 3.38 (Euler). Suppose v

makes angle ψ with the ﬁrst principal curvature direction. Then

ν(v

) = k

cos

ψ + k

sin

In particular, the principal curvatures are the extremes of normal curvature: if k

≤ k

, then

≤ ν(v

) ≤ k

where the bounds are realized precisely when v

points in a curvature direction.

Proof. Choose axes so the curvature directions at P are i, j, and n

= k. The surface is locally a graph

z = f (x, y). If (r, ψ) are polar co-ordinates in the (x, y)-plane, Theorem 3.34 says that

z = f (x, y) ≈

(r cos ψ)

(r sin ψ)

+ ··· =

( k

cos

ψ + k

sin

ψ)r

+ ···

Fix ψ and let v



cos ψ

sin ψ



(assume unit length since only the direction matters). Our curve of interest

y ⊆ S ∩Span{v

, n

} may be parametrized

y(r) = rv

+ f (r cos ψ, r sin ψ) n





r cos ψ

r sin ψ

f (r cos ψ, r sin ψ)









r cos ψ

r sin ψ

νr

+ ···





The last equality used observation (∗), where ν is the normal curvature. For the ﬁrst result, simply

compare the z-expressions in the displayed equations. For the ﬁnal observation, note that

ν(v

) = k

(1 −sin

ψ) + k

sin

ψ = k

+ ( k

−k

) sin

ψ ∈ [k

, k

]

and that the bounds are achieved precisely when ψ = 0,

, when v

is a curvature direction.

Examples 3.39. 1. If P is a planar point (k

= k

= 0), all normal curvatures are zero and all

directions are asymptotic.

2. (Example 3.30.1) All points of the paraboloid z = r

are elliptic (everywhere k

, k

> 0). The

surface has no asymptotic directions at any point, indeed the normal curvature in the direction

= (cos ψ, sin ψ) at P = (r cos θ, r sin θ, r

) is

ν(v

) = k

cos

ψ + k

sin

ψ =

(1 + 4r

)

3/2



cos

ψ + (1 + 4r

) sin



> 0

3. If k

= 0, then v



cos ψ

sin ψ



is asymptotic ⇐⇒ tan ψ = ±

−

The Second Fundamental Form and the Local Shape of a Surface

Our standard approach is to transfer calculations about surfaces back to the parametrization space.

With this in mind, we consider special tangent vectors with respect to the second fundamental form.

Deﬁnition 3.40. Let x : U → E

be an oriented surface and



∈ T

1. A tangent vector



=



0 is asymptotic if I(



) = 0.

2. The Dupin indicatrix at p ∈ U is the set of tangent vectors



such that I(



) = ±1.

Theorem 3.41. The notions of asymptotic in Deﬁnitions 3.37 & 3.40 coincide:

= dx(



) ∈ T

S is asymptotic ⇐⇒



∈ T

is asymptotic

The proof is an exercise. Recalling Theorem 3.20, a direction is asymptotic if and only if the normal

acceleration in said direction is zero.

The Dupin indicatrix turns out to precisely describe level curves near a point. To see this, write



= a



+ b



where



are orthonormal curvature directions;

the indicatrix at p has equation



) =



a b





0 k





= k

+ k

= ±1

This deﬁnes a conic in the tangent space T

whose type depends on the signs of the principal cur-

vatures. In essence, the Dupin indicatrix indicates the level curve obtained by taking the intersection

S ∩ (cn

+ T

S) for inﬁnitesimal c. We summarize all possibilities in a table using the point-types

introduced in Deﬁnition 3.36:

type of point # asymptotic directions Dupin indicatrix

elliptic 0 ellipse

hyperbolic 2 two hyperbolæ

parabolic 1 two parallel lines

planar ∞ empty

With respect to



, the matrices of the fundamental forms at p are [I

] =



1 0

0 1



and [I

] =



0 k



Examples 3.42. For a parametrized surface x at a given point p = (u

, v

), write



= a

∂

∂u



+ b

∂

∂v



1. (Exercise 3.2.6) The tangent developable of the unit-speed helix has



) = −

(



) = −

The single asymptotic direction is



∂

∂v



. The Dupin indicatrix is a pair of parallel lines

−

= ±1 =⇒



= ±

∂

∂u



+ b

∂

∂v



(b is arbitrary!)

2. In its usual parametrization, the surface z = x

y has

I =

√

1 + 4u

+ u



v du

+ 2u dudv



At p = (−1, 2) (i.e., x(p) = (−1, 2, 2)) we see that



) =

√



−2ab



√

a(a −b)

The point is hyperbolic with asymptotic directions

∂

∂v



and

∂

∂u



−

∂

∂v



(a = 0 and a −b = 0). The indicatrix comprises two hyperbolæ a(a −b) = ±

√

Exercises 3.4. 1. Consider the graph of the function z = x

−3y

+ 7xy

+ 9y

(a) Find the Gauss and mean curvatures at the origin.

(Hint: use Theorem 3.34)

(b) Find the normal curvature at the origin for the curve in the surface described by x = y.

2. As in Example 3.35, mk. III (page 70), ﬁnd the asymptotic directions at the origin for the surface

z = y

−3x

3. For the elliptic paraboloid z = x

+ y

, let P = (1, 2, 5) be a ﬁxed point.

(a) Find the maximum and minimum values for the normal curvature at P.

(b) Find the Dupin indicatrix at P.

4. For the hyperbolic paraboloid z = x

− y

, let p = (u

, v

) and P =



, v

, u

− v



. If c = 0,

prove that the intersection of the parallel plane cn

+ T

S and the paraboloid may be expressed

(x −u

)

−( y −v

)

= constant, z = x

−y

That is, the level curves really are hyperbolæ.

5. Consider the graph of the surface z = x

+ y

(a) Compute the Gauss curvature and classify all points according to Deﬁnition 3.36.

(b) Sketch the level curves z = 1,

100

and

10000

and compare to the Dupin indicatrix at (0, 0).

6. Prove Theorem 3.41 by considering the normal acceleration of the curve S ∩Span{v

, n

3.5 Adaptive Frames & Gauss’ Remarkable Theorem

In this section we repurpose the idea of a moving frame ﬁrst encountered when studying curves.

Deﬁnition 3.43. Let x : U → E

parametrize a surface S. A moving frame for S is a triple of smooth

functions e

, e

on U such that, for each p ∈ U,



(p), e

(p)



is a positively oriented orthonormal basis of T

x(p)

When S is oriented, we say that a moving frame is adaptive if e

= n is the unit normal ﬁeld.

For an adaptive frame, the tangent plane at each point is T

x(p)

S = Span{e

(p), e

(p)}.

We will often refer to the matrix-valued function E =





: U → SO

(R ) as the frame.

Examples 3.44. We’ll repeatedly analyze three examples through this section.

1. The parabolic cylinder x(u, v) =



u, v,



has an adaptive frame

√

1 + u

















√

1 + u





−u





2. The sphere of radius R in spherical polar co-ordinates x(ψ, ϕ) has an adaptive frame





−sin ψ

cos ψ









−cos ψ sin ϕ

−sin ψ sin ϕ

cos ϕ





= x =





cos ψ cos ϕ

sin ψ cos ϕ

sin ϕ





We use ψ instead of θ since we’ll need the latter for something else momentarily. . .

3. The paraboloid x(r, ψ) =



r cos ψ, r sin ψ,



has an adaptive frame

√

1 + r





cos ψ

sin ψ









−sin ψ

cos ψ





√

1 + r





−r cos ψ

−r sin ψ





In the pictures we’ve reduced the lengths of the frame vectors for clarity.

In each case e

, e

were obtained by differentiating with respect to the co-ordinates (and normalizing

if necessary). This works because the co-ordinate systems for all three examples are orthogonal.

As with the Frenet frame approach to curves, our strategy is to analyse a surface x : U → E

two

stages:

1. Describe how x moves with respect to the frame E.

2. Describe how the frame E moves (with respect to itself).

We describe inﬁnitesimal changes using 1-forms, following an approach pioneered by

Elie Cartan

around 1899.

Deﬁnition 3.45. Let x : U → E

be a smooth map and E = (e

) a moving frame. The metric

forms θ

and connection forms ω

are the 1-forms on U deﬁned by

:= e

·dx, ω

= e

·de

where j, k ∈ {1, 2, 3}.

Since e

, e

are orthonormal, these forms are nothing more than the co-ordinates of dx, de

, de

and de

with respect to the moving frame:

dx =

∑

j=1

( e

·dx)e

= e

+ e

, de

∑

j=1

(∗)

The frame is adaptive if and only if θ

= 0. Moreover, as the next result shows, for any frame there

are only three independent connection forms (compare this with Theorem 1.29).

Lemma 3.46. For all j, k, we have ω

= −ω

. In particular ω

= 0.

Proof. Take the exterior derivative of the identity e

·e

= 0 or 1, to obtain

0 = de

·e

+ e

·de

= ω

+ ω

If (∗) are arranged in matrix format, the subscripts follow the usual row/column convention:

dx =













= EΘ, dE =









0 ω

−ω

0 ω

−ω





= Eω

The second expression should remind you of the Frenet–Serret equations for a curve! The metric

forms get their name because they measure small changes on the surface. The connection forms tell

us how nearby frames are related (connected): abusing notation a little, if



∈ T

, then

E(p +



) −E(p) ≈ dE(



) = E(p)ω(



)

The fundamental forms of x can be written in terms of Θ and ω; in an adaptive frame this is partic-

ularly simple.

Lemma 3.47. In an adaptive frame

I = dx ·dx = θ

+ θ

and I = −dx ·de

= −θ

−θ

Examples (3.44, mk. II). You needn’t compute all exterior derivatives de

: use the skew-symmetry

of ω to help; also consider which frame ﬁelds are easier to differentiate! The expressions for the

fundamental forms should be a sanity check since we know how to compute them already.

1. The parabolic cylinder has

dx =









du +









dv =

1 + u

du + e

dv =⇒ θ

1 + u

du, θ

= dv

=⇒ I = (1 + u

) du

+ dv

Since e

is constant, we have de

= 0 from which

= e

·de

= 0, ω

= −ω

= −e

·de

= 0

The ﬁnal connection form requires a derivative:

= e

·de

√

1 + u













−u

(1 + u

)

3/2





−u





√

1 + u





−1









−1

1 + u

Putting it together, we have

ω =

1 + u





0 0 −1

0 0 0

1 0 0





du and I = −θ

−θ

= du

2. For the sphere of radius R, dx = R cos ϕe

dψ + Re

dϕ, whence

= R cos ϕ dψ, θ

= Rdϕ =⇒ I = R



cos

ϕ dψ

+ dϕ







−cos ψ

−sin ψ





dψ =⇒

(

= −e

·de

= −sin ϕ dψ

= −e

·de

= cos ϕ dψ

= e

·de





−cos ψ sin ϕ

−sin ψ sin ϕ

cos ϕ













−sin ψ cos ϕ

cos ψ cos ϕ





dψ +





−cos ψ sin ϕ

−sin ψ sin ϕ

cos ϕ





dϕ





= dϕ

=⇒ I = −θ

−θ

= −R



cos

ϕ dψ

+ dϕ



3. For the paraboloid,

dx =





cos ψ

sin ψ





dr + r





−sin ψ

cos ϕ





dψ =

1 + r

dr + re

dψ

=⇒ θ

1 + r

dr, θ

= r dψ =⇒ I = (1 + r

)dr

+ r

dψ

The connection forms are comparatively ugly. The low-hanging fruit is de



−cos ψ

−sin ψ



dψ,

which quickly yields two of them:

= e

·de

= −

dψ

√

1 + r

, ω

= −ω

= −e

·de

−r dψ

√

1 + r

The last connection form requires a nastier differentiation, though only one of the three terms

in de

provides a non-zero result when dotted with e

= e

·de

√

1 + r





cos ψ

sin ψ









···+

√

1 + r





−cos ψ

−sin ψ









−dr

1 + r

We therefore obtain the connection form matrix

ω =

√

1 + r







0 −dψ

−1

√

1+r

dψ 0 −r dψ

√

1+r

dr r dψ 0







and second fundamental form

I = −

1 + r

−dr

1 + r

−r dψ

√

1 + r

√

1 + r



+ r

dψ



The Structure Equations for a Moving Frame

The metric and connection forms satisfy matrix equations dx = EΘ and dE = Eω. Since d

= 0,

something nice happens when we take the exterior derivatives of these expressions:

0 = d

x = d(dx) = d( E Θ) = dE ∧Θ + E dΘ = E(ω ∧Θ + dΘ)

0 = d

E = d(dE) = d(Eω) = dE ∧ω + Edω = E



ω ∧ω + dω



The notation ω ∧Θ means matrix multiplication using the wedge product of forms to evaluate each

entry.

Since each E(p) is an invertible matrix, we conclude two identities.

Theorem 3.48. The metric and connection forms satisfy the structure equations; each amounts to

three separate equations after multiplying out the matrix expressions.

1. dΘ + ω ∧ Θ = 0, equivalently dθ

∑

k=j

∧θ

= 0 for each j = 1, 2, 3

2. dω + ω ∧ω = 0, equivalently dω

+ ω

∧ω

= 0 where i, j, k are distinct.

These are easy to remember if you pay attention to the indices! In an adaptive frame (θ

= 0), things

are a little simpler and some of the equations get special names:

First structure equations



dθ

+ ω

∧θ

= 0

dθ

+ ω

∧θ

= 0

Symmetry equation ω

∧θ

+ ω

∧θ

= 0

Gauss equation dω

+ ω

∧ω

= 0

Codazzi equations



dω

+ ω

∧ω

= 0

dω

+ ω

∧ω

= 0

Be careful not to reverse the order: Θ ∧ ω makes no sense since the dimensions of the matrices are incompatible!

Similarly, ω ∧ ω is unlikely to be zero. . .

Examples (3.44, mk. III). 1. For the parabolic cylinder, Θ =



√

1+u



and ω =

1+u



0 0 1

0 0 0

−1 0 0



du,

so all the structure equations are trivial:

dΘ = 0 = −ω ∧Θ, dω = 0 = −ω ∧ω

2. For the sphere, Θ = R



cos ϕ dψ

dϕ



and ω =



0 sin ϕ dψ −cos ϕ dψ

−sin ϕ dψ 0 dϕ

cos ϕ dψ −dϕ 0



, from which

dΘ = R





−sin ϕ





dϕ ∧dψ = −ω ∧Θ

dω =





0 cos ϕ sin ϕ

−cos ϕ 0 0

−sin ϕ 0 0





dϕ ∧dψ = −ω ∧ω

3. For the paraboloid, Θ =



√

1+r

r dψ



and ω =

√

1+r





0 −dψ −

√

1+r

dψ 0 −r dψ

√

1+r

r dψ 0





The ﬁrst equations aren’t too bad to check:

dΘ =









dr ∧dψ = −ω ∧Θ

The second are a little nastier: you should check that

dω =

(1 + r

)

3/2





0 r 0

−r 0 −1

0 1 0





dr ∧dψ = −ω ∧ω

Gauss’ Remarkable Theorem

Suppose we have an adaptive frame for an oriented local surface x. If θ

, θ

were linearly dependent

at p, then the differential dx = e

+ e

: T

→ T

x(p)

S = Span{e

(p), e

(p)} would have

rank ≤ 1 and thus not be a bijection. We conclude that {θ

, θ

} forms a basis of the space of 1-forms

at p, and that any other 1-form may be written as a linear combination thereof. . .

Lemma 3.49. There exist unique functions a, b, c such that

= aθ

+ bθ

, ω

= bθ

+ cθ

With respect to these functions, the second fundamental form, Gauss and mean curvatures are

I = −aθ

−2bθ

−cθ

, K = ac −b

, H = −

(a + c)

Proof. That ω

= aθ

+ bθ

and ω

bθ

+ cθ

are linear combinations of θ

, θ

is the above discus-

sion. By the symmetry equation and the fact that θ

∧θ

= 0,

0 = ω

∧θ

+ ω

∧θ

= (−b +

b) θ

∧θ

=⇒

b = b

The formula for I follows from Lemma 3.47.

Moreover, if



and



are the dual vector ﬁelds to θ

, θ

, then the matrices of I, I with respect to

these ﬁelds are the identity matrix

and B =



−a −b

−b −c



. The Gauss and mean curvatures are the

determinant and half the trace of B (Theorem 3.29).

Now consider the ﬁnal connection form ω

. Since θ

, θ

form a basis at each point, we may write

= f θ

+ gθ

for some functions f , g : U → R. Applying the 1

structure equations,

dθ

= −ω

∧θ

= −f θ

∧θ

dθ

= −ω

∧θ

= −θ

∧ω

= −gθ

∧θ

whence f , g (and ω

) are determined by θ

, θ

. This brings us to the capstone result of these notes.

Theorem 3.50 (Gauss’ Theorem Egregium). The Gauss curvature depends only on the ﬁrst funda-

mental form.

Proof. By the above discussion, ω

(and thus dω

) depends only on θ

, θ

, which may be recovered

from I = θ

+ θ

by writing it as a sum of squares. But now the Gauss equation reads

dω

= ω

∧ω

= (aθ

+ bθ

) ∧(bθ

+ cθ

) = (ac − b

) θ

∧θ

= Kθ

∧θ

An explicit formula for K as a function of the coefﬁcients E, F, G of I can be found; see Exercise 9.

Egregium (Latin for remarkable/outstanding) is the (modest!) term Gauss applied after proving his

result in 1827. Why did he consider it so remarkable? The original deﬁnition of K relied on the nor-

mal ﬁeld; an object outside the surface which helps describe its position/orientation in E

. Gauss’

Theorem, however, says that K is intrinsic to the surface: it depends only on the metric (ﬁrst fun-

damental form) which may be understood by an occupant of the surface with no ability to escape

(travel outside the surface) in to view its shape. By contrast, the second fundamental form and the

mean curvature depend on how a surface is embedded; these are extrinsic quantities.

As a nice side-effect, the result provides what is often a faster method for calculating K.

1. Compute the ﬁrst fundamental form I = dx ·dx and express it as a sum of squares I = θ

+ θ

2. Write ω

= f θ

+ gθ

and compute f , g using the 1

structure equations.

3. Use the Gauss equation to ﬁnd K.

We need only calculate 1-forms θ

, θ

, ω

that are related to the tangent part of the moving frame.

The unit normal e

needn’t be considered or calculated.

(



) = δ

(

1 j = k

0 j = k

implies that dx(



) = e

and dx(



) = e

are orthonormal.

Examples (3.44, mk. IV). We return to our examples one last time. Even though we’ve already

calculated the connection forms, the goal is to see that ω

= f θ

+ gθ

and thus K may be found

directly from I.

1. The parabolic cylinder has I = (1 + u

) du

+ dv

so the natural choice is

1 + u

du and θ

= dv

Since dθ

= 0 = dθ

we see that f = g = 0. We conclude that

= 0 =⇒ dω

= 0 =⇒ K = 0

2. For the sphere I = R

(cos

ϕ dψ

+ dϕ



so we choose θ

= R cos ϕ dψ and θ

= R dϕ. Certainly

0 = dθ

= −gθ

∧θ

=⇒ g = 0. Moreover,

dθ

= −f θ

∧θ

=⇒ R sin ϕ dψ ∧dϕ = −f R

cos ϕ dψ ∧dϕ =⇒ f = −R

−1

tan ϕ

We conclude that ω

= −R

−1

tan ϕ θ

= −sin ϕ dψ, from which

dω

= cos ϕ dψ ∧dϕ =

∧θ

=⇒ K =

3. For the paraboloid, I = (1 + r

)dr

+ r

dψ

so we choose θ

√

1 + r

dr and θ

= r dψ. This

time dθ

= 0 =⇒ f = 0 and

dθ

= −gθ

∧θ

=⇒ dr ∧dψ = −gr

1 + r

dr ∧dψ =⇒ g = −

√

1 + r

We conclude that ω

= −

√

1+r

= −

√

1+r

dψ, from which

dω

(1 + r

)

3/2

dr ∧dψ =

(1 + r

)

∧θ

=⇒ K =

(1 + r

)

Since K depends only on the metric, it is invariant under isometric transformations of the surface. This

helps explain why the Gauss curvature of a cylinder and a cone are both zero: both may constructed

by rolling up a ﬂat plane without other distortion.

The contrapositive of Gauss’ Theorem is also important: surfaces with distinct Gauss curvatures

cannot be isometric. Since the metric I determines how we measure angle and length, this explains

why a perfect ﬂat map (K = 0) of any part of the Earth (K =

) is impossible to achieve. The holy

grail of map-making would be a map free of direction, angle and length/area distortion:

1. Straight lines on the map should correspond to paths of shortest distance on the Earth.

2. Angles on the map should equal corresponding angles on the Earth’s surface.

3. Areas on the map and the Earth should be in constant ratio.

Gauss’ Theorem implies that you cannot have all these properties in one map. In fact, at most one of

these properties is possible in a single map.

Riemannian Geometry

We can even employ the method when there is no surface! The idea is to equip a domain with an

abstract ﬁrst fundamental form and use it to compute lengths, angles, area, geodesics, curvature, etc.

Example 3.51. The Poincar

e disk model of hyperbolic space is the disk



(x, y) ∈ R

: x

+ y

< 1



equipped with the metric (ﬁrst fundamental form)

I =

4( dx

+ dy

)

(1 − x

−y

)

4( dr

+ r

dψ

)

(1 −r

)

As one approaches the boundary of the disk, the idea is that measured distance gets larger: the

boundary circle is in fact inﬁnitely far from any point inside the disk. To express I as a sum of squares,

a natural choice is θ

2dr

1−r

and θ

2r dψ

1−r

, from which dθ

= 0 =⇒ f = 0 and

dθ

= −gθ

∧θ

=⇒

2( 1 + r

)

(1 −r

)

dr ∧dψ = −

4gr

(1 −r

)

dr ∧dψ =⇒ g = −

1 + r

from which

dω

= d(gθ

) = −d



1 + r

1 −r



∧dψ =

−4r

(1 −r

)

dr ∧dψ = −θ

∧θ

=⇒ K = −1

Hyperbolic space is the canonical example of a negatively curved geometry. There is no surface here, no

second fundamental form, and no mean curvature! Since there is no surface, it is harder to visualize

what K means in this context (e.g. Section 3.4).

The Gauss curvature of a surface is the simplest avatar of a more general object called the Riemann

curvature tensor. As an example of how this is applied, in general relativity

mass is construed as

changing the metric of spacetime (i.e. I); it can be seen that this metric is compatible with unique

connection (essentially ω) from which the curvature (dω + ω ∧ ω) may be computed. When a

physicist says spacetime is curved, this is what they mean: there is no exterior to spacetime from which

we can measure curvature, so everything is computed intrinsically.

The Fundamental Theorem of Surfaces

Recall the equivalence of spacecurves up to rigid motions (Theorem 1.38) and the Fundamental The-

orem of Biregular Spacecurves (Corollary 1.42). A similar discussion is available for surfaces once we

replace curvature and torsion with the fundamental forms I, I.

The equivalence problem is almost identical. Suppose x : U → E

is an oriented surface, and that

A ∈ O

(R ) and b = E

are constants. Then y := Ax + b is a new surface, the result of applying an

isometry to x. A moving frame for x is transformed to a frame for y via



± Ae



where ±1 = det A

The most famous consequence concerns angle-sums of geodesic triangles: A + B + C = π +

△

K. If K < 0, the angle-

sum of a geodesic triangle is less than 180°. When K > 0 (e.g., a sphere), the angle sum is greater than 180°. This topic, the

related Gauss–Bonnet Theorem, and other consequences, are a matter for another course.

Really this is pseudo-Riemannian geometry, since I is not positive-deﬁnite.

The upshot is that n

= (det A)An

, and I, I transform exactly as κ, τ:

= dy ·dy = (Adx) · (Adx) = dx ·dx = I

= −dy ·dn

= −(det A)(Adx) ·(Adn

) = (det A)I

As with curves, we may ask the question in reverse. If we know the fundamental forms, can we also

recover the surface up to a rigid motion? The answer is yes, though with a caveat: unlike κ, τ for

spacecurves, the fundamental forms cannot be chosen independently.

Theorem 3.52 (Bonnet). Suppose I and I are symmetric bilinear forms where I is positive-deﬁnite.

Provided the Gauss–Codazzi equations are satisﬁed, there exists a local parametrized surface with

these fundamental forms, which is moreover unique up to rigid motions.

Everything ultimately depends on a generalization of the existence/uniqueness theorem for ODE

(another version of the Frobenius Theorem (3.33)). Here is a rough sketch of how the process works.

1. Suppose we are given I, I on U, and initial conditions at some p ∈ U (for the surface x(p) = x

and frame E(p) = E

2. Since I is positive-deﬁnite, it may be written I = θ

+ θ

3. The ﬁrst structure equations determine ω

and I determines ω

and ω

(Lemma 3.49).

4. The Frobenius Theorem shows that the initial value problem

dE = Eω E(p) = E

(∗)

has a unique local solution provided the Gauss–Codazzi equations dω + ω ∧ ω = 0 are sat-

isﬁed. The solution E is SO

(R )-valued and supplies an adapted frame (compare Corollary

1.41).

5. To ﬁnd the surface, solve a second initial value problem

dx = EΘ x(p) = x

Frobenius says this has a unique solution provided dΘ + ω ∧ Θ = 0. Since this is precisely

what we used to determine ω in step 2, we don’t need to check this condition.

6. Any different choice of metric forms in step 2 merely rotates E around n = e

and does not

affect the resulting surface.

It is a little easier to understand the integrability condition when written in co-ordinates: (∗) is a

linear system of eighteen PDE in nine unknowns

(

∂E

∂u

= EP

∂E

∂v

= EQ

where P = ω



∂

∂u



, Q = ω



∂

∂v



are skew-symmetric matrix functions

The Gauss–Codazzi equations are essentially the fact that mixed partial derivatives commute:

0 = E

−E

= E

P + EP

−E

Q + EQ

= E



− Q

−[P, Q]



− Q

−[P, Q] =

∂

∂v



∂

∂u



−

∂

∂u



∂

∂v



−





∂

∂u



, ω



∂

∂v





dω + ω ∧ω





∂

∂v

∂

∂u



[P, Q] = PQ −QP and dω is evaluated as in Exercise 2.3.10.

The part that requires some proof is that the integrability condition (P

−Q

= [P, Q]) is sufﬁcient for

a solution. This is not as hard as it sounds; here is another sketch:

1. If p = (u

, v

), use Picard’s ODE existence/uniqueness theorem to

solve an initial value problem on the horizontal line v = v

EP(u, v

E(u

, v

) = E

2. For each u

, apply the ODE theorem to solve another IVP on the

vertical line u = u

= EQ(u

, v), E(u

, v

) =

E(u

, v

)

3. Finally, one shows that the resulting E is differentiable with respect to u, and uses the integra-

bility condition to check that E

= EP as required.

The ﬁrst two steps may be accomplished approximately using a numerical method to desired accu-

racy, so this amounts to an algorithm for the approximation of E. The same approach can then be

followed to approximate the surface.

The Gauss–Codazzi equations in curvature-line co-ordinates Suppose (u, v) are curvature-line

co-ordinates. Then the fundamental forms are

I = E du

+ G dv

, I = k

E du

+ k

Gdv

(†)

where E, G are positive functions and k

, k

are the principal curvatures. We therefore choose metric

forms θ

√

E du and θ

√

G dv. In the language of Lemma 3.49,

a = −k

, b = 0, c = −k

, ω

= −k

√

E du, ω

= −k

√

G dv

The ﬁrst structure equations determine

√

(

du −G

)

(see Exercise 9). Moreover, the Gauss–Codazzi equations are equivalent to

dω

+ ω

∧ω

= 0 ⇐⇒



√





√



= −2k

√

dω

+ ω

∧ω

= 0 ⇐⇒ 2(k

)

E = (k

−k

dω

+ ω

∧ω

= 0 ⇐⇒ 2(k

)

G = (k

−k

These equations show the relationship between I and I: we cannot independently choose the metric

(E, G) and the curvatures (k

, k

). However, if E, G, k

, k

satisfy these equations, Bonnet’s theorem

guarantees the existence of a surface with fundamental forms (†) , unique up to rigid motions.

While I, I cannot be chosen independently, Bonnet’s result is considered the best description of the

minimal data for a surface. You might suspect/hope that knowledge of K, H would be enough to

determine a surface up to rigid motions, but Exercise 10 shows such to be vain!

Exercises 3.5. 1. The unit cylinder x(ϕ, v) =



cos ϕ, sin ϕ, v



has adaptive frame





−sin ϕ

cos ϕ





, e









, e

= e

×e





cos ϕ

sin ϕ





(a) Directly compute the metric forms θ

and connection forms ω

(b) That the six structure equations are satisﬁed should be obvious from your answers to (a):

why?

2. For a general regular surface, explain why we cannot, in general, ﬁnd co-ordinates u, v for

which I = du

+ dv

3. For the paraboloid example (3.44.3) verify the Gauss–Codazzi equations dω + ω ∧ω = 0.

(Hint: this is easier if you treat the three equations separately!)

4. Verify that the metric I =

+dy

on the upper half-plane y > 0 has curvature K = −1.

(Hint: Recall Example 3.51 and Exercise 3.2.9)

5. Consider the catenoid x(u, v) =



cos u cosh v, sin u cosh v, v



obtained by revolving the cate-

nary x = cosh z around the z-axis.

(a) Show that there exists a moving frame for which the metric forms are

= cosh v du, θ

= cosh v dv

(b) Show that ω

= tanh v du =

sinh v

cosh v

du and use it to prove that the Gaussian curvature of

the catenoid is

K = −

cosh

6. We re-prove Exercise 3.3.12 using our new language.

(a) Suppose a surface x is totally umbilic: I = λI, where λ is some function. Explain why

= −λθ

and ω

= −λθ

(b) Use the 1

structure equations and the Codazzi equations to prove that dλ = 0.

(d) If a = 0, deﬁne c := x −

. Prove that dc = 0 and hence conclude that the surface is

(part of a) round sphere.

7. Suppose E =





is an adaptive frame for a surface. Any other adaptive frame (with the

same orientation) is obtained by rotating around e

: that is

E =





where

= cos φ e

+ sin φ e

= −sin φ e

+ cos φ e

for some smooth function φ : U → R.

(a) Compute θ

, θ

in terms of

and conclude that

∧

= θ

∧θ

(b) Use Deﬁnition 3.45 to compute

in terms of ω

and φ. Verify that d

= dω

so that

the Gauss equation is identical for the new moving frame.

8. Suppose I is the 1

fundamental form of a surface. Suppose I = θ

+ θ

for some 1-forms θ

, θ

Prove that there exists a moving frame E = (e

) for which dx = e

+ e

(Hint: consider the dual vector ﬁelds to θ

, θ

)

9. Suppose u, v are orthogonal co-ordinates so that θ

√

E du and θ

√

G dv.

(a) Use the structure equations to prove that

√

(

du −G

)

(b) Hence deduce an explicit formula for the Gauss curvature in terms of the coefﬁcients of

the 1

fundamental form:

K = −

√



∂

∂u

√

∂

∂v

√



This can be multiplied out to remove the square roots, though you’ll get more terms. A nastier

expression (the Brioshi formula) may be found for general co-ordinates with F = 0.

10. In Exercise 3.3.5 we saw that the tangent developable x(u, v) = y( u) + vy

′

( u) of a unit-speed

curve has curvatures K = 0, H = −

2vκ

. Use this to describe two surfaces with the same curva-

ture functions which are not related by a direct isometry.

11. Show that the surfaces parametrized by

x(u, v) =



u cos ϕ, u sin ϕ, ln u



, y(u, v) =



u cos ϕ, u sin ϕ, ϕ



have the same Gauss curvature but distinct ﬁrst fundamental forms I

= I

. To do this properly,

you should argue that there is no reparametrization of y so that K

= K

and I

= I

(Gauss’ Theorem isn’t biconditional: surfaces can have the same K without being locally isometric)

12. Consider the family of surfaces

( u, v) = cos t





sin u sinh v

−cos u sinh v





+ sin t





cos u cosh v

sin u cosh v





, t ∈ [0,

]

When t = 0 this is a helicoid. When t =

this is the catenoid from Exercise 5.

(a) Compute the ﬁrst fundamental form of x

and show that it is independent of t (the family

is therefore isometric).

(b) Show that the unit normal of x

is also independent of t:

cosh v





cos u

sin u

−sinh v





Hence compute the second fundamental form of x

for each t.

. What is special about this family?

Relate this to Gauss’ Theorem.