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Lecture Notes on Differential Geometry

Name: Lecture Notes on Differential Geometry
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Author: Dr./Ph.D. Prakash Dabhi
ISBN: 978-3-668-01573-9

Textbook, 2015

77 Pages

Dr./Ph.D. Prakash Dabhi (Author)

Excerpt

Dedicated to:

The Lecture Notes is dedicated to my students

Preface

This is a Lecture Notes on a one semester course on Differential Geometry taught as a basic

course in all M.Sc./M.S. programmes in Mathematics. This consists normally of curve theory leading

up to fundamental theorem of space curves as well as the Gauss theory of surfaces covering first fun-

damental form, second fundamental form, Gaussian curvature, geodesic and Gauss Bonnet theorem.

This Lecture Notes is based on lectures I have given to M.Sc. Mathematics students of Sardar Patel

University, Vallabh Vidyanagar, India.

Here are the salient features of the Lecture Notes. Proofs of all assertions are completely given

in a lucid student friendly manner. A large number of solved exercises are included. All these are to

facilitate self study by the students. I have also adopted the modern approach to develop the classical

topics treated here. The Lecture Notes is highly influenced by the approach adopted in Elementary

Differential Geometry by Andrew Pressley and Differential Geometry of Curves and Surfaces by

Manfredo P. do Carmo. I am indebted to these authors whose work have influenced my learning of

the subject as well as the preparation of this Lecture Notes.

I hope this little book would invite the students to the subject of Differential Geometry and

would inspire them to look to some comprehensive books including those mentioned above.

Place : Vallabh Vidyanagar

Date : 7 July 2015

Prakashkumar A. Dabhi

Department of Mathematics

Sardar Patel University

iii

Contents

Preface

iii

Chapter 1. Curve Theory

1. Curves

2. Regular curves

3. Curvature and Torsion

4. Fundamental Theorem of Space Curves

5. Isoperimetric Inequality and Four Vertex Theorem

Chapter 2. Surfaces in R

6. Surfaces

7. Calculus on Surface

8. First Fundamental Form

9. Local isometries and conformal maps

Chapter 3. Curvature of a surface

10. Second Fundamental Form

11. Gaussian, Mean and Principal Curvatures

Chapter 4. Geodesics and some fundamental results

12. Christoffel's Symbols and tangent vector fields

13. Geodesics

14. Codazzi - Mainardi and Gauss equations

15. The Gauss-Bonnet Theorem

Bibliography

CHAPTER 1

Curve Theory

1. Curves

Definition 1.1.1. Let I = (a, b) be an interval, - a < b . A continuous map : I R

is called a parametrized curve in R

Example 1.1.2. Parametrization of Cartesian curves.

(i) A parametrization of a parabola y = x

is (t) = (t, t

), t R. The curve

(t) = (t

, t

), t

R is not a parametrization of y = x

. The curve (t) = (t

, t

), t R is a parametrization of

y = x

A curve may have more than one parametrization.

(ii) The parametrized curve (t) = (a cos t, b sin t), t R is a parametrization of the ellipse

= 1.

(iii) The curve (t) = (a sec t, b tan t), t (-

) is a parametrization of the hyperbola

1, x > 0.

(iv) The curve (t) = (a cos

t, a sin

t), t R is a parametrization of the astroid x

+ y

= a

Example 1.1.3.

(i) Let : R R

be defined by (t) = (a cos t, b sin t), where a, b R\{0}. Then the image

of is an ellipse in R

. Its Cartesian equation is

= 1. In particular when a = b, then

the image of is a circle in R

(ii) Let : R R

be defined by (t) = (t, t

). Then the image of is a parabola in R

. Its

Cartesian equation is y = x

(iii) Let : R R

be defined by (t) = (a cosh t, b sinh t), where a, b R\{0}. Then the image

of is a part of the hyperbola namely all those points (x, y) on the hyperbola

= 1 such

that x > 0.

(iv) Let : R R

be defined by (t) = (e

cos t, e

sin t). Then the image of is a logarithmic

spiral in R

(v) Let : R R

be defined by (t) = (a + lt, b + mt, c + nt), where a, b, c, l, m, n R and

+ m

+ n

= 0. Then the image of is a line in R

passing through (a, b, c) having direction

(l, m, n). Its Cartesian equation is

x-a

y-b

z-c

1. CURVE THEORY

(vi) Let : R R

be defined by (t) = (a cos t, b sin t, ct), where a, b, c R\{0}. Then the

image of is a helix in R

. It is a helix with the base ellipse

= 1. When a = b, it is

called circular helix.

(vii) Let (t) = (cos

t, sin

t), t R. Then the Cartesian equation for the image of is x + y = 1,

0 x 1 and 0 y 1.

(viii) Let (t) = (e

, t

). Then the Cartesian form of the curve is y = (ln x)

Definition 1.1.4. A parametrized curve in R

is called planar if it is contained in some plane of R

It is called non-planar or twisted if it is not planar.

Let : (a, b) R

be a parametrized curve. Then = (

, . . . ,

), where each

is a

mapping from (a, b) to R. The symbol

is the derivative of and it means

= (

, . . . ,

Throughout it is assumed that all curves are smooth, i.e., all the derivatives of exist.

We note that a parametrized curve = (

) : (a, b) R

is planar if and only if there

are real numbers , , , R with

= 0 such that

(t) +

(t) = for

all t (a, b).

Example 1.1.5.

(i) The curve (t) = (t, t

, t

) is not planar.

Suppose that is planar. Then the exist a, b, c, d R, a

= 0, such that at+bt

+ct

d for all t. It follows from Fundamental Theorem of Algebra that a = b = c = 0. This

contradicts the fact a

+ b

+ c

= 0.

(ii) The curve (t) = (cos t, sin t, 3 sin t + 4 cos t) is planar.

One can see that the coordinates of satisfy the equation of the plane z = 4x + 3y. Hence

is planar.

(iii) The curve

(t) =

cos t, 1 - sin t, -

cos t) is a plane curve.

Definition 1.1.6. Let : (a, b) R

be a parametrized curve, and let t (a, b). The vector

(t) is

called the tangent vector to at the point (t).

(t) = 0, then the equation of the tangent to at the point (t) is R - (t) = u (t), u R.

Exercise 1.1.7.

(i) Find the equation of the tangent to the following curves.

(a) (t) = (a cos t, a sin t, bt), (a, 0, 2b).

(b) (t) = (t, t

, t

), (1, 1, 1).

t, sin

t), (

(d) (t) = (e

, t

), (1, 0).

(ii) Sketch the astroid (t) = (cos

t, sin

t). Calculate its tangent vector at each point. At which

point is the tangent vector zero?

2. Regular curves

Example 1.1.8. Consider the logarithmic spiral (t) = (e

cos t, e

sin t). We show that the angle

between (t) and

(t) is independent of t.

Here (t) = (e

cos t, e

sin t) and (t) = (e

cos t - e

sin t, e

sin t + e

cos t). Let (t) be the

angle between (t) and

(t). Then

(t) = cos

-1

(t) (t)

(t)

= cos

-1

= cos

-1

Therefore the angle between (t) and

(t) is independent of t.

Proposition 1.1.9. Let be a parametrized curve in R

. If = a for some nonzero vector a R

then is part of a line.

. Since

= a, we have (t) = at + b. Hence is part of a line.

2. Regular curves

Definition 1.2.1. Let : (a, b) R

be a parametrized curve. Then the arc-length s(t) of a curve

starting at the point (t

) is given by

s(t) =

(u) du (t (a, b)).

Since

is a nonnegative continuous function, it follows that s(t

) = 0, s(t) 0 if t > t

and

s(t) < 0 if t < t

Example 1.2.2.

(i) Parametrize the graph y = f (x), x [a, b], where f is a smooth map, and we

show that its arc-length is given by the formula

1 + f (x)

dx.

We may parametrize y = f (x) by (t) = (t, f (t)), t [a, b]. Then

(t) = (1, f (t)). Thus

(u) du =

1 + f (u)

du.

(ii) Compute the arc-length of the logarithmic spiral (t) = (e

cos t, e

sin t) starting at the point

(1, 0).

We have

(t) = (-e

sin t + ke

cos t, e

cos t + ke

sin t). Therefore

(t) = e

(- sin t + k cos t)

+ (cos t + k sin t)

= e

1 + k

1. CURVE THEORY

Since (0) = (1, 0), the arc-length of starting at the point (1, 0) is

s(t) =

1 + k

du =

1 + k

- 1).

Definition 1.2.3. Let : (a, b) R

be a parametrized curve, and let t (a, b). Then is called

regular at (t) or (t) is called a regular point of if (t) = 0 (or (t) > 0). A point (t) of the

curve is called a singular point if is not regular at that point. A curve is called to be regular if

all its points are regular points.

Definition 1.2.4. Let : (a, b) R

be a parametrized curve. Then the scalar

(t) is called the

speed of at the point (t). The curve is called a unit-speed curve if (t) = 1 for all t.

It follows from the definitions that every unit-speed curve is regular. The converse is not true.

For example (t) = (cos 2t, sin 2t) is regular but not unit-speed.

Exercise 1.2.5.

(i) Calculate the arc-length of the catenary (t) = (t, cosh t) starting at the point (0, 1).

(ii) Show that the following curves are unit-speed:

(a) (t) = (

(1 + t)

3/2

(1 - t)

3/2

)

(b) (t) = (

cos t, 1 - sin t, -

cos t)

(iii) Which of the following curves are regular?

(a) (t) = (cos

t, sin

t) for t R.

(b) The same curve as in (i), but with 0 < t < /2.

Lemma 1.2.6. Let : (a, b) R

be a parametrized curve. If (t) = 1 for all t, then (t) and

(t)

are perpendicular for all t. In particular, if is a unit-speed curve, then (t) ¨

(t)

for all t.

. Since

(t)

= 1 for all t, we have (t)(t) = 1. Hence (t) (t) = 0 for all t, i.e.,

(t)

(t) for all t.

It follows that if is a unit-speed curve, then

(t) ¨

(t) for all t as (t) = 1 for all t.

Lemma 1.2.7. Let : (a, b) R

be a regular curve. Then the arc-length of , starting at any

point of the curve, is a smooth map.

. We have s(t) =

(u) du. Then

. Let = (

, . . . ,

). Since is regular,

it follows that

+ · · · +

> 0. Define f : (0, ) R and g : (a, b) R by f (t) =

and g(t) =

(t) + · · · +

(t). Then both f and g are smooth maps. Since the range of g is

contained in (0, ), the domain of f , it follows that the map f g is a smooth map. But f g =

Therefore

is a smooth map and hence s is a smooth map.

2. Regular curves

Definition 1.2.8. Let : (a, b) R

be a parametrized curve. A parametrized curve : (~

a, ~

b) R

is said to be a reparametrization of if there is a bijective smooth map : (~

a, ~

b) (a, b), whose

inverse is also smooth, such that = . The map is called the reparametrization map for the

above reparametrization.

If is a reparametrization of with the reparametrization map , then =

-1

. Hence

is a reparametrization of with a reparametrization map

-1

If and are reparametrizations of each other, then they have the same trace.

Exercise 1.2.9. Let C be the collection of all parametrized curve in R

. Let , C. We say

if is a reparametrization of . Show that the above relation on C is an equivalence relation.

Proposition 1.2.10. A reparametrization of a regular curve is regular.

. Let : (~

a, ~

b) R

be a reparametrization of a regular curve : (a, b) R

. Then there

is a bijective smooth map : (~

a, ~

b) (a, b), whose inverse is also smooth, such that = . Let

: (a, b) (~

a, ~

b) be the inverse of . Then (t) = t for all t (a, b). Therefore

= 1

and hence

t) = 0 for any ~t (~a, ~b). Since = , we have

. Since is regular

= 0. Since

is never vanishing, it follows that

is never zero (vector), i.e., is regular.

The following is a characterization of regular curves.

Proposition 1.2.11. Let : (a, b) R

be a parametrized curve. Then is regular if and only if

it has a unit-speed reparametrization.

. Let : (a, b) R

be a unit-speed reparametrization of the curve . Since is a reparametriza-

tion of , is a reparametrization of . Since is regular (as it is unit-speed) and any reparametrization

of a regular curve is regular, it follows that is regular.

Conversely, assume that be regular. Let s be the arc-length of , starting at the point (t

), i.e.,

s(t) =

(u) du. We note that s is a smooth map. Since is regular,

(t) =

(t) > 0 for all

t, and hence s is strictly increasing and so it is one-one. Let (a, b) be the range of s. Since

(t) = 0

for all t, it follows from the Inverse Function Theorem that s

-1

: (~

a, ~

b) (a, b) is smooth. Consider

the reparametrization : (~

a, ~

b) R

given by = s

-1

, i.e., s = . Now

. Then

, i.e.,

= 1. Hence has a unit-speed reparametrization.

Corollary 1.2.12. Let be a regular curve, and let be a unit-speed reparametrization of given

by u = , where u is a smooth map. Let s be the arc-length of starting at any point of .

Then u = ±s + c for some constant c. Conversely, if u = ±s + c for some constant c, then is a

unit-speed reparametrization of .

. Since u = , we have

. This implies

. Therefore

as is unit-speed. Since both

and

are strictly monotonic continuous functions, it follows that

(t) =

(t) for all t or

(t) = -

(t) for all t. Hence u = ±s + c for some constant c.

1. CURVE THEORY

Conversely, assume first that u = -s + c. Clearly u is bijective, smooth and its inverse is also

smooth. We have (-s+c) = . Hence -

. This implies

. Therefore

= 1, i.e., is unit-speed reparametrization of . Similarly, if u = s + c, then is unit-speed.

Example 1.2.13.

(i) Let : (a, b) R

be a regular curve which does not pass through the origin. If (t

) is the

point on the trace of closest to the origin and

) = 0 for t

(a, b), then show that (t

)

is perpendicular to

Define f : (a, b) R by f (t) = (t)

= (t)(t). Then f is a smooth map. Since (t

) is

point closest to the origin, f has its minimum at t

. Therefore

f (t

) = 0, i.e., (t

) (t

) = 0

(ii) Let : (a, b) R

be a regular curve and v R

be a fixed vector. Assume that

(t) is

perpendicular v for all t and (0) is perpendicular to v. Prove that (t) is perpendicular to v

for all t.

Here

(t)v = 0 for all t (a, b). Integrating we get (t)v = c for all t and for some constant

c R. Since (0)v = 0, we have c = 0. Thus (t)v = 0 for all t.

(iii) Let : (a, b) R

be a regular curve. Show that (t) is a nonzero constant if and only if

(t) is perpendicular to (t) for all t.

Assume that

(t)

= c for all t. Then (t)(t) = c

for all t. Differentiating it we get

(t)(t) = 0 for all t. Conversely, assume that (t)(t) = 0 for all t. Integrating we get

(t)

= c for all t for some constant c. Since is regular, it follows that c > 0. Hence

(t) =

c for all t.

(iv) Let : I R

be a regular curve, and I is an interval. Let [a, b] I. Let (a) = p and

(b) = q.

(a) If v R

with v = 1, then

(q - p) · v =

(t).v dt

(t) dt.

(b) Set v =

q - p

and show that

(b) - (a)

(t) dt.

Deduce that the curve of the shortest length from (a) to (b) is the straight line joining

these points.

It follows from Fundamental Theorem of Calculus that

(t)vdt = ((b) - (a))v. Also

(t)vdt

(t) v dt =

(t) dt. Thus

(t)vdt

(t) dt.

3. Curvature and Torsion

Let v =

q-p

q - p . Then it follows from above inequality that

q - p

(b) - (a)

(t) dt. Since

(t) dt is the arc length of the curve between points (a) and (b), it

follows that among all curves passing through p and q, the line segment is of shortest length.

3. Curvature and Torsion

Definition 1.3.1. Let : (a, b) R

be a unit-speed curve. Then the curvature of (t) of at the

point (t) is defined by (t) = ¨

(t) .

Example 1.3.2.

(i) Let a, b R

, and let a

= 1. Then the curvature of the unit-speed curve (t) = ta + b is

zero everywhere.

(ii) Let (x

, y

) R

, and let R > 0. Then the curvature of the unit-speed curve (t) = (x

R cos(

), y

+ R sin(

)), which is a circle with center (x

, y

) and radius R, is

everywhere.

(iii) Let a, b R, and let a

+ b

= 0. Let (t) = (a cos(

), a sin(

). Then

easy computations show that (t) =

|a|

for all t.

Note that all the three curves above have constant curvatures.

If is a unit-speed curve, then its curvature is given by (t) =

(t) for all t. The following

gives a formula to compute the curvature of a regular curve which is not necessarily unit-speed.

Proposition 1.3.3. If is a regular curve in R

, then its curvature is given by the formula

. Let be a unit-speed reparametrization of given by s = , where s is the arc-length of

starting at any point of . We note that the curvature of at (s(t)) is same as the curvature of

at (t). We have

= ¨

, and hence

= ¨

. Let denote the derivative of

with respect to s. Since s = ,

= , i.e., =

. Differentiating it again, we have

( )¨

- ( ¨

)

× (¨ × )

Hence =

× (¨ × )

(¨ × )

. The last step follows because

× .

Example 1.3.4.

1. CURVE THEORY

(i) Let : (a, b) R

be a regular parametrized curve. Assume that there exists t

(a, b)

such that the distance |(t)| from the origin to the trace of is maximum at t

. We prove that

|(t

)| >

) .

We may assume that is unit-speed. Define f : (a, b) R by f (t) =

(t)

= (t)(t).

Then f is a smooth map. Since (t

) is farthest from origin, the function f has maximum at

. Thus f (t

) = 0 and f (t

) 0. Since f (t

) 0, we have (t

)¨

) + (t

)(t

) 0.

As is unit-speed, (t

)¨

) -1. This gives |(t

)¨

)| 1. Now

1 |(t

)¨

)| (t

)

) = (t

) (t

Thus |(t

)| >

) .

(ii) We compute the curvature of the curve (t) = (a cos t, a sin t, bt), where a, b > 0.

We have

(t) = (-a sin t, a cos t, b), ¨

(t) = (-a cos t, -a sin t, 0). Therefore (t) =

+ b

and ¨

(t)× (t) = (-ab sin t, ab cos t, -a

). Therefore ¨

(t) × (t) = a

+ a

Thus (t) =

)

Exercise 1.3.5.

(i) Compute the curvature of the following curves.

(a) (t) = (e

cos t, e

sin t).

(b) (t) = (t, cosh t). Also determine a point having maximum curvature.

, t

(ii) Show that, if the curvature of a regular curve is positive everywhere, then is a smooth

function. Give an example to show that this may not be the case without the assumption that

> 0.

Now we define a special type of curvature for plane curves only.

Definition 1.3.6. Let : (a, b) R

be a unit-speed curve. Let t be the unit tangent of , i.e., t =

The unit vector n

obtained by rotating the tangent vector t anti-clockwise by an angle

is called

the signed unit normal of . Since is unit-speed, ¨

is perpendicular to t and hence ¨

and n

are

linearly dependent. Therefore there exists a map

: (a, b) R such that ¨

. The map

is called the signed curvature of .

Since =

, it follows that = |

|, i.e., the absolute value of the signed curvature is the

curvature

Let R

: R

be a rotation operator (it rotates a vector anti-clockwise by an angle ). We

note that R

is linear and R

(1, 0) = (cos , sin ) and R

(0, 1) = (- sin , cos ). Let (x, y) R

Then

(x, y) = R

(x(1, 0) + y(0, 1))

3. Curvature and Torsion

= xR

(1, 0) + yR

(0, 1)

= x(cos , sin ) + y(- sin , cos ).

Hence

(x, y) = (x cos - y sin , x sin + y cos ).

In fact, R

is an isometry of R

, i.e., it satisfies

, y

) - R

, y

) = (x

, y

) - (x

, y

)

((x

, y

), (x

, y

) R

In particular, when =

, R

(x, y) = (-y, x).

Thus if : (a, b) R

is a unit-speed curve, then the signed unit normal of can be computed

by the formula n

(t) = R

(t(t)) for all t.

Fix a vector a R

. Define T

: R

by T

(x) = x + a. Then T

is called a translation

by a vector a (it translates given vector by a) and it is an isometry.

Definition 1.3.7. An isometry of R

of the form T

is called a direct isometry, where a R

and [0, 2).

Example 1.3.8. Compute the signed unit normal of the curve (t) = (e

cos t, e

sin t).

The curve is not unit-speed. The unit tangent will be the unit vector in the direction of the tangent

vector, i.e., t(t) =

(t)

(t) . Now

(t) = (-e

sin t + ke

cos t, e

cos t + ke

sin t) and (t) =

1 + k

. Hence t(t) =

1+k

(- sin t + k cos t, cos t + k sin t). Since n

(t) = R

(t(t)), we

have

(t) =

1 + k

(- cos t - k sin t, - sin t + k cos t)

Exercise 1.3.9. If : (a, b) R

is a unit-speed curve, then show that both signed unit normal and

signed curvature are smooth maps.

Proposition 1.3.10. Let : (a, b) R

be a unit-speed curve, and let s

(a, b). Let

R be

such that (s

) = (cos

, sin

)

. Then there exists a unique smooth map : (a, b) R such

that (s) = (cos (s), sin (s)) for all s (a, b) and (s

) =

. Since

: (a, b) R

is a smooth map, there are smooth maps f, g : (a, b) R such

that

(s) = (f (s), g(s)) (s (a, b)). As (s

) = (cos

, sin

), we have f (s

) = cos

and

g(s

) = sin

. Since is unit-speed, f

+ g

= 1 and hence f

f + g g = 0. Define : (a, b) R

(s) =

(f g -

f g).

1. CURVE THEORY

Since f and g are smooth, is a smooth map and (s

) =

Let F = f cos + g sin and G = f sin - g cos . Then

f cos - f sin

+ g sin + g cos

= (

f + g

) cos + ( g - f

) sin .

Now,

f + g

f + g(f g -

f g) =

f (1 - g

) + f g g =

f f

+ f g g = f (f

f + g g) = 0.

By similar arguments it follows that

g - f

= 0. Hence

F = 0, i.e., F is a constant map. Therefore

F (s) = F (s

) for all s. Now

F (s

) = f (s

) cos (s

) + g(s

) sin (s

) = cos

+ sin

= 1.

This shows that F = 1. Similarly, we get G = 0. Hence f cos +g sin = 1 and f sin -g cos -

0. Solving these equations we get f = cos and g = sin . Therefore (s) = (cos (s), sin (s))

for all s (a, b) and (s

) =

For the uniqueness, let : (a, b) R be a smooth map satisfying (s

) =

and

(s) =

(cos (s), sin (s)) for all s. Then there is a map n : (a, b) Z such that (s) - (s) = 2n(s)

for all s. Since both and are smooth, n is smooth and hence it is continuous. Since (a, b) is

connected, n is continuous and non-empty connected subsets of Z are singletons only, it follows that

n is a constant map, say, c. Therefore (s) - (s) = 2c for all s. Since (s

) = (s

) =

, we

get c = 0. Hence (s) = (s) for all s.

Definition 1.3.11. Let : (a, b) R

be a unit-speed curve. Then there is a smooth map :

(a, b) R such that (s) = (cos (s), sin (s)) for all s, which is determined uniquely by the

condition (s

) =

. The map is called the turning angle of .

Proposition 1.3.12. Let : (a, b) R

be a unit-speed curve, and let be the turning angle of .

Then the signed curvature of can be determined by the formula

. Since is the turning angle of , we have

(s) = (cos (s), sin (s)) for all s. Hence

= (- sin , cos )

. Now n

= R

( ) = (- sin , cos ). Therefore ¨

and hence the

signed curvature

of is

, i.e.,

It follows from the Proposition 1.3.12 that the signed curvature is the rate of rotation of the

tangent vector.

Example 1.3.13. If : (a, b) R

is a unit-speed curve, then its signed curvature and signed unit

normal are smooth maps.

The map n

: (a, b) R

is given by n

(t) = R

(t(t)). The map R

: R

defined

by R

(x, y) = (-y, x) is a smooth map and the map t = is a smooth map and hence the map

t = n

is a smooth map, i.e., the signed unit normal is a smooth map.

3. Curvature and Torsion

Now the map

: (a, b) R is given by ¨

(t) =

(t)n

(t). Therefore

(t) = ¨

(t)n

(t).

Since both ¨

and n

are smooth maps, it follows that

is a smooth map.

We saw that if is unit-speed plane curve, then its signed curvature is a smooth map. If k :

(a, b) R is a smooth map, does there exist a unit-speed plane curve whose signed curvature is k?

If yes, will it be unique? The following answers these questions.

Theorem 1.3.14. If k : (a, b) R is a smooth map, then there is a unit-speed curve : (a, b) R

whose signed curvature is k. Moreover, if : (a, b) R

is a unit-speed curve whose signed

curvature is k, then there is a direct isometry M of R

such that = M .

. Let s

(a, b). Define : (a, b) R by (s) =

k(u)du. Since k is smooth, is a

smooth map. Note that (s

) = 0. Define : (a, b) R

(s) =

cos (u)du,

sin (u)du

Then

(s) = (cos (s), sin (s)) for all s. Therefore is unit-speed curve. It follows from the

Proposition 1.3.10 that is the turning angle of satisfying (s

) = 0 and hence the signed curvature

of is

. But

= k. Hence the signed curvature of is k.

Let : (a, b) R

be a unit-speed curve with signed curvature k. Let be the turning angle

of . Then

= k. But then (s) =

k(u)du + (s

) = (s) + , where = (s

) is a constant.

So we have = + . Since is the turning angle of ,

(s) = (cos (s), sin (s)) for all s. Let

a = (s

). Integrating the above equation, we have

(s) =

cos du,

sin du

+ a = T

cos du,

sin du

= T

cos( + )du,

sin( + )du

= T

(cos cos - sin sin )du,

(sin sin + cos cos )du

= T

cos

cos du - sin

sin du, cos

cos du + sin

sin du

= T

cos du,

sin du

= T

((s)).

1. CURVE THEORY

Therefore (s) = M ((s)) for all s, where M = T

is a direct isometry. Hence = M .

Exercise 1.3.15.

(i) If is a unit-speed plane curve, then show that

= -

(ii) Let and be two plane curves. Show that, if is obtained from by applying an isometry M

of R

, the signed curvatures

and

of and are equal if M is direct but that

= -

M is opposite (in particular, and have the same curvature). Show, conversely, that if and

have the same nowhere-vanishing curvature, then can be obtained from by applying an

isometry of R

Example 1.3.16.

(i) Let be a regular plane curve and let be a constant. The parallel curve

of is defined

(t) = (t) + n

(t). If

(t) = 1 for all values of t, then we prove that

is a regular

curve and that its signed curvature is

/|1 -

Here

- 1 = 0. Since

- 1 is a continuous, it follows that

(t) - 1 > 0 for all t or

(t) - 1 < 0 for all t.

Assume that

(t) - 1 < 0 for all t. Let =

. Let us denote the derivative of with

respect to its arc-length parameter s by (dash). Here (t) = (t) + n

(t). Differentiating it

with respect to s we get = (t +

)

= (t -

= (1 -

. As

= 1, we

have

|1-

because 1 -

> 0. Thus = t. Now =

= ¨

. Let n

be the signed unit normal of . As unit-tangents to and are same, they have

the same signed unit normal, i.e., n

= n

. Hence =

. This shows that the signed

curvature of is

Similarly, if

(t) - 1 > 0 for all t, then the signed curvature of will be

-1

. Hence the

result.

(ii) Let : (a, b) R

be a regular curve with constant signed curvature. Show that is (part of)

a circle.

We may assume that is unit-speed. Let k be the curvature of . Let k < 0. Then k = -

for some

> 0. Let : (a, b) R

be defined by (t) = (

cos( t), -

sin( t)). Then

is a unit-speed and t(t) = (t) = (- sin( t), - cos( t)), ¨

(t) = (- cos( t), sin( t)).

Hence n

(t) = (cos( t), - sin( t)). Therefore ¨

(t) = - n

(t). This implies that the signed

curvature of is - = k. Since and have the same signed curvature, there exists a direct

isometry M of R

such that = M . Since N is (part of) a circle for any direct isometry

N of R

, it follows that is also a circle.

Similarly one can show that is a circle when k > 0.

Example 1.3.17.

(i) Compute the signed curvature of (t) = (t, cosh t).

3. Curvature and Torsion

We have

(t) = (1, sinh t). Let (t) be the angle between positive x-axis and (t). Then

cos (t) =

(1, 0)(1, sinh t)

1 + sinh

cosh t

Therefore sin (t) = tanh t and tan (t) = sinh t. Let s be the arc-length of , i.e., s(t) =

(u) du = sinh t. Differentiating tan (t) = sinh t = s with respect to s we have

sec

= 1, i.e, (1 + tan

)

= 1. Therefore

1+tan

1+s

1+sinh

cosh

Hence

(t) =

cosh

(ii) Give an example of a plane curve having signed curvature

1+s

It is

(s) =

cos(tan

-1

u)du,

sin(tan

-1

u)du

See the proof of Theorem 1.3.14.

Definition 1.3.18. Let : (a, b) R

be a unit-speed curve with nowhere vanishing curvature.

Then the function n =

is called the unit normal or principal normal of .

It is clear that n(t) = 1 for all t. Since t(t) = 1 for all t,

t t = 0, i.e., ¨ t = 0. This means

t n. The map b defined by b = t × n is called the unit binormal of .

Thus at each point (t) of a unit-speed curve , we have a right handed system of orthonormal

basis {t(t), n(t), b(t)}, i.e, t(t) × n(t) = b(t), n(t) × b(t) = t(t), b(t) × t(t) = n(t), t(t) t(t) =

n(t) n(t) = b(t) b(t) = 1.

Since b(t) b(t) = 1 for all t,

b b = 0, i.e,

b b. Also, b = t×n. Therefore

b = t×n+t×

n =

(n) × n + t ×

n = t ×

n. This implies that

b t. Hence

b and n are linearly dependent. Therefore

there is a map such that

b = - n. The map is called the torsion of .

Since n = b × t, we have

n =

b × t + b × t = - (n × t) + (b × n) = b - t.

Proposition 1.3.19. Let : (a, b) R

be a regular curve with nowhere vanishing curvature.

Then the torsion of is given by

( × ¨

)

...

× ¨

. First assume that is unit-speed. Then

= t, ¨

= n and

...

= n +

n = n + ( b - t). Now

( × ¨

)

...

= (t × n)( n + ( b - t)) = b( n + ( b - t)) =

and

× ¨

= b

. Hence

(

×¨

)

...

× ¨

= .

Let be a regular curve with nowhere vanishing curvature (not necessarily unit-speed). Let

be its unit-speed reparametrization given by s = , where s is the arc-length of starting

1. CURVE THEORY

at any point of . Note that the torsion of at (s(t)) is same as the torsion of at the point

(t). Since s = , we have

= , i.e,. =

. Therefore (

)

= ¨

and

(

)

+ 2

...

. It follows that × ¨

= ( × )(

)

and (

× ¨

)

...

( × ) (

)

. Also

× ¨

= ×

(

)

. Hence

(

×¨

)

...

× ¨

( × )

= , as is

a unit-speed curve.

Exercise 1.3.20.

(i) Prove that the curvature of the plane curve y = f (x) is given by =

|f |

(1+f

)

(ii) Compute curvature and torsion of the following curves.

(a) (t) = (t, t

, t

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

cos t, e

sin t, e

)

(d) (t) = (

cos t, 1 - sin t,

cos t)

(e) (t) = (

(1 + t)

3/2

(1 - t)

3/2

)

(f) (t) = (

1 + t

, ln(t +

1 + t

))

Example 1.3.21.

(i) If : (a, b) R

is a unit-speed curve with nowhere vanishing curvature, then the unit normal,

unit binormal, curvature and torsion are smooth maps.

We know that n, b : (a, b) R

and , : (a, b) R. By definition (t) =

(t) . Let

= (

). Since is nowhere vanishing,

(t) +

(t) > 0 for all t. Define

f : (0, ) R by f (t) =

t and g : (a, b) R by g(t) =

(t) +

(t). Then

both f and g are smooth maps. Since g(t) > 0 for all t, f g is a smooth map. But f g =

and hence is a smooth map.

Since is a smooth map and (t) > 0 for all t,

is a smooth map. Therefore

= n is a

smooth map.

By definition b = t × n and both t and n are smooth maps Therefore b is a smooth map.

Now

b = - n. It means = -

bn. Since both b (so is

b) and n are smooth maps, is a

smooth map.

(ii) Let be a regular curve in R

with nowhere vanishing curvature. Then is planar if and only

if the torsion of is identically zero.

We may assume that is unit-speed. Suppose that is planar. Then there exist a unit vector

a R

and a constant d R such that a = d. Differentiating, we get ta = 0. Differentiating

it again, we have

ta = 0, i.e., (n a) = 0. This gives n a = 0, as is nowhere vanishing.

Therefore a t and a n. It means b = ±a. Since b is continuous, it follows that either

b(t) = a for all t or b(t) = -a for all t. So, in either case,

b = 0. Since

b = - n, it follows

that = 0.

3. Curvature and Torsion

Conversely, assume that = 0. Since

b = - n, we have

b = 0, i.e, b is a constant unit vector.

Now

( b) = b = t b = 0. This means b is a constant function, say, d, i.e., b = d.

Hence is lying on the plane R b = d, i.e., is planar.

(iii) Let be a regular curve in R

with constant curvature and zero torsion. Then is (part of) a

line or (part of) a circle.

Assume that the curvature = 0. Since ¨

= n, ¨

= 0. Therefore is (part of) a line.

Let > 0. Note that = 0. We see that

( +

n) = t +

n = t +

( b - t) = 0.

Therefore +

n = a for some constant vector a R

. But then - a =

, i.e., is lying

on the sphere R - a =

. Since = 0, is lying on some plane. Therefore is lying on a

sphere as well as a plane and hence it is (part of) a circle.

(iv) Let be a regular curve in R

with nowhere vanishing curvature. Show that is spherical if

and only if

We may assume that is unit-speed. Assume that is spherical. Then there exist a constant

vector a R

and a constant R R such that ( -a)( -a) = - a

= R

. Differentiating

it we get ( - a)t = 0. Differentiating it again we have ( - a)n = -

. Again differentiation

gives ( - a)b =

. Since {t, n, b} is an orthonormal basis of R

, - a = (( - a)t)t +

(( - a)n)n + (( - a)b)b. Therefore

- a = -

n +

But then R

= - a

. Differentiating it we obtain

)

= 0, i.e.,

-2

= 0, i.e.,

Conversely, assume that

. Let a = +

n -

b. Then

a = t -

n +

n -

b -

= t -

n +

( b - t) -

b -

(- )n

b = 0.

Hence a is a constant vector. Now

- a

. Again using

, it

follows that

is a positive constant, say, R

. Therefore - a

= R

, and hence

is spherical.

Let =

and =

. Assume that is spherical. Then there exists a constant vector a R

and a constant R > 0 such that ( - a)( - a) =

- a

= R

. Differentiating it we get

( - a)t = 0. Differentiating it again we have ( - a)n = -

= -. One more differentiation

1. CURVE THEORY

gives ( - a)b = -

. Now

- a = (( - a)t)t + (( - a)n)n + (( - a)b)b

= -n -

Therefore R

- a

- n - b

+ (

)

. Differentiation of the last

equation gives

+ (

)

= 0, which is required.

Conversely, assume that

+ (

)

= 0. Then

+ (

)

= d for some constant d > 0(???).

Let a = + n +

b. Then a = t + n + ( b - t) + ( )

b -

n = (

+ (

)

)b = 0.

Therefore a is a constant vector. Now - a

- n - b

+ (

)

= d. Hence

is spherical.

(v) Let be a unit-speed curve in R

with nowhere vanishing curvature. Show that the curve = t

is a regular curve and find the curvature and torsion of .

Here = t. Therefore

= t = ¨

. Since the curvature = ¨

> 0, the curve is regular.

Now

= t = n, ¨

= n +

n = n + ( b - t) and

...

= ¨

n + ( b - t) + b +

b -

n - 2 t -

n. Now × ¨

t +

b, ( × ¨

)

...

= -

, =

and

× ¨

. Let

be the curvature of and

be the torsion of . Then

× ¨

and

(

× ¨

)

...

× ¨

( - )

(vi) Let be a unit-speed curve in R

such that all the tangent to pass through a fixed point. We

show that is part of a line.

The equation of tangent to at a point (t) is given by R = (t) + ut(t), u R. Since

every tangent to pass through a fixed point, say, c, for every t there is u(t) R such that

c = (t) + ut(t). This implies that u(t) = (c - (t))t(t) and hence u is a smooth map.

Differentiating c = (t) + ut(t), we get 0 =

+ ut + ut = t + ut + un = (1 + u)t + un.

Since t and n are linearly independent,

u + 1 = 0 and u = 0. Thus u(t) = t + c for some

constant c. Since u = 0 and is continuous, we have = 0, i.e., ¨

= 0. Integrating twice

= 0, we have (t) = at + b for some constant vectors a and b in R

. Since is regular,

a = 0. Thus is part of a line.

(vii) Let be a unit-speed curve in R

with nonzero curvature. The normal line to at (t) is the

line through (s) with direction vector n(t). Suppose all the normal lines to pass through a

fixed point. We show that is part of a circle.

A normal at a point (t) is given by R = (t) + un(t), u R. Assume that all the normal lines

pass through a point c. Then for each t, there is u(t) R such that c = (t)+u(t)n(t). But then

4. Fundamental Theorem of Space Curves

u(t) = (c-(t))n(t). This implies that u is a smooth map. Differentiating c = (t)+u(t)n(t),

we get 0 =

+ un + u

n = t + un + u( b - t) = (1 - u)t + un + u b. Since t, n and b

are linearly independent, we get 1 - u = 0,

u = 0 and u = 0. Since u = 0, u is a constant

map. Since is regular (as it is unit-speed), u = a for some nonzero a R. As u = 0 and u

is a nonzero constant map, is identically 0. Therefore is a plane curve. Now the equation

1 - u = 0 implies that =

. Thus the curvature of is constant. Since is a plane curve

with (nonzero) constant curvature, it follows that is part of a circle.

Theorem 1.3.22 (Frenet-Serret). Let be a unit-speed curve in R

. Let t, n and b be its unit

tangent, unit normal and unit binormal respectively. Let and be the curvature and torsion of

respectively. Then

t =

n = -t

b =

- n

The equations in Theorem 1.3.22 are called Frenet-Serret equations or Serret-Frenet equations.

Example 1.3.23. Verify the Serret-Frenet equations for the curve

(a) =

a cos

+ b

, a sin

+ b

4. Fundamental Theorem of Space Curves

We have seen that if : (a, b) R

is a unit-speed curve with nowhere vanishing curvature,

then its curvature and torsion are smooth maps. If k, t : (a, b) R are smooth maps with k(x) > 0

for all x, does there exist a unit-speed curve : (a, b) R

whose curvature is k and torsion is t? If

yes, is the curve unique? The following answers these.

Theorem 1.4.1. (Existence) If k, t : (, ) R are smooth maps with (s) > 0 for all s, then

there exists a unit-speed curve : (, ) R

whose curvature is k and torsion is t.

. Fix s

(, ). It follows from the theory of Differential Equations that the equations

T = kN,

N = -kT + tB and

B = -tN have a unique solution T, N and B such that

{T(s

), N(s

), B(s

)} is a standard basis of R

. Since the matrix

-k

-t 0

expressing

N and

B in terms of T, N and B is skew-symmetric, it follows that T, N and B are orthogonal

unit vectors. Since B is a smooth map orthogonal to both T and N, there exists a smooth map

: (, ) R such that B = T × N. Note that (s) is 1 or -1. Since is continuous, either

1. CURVE THEORY

(s) = 1 for all s or (s) = -1 for all s. Since B(s

) = T(s

) × N(s

), we have (s) = 1 for all

s, i.e., B = T × N. Define : (, ) R

(s) =

T(u)du.

Then

= T. Therefore is a unit-speed curve and T is the unit tangent of . Now ¨

T = kN.

Therefore

= k, i.e., k is the curvature of . It follows from ¨

T = kN that N is the unit

normal of . Since B = T × N, B is the unit binormal of . It follows from the equation

B = -tN

that t is the torsion of .

Theorem 1.4.2. (Uniqueness) If , : (, ) R

are unit-speed curves with same curvature and

same torsion, then there is an isometry M of R

such that = M .

. Let t, n and b be the unit tangent, unit normal and unit binormal of , and let t, n and b

be those of . Fix s

(, ). Since {t(s

), n(s

), b(s

)} and {t(s

), n(s

), b(s

)} are both right

handed orthonormal basis of R

, there is a rotation about origin that maps t(s

), n(s

) and b(s

)

to t(s

), n(s

) and b(s

) respectively. Further, there is a translation which takes (s

) to (s

)

(and this has no effect on t, n and b). By applying rotation followed by translation, we can therefore

assume that (s

) = (s

), t(s

) = t(s

), n(s

) = n(s

) and b(s

) = b(s

). Hence to prove the

theorem we need to prove = . Define A : (, ) R by

A(s) = t(s)t(s) + n(s)n(s) + b(s)b(s).

Then A is a smooth map. Note that A(s) 3 for all s and A(s

) = 3. Also observe that

A(s) = 0

for all s (you need to verify this). Therefore A is a constant map. Since A(s

) = 3, we have A(s) = 3

for all s. Hence it follows that t = t, n = n and b = b. The equation t = t implies that = + c.

Since (s

) = (s

), we get c = 0. Hence = . This proves the theorem.

Example 1.4.3. Describe all curves in R

which have constant curvature > 0 and constant torsion

Let : (a, b) R

be a unit-speed curve with constant curvature > 0 and constant torsion

. Let a =

and b =

. Define : (a, b) R

(t) =

a cos

+ b

, a sin

+ b

, b

+ b

Then is a unit-speed curve. The curvature of is

= and its torsion is

= . Therefore

, : (a, b) R

are unit-speed curves with same curvature and same torsion. So, there is an

isometry M of R

such that = M . But is a helix and hence M is a helix. This implies

that is a helix. Hence any curve with constant (non-zero) curvature and constant torsion is a helix.

5. Isoperimetric Inequality and Four Vertex Theorem

Exercise 1.4.4.

(i) A regular curve in R

with a positive curvature is called a generalized helix if its tangent

vector makes a fixed angle with a fixed unit vector a. Let be a regular curve with positive

curvature everywhere. Show that is a generalized helix if and only if

is constant

(ii) Let P be an n × n orthogonal matrix and let a R

, so that M (v) = P v + a is an isometry of

. Show that, if is a unit-speed curve in R

, the curve = M () is also unit-speed. Show

also that, if t, n, b and T, N, B are the tangent vector, principal normal and binormal of and

, respectively, then T = P t, N = P n and B = P b.

5. Isoperimetric Inequality and Four Vertex Theorem

Definition 1.5.1. Let a R be a positive constant. A simple closed curve in R

is a (regular) curve

: R R

such that (t) = (t ) if and only if t - t = ka for some k Z.

Thus, the point (t) returns to its starting point when t is increased by a, but not before that.

By Jordan Curve Theorem any simple closed curve in R

has an `interior' and `exterior'. The

following defines interior and exterior of a simple closed curve. The set of points in R

that are not

on the curve is the disjoint union of two subsets of R

, denoted by int() and ext(), with the

following properties.

(i) int() is bounded,

(ii) ext() is unbounded,

(iii) both int() and ext() are connected, but any curve joining a point of int() to a point of ext()

crosses .

Proposition 1.5.2. Let be a simple closed curve with period a. Then its unit-speed reparametriza-

tion is ()- periodic, where () is the length of .

. Since is a- periodic, the length, (), of is

() =

(u) du.

Let be a unit-speed reparametrization of given by s = , where s is the arc-length of starting

at the point (0). Let t, t R, and let t t . Then there is k N {0} such that t - t = ka + ,

where 0

< a. Now

s(t ) - s(t) =

(u) du -

(u) du =

(u) du

1. CURVE THEORY

t+a

(u) du +

t+2a

t+a

(u) du + · · · +

t+ka

t+(k-1)a

(u) du +

t+ka+

t+ka

(u) du

= k () +

(u) du.

Hence s(t ) - s(t) = k () if and only if

(u) du = 0 if and only if = 0 if and only if

t - t = ka. Now (s(t)) = (s(t )) if and only if (t) = (t ) if and only if t - t = ka for some

k Z if and only if s(t ) - s(t) = k (). Therefore is ()- periodic.

Let be a simple closed curve in R

. Let A(int()) be the area of the interior of . Then

A(int()) =

int()

dxdy.

Definition 1.5.3. A simple closed curve in R

is called positively oriented if its unit signed normal

points towards the interior of the curve at ach point of the curve.

Theorem 1.5.4. (Green's Theorem) Let be a simple closed curve in R

, and let be positively

oriented. If f and g are continuous on the closure of int() and smooth on int(), then

int()

dxdy =

[f (x, y)dx + g(x, y)dy].

Proposition 1.5.5. Let (t) = (x(t), y(t)) be a simple closed curve in R

with period a. Let be

positively oriented. Then

A(int()) =

y - y x)dt.

. Take f (x, y) = -

y and g(x, y) =

x. Then both f and g are smooth functions on R

. It

follows from Green's theorem that

int()

(

- (-

)dxdy =

[

xdy -

ydx]. This implies

A(int()) =

[

xdy -

ydx] =

y - y x)dt.

Example 1.5.6. If f : [a, b] R is a nonnegative continuous function with

f (x)dx = 0, then

f 0.

If f is not identically zero, then there is x

[a, b] such that f(x

) > 0. Take =

f (x

)

Assume that x

= a. Then there is > 0 such that |f (x) - f (x

)| <

for every x [a, a + ).

This implies that f (x)

f (x

)

for all x [a, a + ). Now 0 =

f (x)dx =

f (x)dx +

f (x)dx

f (x

)

dx =

f (x

)

> 0. This is a contradiction.

5. Isoperimetric Inequality and Four Vertex Theorem

Assume that a < x

< b. By continuity of f , there is > 0 such that |f (x) - f (x

)| <

for every x [x

- , x

+ ). It follows that f (x)

f (x

)

for all x (x

- , x

+ ). Again

0 =

f (x)dx

f (x

)

dx = f (x

) > 0, which is a contradiction.

Similar arguments shows that x

= b is also not possible.

We use this example to prove the following.

Proposition 1.5.7 (Wirtinger's Inequality). Let F : [0, ] R be a smooth map satisfying F (0) =

F () = 0

. Then

and the equality holds if and only if there is a constant A such that F (t) = A sin t for all t.

. Define G : [0, ] R by G(t) =

F (t)

sin t

. Note that G is a smooth map. Then

dt =

(

G sin t + G cos t)

sin

tdt + 2

G sin t cos tdt +

cos

tdt

sin

tdt -

(cos

t - sin

t)dt +

cos

tdt

sin

tdt +

sin

tdt =

dt +

sin

tdt.

Hence

dt -

dt =

sin

tdt 0.

(1.5.7.1)

By equation (1.5.7.1) it follows that

dt =

dt if and only if

sin

tdt = 0 if and only

sin

t = 0 if and only if

G sin t if and only if

G = 0. But

G = 0 implies G(t) = A for all t, for

some constant A, i.e., F (t) = A sin t for all t.

1. CURVE THEORY

Example 1.5.8. The length and the area of interior of a simple closed are invariant under isometry.

Let : R R

be a simple closed curve with period , and let M : R

be an isometry

Then there is an orthogonal matrix A and a R

such that M (x) = A(x) + a (x R

). Also note

that DM (x) = DA(x) = A (x R

) Let : R R

be = M . We need to prove that the

lengths and the area of interiors of and are same. Let t, t R. Then (t ) = (t) if and only if

M ((t )) = M ((t)) if and only if (t ) = (t) if and only if t - t = k for some k Z. Therefore

is -periodic. Now

() =

(u) du =

DM ((u)) du

DA((u)) (u) du

A( (u)) du =

(u) du = ().

Since M is an isometry, there are a, b, c, d, e, f R

such that a

+ b

= c

+ d

= a

+ c

= 1, ac+bd = ab+cd = 0 and M(x, y) = (ax+by+e, cx+dy+f ). Let (t) = (x(t), y(t)).

Then (t) = (ax(t) + by(t) + e, cx(t) + dy(t) + f ). Now

A(int()) =

((ax + by + e)(cx + dy + f )

- (ax + by + e)

(cx + dy + f ))dt

((ax + by + e)(c x + d

y) - (a x + b

y)(cx + dy + f ))dt

(ad - bc)

y - xy)dt + (ec - af )(x( ) - x(0)) + (ed - bf )(y( ) - y(0))

y - xy)dt = A(int()),

as ad - bc = ±1, x(0) = x( ) and y(0) = y( ) (???).

We know that N : R

is an isometry if and only if there are a, b, c, d, e, f R

such that a

+ b

+ d

= a

+ c

= b

+ d

= 1

, ac + bd = ab + cd = 0 and N (x, y) = (ax + by, cx + dy) + (e, f ) OR N is an

isometry of R

if and only if there is a 2 × 2 orthogonal matrix A and a R

such that N (x) = Ax + a

5. Isoperimetric Inequality and Four Vertex Theorem

Theorem 1.5.9 (The Isoperimetric Inequality). Let be a simple closed curve, let

be its length,

and let A be the area of interior of . Then

4A. Moreover,

= 4A

if and only if is a

circle.

. We may assume that is unit-speed. Then the length of is its period, i.e., is - periodic.

Consider the curve (t) = (x(t), y(t)) defined by (t) =

, i.e., = s or s

-1

= ,

where s(t) =

. [Note that is a unit-speed reparametrization of with the parametrization map

-1

(t) =

, i.e., s(t) =

.] Then is a simple closed curve with the same length

and period .

Also the area of interior of is same as the area of interior of . Since the length and the area of a

curve are invariant under isometry, we may assume that (0) = (0) = (0, 0). Let x = r cos and

y = r sin . Note that r, : [0, ] R are smooth maps. Since is - periodic, (0) = (). This

gives r(0) = r() = 0 as (0) = (0, 0). Now

= r

+ r

and x

y - y x = r

. Since

= s, we have =

. But then

+ r

= x

(

)

. We also

note that A = A(int()) =

y - y x)dt =

. Now,

- A =

( r

+ r

)dt -

( r

+ r

- 2r

)dt

(

- 1)

dt +

( r

- r

)dt 0 [

Wirtinger's Inequality].

It follows from the above equation that

-A = 0 if and only if

(

-1)

dt+

( r

-r

)dt = 0

if and only if

(

- 1)

dt =

( r

- r

)dt = 0 if and only if = 1 and r = A sin t for some A if

and only if = t + for some constant and r = A sin t for some A if and only if r = A sin( - ).

But r = A sin( - ) is a circle. Hence

- 4A = 0 if and only if is a circle.

Example 1.5.10. Let a and b be positive reals. By applying the isoperimetric inequality to the ellipse

= 1, we prove that

sin

t + b

cos

t dt 2

ab.

We also show that equality holds if and only if a = b.

1. CURVE THEORY

Consider (t) = (a cos t, b sin t). Then the trace of is the ellipse

= 1. Note the is

2-periodic. Therefore length of is

(t) dt =

sin

t + b

cos

t dt.

Also, the area A of the interior of is

A =

y - y x)dt =

ab(cos t cos t + sin t sin t)dt =

2 = ab.

It follows from the Isoperimetric inequality that

sin

t + b

cos

t dt

4ab = 4

ab,

i.e.,

sin

t + b

cos

t dt 2

ab.

Equality holds in above inequality if and only if is a circle if and only if a = b. [Because

(t) = (a cos t, b sin t) is a circle if and only if a = b.]

Definition 1.5.11. A simple closed curve in R

is called convex if its interior is a convex subset of

Definition 1.5.12. Let be a regular plane curve. A point (t) on is called a vertexif

(t) = 0.

Theorem 1.5.13 (Four Vertex Theorem). Every convex, simple closed curve in R

has at least four

vertices.

Example 1.5.14.

(i) The ellipse

= 1 (or the curve (t) = (a cos t, b sin t)) is a convex curve.

Let (x, y), (z, w) int(), i.e.,

< 1 and

< 1. Let X =

and Y =

Then X

< 1 and Y < 1. Let t [0, 1]. Then tX + (1 - t)Y t X + (1 - t) Y <

t + 1 - t = 1. Now

(tx + (1 - t)z)

(ty + (1 - t)w)

= tX + (1 - t)Y

< 1.

Therefore t(x, y) + (1 - t)(z, w) int() and hence is convex.

5. Isoperimetric Inequality and Four Vertex Theorem

(ii) Find the vertices of the curve (t) = (a cos t, b sin t), where a, b > 0 and a = b.

Let s be the arc-length of starting at some point of . Denoting the derivative of with respect

to s by , we have (t) = (-a sin t, b cos t)

. Since

= 1,

sin

t+b

cos

Therefore

t(t) = (t) =

sin

t + b

cos

(-a sin t, b cos t).

Differentiating with respect to s, we get

(t) =

sin

t + b

cos

(-b cos t, -a sin t).

Now n

(t) = R

(t(t)) =

sin

t+b

cos

(-b cos t, -a sin t). Hence

sin

t + b

cos

This implies that

(t) =

sin

t+b

cos

. Now

(t) =

-3ab(a

-b

) sin t cos t

sin

t+b

cos

. Since a, b > 0

and a = b, we have

(t) = 0 if and only if sin t cos t = 0 if and only if sin 2t = 0 if and

only if t =

for some n Z. Since is 2- periodic, it follows that the vertices of are

(0) = (a, 0), (/2) = (0, b), () = (0, -a) and (3/2) = (0, -b) only.

Exercise 1.5.15. If is a simple closed curve, then show that the length of () and the area of interior

of are unchanged by applying an isometry to .

CHAPTER 2

Surfaces in R

6. Surfaces

Definition 2.6.1. A subset S of R

is called a surface if for every p S there is an open subset W

of R

containing p and an open subset U

of R

such that S W

and U

are homeomorphic.

Thus if S is a surface, then for every p S there is homeomorphism

: U

S W

. These

homeomorphisms are called surface patches or parametrizations of a surface. The collection of all

these surface patches is called an atlas of the surface.

6.1. Examples.

(i) The set S = {(x, y, z) R

: z = x

+ y

} is a surface.

Let W = R

and U = R

. Then W is open in R

and U is open in R

. Consider the map

: U S W defined by (u, v) = (u, v, u

+ v

). Then is a bijection and continuous.

Since

-1

is the restriction on S of the continuous map : R

, (x, y, z) = (x, y), the

map

-1

is continuous. Hence is a homeomorphism between S W and U . This shows that

S is a surface.

In this example same W and same U work for all points of the surface. The above surface is

an elliptical paraboloid.

(ii) The set S = {(x, y, z) R

: z = y

- x

} is a surface.

The proof is same as above by considering (u, v) = (u, v, v

- u

(iii) The set S = {(x, y, z) R

: z

= x

+ y

, z 0} is a surface.

Consider (u, v) = (u, v,

+ v

(iv) The set S = {(x, y, z) R

: x

+ y

+ z

= 1} is a surface.

Let W

= {(x, y, z) R

: x < 0} {(x, y, z) R

: y = 0}, W

= {(x, y, z) R

: x >

0} {(x, y, z) R

: z = 0} and U = (0, 2) × (-

). Then both W

and W

are open

in R

and U is open in R

. For i = 1, 2, define

: U S W

(u, v) = (cos u cos v, cos u sin v, sin v)

and

(u, v) = (- cos u cos v, - sin v, - cos u sin v).

Then both

and

are homeomorphisms. Since

(U )

(U ) = S, it follows that S is a

surface.

2. SURFACES IN R

We now see another parametrization of S. Let U = {(u, v) R

: x

+ y

< 1}. Let

= {(x, y, z) R

: z > 0}, W

= {(x, y, z) R

: z < 0}, W

= {(x, y, z)

: y > 0}, W

= {(x, y, z) R

: y < 0}, W

= {(x, y, z) R

: x > 0} and

= {(x, y, z) R

: x < 0}. Then U is open in R

and each W

is open in R

. For

i = 1, . . . , 6, define

: U S W

(u, v) = (u, v,

1 - u

- v

(u, v) = (u, v, -

1 - u

- v

(u, v) = (u,

1 - u

- v

, v),

(u, v) = (u, -

1 - u

- v

, v),

(u, v) = (

1 - u

- v

, u, v) and

(u, v) = (-

1 - u

- v

, u, v).

Then each of

is a homeomorphism. Since

i=1

(U ) = S, the set S is a surface.

We now give another parametrization of the sphere S called the stereographic projection. Let

= R

and W

= R

\ {(0, 0, z) : z 0}. Then W

is open in R

. Define : U

S W

(u, v) =

+ v

+ 1

+ v

+ 1

+ v

- 1

+ v

+ 1

Then is a homeomorphism. We note that (U

) = S \ {(0, 0, 1)}. Now take U

= R

and

= R

\ {(0, 0, z) : z 0}. Then W

is open in R

. Define : U

S W

(u, v) =

+ v

+ 1

+ v

+ 1

1 - u

- v

+ v

+ 1

Then is a homeomorphism and that (U

) = S \ {(0, 0, -1)}. Thus (U

) (U

) = S.

Hence S is a surface.

(v) The set S = {(x, y, z) R

: x

+ y

= 1} is a surface.

Let W

= {(x, y, z) R

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

: y = 0}, W

= {(x, y, z) R

: x <

0} {(x, y, z) R

: z = 0}, U

= (0, 2) × R and U

= (-, ) × R. Then both W

and

are open in R

and both U

and U

are open in R

. For i = 1, 2, define

: U

S W

(u, v) = (cos u, sin u, v).

Then both

and

are homeomorphisms. Since

)

) = S, the set S is a surface.

Let U = {(u, v) :

+ v

} and W = R

. Then U is open in R

and W is open in

. Define : U S W by

(u, v) =

+ v

, tan(

+ v

- ) .

6. Surfaces

Then is a homeomorphism. Hence S is a surface.

(vi) Let a, b, c, d R, and let a

= 0. Then the set S = {(x, y, z) R

: ax+by+cz = d}

is a surface.

Here S is a plane. We may assume that c = 0. Let U = R

and W = R

. Then U is open in

and W is open in R

. Define : U S W by

(u, v) = (u, v,

(d - au - bv)).

Then is a homeomorphism (what is its inverse?). Since (U ) = S, it follows that S is a

surface.

(vii) The set S = {(x, y, z) : x

+ y

- z

= 1} is a surface.

Let U = {(u, v) :

+ v

} and W = R

. Then U is open in R

and W is open in

. Define : U S W by

(u, v) =

tan

(r - ) + 1

tan

(r - ) + 1

, tan(r - )

where r =

+ v

. Then is a homeomorphism. Hence S is a surface.

(viii) The set S = {(x, y, z) R

: z

= x

+ y

} is not a surface.

Suppose that S is a surface. Then there exists an open subset W of R

containing (0, 0, 0)

such that S W is homeomorphic to some open subset U of R

. We may assume that W is

an open ball around origin (why?). Since S W is connected, U is connected (why?). If we

remove origin from S W , we get two components. While removal of any point from U gives

connected subset. This is not possible (again why?). Hence our assumption that S is surface is

false.

(ix) Let : (a, b) R

be a curve given by (v) = (0, f (v), g(v)), i.e, is a curve in yz-

plane. We also assume that the curve does not intersect the z- axis and the curve does not

intersect itself. The surface obtained by rotating about the z- axis is called the surface of

revolution. Let p = (x, y, z) be a point on this surface S. Then the distance of p from the

z- axis is f (v) for some v (a, b). Let q be the projection of p on the xy- plane. Let u

be the angle between the segment joining q with the origin. Then the coordinates of q are

(f (v) cos u, f (v) sin u, 0). Since the point p is at a distance g(v) from the xy- plane, the

coordinates of p are (f (v) cos u, f (v) sin u, g(v)). Now we prove that S is a surface. Let

: (0, 2) × (a, b) and U

= (-, ) × (a, b), W

= {(x, y, z) R

: x > 0} {(x, y, z)

: y = 0} and W

= {(x, y, z) R

: x < 0} {(x, y, z) R

: z = 0}. Then U

are open in R

and W

are open in R

. For i = 1, 2, define

: U

S W

(u, v) = (f (v) cos u, f (v) sin u, g(v)). Then both

and

are homeomorphisms and

)

) = S. Hence S is a surface.

(x) Let b > a > 0. Let C be the circle in xy-plane with centre (b, 0, 0) and radius a. The surface

obtained by revolving C about z- axis is called the torus. We show that torus is a surface.

2. SURFACES IN R

Let S be the torus. Let U

= (0, 2)×(0, 2), U

= (0, 2)×(-, ), U

= (-, )×(0, 2)

and U

= (-, ) × (-, ). The each U

is open in R

. Let

= R

\ ({(x, y, z) : x

+ y

= (a + b)

} {(x, y, 0) : 0 x a + b}).

= R

\ ({(x, y, z) : x

+ y

= (a - b)

} {(x, y, 0) : 0 x a + b}).

= R

\ ({(x, y, z) : x

+ y

= (a + b)

} {(x, y, 0) : -(a + b) x 0}), and

= R

\ ({(x, y, z) : x

+ y

= (a - b)

} {(x, y, 0) : -(a + b) x 0}). Then each

is open R

. For i = 1, 2, 3, 4, define

: U

S W

(u, v) = ((b + a cos v) cos u, (b + a cos v) sin u, a sin v).

Then each

is a homeomorphism and

i=1

) = S. Therefore S is a surface.

Example 2.6.3. The surface S = {(x, y, z) R

: x

+ y

+ z

= 1} cannot be covered by a single

patch.

Suppose that S is covered by a single patch. Then there is an open subset W of R

and an open

subset U of R

such that S W = S and U are homeomorphic. Clearly, U = . Let : U S

be a homeomorphism. Since S is compact, U is compact. By Heine - Borel Theorem U is closed

and bounded. Since U is open, it is both open and closed. Since U is bounded, U

. This is a

contradiction as nonempty clopen subset of R

is R

only.

In fact, the same proof works for any compact surface.

7. Calculus on Surface

If : U S W R

is a surface patch of a surface S, then we denote the partial derivative

of with respect to the first variable and second variable by

and

respectively.

Definition 2.7.1. A surface patch : U R

is called a regular if is a smooth map and

and

are linearly independent for all (u, v) in U .

A surface S is called smooth if it has an atlas consisting of regular patches.

Exercise 2.7.2. Check that the surfaces in examples (i), (ii), (iv), (v), (vi), (vii) in section 6.1 are smooth

surfaces.

Example 2.7.3.

(i) Show that the unit sphere is a smooth surface.

We know that the unit sphere has an atlas consisting of two surface patches namely

: U

S W

, i = 1, 2,

(u, v) = (cos u cos v, cos u sin v, sin v)

and

(u, v) = (- cos u cos v, - sin v, - cos u sin v),

where W

= {(x, y, z) R

: x < 0} {(x, y, z) R

: y = 0}, W

= {(x, y, z)

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

: z = 0} and U = (0, 2) × (-

). Clearly, both the

7. Calculus on Surface

maps

and

are smooth maps. We show that (

)

and (

)

are linearly independent. We

have (

)

= (- sin u cos v, cos u cos v, 0) and (

)

= (- cos u sin v, - sin u sin v, cos v).

Therefore (

)

× (

)

= (cos u cos

v, sin u cos

v, cos v sin v) gives

(

)

× (

)

= cos

v > 0 as v (-

Therefore

is a regular surface patch. Similarly, it follows that

is a regular surface patch.

(ii) Show that translations and invertible linear transformations of R

take smooth surfaces to

smooth surfaces.

Let S be a smooth surface and let { : U

: } be an atlas of S containing regular

surface patches. Fix a R

. We want to show that S + a = {s + a : s S} is a smooth

surface. Let T

: R

be T

(x) = x + a (x R

). Then T

is a homeomorphism. Let

A = {

= T

: U

: }. Then

) =

)(U

) = T

)

= S + a.

Since U

) and

)

- T

(

)) are homeomorphism, U

- T

(

))

is a homeomorphism. Therefore A is an atlas of S + a. Let

A. Since

and the constant

map a are smooth maps,

is a smooth map. Note that (

)

and (

)

. Since

and

are linearly independent, it follows that (

)

and (

)

are linearly independent.

Hence A is an atlas of S + a having regular patches, i.e., S + a is a smooth surface.

Let T : R

be an invertible linear map. Then A = {T

: U

: } is an

atlas of T (S) consisting of regular patches (the proof is as above).

(iii) Show that every open subset of a smooth surface is a smooth surface.

Let W be an open subset of a smooth surface S. Let {

: U

: } be an

atlas of S such that each

is a regular patch. Let = { :

) W = }.

Let A = {

-1

(W )

: U

-1

(W ) R

: }. We show that A is an

atlas of W and each

in A is a regular patch. Clearly,

-1

(W )) W . Let

q W S =

). Then q

) for some . But then q W

), i.e.,

. It also follows that q

-1

(W )). Therefore W

-1

(W )).

Since

is a restriction of

on an open subset of U

, the map

is a smooth map and

(

)

= (

)

and (

)

= (

)

. Therefore (

)

and (

)

are linearly independent.

Hence W is a smooth surface.

Let S be a smooth surface, and let be a curve lying on a surface patch : U R

of S.

Then has the form (t) = (u(t), v(t)), where u and v are smooth (real) functions.

Definition 2.7.4. Let p be a point on a surface S. A vector w R

is called tangent vector to S at p

if there is a curve in S passing through p whose tangent vector at p is w.

2. SURFACES IN R

The tangent space at a point p of a surface S is the set of tangent vectors at p . We denote the

tangent space of S at p S by T

Proposition 2.7.5. Let : U R

be a patch of a surface S containing a point p of S, i.e.,

p = (u

, v

)

for some (u

, v

) U

. Then the tangent space to S at p is a vector subspace of R

spanned by the vectors

, v

)

and

, v

)

. Let be a curve in S passing through p. Then (t) = (u(t), v(t)) and (t

) = (u(t

), v(t

)) =

, v

) = p for some t

. Then the tangent vector to at p = (t

) will be

) =

(u(t

), v(t

)) u(t

) +

(u(t

), v(t

)) v(t

)

, v

) u(t

) +

, v

) v(t

Hence any tangent vector to S at p is a linear combination of

, v

) and

, v

), i.e., T

span{

, v

)}.

Conversely, let a span{

, v

)}. Then a =

, v

) for some

, R. Consider the curve (t) = (u

+ t, v

+ t). Then is lying on the patch and and

(0) = (u

, v

) = p. Now the tangent vector to at 0 will be (0) =

, v

) = a.

This shows that a is tangent vector to some curve in S passing through p, i.e., a T

S. Therefore

span{

, v

)} T

Definition 2.7.6. Let S be a surface, and let p be a point in the patch : U R

with (u, v) = p.

Then the plane passing through p parallel to the tangent space at p is called the tangent plane to S

at p. The vector

N(u, v) =

(u, v) ×

(u, v)

(u, v) ×

(u, v)

is called the unit normal to S at p.

Definition 2.7.7. Let S

and S

be smooth surfaces with surface patches

: U

and

(i) A map f : S

is called a smooth map if the map

-1

: U

is a smooth

map.

(ii) A map f : S

is called a diffeomorphism if f is bijective and both f and f

-1

are smooth

maps.

(iii) A map f : S

is called a local diffeomorphism if for every p S

there exists an open

subset O of S

containing p such that f

: O f (O) is a diffeomorphism.

Example 2.7.8. Let a, b, c > 0. We construct a diffeomorphism between the ellipsoid

= 1

and the unit sphere x

+ y

+ z

= 1.

Let S

be the ellipsoid and S

be the unit sphere. Define f : S

by f (x, y, z) = (

Clearly, f is bijective. Check that f and f

-1

are smooth maps.

7. Calculus on Surface

Exercise 2.7.9. Let C be a collection of smooth surfaces. Two smooth surfaces in C are said to be

related, S

, if S

and S

are diffeomorphic. Show that the relation on C is an equivalence

relation.

Let f : S S be a smooth map, and let p S. Let w T

S, i.e., w is a tangent vector to S

at p. Then, by definition, there exists a curve in S through p such that (t

) = p and (t

) = w

for some t

. Then = f is a curve in S passing through f (p) ((t

) = f ((t

)) = f (p)). Let

w be the tangent vector to at the point f (p), i.e., w = (t

). (Hence we have a map which maps

w T

S to w T

f (p)

S.) The map D

f : T

S T

f (p)

S given by D

f (w) = w is called the

derivative of f , D

f , at p.

We now show that the definition of D

f (w) depends on f , p and w.

Let : U R

be a surface patch containing p, say, p = (u

, v

), and let , be smooth

functions such that f ((u, v)) = ((u, v), (u, v)). Let w =

+ µ

be a tangent vector at

p of a curve (t) = (u(t), v(t)), where u and v are smooth functions such that u(t

) = and

v(t

) = µ. Since the corresponding curve on S is (t) = ((u(t), v(t)), (u(t), v(t))), we have

f (w) =

( u

+ v

) +

( u

+ v

the derivatives of u and v are evaluated at t

. Thus

f (w) =

(

+ µ

) +

(

+ µ

The righthand side depends only on p, f , µ (hence on w).

Proposition 2.7.10. Let S and S be smooth surfaces. Let f : S S be a smooth map, and let

p S

. Then the derivative D

f : T

S T

f (p)

S is a linear map.

. Let w

+ µ

and w

+ µ

be in T

S, and let R. Then

f (w

+ w

) = D

f ((

)

+ (µ

+ µ

)

= [(

)

+ (µ

+ µ

)

]

+ [(

)

+ (µ

+ µ

)

]

= [(

+ µ

)

+ (

+ µ

)

]

+ µ

)

+ (

+ µ

)

= D

f (w

) + D

f (w

Hence the map D

f is linear.

Note that the matrix of the linear map D

f with respect to the basis {

} of T

S and

{

} of T

f (p)

S is

Proposition 2.7.11. Let S

, S

and S

be smooth surfaces.

(i) If p S

, then the derivative at p of the identity map from S

to itself is the identity map

from T

to itself.

2. SURFACES IN R

(ii) (Chain rule) If f : S

and g : S

are smooth maps, then for all p S

(g f ) = D

f (p)

g D

(iii) If f : S

is a diffeomorphism, then for all p S

, the linear map D

f : T

f (p)

is invertible.

. (i) Here I : S

is the identity map, i.e., I(x) = x for all x S

. Let p be a point

on S

, and let w T

S. Let be a curve in S

passing through p whose tangent vector at p is

w. Since I is the identity map I = . Hence the tangent vector to I at p is w itself. This

shows that D

I(w) = w. Since w is an arbitrary point of T

, then map D

I is the identity map, i.e.,

I(w) = w for all w T

(ii) Let p be a point on S

, and let w T

. Let be a curve in S

passing through p whose

tangent vector at p is w. Let

= f . Then

is a curve in S

passing through f (p). Let w

be the

tangent vector to

at f (p). Then D

f (w) = w

. Let

= g

= g f . Then

is curve in

passing through g(f (p)). Let w

be the tangent vector to

at g(f (p)). Then D

f (p)

g(w

) = w

i.e, D

f (p)

g(D

f (w)) = w

, i.e., (D

f (p)

g D

f )(w) = w

. Also, w

is the tangent vector to (g f )

at g f (p). Therefore D

(g f )(w) = w

. This shows that D

(g f )(w) = (D

f (p)

g D

f )(w).

Since w is an arbitrary point of T

, we get D

(g f ) = D

f (p)

g D

f .

(iii) Since f is a diffeomorphism, the map f

-1

is smooth and the map f

-1

f : S

is the

identity map. Let p S. Then I = D

-1

f )

Proposition 2.7.12. Let S

and S

be smooth surfaces and let f : S

be a smooth map. Then

is a local diffeomorphism if and only if the linear map D

f : T

f (p)

is invertible for

all p S

. Assume that f is a local diffeomorphism. Let p S

. Since f is a local diffeomorphism, there

exists an open subset O of S

containing p such that f

: O f (O) is a diffeomorphism. It follows

from Proposition 2.7.11 (ii) that D

f is invertible.

The converse follows from the Inverse Function Theorem.

Example 2.7.13.

(i) Let f : S

be a local diffeomorphism, and let be a regular curve in

. We show that f is a regular curve in S

We note that (f )

(t) is a tangent vector to the curve f at f (t). Hence by the definition

of the derivative D

(t)

f ( (t)) = (f )

(t) for all t. Since f is a local diffeomorphism, D

is invertible for all t. In particular, D

(t)

f is invertible for all t. Fix t. Since (t) = 0, and the

linear map D

(t)

f is invertible, we have (f )

(t) = D

(t)

f ( (t)) = 0. Since t is arbitrary,

it follows that, (f )

(t) > 0 for all t, i.e., f is regular.

(ii) Let S be the half-cone x

+ y

= z

, z > 0. Define a map f from the half-plane {(0, y, z)|y >

0} to S by f (0, y, z) = (y cos z, y sin z, y). Then f is a local diffeomorphism but not a diffeo-

morphism.

Since f (0, 1, 0) = f (0, 1, 2), the map is not one one and hence not a diffeomorphism. Let

= {(0, y, z)|y > 0} and S

= {(x, y, z) : x

+ y

= z

, z > 0}. Then S

and S

8. First Fundamental Form

are smooth surfaces. Let U

= {(u, v) : u > 0}, U

= R

\{0} and W = R

. Consider

: U

W S

defined by

(u, v) = (0, u, v) and : U

W S

defined by

(u, v) = (u, v,

+ v

). Then

is a surface patch of S

and is a surface patch of S

Also note that (U

) = S

and (U

) = S

. Since f (U

) is a subset of S

= (U

), there ex-

ists smooth maps , : U

R such that f (u, v) = ((u, v), (u, v)), i.e., f(0, u, v) =

(u cos v, u sin v, u) = ((u, v), (u, v),

(u, v)

+ (u, v)

). This gives (u, v) = u cos v

and (u, v) = u sin v. Fix p = ((u

, v

)) S

. Then the matrix of D

f with respect to the

basis {

, v

)} of T

and the basis {

((u

, v

), (u

, v

)),

((u

, v

), (u

, v

))}

, v

)

, v

)

, v

)

, v

)

cos v

-u

sin v

cos v

. Now D

f is invertible if and only if

the above matrix is invertible if and only if u

= 0. But u

> 0. Hence D

f is invertible. Since

p is an arbitrary point of S

, it follows that D

f is invertible for every p S

. Hence f is a

local diffeomorphism.

8. First Fundamental Form

Definition 2.8.1. Let S be a smooth surface, and let p S. Then the first fundamental form of S at

p is a map I

: T

S × T

S R given by

(v, w) = v, w

= vw

(v, w T

S).

Suppose that : U R

be a surface patch of a surface S. Fix p S. Then any tangent

vector to S at p is a linear combination of

and

. Define du, dv : T

S R by du(w) = and

dv(w) = µ if w =

+ µ

. Then both du and dv are linear maps. Now

I(w, w) = w, w =

+ 2µ

+ µ

Let E =

, F =

and G =

. The functions E, F, G : U R

are called the first order magnitudes of the surface S. Hence I(w, w) =

w, w = Edu(w)

2F du(w)dv(w) + Gdv(w)

·, · = Edu

+ 2F dudv + Gdv

(2.8.1.1)

The expression Edu

+ 2F dudv + Gdv

is known as the first fundamental form on the surface patch

. Note that the first fundamental form on the surface does not depend on the choice of surface patch,

but it depends on S and p.

Let : (, ) R be a curve lying on a surface patch (u, v) of a surface S. Then (t) =

(u(t), v(t)). Therefore = u

+ v

and hence

= ( , )

1/2

F u

+ 2F u v + G v

. If

p = (t

) and q = (t

), then the length of the curve (on the surface patch ) is

F u

+ 2F u v + G v

dt.

Thus the first fundamental form is helpful in defining metric on the surface.

Example 2.8.2. Compute the first fundamental form on the cylinder x

+ y

= 1.

2. SURFACES IN R

We know that the surface patch of the above surface is given by (u, v) = (cos u, sin u, v).

Now

= (- sin u, cos u, 0) and

= (0, 0, 1). Therefore E =

= 1, F =

= 0 and

G =

= 1. Hence the first fundamental form on the surface is du

+ dv

Exercise 2.8.3. Compute the first fundamental form on the following surfaces.

(i) (u, v) = (f (u) cos v, f (u) sin v, g(u))

(ii) (u, v) = (cos u cos v, cos u sin v, sin u)

(iii) (u, v) = (sinh u sinh v, sinh u cosh v, sinh u)

(iv) (u, v) = (u - v, u + v, u

+ v

)

(v) (u, v) = (cosh u, sinh u, v)

(vi) (u, v) = (u, v, u

+ v

)

Example 2.8.4. When applying an isometry of R

to a surface, it does not change its first fundamental

form. We shall see the effect of applying dilation on the surface. (Given a = 0, a dilation is a map

from R

to R

given by v av).

Since the first fundamental form of a surface depends only on the surface and a point, we may

consider any parametrization of the surface containing that point. Let : U R

be a surface

patch of a surface S, and let T : R

be a linear isometry. Then : U R

is a surface

patch of the surface T (S). It follows that

= T (

) and

= T (

) (check!!!). Let E, F and G

be the first order magnitudes of S, and let E, F and G be those of T (S). Then E =

T (

), T (

) =

= E and similarly F = F and G = G. Hence the first fundamental

forms of S and T (S) (at corresponding points) are same.

Let : U R

be a surface patch of S. Let T (v) = av be a dilation, where a is a nonzero

real. Then T = : U R

is a surface patch of T (S). We see that = T = a. Let E, F

and G be the first order magnitudes of T (S). Then E =

= a

E. Similarly, we get

F = a

F and G = a

G. Thus the first fundamental form on T (S) is a

times the first fundamental

form on S.

9. Local isometries and conformal maps

Definition 2.9.1. Let S

and S

be smooth surfaces. A smooth map f : S

is called a local

isometry if it takes any curve in S

to a curve of same length in S

A local isometry that is a diffeomorphism is called an isometry.

Thus if f : S

is a local isometry, then for any curve on S

, the curves and f

have the same length.

Example 2.9.2.

(i) Let S

, S

and S

be surfaces. Let f : S

and g : S

be local isometry. Then

g f : S

is a local isometry (i.e., composition of local isometries is a local isometry).

9. Local isometries and conformal maps

Let be a curve in S

. Since f is a local isometry f is a curve in S

of the same length.

Since g is a local isometry, g f is a curve in S

of the same length. Hence if is a curve

, then and g f have the same length, i.e., g f is a local isometry.

(ii) The inverse of an isometry is an isometry.

Let f : S

be an isometry. Then by definition f

-1

: S

is a diffeomorphism. Let

be a curve in S

. Since f is a (local) isometry, the lengths of f

-1

and f f

-1

= are

same. Hence f

-1

is a local isometry. Therefore f

-1

is an isometry.

(iii) Is the map from the circular half-cone x

= z

, z > 0, to the xy-plane given by (x, y, z)

(x, y, 0) a local isometry? Justify.

Consider : (1, 2) R defined by (t) = (t, 0, t). Then is a curve on the cone and the

length of the curve is

(u) du =

2. Now f : (1, 2) R

will be (f )(t) = (t, 0, 0).

It is on the xy-plane and its length is 1. Hence the length of is not preserved under f . Therefore

f is not a local isometry.

Definition 2.9.3. Let f : S

be a smooth map, and let p S

. Define f

: T

× T

v, w

= D

f (v), D

f (w)

f (p)

(v, w T

Example 2.9.4. The map f

is a symmetric bilinear map.

Let u, v, w T

and , µ R. Then

u, v

f (u), D

f (v)

f (p)

= D

f (u)D

f (v) = D

f (v)D

f (u)

f (v), D

f (u)

f (p)

= f

v, u

Therefore f

is symmetric. Now

u + µv, w

f (u + µv), D

f (w)

f (p)

= D

f (u) + µD

f (v), D

f (w)

f (p)

= D

f (u), D

f (w)

f (p)

+ µ D

f (v), D

f (w)

f (p)

= f

u, w

+ µf

v, w

Hence f

is bilinear.

The following is an important fact. Let f : S

be a smooth map. Let be a curve in

. Then

) is a tangent to S

at the point (t

) = p. Now = f is a curve in S

passing

through f ((t

)) = f (p). Therefore by definition D

f ( (t

)) = (t

). Since t

arbitrary, we have

f ( ) =

Example 2.9.5. Let T : R

× R

R be a symmetric bilinear map. If T (x, x) = 0 for all x R

then T = 0.

Let x, y R

. Then

0 = T (x + y, x + y) = T (x, x) + 2T (x, y) + T (y, y) = 2T (x, y).

2. SURFACES IN R

Therefore T (x, y) = 0, i.e., T 0.

It follows from the example that, if S and T are symmetric bilinear maps on R

and S(x, x) =

T (x, x) for all x, then S = T .

Theorem 2.9.6. Let S

and S

be smooth surfaces, and let f : S

be a smooth map. Then

is a local isometry if and only if the symmetric bilinear maps ·, ·

and f

·, ·

on T

are

equal for every p S

. If is curve in S

, then the length of the part of with endpoints (t

) and (t

) is

1/2

dt.

Then = f is a curve in S

. The length of between the points (t

) and (t

) is

1/2

dt =

f ( ), D

f ( )

1/2

dt =

1/2

(2.9.6.1)

Assume that the symmetric bilinear maps ·, ·

and f

·, ·

are equal for every p S

. Then it

follows from the equation (2.9.6.1) that any curve on S

and the corresponding curve = f

have the same length. Therefore f is a local isometry.

Conversely, assume that f is a local isometry. Then for any curve and the corresponding

curve = f in S

have the same length, i.e.,

1/2

dt =

1/2

dt.

Suppose that f

·, ·

= ·, ·

for some p S

. Then there is v T

such that f

v, v

. Let : (a, b S

such that (s) = p and

(s) = v for some s (a, b). Then

(s), (s)

(s)

= f

(s), (s)

(s)

. We may assume that

(s), (s)

(s)

-f

(s), (s)

(s)

> 0.

By continuity of the map g(u) =

(u), (u)

- f

(u), (u)

(u (a, b)) there exists > 0

and c > 0 such that

(u), (u)

(u)

- f

(u), (u)

(u)

> c for all u (s - , s + ). But

then

( (u), (u)

(u)

- f

(u), (u)

(u)

)du 2c > 0. i.e.,

(u), (u)

(u)

du >

(u), (u)

(u)

du. It means the length of the curve between the points (s- ) to (s+ )

is not same as the length of curve = f between the points (s - ) to (s + ), which is not

possible as f is a local isometry. Hence ·, ·

= f

·, ·

for every p S

Thus a smooth map f : S

is a local isometry if and only if for every p S

f (v), D

f (w)

f (p)

= v, w

(v, w T

Exercise 2.9.7. Let H be a finite dimensional Hilbert space. A linear map T : H H is called an

isometry if T x = x for all x H.

Prove the following statements.

9. Local isometries and conformal maps

(i) The map T is an isometry if and only if T x, T y = x, y for all x, y H.

(ii) Every linear isometry on H is invertible.

Corollary 2.9.8. Let S

and S

be smooth surfaces, and let f : S

be a smooth map. Then

is a local isometry if and only if for every p S

, the linear map D

f : T

f (p)

is an

isometry.

. Assume that f is a local isometry. Fix p S

. Since f is a local isometry, f

u, v

= u, v

for all u, v T

, i.e., D

f (u), D

f (v)

= u, v

for all u, v T

. This shows that D

f is

an isometry. Since p S

is arbitrary, we are done.

Conversely, assume that D

f is isometry for all p S

. Fix p S

. Since D

f is an isometry,

it follows that D

f (u), D

f (v)

= u, v

for all u, v T

, i.e., f

u, v

= u, v

for all

u, v T

. This shows that f

·, ·

= ·, ·

. Since p S

is arbitrary, it follows that f is a local

isometry.

Let : U R

be a surface patch. Fix (u

, v

) in U . Define

and

(v) = (u

, v)

and

(u) = (u, v

). Then both

and

are curves on passing through (u

, v

) (

) =

, v

) =

)). Now

) =

, v

) and

) =

, v

). Therefore

, v

)

and

, v

) are tangent vectors at (u

, v

). The curves

and

are called parameter curves

to the surface patch .

Let f : S

be a smooth map. Let : U R

be a surface patch of S

. Let = f :

U R

. Let (u, v) U . Then

(u, v) =

lim

f (u + h, v) - f (u, v)

lim

(u + h) - f

(u)

= (f

)

(u)

Since

(u) =

(u, v), by definition of derivative D

f of f at p = (u, v) we have D

f (

(u)) =

)

(u), i.e., D

f (

(u, v)) =

(u, v) or D

f (

) =

. Similarly, we get D

f (

) =

Corollary 2.9.9. Let S

and S

be smooth surfaces, and let f : S

be a local diffeomorphism.

Then f is a local isometry if and only if for any surface patch of S

, the patches and f of

and S

, respectively, have the same first fundamental form.

. Let E, F and G be the first order magnitudes of and E

, F

and G

be those of f . Assume

that f is a local isometry. Then to prove both and f have the same fundamental form, we need to

prove E = E

, F = F

and G = G

. Since f is a local isometry f

v, w

= v, w

for all p S

In particular, f

u p

, f

v p

and f

v p

. Now,

E =

u p

= f

u p

= D

f (

), D

f (

)

f (p)

u f (p)

= E

2. SURFACES IN R

Similar arguments shows that F = F

and G = G

. Hence the first fundamental forms of and

f are same.

Conversely, assume that the first fundamental forms of and f are same. Then E = E

F = F

and G = G

. Fix p = (u, v). Let w =

+ µ

, z =

+ µ

S. Then

w, z

f (w), D

f (z)

f (p)

= D

f (

+ µ

), D

f (

+ µ

)

f (p)

f (

), D

f (

)

f (p)

+ (

) D

f (

), D

f (

)

f (p)

+µ

f (

), D

f (

)

f (p)

u p

+ (

)

v p

+ µ

v p

+ µ

v p

= w, z

Hence f

·, ·

= ·, ·

. Since p S

is arbitrary, f

·, ·

= ·, ·

for every p S

, i.e., f is a

local isometry.

Definition 2.9.10. Let S

and S

be smooth surfaces. A local diffeomorphism f : S

is called

a conformal map if it preserves angle between curves, i.e., if and

are any two curves on S

intersecting at a point p S

and if

and

are their images under f , then the angle between

and

at p is equal to the angle between

and

at f (p).

If : U R

is a surface patch, then may be viewed as a map from an open subset of the

plane (namely U ) to the image of . We say that is a conformal parametrization or conformal

surface patch if the map is conformal.

Exercise 2.9.11. Composition of two conformal maps is a conformal map.

Theorem 2.9.12. Let S

and S

be smooth surfaces, and let f : S

be a local diffeomorphism.

Then f is conformal if and only if there is map : S

R such that f

v, w

= (p) v, w

for

all p S

and for all v, w T

. Let and be two curves on S

that intersects at p S

. If is an angle between these two

curves, then

cos =

1/2

Let be the corresponding angle of intersection between the curves f and f on S

at f (p).

Then

cos =

f ( ), D

f ( )

f (p)

f ( )

f ( ), D

f ( )

f (p)

f ( , D

f ( )

1/2

f (p)

f ( ), D

f ( )

1/2

f (p)

9. Local isometries and conformal maps

1/2

Suppose that there is a map : T

R such that f

v, w

= (p) v, w

for all p S

and for

all v, w T

. Then

cos =

1/2

(p) ,

(p)

1/2

(p)

1/2

= cos ,

which proves that the map f is conformal.

Conversely, assume that f is conformal. Then

1/2

(2.9.12.1)

Since every tangent vector to S

is a tangent vector to a curve in S

, the equation (2.9.12.1) implies

that

v, w

v, v

1/2

w, w

1/2

v, w

v, v

1/2

w, w

1/2

for all p S

and v, w T

. Fix p S

. Let {x, y} be an orthonormal basis of T

. Let

= f

x, x , µ = f

x, y and = f

y, y . Let w = cos x + sin y T

. Then w makes an

angle with x. Therefore

cos =

x, w

x, x

1/2

w, w

1/2

cos + µ sin

( cos

+ 2µ sin cos + sin

)

Take =

. Then µ = 0. Hence we get = cos

+ sin

, which gives = µ, say, (p). Let

v =

x +

y, w =

x +

y T

. Then

v, w

= f

x +

x, x

+ (

x, y

y, y

= (p)(

) = (p)(v · w)

= (p) v, w

2. SURFACES IN R

Therefore f

v, w

= (p) v, w

for all v, w T

. Since p S

arbitrary, for every p S

there exists (p) R (hence there is a map : S

R) such that f

v, w

= (p) v, w

for all

v, w T

. Hence the theorem.

Corollary 2.9.13. A local diffeomorphism f : S

is conformal if and only if for any surface

patch of S

, the first fundamental forms of patches of S

and f of S

are proportional.

. Let E, F and G be the first order magnitudes of and E

, F

and G

be those of f . Assume

that f is conformal. Then to prove both and f have proportional fundamental form, we need

to prove E = E

, F = F

and G = G

. Since f is conformal, there is a map : S

such that f

v, w

= (p) v, w

for all p S

. Fix p S

. Then f

u p

= (p)

u p

v p

= (p)

v p

and f

v p

= (p)

v p

. Now,

(p)E = (p)

u p

= f

u p

= D

f (

), D

f (

)

f (p)

u f (p)

= E

Similar arguments shows that (p)F = F

and (p)G = G

. Hence the first fundamental forms of

and f are proportional.

Conversely, assume that the first fundamental forms of and f are proportional. Fix p S

Then (p)E = E

, (p)F = F

and (p)G = G

. It means that (p)

u p

= (f )

, (f

)

u p

= f

u p

, (p)

v p

= (f )

, (f )

v p

= f

v p

and (p)

v p

(f )

, (f )

v p

= f

v p

. Let w =

+ µ

, z =

+ µ

S. Then

w, z

f (w), D

f (z)

f (p)

= D

f (

+ µ

), D

f (

+ µ

)

f (p)

f (

), D

f (

)

f (p)

+ (

) D

f (

), D

f (

)

f (p)

+µ

f (

), D

f (

)

f (p)

u p

+ (

v p

+ µ

v p

= (p)(

u p

+ (

)

v p

+ µ

v p

)

= (p)

+ µ

v p

= (p) w, z

Hence f

·, ·

= (p) ·, ·

. Since p S

is arbitrary, f

·, ·

= (p) ·, ·

for every p S

, i.e.,

f is a conformal map.

In particular, a surface patch is conformal parametrization if and only if its first fundamental

form is of the form (du

+ dv

) for some smooth map .

Example 2.9.14. Let : U V be a diffeomorphism between open subsets of R

. Let (u, v) =

(f (u, v), g(u, v)), where f and g are smooth functions. We show that is conformal if and only if

either (f

= g

and f

= -g

) or (f

= -g

and f

= g

We have (u, v) = (f (u, v), g(u, v)). Therefore

= (f

, g

) and

= (f

, g

). This gives

E = f

+ g

, F = f

+ g

and F = f

+ g

. Assume that is conformally parametrized.

Then its first fundamental form is of the form (du

+ dv

), where is a smooth function of u and

9. Local isometries and conformal maps

v. This gives F = f

+ g

= 0 and f

+ g

= E = G = f

+ g

. It means the vectors (f

, g

)

and (f

, g

) are orthogonal and have the same norm. Therefore (f

, g

) can be obtained by rotating

, g

) by an angle /2 or 3/2. Hence (f

, g

) = (g

, -f

) or (f

, g

) = (-g

, f

). Therefore

= -g

and f

= g

) or (f

= g

and f

= -g

Conversely, if (f

= -g

and f

= g

) or (f

= g

and f

= -g

), then F = 0 and E =

G = f

+ g

. Therefore the first fundamental form of is (f

+ g

)(du

+ dv

) and hence is

conformally parametrized.

Definition 2.9.15. Let : U R

be a surface patch, and let R U . Then the surface area, A

(R),

of the part (R) of (U ) is

(R) =

dudv.

Proposition 2.9.16. Let : U R

be a surface patch. Then

EG - F

. We know that if a , b, c, d R

, then (a × b)(c × d) = (a c)(b d) - (a d)(bc). Now

= (

)(

) - (

)

= EG - F

. Therefore

EG - F

Example 2.9.17.

(i) Compute the surface area of sphere of radius r.

We know that a parametrization of sphere of radius r is given by

(u, v) = (r cos v cos u, r cos v sin u, r sin v) (u (0, 2), v (-

)).

Now

= (-r cos v sin u, r cos v cos u, 0),

= (-r sin v cos u, -r sin v sin u, r cos v). There-

fore E =

= r

cos

v, F =

= 0 and G = r

, and EG - F

= r

cos

v. Now the

surface area, A, of the sphere is

A =

(0,2)×(-

)

cos v dudv

= r

-/2

cos v dudv

= r

-/2

cos vdv

= 4r

(ii) Compute the area of {(x, y, z) : x

+ y

= 1, |z| 1}.

We know that the cylinder may be parametrized by (u, v) = (cos u, sin u, v), where u

(0, 2) and v R). We have E = G = 1 and F = 0. We want the surface area of (R),

2. SURFACES IN R

where R = {(u, v) : u (0, 2), |v| 1}. Therefore the surface area A of (R) is

A =

(0,2)×[-1,1]

1dv =

-1

= 4.

(iii) Let a > 0 be fixed. Let : (, ) R

be a unit-speed curve in R

whose curvature is

strictly less than a

-1

. Consider the surface patch of the tube of radius a around given by

(t, ) = (t) + a(cos n(t) + sin b(t)), t (, ), (0, 2), where n and b are the unit

normal and unit binormal of respectively. We want to compute the surface are of this tube.

(Obviously the answer should be 2a( - )).

We have

= t + a(cos

n + sin

b) = (1 - a cos )t - a sin n + a cos b and

-a sin n + a cos b. Therefore E =

= (1 - a cos )

+ a

, F =

= a

and

G =

= a

. Now EG - F

= ((1 - a cos )

+ a

- a

= a

(1 - a cos )

Now

(,)×(0×2)

EG - F

dtd

a(1 - a cos )dtd

= 2a( - ).

Exercise 2.9.18.

(i) Consider the catenoid (u, v) = (a cosh u cos v, a cosh u sin v, au), u R, 0 < v < 2,

where a > 0 is fixed. Compute the surface area of the portion of the catenoid given by |u|

(ii) Compute the surface area of a portion of the helicoid (u, v) = (u cos v, u sin v, bv), 1 < u <

3, 0 < v < 2.

CHAPTER 3

Curvature of a surface

10. Second Fundamental Form

Let : U R

be a surface patch with the unit normal N. Fix (u, v) U . As the parameter

(u, v) changes to (u + u, v + v), the surface moves away from the tangent plane at (u, v) by a

distance ((u + u, v + v) - (u, v))N. Let u and v be very small. By Taylor's theorem

(u + u, v + v) - (u, v) =

u +

v +

[

(u)

+ 2

uv +

(v)

] +

u +

v +

[

(u)

+ 2

uv +

(v)

Hence ((u + u, v + v) - (u, v))N =

[L(u)

+ 2M (u)(v) + N (v)

], where L =

M =

N and N =

N. The expression Ldu

+ 2M dudv + N dv

is known the second

fundamental form on a surface. The functions L, M and N are called the second order magnitudes.

Exercise 3.10.1. Calculate the second fundamental forms on sphere, cylinder, plane, torus, surface of

revolution, hyperboloid with one sheets and two sheets, hyperbolic cylinder.

Let S

denote the unit sphere in R

, i.e., S

= {(x, y, z) : x

+ y

+ z

= 1}.

Definition 3.10.2. Let S be a smooth surface. The map G : S S

defined by G(p) = N

, where

is the unit normal at p S, is called the Gauss map.

An oriented surface is a smooth surface S together with a smooth choice of unit normal N at

each point, i.e., a smooth map N : S R

(It means that each of the three components of N is a

smooth function S R) such that, for all p S, N(p) is a unit vector perpendicular to T

S. Al most

all surfaces which we have seen are oriented and we will consider them only. There are surfaces

which are not oriented for example Mobius band is not oriented.

Let S be an oriented surface. The map G is a smooth map. Therefore, for p S, D

G : T

G(p)

is a linear map. Note that T

S = T

G(p)

. Hence D

G : T

S T

S is a linear map.

Example 3.10.3. Find the image of the Gauss map for (u, v) = (u, v, u

+ v

), u, v R.

We have

= (1, 0, 2u) and

= (0, 1, 2v). Therefore

= (-2u, -2v, 1). This gives

N =

1+4u

+4v

(-2u, -2v, 1). Hence the image of the Gauss map is {

1+4u

+4v

(-2u, -2v, 1) :

u, v R}.

3. CURVATURE OF A SURFACE

Definition 3.10.4. Let p be a point on an oriented surface S.

(i) The map W

: T

S T

S defined by W

= -D

G is called the Weingarten map.

(ii) The second fundamental form on a surface S at p S is a bilinear map

: T

S × T

S R defined by

v, w

= W

(v), w

(v, w T

S).

Since W

= -D

G and D

G is a linear map, the map W

: T

S T

S is a linear map.

Example 3.10.5. Let : U R

be a regular patch of an oriented surface. Then D

) = N

and D

) = N

Fix a point p = (u, v) on the surface. Consider the parametric curves

(t) = (t, v) and

(t) = (u, t). Then

(u) = (u, v) = p =

(v). Now

(u) =

(u, v) and

(v) =

(u, v).

Now G(

)(t) = N

(t) = N(t, v). Hence tangent to the curve G

at G (u, v) (or the tangent

to G

at G(p)) is (G

)

(u) = N

(u, v). Hence D

(u, v)) = N

(u, v) or D

) = N

One may similarly show that D

) = N

Lemma 3.10.6. Let be a surface patch of an oriented surface with the unit normal N. Then

= -L

, N

= -M = N

and N

= -N

. Since

and

generates the tangent space and N is orthogonal to the tangent space, we have

= N

= 0. Differentiating N

= 0 partially with respect to u, we obtain N

+ N

= 0.

But N

= L. Hence N

= -L. By differentiating N

= 0 partially with respect to v, we get

= -M . Similarly, by differentiating N

= 0 partially with respect to u and v, we get the

relations N

= -M and N

= -N respectively.

Proposition 3.10.7. Let be a surface patch of an oriented surface S containing a point p of S,

say, p = (u, v). Then

w, x = Ldu(w)du(x) + M (du(w)dv(x) + du(x)dv(w)) + N dv(w)dv(x)

for all w, x T

where du, dv : T

S R are defined by du(w) = and dv(w) = µ when w =

+ µ

. Since D

) = N

and D

) = N

, we have W

(

) = -N

and W

(

) = -N

Let w =

+ µ

and x =

+ µ

be in T

S. Then

w, x

(w), x

(

+ µ

v p

(

) + µ

(

+ µ

v p

- µ

+ µ

v p

= -

- µ

= Ldu(w)du(x) + M (du(w)dv(x) + du(x)dv(w)) + N dv(w)dv(x).

This proves the result.

11. Gaussian, Mean and Principal Curvatures

A bounded linear map T on a Hilbert space H is self-adjoint if T x, y

x, T y for all

x, y H. When the Hilbert space is real (i.e., when the field is R), self-adjoint map is often called

symmetric.

Corollary 3.10.8. The second fundamental form at a point on an oriented surface is a symmetric

bilinear map. Equivalently, the Weingarten map is a self adjoint map.

. One can check (means you) that II

is linear in both the variable (by fixing one), i.e., II

bilinear. It follows from the Proposition 3.10.7 that II

w, x

= II

x, w

for all x, w T

S,i.e.,

is symmetric. Hence the second fundamental form II

is a symmetric bilinear map.

Let x, w T

S. Since II

w, x

= II

x, w

, we have W

(w), x

= W

(x), w

w, W

(x)

, i.e, W

(w), x

= w, W

(x)

. Hence W

is symmetric.

11. Gaussian, Mean and Principal Curvatures

Definition 3.11.1. Let W

be the Weingarten map on an oriented surface S at p S. Then the

Gaussian curvature, K

, and the mean curvature, H

, at p are defined by

= det(W

) and H

trace(W

Proposition 3.11.2. Let : U R

be a surface patch of an oriented surface S, and let p be in

the image of , i.e., p = (u, v) for some (u, v) U . Then the matrix of W

with respect to the

basis {

} of T

S is

-1

. Since N is a unit vector, N N = 1. This will imply that N

N = 0 and N

N = 0. Hence N

and N

(and so -N

and -N

) are in the tangent space T

S. Therefore there exist , , , R

such that -N

and -N

. Since W

(

) = -N

and W

(

) = -N

we have W

(

) =

and W

(

) =

. Therefore the matrix of W

with respect

to basis {

} of T

S is

. Taking dot product with

and

on both the sides of the

equations -N

and -N

, we obtain L = E + F , M = F + G,

M = E + F and N = F + G. The last four equations can be written as

This gives

-1

which proves the result.

3. CURVATURE OF A SURFACE

We have

-1

EG-F

-F

Corollary 3.11.3. Let be a surface patch of an oriented surface S containing point p = (u, v).

Then the Gaussian curvature K

and the mean curvature H

are

LN - M

EG - F

and H

LG - 2M F + N E

2(EG - F

)

. Since K

= det(W

), we have

= det

-1

= det

-1

det

= det

-1

det

LN - M

EG - F

Now

-1

EG-F

LG - M F

M G - N F

M E - LF

N E - M F

, and hence H

trace(W

) =

LG-2M F +N E

2(EG-F

)

Example 3.11.4.

(i) We shall compute the Gaussian curvature and the mean curvature of the surface z = f (x, y),

where f is a smooth map.

The surface z = f (x, y) may be parametrized as (u, v) = (u, v, f (u, v)). Then

(1, 0, f

= (0, 1, f

= (0, 0, f

) and

= (0, 0, f

). Therefore

E = 1 + f

, F = f

and G = 1 + f

. Now N =

1+f

(-f

, -f

, 1).

So, L =

N =

1+f

, M =

N =

1+f

and N =

N =

1+f

. Now

EG - F

= 1 + f

+ f

and LN - M

-f

1+f

. Thus

K =

LN - M

EG - M

- f

(1 + f

+ f

)

also

H =

LG - 2M F + N E

2(EG - F

)

(1 + f

- 2f

+ (1 + f

)

(1 + f

+ f

)

11. Gaussian, Mean and Principal Curvatures

(ii) We shall now compute the Gaussian curvature and the mean curvature of the surface of revo-

lution (u, v) = (f (u) cos v, f (u) sin v, g(u)), where f

+ g

= 1.

We have

= (f cos v, f sin v, g ),

= (-f sin v, f cos v, 0),

= (f cos v, f sin v, g ),

= (-f sin v, f cos v, 0) and

= (-f cos v, -f sin v, 0). Therefore E = 1, F = 0

and G = f

. Now N = (-g cos v, -g sin v, f ). Thus L = -f g , M = 0 and N = f g .

This gives K =

-f f (g )

= -

f (g )

and H =

-f

f g +f g

g (1-f f )

(iii) Let : (a, b) R

be a unit-speed curve with nowhere vanishing curvature. Define :

(a, b) × (0, ) R

by (u, v) = (u) + vt(u). We shall compute the Gaussian curvature

and the mean curvature of this surface.

Here

= t + v t = t + vn,

= t,

= t + v n + v

n = n + v n + v( b - t),

= n and

= 0. Thus E = 1 + v

, F = 1 and G = 1. Also,

= -vb gives

N = -b. Now L = -v , M = N = 0. Thus LN - M

= 0 implies that the Gaussian

curvature is zero everywhere. Also, H =

-v

= -

Exercise 3.11.5.

(i) Calculate the Gaussian curvature and mean curvature for sphere, cylinder, plane, torus, surface

of revolution, hyperbolic paraboloid, . . ..

(ii) Show that the Gaussian curvature of the parametrized surfaces (u, v) = (u cos v, u sin v, v)

and (u, v) = (u cos v, u sin v, ln u) are same for each (u, v) and yet the first fundamental

forms do not agree.

The following is analogous to the result which says that the derivative of a turning angle of a

plane curve is the signed curvature (see [1, page 182]).

Theorem 3.11.6. Let : U R

be a regular surface patch, and let (u

, v

) U

. For > 0, let

= {(u, v) : (u - u

)

+ (v - v

)

}. Let K be the Gaussian curvature of the surface at p.

Then lim

)

= |K|

. Since (u

, v

) U and U is open, there is

> 0 such that B

U . We have N

× N

LN -M

EG-G

(

) = K(

). If <

, then

)

× N

dudv

|K|

dudv

Let

> 0. Since K is continuous, there exists > 0 ( <

) such that |K(u

, v

)| - < |K(u, v)| <

|K(u

, v

)| + for all (u, v) B

. Let 0 < < . Then

(|K(u

, v

)| - )

dudv

|K|

dudv

3. CURVATURE OF A SURFACE

(|K(u

, v

)| + )

dudv.

This gives

|K(u

, v

)| -

|K|

dudv

|K(u

, v

)| + ,

i.e.,

|K(u

, v

)| -

)

|K(u

, v

)| + .

Hence

)

- |K(u

, v

)| <

for every 0 < < ,

i.e., lim

)

= |K(u

, v

)|.

Example 3.11.7. For a plane, the image of the unit normal N is a point only. Hence A

) = 0 for

every > 0. Therefore |K| = 0, i.e., K = 0 at every point.

For a cylinder, the image of the unit normal N is a circle. Hence A

) = 0 for every > 0.

Therefore |K| = 0, i.e., K = 0 at every point.

For a unit sphere, the unit normal is the identity map. Hence A

) = A

) for every

> 0. Hence K = 1 at every point.

Proposition 3.11.8. Let p be a point on an oriented surface S. Then there are real numbers

and a basis {t

, t

} of T

S such that W

) =

for i = 1, 2. Further, if

, then the

vectors t

and t

are orthogonal.

. Since W

is symmetric, both its eigen values are real, say,

and

. Let t

and t

corresponding eigen vectors. Then W

) =

for i = 1, 2. If

, then W

I. But

then every element of T

S is an eigen vector. So we may choose t

and t

so that they are linearly

independent and hence a basis of T

S. Let

. Then

, t

= W

), t

, W

) = t

, t

. Since

, it follows that t

, t

= 0, i.e, t

and t

are

orthogonal and hence {t

, t

} is a basis of T

If x is an eigen vector of a linear map T , then any non-zero scalar multiple of x is also an eigen

vector of T . Hence we may assume that the vectors t

and t

in above proposition are unit vectors.

Definition 3.11.9. Let p be a point on an oriented surface S. Then there exist real numbers

and an orthonormal basis {t

, t

} of T

S such that W

) =

for i = 1, 2. The scalars

and

are called the principal curvatures of S at p and the vectors (directions) t

and t

are called the

principal vectors (directions) of S at p.

A point of a surface at which both the principal curvatures are equal is called an umbilical

point.

11. Gaussian, Mean and Principal Curvatures

It is known that every real symmetric matrix is diagonalizable, i.e., if A M

(R) is symmetric,

then there is an invertible matrix C M

(R) such that C

-1

AC is a diagonal matrix and the entries

on the main diagonal of the matrix are precisely the eigen values of A. Consider W

as a matrix.

Then W

is symmetric as W

= W

. Therefore there is an invertible matrix C M

(R) such that

-1

C =

Hence it follows that

if and only if C

-1

C =

I if and only if

Hence we have the following result. A point p on a surface S is umbilic if and only if W

is a

scalar multiple of the Identity.

Proposition 3.11.10. Let p be a point on an oriented surface S. Let

and

be principal curva-

tures of S at p. Then the Gaussian curvature K

and the mean curvature H

at p are given by

and H

(

)

. Let t

and t

be the principal vectors. Then W

) =

for i = 1, 2. Then the matrix

of W

with respect to basis {t

, t

} of T

S is

. Hence K

= det

and

trace

(

)

Proposition 3.11.11. Let : U R

be a regular surface patch of an oriented surface. Then the

principal curvatures are the roots of the equation

L - E

M - F

N - G

= 0,

and the corresponding principal vectors are t =

, where

L - E

M - F

N - G

. We know that the principal curvatures at p are the eigen values of W

, i.e., the principal

curvatures are the roots of the characteristic equation of W

. Since the matrix of W

with respect to

basis {

} of T

S is

-1

, the characteristic equation will be

-1

1 0

0 1

= 0, i.e.,

L - E

M - F

N - G

= 0. If w =

is a principal direction, then it is an eigen vector of W

corresponding to some eigen value, say, .

3. CURVATURE OF A SURFACE

Therefore

-1

, i.e.,

, which

will give the equation

L - E

M - F

N - G

Example 3.11.12. We shall compute the principal curvatures on the surface (u, v) = (u, v, uv).

We also compute the Gaussian curvature and the mean curvature of the surface and the principal

directions at (0, 0, 0).

We see that

= (1, 0, v),

= (0, 1, u),

= (0, 0, 0),

= (0, 0, 1) and

= (0, 0, 0).

Therefore E = 1 + v

, F = uv and G = 1 + u

. Also, N =

1+u

(-v, -u, 1). Thus L = 0,

M =

1+u

and N = 0. To find the principal curvatures we need to solve the determinant

L - E

M - F

N - G

= 0 for . Now

0 =

L - E

M - F

N - G

-(1 + v

)

1+u

- uv

1+u

- uv

-(1 + u

)

Simplifying this equation we get

(1 + u

+ v

) +

2uv

1+u

= 0 and hence =

2uv

1+u2+v2

4u2v2

1+u2+v2

2(1+u

)

-uv±

(1+u

)(1+v

)

(1+u

)

. Thus principal curvatures at a point (u, v) of the

surface are

-uv+

(1+u

)(1+v

)

(1+u

)

and

-uv-

(1+u

)(1+v

)

(1+u

)

. Now

K =

-uv -

(1 + u

)(1 + v

)

(1 + u

+ v

)

-uv +

(1 + u

)(1 + v

)

(1 + u

+ v

)

- (1 + u

)(1 + v

)

(1 + u

+ v

)

= -

(1 + u

+ v

)

and

-uv-

(1+u

)(1+v

)

(1+u

)

-uv+

(1+u

)(1+v

)

(1+u

)

= -

(1 + u

+ v

)

Since (0, 0) = (0, 0, 0), by above computations

= 1 and

= -1. Let (0, 0) +

(0, 0) = (1, 0, 0) + (0, 1, 0) be a principal direction corresponding to 1. Then

11. Gaussian, Mean and Principal Curvatures

L(0, 0) - 1E(0, 0)

M (0, 0) - 1F (0, 0)

N (0, 0) - 1G(0, 0)

, i.e.,

-1

. This

will give - + = 0 and - = 0. Taking = =

, we get the principal direction corresponding

to 1 is (

, 0). Similar computations shows that a principal direction at (0, 0, 0) corresponding

to the principal curvature -1 is (

, -

, 0).

Exercise 3.11.13.

(i) Compute the principal curvatures and principal vectors to sphere, cylinder, hyperbolic parabo-

loid, torus, surface of revolution.

(ii) Compute the principal curvatures on the surface (u, v) = (u, v, v

- u

). Also compute

the Gaussian curvature and the mean curvature of the surface and the principal directions at

(0, 0, 0).

Let be a unit-speed curve on a surface S, then

is a unit vector and by definition it is a tangent

vector to S. Therefore it is orthogonal to the unit normal N

. Since ¨

is perpendicular to , ¨

is a

linear combination of N

and N

× . Therefore there exists functions

and

(of t) such that

× ).

Definition 3.11.14. Let be a unit-speed curve on an oriented surface S. Then the functions

and

(of t) satisfying ¨

× ) are called the normal curvature and the geodesic

curvature of respectively.

Proposition 3.11.15. Let be a unit-speed curve on an oriented surface S. Then

= ¨

× ),

= cos

and

= ± sin

, where is the curvature of

and is the angle between N

and the principal normal n of .

. Since N

and N

× are orthogonal vectors, by taking dot product with N

and N

× both

the sides of the equation ¨

× ) we obtain

= ¨

and

= ¨

× ). Since

and N

× are orthogonal unit vectors,

= ¨

. Since is the angle between N

and n, cos = N

n. Therefore cos = (n)N

= ¨

. Since

, we have

= ± sin .

Definition 3.11.16. A normal section of a surface S is a curve on S such that the unit normal n of

and the unit normal N

are linearly dependent at each point of the curve .

Corollary 3.11.17. The curvature , the normal curvature

and the geodesic curvature

of a

normal section of a surface are related by

= ±

and

= 0

Proposition 3.11.18. If is a unit-speed curve on an oriented surface S, then its normal curvature

is given by the formula

= II

, = W

( ),

. If is a surface patch of a surface S and

if (t) = (u(t), v(t)) is a unit-speed curve on , then

= L u

+ 2M u v + N v

3. CURVATURE OF A SURFACE

. Since

is a tangent vector to S, N

(t)

(t) = 0 for all t, i.e., (G (t)) (t) = 0 for all

t. This gives (G (t))¨

= - (G (t))

= - D

G( (t)) = W

( (t)) for all t. Hence

= N

= W

( (t)) = W

( ), .

Let (t) = (u(t), v(t)). Then

u +

v. Now

( ), = W

(

u +

v),

u +

(

) + vW

(

u +

u(-N

) + v(-N

u +

(-N

) + u v(-N

) + v

(-N

)

= L u

+ 2M u v + N v

This completes the proof

Let p be point on an oriented surface S, and let v be a unit vector in T

S. Then there is a

unit-speed curve in S such that (t) = p and

(t) = v. Then by above the normal curvature of

at the point (t) is W

(t)

( (t)), (t) = W

(v), v = II

v, v .

Let t

be a principal direction corresponding to the principal curvature

at p. Then the normal

curvature of the surface at p in the unit direction t

= W

), t

, t

. Hence the principal curvatures are the normal curvatures.

Theorem 3.11.19. (Meusnier's Theorem) Let p be a point on an oriented surface S, and let v be a

unit tangent vector at p. Let

be a plane containing a line through p parallel to v and making

an angle with the tangent plane at p and is not the tangent plane at p. Suppose that

intersects

S in a curve with curvature

(at p). Then

sin

is independent of .

. Let

be a unit-speed reparametrization of the curve of the intersection of

and the surface

S. Then at p,

= ±v, so ¨

is perpendicular to v and parallel to

. Thus the unit normal n to

the curve

at p makes an angle

± with the unit normal N of the surface at p. Note that the

normal curvature at p in the direction v is W

(v), v . Also, the normal curvature

(p) of

at p is

(p) = W

(

. Now

(v), v

(±

), ±

= W

(

(p) =

cos(

± ) =

sin .

Since W

(v), v is independent of , it follows that

is independent of .

Theorem 3.11.20. (Euler's Theorem) Let p be point on an oriented surface S, and let

and

the principal curvatures of S with principal vectors t

and t

. Let v be a unit vector in T

S. Then

the normal curvature of S at p in the direction v is given by

cos

sin

, where is

the oriented angle t

11. Gaussian, Mean and Principal Curvatures

. By replacing t

by -t

(if needed) we may assume that t

. Also we may assume that

both t

and t

are unit vectors. Since is an oriented angle between t

and v, v = cos t

+ sin t

Now

(v), v

(cos t

+ sin t

), cos t

+ sin t

cos t

sin t

, cos t

+ sin t

cos

sin

which proves the theorem.

Corollary 3.11.21. The principal curvatures at a point of a surface are maximum and minimum

values of the normal curvatures. Moreover, the principal vectors (directions) are the directions

giving these maximum and minimum values.

. Let

and

be the principal curvatures at a point p on the surface S. We may assume that

. Let t

and t

be the corresponding principal directions. We may assume that t

. Let

v be a unit tangent at p. Then the normal curvature at p in the direction v is

cos

sin

where is the oriented angle t

v. Now

cos

sin

cos

sin

cos

sin

. Hence

lies between

and

. Note that the normal curvature at p in

the direction of t

and the normal curvature at p in the direction of t

. Hence the principal

vectors are the directions at p giving these maximum and minimum values.

Example 3.11.22. Let p be an oriented surface. Let t

and t

be principal directions at p, and let

and

be corresponding principal curvatures at p. If

() is a normal curvature at p in the direction

making an oriented angle with the principal direction, then show that

() d = H

We may assume that t

. It follows from Euler's theorem that

() =

cos

sin

. Now, the calculations give

() d = H

Exercise 3.11.23. With the notations of the above example prove the following.

(i) H

(

() +

( +

)) for any .

(ii) Let m 3. Then H

() +

+ · · · +

2(m-1)

Proposition 3.11.24. Let : U R

be a connected surface patch of an oriented surface such

that every point of (U ) is umbilic. Then (U ) is (part of) a plane or sphere.

. Since every point is an umbilical point, both the principal curvatures are same, say, and

W = I. Since W

(

) = -N

and W

(

) = -N

, we have N

= -

and N

= -

Now (

)

= -(N

)

= -(N

)

= (

)

. Therefore

, i.e,

. Since

and

are linearly independent,

= 0, i.e., is a constant map.

3. CURVATURE OF A SURFACE

Suppose that = 0. Since N

= -

and N

= -

, we get N

= N

= 0, i.e., N is a

constant unit vector. Now (N)

N = 0 and (N)

N = 0. Hence N = d for some

constant d R. This shows that is lying on a plane R N = d, i.e., S is (part of) a plane.

Suppose that = 0. Then ( +

= 0 and similarly

( +

= 0. Hence +

N is a constant vector, say, a. Therefore - a =

and hence

is lying on the sphere R - a =

, i.e., is (part of) a sphere.

Let be a surface patch of a surface S, and let p be in . By applying a rigid motion to S we

may assume that p = (0, 0, 0) and the principal vectors are t

= (1, 0, 0) and t

= (0, 1, 0). By

Taylors theorem

(s, t) (0, 0) + s

+ t

+ 2st

+ t

If (s, t) = (x, y, z), then z =

+2st

]N =

(Ls

+2M st+N t

). If t = s

then Ls

+2M st+N t

= W (t), t . Now t = xt

+yt

. Therefore W (t) = xW (t

)+yW (t

) =

= (

y, 0). This gives Ls

+2M st+N t

= (

y, 0)(x, y, 0) =

Hence we obtain

z =

(

(3.11.24.1)

(i) If both

and

are positive of both negative (i.e., K > 0), then equation (3.11.24.1) represents

an elliptic paraboloid. In this case p is called an elliptic point.

(ii) If both

and

are of opposite signs, both non-zero, (i.e., K < 0), then equation (3.11.24.1)

represents a hyperbolic paraboloid. In this case p is called an hyperbolic point.

(iii) If one of

and

is zero and the other is non-zero, (i.e., K = 0), then equation (3.11.24.1)

represents a parabolic cylinder. In this case p is called an parabolic point.

(iv) If both

and

are zero, (i.e., K = 0), then equation (3.11.24.1) represents a plane. In this case

p is called an flat point or a planar point.

Definition 3.11.25. A point p on an oriented surface S is called

(i) an elliptic point if the Gaussian curvature at p is positive.

(ii) a hyperbolic point if the Gaussian curvature at p is negative.

(iii) a parabolic point if the Gaussian curvature at p is zero but one of the principal curvature at p

is non-zero.

(iv) a flat point or a planar point if both the principal curvatures at p are zero.

Exercise 3.11.26. Determine the points on (u, v) = ((b+a cos v) cos u, (b+a cos v) sin u, a sin v),

where b > a, which are parabolic, elliptical and hyperbolic.

11. Gaussian, Mean and Principal Curvatures

We note that the principal curvatures of are

and

cos v

b+a cos v

. Thus no point on

torus is a flat point. Note that the Gaussian curvature K will be

K(u, v) =

cos v

a(b + a cos v)

and

a(b+a cos v)

> 0. Thus K(u, v) will be positive, negative or zero if and only if cos v is positive,

negative or zero respectively. Thus

the set of ellptic points of is {(u, v) : v (0,

) (

, 2)},

the set of parabolic points of is {(u, v) : v {

}},

and the set of hyperbolic points of is {(u, v) : v (

)}.

CHAPTER 4

Geodesics and some fundamental results

12. Christoffel's Symbols and tangent vector fields

Definition 4.12.1. Let : U R

be a regular surface patch. Since

and N are linearly

independent, there exist function

: U R such that

+ M N

N N

(4.12.1.1)

The functions

are called the Christoffel symbols of second kind.

Example 4.12.2. Show that

= F

and

We have E

= (

)

= 2

. So,

Now F

= (

)

(

)

implies that

= F

Also, F

= (

)

= 2

implies that

Similarly, G

= (

)

= 2

implies that

Now F

= (

)

(

)

gives

= F

Also, G

= (

)

= 2

gives

4. GEODESICS AND SOME FUNDAMENTAL RESULTS

Proposition 4.12.3. Let be a regular surface patch. Then

-2F F

+F E

2(EG-F

)

2EF

-EE

-F E

2(EG-F

)

-F G

2(EG-F

)

-F E

2(EG-F

)

2GF

-GG

-F G

2(EG-F

)

and

-2F F

+F G

2(EG-F

)

. We know that

+ LN. Taking dot product with

and

, we get

= E

+ F

and

= F

+ G

. Putting the values of

and

, we have

= E

+ F

and F

= F

+ G

. Solving the last two equations for

and

, we obtain

-2F F

+F E

2(EG-F

)

and

2EF

-EE

-F E

2(EG-F

)

. For the values of

consider the equations

+ M N and

+ N N and take their

dot products with

and

and use the last example.

Example 4.12.4. We shall compute Christoffel's symbols of second kind for the circular cylinder

(u, v) = (cos u, sin u, v).

We have

= (- sin u, cos u, 0),

= (0, 0, 1),

= (- cos u, - sin u, 0),

= (0, 0, 0)

and

= (0, 0, 0). Therefore E = 1, F = 0, G = 1. Also, N = (cos u, sin u, 0), L = -1,

M = N = 0. We see that

= (- cos u, - sin u, 0) = 0

+ 0

+ (-1)N. Since

and

N are linearly independent, it follows that

= 0. Similar observation gives

= 0. Hence all the Christoffel's symbols are 0 for circular cylinder.

Exercise 4.12.5. Calculate Christoffel's symbols of second kind (either use above formula or directly).

(i) (u, v) = (u cos v, u sin v, u)

(ii) (u, v) = (u cos v, u sin v, v)

(iii) (u, v) = (f (u) cos v, f (u) sin v, g(u)), where f

+ g

= 1.

Definition 4.12.6. Let : (, ) R

be a curve on a smooth surface S. A tangent vector field

along is a smooth map from an interval (, ) to R

such that v(t) T

(t)

S for all t (, ).

Let v be a tangent vector field along a curve lying on a surface S. Then the component of

along the unit normal N

is (

. Therefore the orthogonal projection of

v on the tangent space

at T

S is

( v) = v - ( vN

Definition 4.12.7. Let be a curve on a surface S, and let v be a tangent vector field along . Then

the covariant derivative of v along is the orthogonal projection

( v) = v - ( vN

v on

the tangent space T

A tangent vector field v along a curve on a surface S is said to be parallel along if

( v) = 0

at every point of .

Proposition 4.12.8. A tangent vector field v is parallel along a curve on a surface S if and only if

is perpendicular to the tangent space of S at every point of (i.e., v is parallel to the unit normal

at every point of ).

. Suppose that v is parallel along . Then

( v) = v - ( vN

= 0, i.e., v = ( vN

Therefore

v and N

are linearly dependent hence it is perpendicular to the tangent space of S at

every point.

13. Geodesics

Conversely, assume that

v is perpendicular to the tangent space of S at every point of . Then

v = N

for some function . Now

( v) = v - ( vN)N

= N

- (N

= N

= 0. Hence v is parallel along .

Proposition 4.12.9. Let (t) = (u(t), v(t))

be a curve on a surface patch , and let w(t) =

(t)

(u(t), v(t)) + (t)

(u(t), v(t))

be a tangent vector field along a curve , where and

are smooth maps of t. Then v is parallel along if and only if

+ (

u +

v) + (

u +

v) = 0

+ (

u +

v) + (

u +

v) = 0

(4.12.9.1)

(Note that these equations involve first order magnitudes of only.)

. Since w =

, we have

w =

+ (

u +

v) +

+ (

u +

+ ((

+ LN

) u + (

+ M N

) v)

+ ((

+ M N

) u + (

+ N N

) v)

= (

+ (

u +

v) + (

u +

v))

+ (

u +

v) + (

u +

v))

+( uL + ( v + v)M + vN )N

Now w is parallel along if and only if

w is parallel to N

if and only if the coefficients of

and

in the expression of

w are 0. Hence w is parallel along if and only if the equations in (4.13.2.1)

hold.

13. Geodesics

Definition 4.13.1. A curve on a smooth surface S is called a geodesic if ¨

(t) is perpendicular to

the tangent space of the surface at (t) for all t (i.e., ¨

and N

are linearly dependent at every point

of ).

Equivalently, is a geodesic if and only if

is a tangent vector field parallel along .

Proposition 4.13.2. Let (t) = (u(t), v(t)) be a curve on a surface patch . Then is a geodesic

if and only if

u + (

u +

v) u + (

u +

v) v = 0

v + (

u +

v) u + (

u +

v) v = 0

(4.13.2.1)

. Here (t) = (u(t), v(t)). Therefore

= u

+ v

. Then is a geodesic if and only if

is parallel along if and only if both the equations hold (See Proposition 4.12.9) with = u and

= v.

Proposition 4.13.3. Any geodesic has a constant speed.

4. GEODESICS AND SOME FUNDAMENTAL RESULTS

. Let be a geodesic on a surface S. We have

( ) = 2 ¨

. Since is a geodesic,

is perpendicular to the tangent space and is therefore perpendicular to . So ¨

= 0 and hence

is constant.

Let be a geodesic on a surface S. Let =

. Take (t) =

. Then is a unit-speed

curve also

. Since ¨

is perpendicular to the tangent space, ¨

is perpendicular to the tangent

space. Therefore is a geodesic. Hence we may now assume that any geodesic is unit-speed.

Proposition 4.13.4. A unit-speed curve on a surface S is a geodesic if and only if its geodesic

curvature is zero every where.

. Here is a unit-speed curve on a surface S. We know that a geodesic curvature of a curve

is given by

= ¨

× ).

Assume that is a geodesic. Then ¨

and N

are linearly dependent. Hence

= ¨

× ) =

(¨

× N

) = 0.

Conversely, assume that

= 0. Then ¨

is perpendicular to N

× . Since , N

and

×N

are

mutually perpendicular and ¨

is perpendicular to , it follows that ¨

and N

are linearly dependent.

Therefore is a geodesic.

Proposition 4.13.5. Any (part of a) line on a surface is a geodesic.

. Let be (part of a) line lying on a surface. Then may be parametrized as (t) = t a + b.

Then ¨

= 0. Hence is a geodesic.

Theorem 4.13.6. Let (t) = (u(t), v(t)) be a curve on a regular surface patch of a surface S.

Then is a geodesic if and only if the following equations hold.

(E u + F v) =

+ 2F

u v + G

)

(F u + G v) =

+ 2F

u v + G

(4.13.6.1)

The equations in (4.13.6.1) are called geodesic equations.

. We know that {

} is a basis of the tangent space of a surface. Therefore is a geodesic

if and only if ¨

is perpendicular to both

and

, i.e., ¨

= 0 and ¨

= 0. Since (t) =

(u(t), v(t)), = u

+ v

and hence ¨

( u

+ v

). It follows that is a geodesic if and

only if

( u

+ v

) = 0 and

( u

+ v

) = 0. Now

(

( u

+ v

)) =

( u

) + ( u

+ v

)( u

+ v

). Therefore

( u

+ v

) =

(

( u

+ v

)) - ( u

+ v

)( u

+ v

)

( uE + vF ) - ( u

+ u v(

) + v

)

( uE + vF ) - ( u

+ u vF

+ v

)

13. Geodesics

( uE + vF ) -

( u

+ u v2F

+ v

Similarly, we get

( u

+ v

) =

( uF + vG) -

( u

+ u v2F

+ v

Hence is a geodesic if and only if both the equations in (4.13.6.1) are satisfied.

Example 4.13.7. Geodesics on the cylinder.

We know that a parametrization of the cylinder S may be given as (u, v) = (cos u, sin u, v).

The first order magnitudes of the cylinder are E = 1, F = 0 and G = 1. Therefore E

= E

= F

= G

= 0. Let (t) = (u(t), v(t)) be a unit-speed curve on S. Then is a

geodesic equation if and only if the geodesic equations hold, i.e.,

( u) = ¨

u = 0 and

( v) = ¨

v = 0.

This implies that u(t) = at + b and v(t) = ct + d for some constants a, b, c and d. Therefore

(t) = (at + b, ct + d) = (cos(at + b), sin(at + b), ct + d).

Since is unit-speed curve, a

+ c

= 1.

If a = 0 and c = 0, then is (part of) a circle (in fact it is a part of the circle which is the

intersection of S with the plane z = d).

If a = 0 and c = 0, then is (part of) a line.

If a = 0 and c = 0, then is (part of) a helix.

Hence the geodesics on S are (part of) lines, circles and helices on S.

Example 4.13.8. Determine the geodesics on a plane.(Ans. (part of) lines on the plane)

We may parametrize a plane by (u, v) = (u, v, au + bv), where a and b are some constants.

Then E = 1 + a

F = ab and G = 1 + b

. Therefore E

= E

= F

= G

= 0.

Let (t) = (u(t), v(t)) = (u(t), v(t), au(t) + bv(t)) be a unit-speed curve on the plane. Then

is a geodesic if and only if

((1 + a

) u + ab v) = 0 and

(ab u + (1 + b

) v) = 0 if and only if

(1+a

) u+ab v = and ab u+(1+b

) v = for some constants and if and only if u(t) = At+B

and v(t) = Ct+D for some constants A and B. Hence (t) = (A, C, Aa+Cc)t+(B, D, Ba+Dc).

Therefore is (part of) a line.

Example 4.13.9. Geodesics on the unit sphere S = {(x, y, z) : x

+ y

+ z

= 1}.

We know that the sphere S may be parameterized as

(u, v) = (cos u cos v, sin u cos v, sin v).

Then

= (- sin u cos v, cos u cos v, 0),

= (- cos u sin v, - sin u sin v, cos v). Therefore

E = cos

v, F = 0 and G = 1,

4. GEODESICS AND SOME FUNDAMENTAL RESULTS

so E

= F

= G

= 0 and E

= -2 cos v sin v. Let (t) = (u(t), v(t)) be a

unit-speed curve on S. Since is a unit-speed curve

1 =

u +

= u

cos

v + v

(4.13.9.1)

Suppose that is a geodesic. Then the geodesic equations imply that

( u cos

v) = 0 and

( v) =

(-2 u

cos v sin v). Therefore u cos

v = , for some constant .

If = 0, then

u = 0, i,e, u is constant. Therefore is a part of meridian.

Let = 0. Since is a unit-speed curve,

= 1 - u

cos

v = 1 -

cos

. Now

cos

- cos

v = cos

-2

cos

v - 1 .

Therefore

±du =

cos v

-2

cos

v - 1

Solving above differential equation we get

±(u - u

) = sin

-1

tan v

-2

- 1

i.e., tan v = ± sin(u - u

)

-2

- 1.

(4.13.9.2)

Take a =

sin u

-2

- 1 and b = ± cos u

-2

- 1. Then

sin u

-2

- 1 cos u cos v ± cos u

-2

- 1 sin u cos v

-2

- 1 cos v(cos u sin u

- cos u

sin u)

= ±

-2

- 1 cos v sin(u - u

)

= sin v.

(

equation (4.13.9.2))

Therefore every point on the curve is lying on the plane Ax+By = z. Hence is on the intersection

of the sphere and the plane Ax + By = z. This shows that is part of a great circle of the sphere.

Hence the geodesics on the sphere are great circles of the sphere. Since every great circle is a normal

section, it follows that the geodesics on the sphere are precisely (part of) great circles of the sphere.

Example 4.13.10. Consider the surface of revolution (u, v) = (f (u) cos v, f (u) sin v, g(u)), where

f > 0 and

= 1. Then

(i) Every meridian is a geodesic, i.e., the curves (t) = (u(t), v

) is a geodesic.

(ii) A parallel u = u

(say) is a geodesic if and only if

) = 0.

14. Codazzi - Mainardi and Gauss equations

Here

= (

cos v,

sin v,

) and

= (-f sin v, f cos v, 0). Therefore E = 1, F = 0

and G = f

, and so E

= E

= F

= G

= 0 and G

= 2f

Let (t) = (u(t), v(t)) be a unit-speed curve on S. Since is unit-speed,

1 =

u +

= u

+ f

(4.13.10.1)

Then is a geodesic if and only if

( u) = f

and

v) = 0.

(4.13.10.2)

(i) Let (t) = (u(t), v

). Since is unit-speed, u = ±1 [see (4.13.10.1)]. Hence the equations in

(4.13.10.2) are satisfied, i.e., is a geodesic.

(ii) Let (t) = (u

, v(t)) be a parallel. Since is unit-speed, v = ±

f (u

)

[see (4.13.10.1)]. Clearly

the second equation in (4.13.10.2) holds and the first equation holds if and only if 0 =

( u) =

f (u

)

), i.e., if and only if

) = 0 as f (u

) > 0. Hence is a geodesic if and only if

) = 0.

14. Codazzi - Mainardi and Gauss equations

Proposition 4.14.1 (Codazzi - Mainardi Equations). Let be a surface patch of an oriented surface

S. Then

- M

= L

+ M (

) - N

- N

= L

+ M (

) - N

. Since

uuv

uvu

, we have

(

+ LN)

= (

+ M N)

We first simplify both the sides of the above equation. Now

(

)

= (

+ LN)

= (

)

+ (

)

+ L

N + LN

= (

)

+ M N + (

)

+ N N + L

N + LN

= ((

)

+ (

)

+ (

)

+(M

+ N

+ L

)N + LN

(4.14.1.1)

Also

(

)

= (

+ M N)

= (

)

+ (

)

+ M

N + M N

= (

)

+ LN + (

)

4. GEODESICS AND SOME FUNDAMENTAL RESULTS

+ M N + M

N + M N

= ((

)

+ (

)

+ (

)

+(L

+ M

)N + M N

(4.14.1.2)

Taking dot product of equations (4.14.1.1) and (4.14.1.2) with N and comparing we get M

+ L

= L

+ M

. Therefore L

- M

= L

+ M (

) - N

Since

uvv

vvu

, we have

(

+ M N)

= (

+ N N)

Now

(

)

= (

+ M N)

= (

)

+ (

)

+ M

N + M N

= (

)

(

+ M N) + (

)

(

+ N N) + M

N + M N

= ((

)

+ ((

)

+ (

)

+(M

+ N

+ M

)N + M N

(4.14.1.3)

Also

(

)

= (

+ N N)

= ((

)

+ (

)

+(L

+ M

+ N

)N + N N

(4.14.1.4)

Taking dot product of equations (4.14.1.3) and (4.14.1.4) with N and comparing we get M

+ M

= L

+ M

+ N

. Therefore M

- N

= L

+ M (

) - N

Proposition 4.14.2. Let be a surface patch of an oriented surface S. Then

EG - F

((M F - LG)

+ (LF - M E)

)

EG - F

((N F - M G)

+ (M F - N E)

. We know that the matrix of W

with respect to the basis {

} is

-1

EG-F

LG - M F

M G - N F

-LF + M E -M F + N E

. Therefore N

= -W

(

) = -

LG-M F

EG-F

-LF +M E

EG-F

i.e., N

EG-F

((M F - LG)

+ (LF - M E)

). Similarly one gets the desired expression for

14. Codazzi - Mainardi and Gauss equations

Proposition 4.14.3 (Gauss Equations). Let be a surface patch of an oriented surface S, and let K

be the Gaussian curvature of the surface. Then

EK =

- (

)

- (

)

+ (

)

F K =

+ (

)

- (

)

F K =

+ (

)

- (

)

GK =

- (

)

+ (

)

- (

)

. We have

(

)

= ((

)

+ (

)

+ (

)

+(M

+ N

+ L

)N + LN

= ((

)

+ (

)

+ (

)

+(M

+ N

+ L

)N +

EG - F

((N F - M G)

+ (M F - N E)

)

(

)

+ (

)

L(N F - M G)

EG - F

+ (

)

L(M F - N E)

EG - F

+(M

+ N

+ L

)N.

(4.14.3.1)

Also

(

)

= ((

)

+ (

)

+ (

)

+(L

+ M

)N + M N

= ((

)

+ (

)

+ (

)

+(L

+ M

)N +

EG - F

((M F - LG)

+ (LF - M E)

)

(

)

M (M F - LG)

EG - F

+ (

)

+ (

)

M (LF - M E)

EG - F

+(L

+ M

)N.

(4.14.3.2)

Since

uuv

uvu

and the vectors

and N are linearly independent, comparing the coefficient

we get

(

)

+ (

)

L(N F - M G)

EG - F

= (

)

M (M F - LG)

EG - F

4. GEODESICS AND SOME FUNDAMENTAL RESULTS

This gives

F K = F

LN - M

EG - F

= (

)

- (

)

- (

)

Comparing the coefficients of

, we have

+ (

)

L(M F - N E)

EG - F

+ (

)

+ (

)

M (LF - M E)

EG - F

This gives

EK = E

LN - M

EG - F

+ (

)

- (

)

- (

)

The remaining equalities may be obtained by using

vvu

uvv

If is a regular surface patch, then EG - F

> 0, i.e., at least one of E, F and G is non-zero at

any point. We also know that

can be expressed by using the first order magnitudes E, F and G

and their first order partial derivatives. Hence it follows from the Gauss equations that the Gaussian

curvature K can be expressed by the first order magnitudes E, F and G and their partial derivatives.

Gauss's Remarkable Theorem says that the Gaussian curvature is an intrinsic surface property,

i.e., it does only depend on the first fundamental form.

Theorem 4.14.4 (Gauss' Theorema Egregium). The Gaussian curvature of a surface is preserved

under local isometries.

. Let f : S

be a local isometry. Then the first fundamental forms (or the first order

magnitudes) of S

and S

are same. It follows from the Gauss equations that the Gaussian curvature

of S

can be expressed by the first order magnitudes E

, F

and G

and their partial derivatives.

Since the first order magnitudes E

, F

and G

are same as E

, F

and G

respectively, the Gaussian

curvature K

of S

is same as the Gaussian curvature K

of S

. Hence the Gaussian curvature of a

surface is preserved under local isometries.

The following result is analogous to the Fundamental Theorem of Space Curves.

Theorem 4.14.5 (Bonnet's Theorem). If : U R

and : U R

are surface patches with

the same first and second fundamental forms, then there is a direct isometry M of R

such that

= M ()

Further, let V be an open subset of R

and let E, F , G, L, M and N be smooth functions on

. Assume that E > 0, G > 0, EG - F

> 0

and that the Mainardi - Codazzi equations and

Gauss equations hold, with K =

LN -M

EG-F

. If (u

, v

) V

, then there is an open set U contained in

and containing (u

, v

)

, and a surface patch : U R

, such that Edu

+ 2F dudv + Gdv

and Ldu

+ 2M dudv + N dv

are the first and second fundamental forms of , respectively.

15. The Gauss-Bonnet Theorem

15. The Gauss-Bonnet Theorem

Theorem 4.15.1 (Gauss-Bonnet Theorem). Let be a positively oriented piecewise smooth curve,

consisting on n smooth arcs with exterior angles

, . . . ,

, on a smooth surface S with Gauss-

ian curvature K. Let int () be simply connected. Then

(s)ds +

int()

K dudv +

i=1

= 2,

(4.15.1.1)

where

is the geodesic curvature of .

Example 4.15.2. The sum of interior angles of a regular n-gon on a plane is (n - 2).

Let be the regular n-gon and that it is positively oriented. Since is made up of lines of the

plane (which are geodesics),

= 0. We also note that the Gaussian curvature K of the plane is 0.

Let

, . . . ,

be the exterior angles of , and let

, . . . ,

be the interior angles of . Then

= (i = 1, 2, . . . , n). Clearly int() is simply connected. Hence by Gauss-Bonnet Theorem

we have

i=1

= 2, i,e,

i=1

( -

) = 2. This implies that

i=1

= (n - 2).

Example 4.15.3. The sum of interior angles of a triangle on a sphere is > .

Let be the triangle and that it is positively oriented. Since is made up of geodesics of the

sphere (which are geodesics),

= 0. We also note that the Gaussian curvature K of the sphere of

radius a is

. Let

be the exterior angles of , and let

be the interior angles of

. Then

= (i = 1, 2, 3). Clearly int() is simply connected. Hence by Gauss-Bonnet

Theorem we have

int()

dudv +

i=1

= 2. Therefore

i=1

= +

int()

dudv > .

Bibliography

[1] Andrew Pressley, Elementary Differential Geometry, SUMSeries, second edition, Springer London Dordrecht

Heidelberg New York, 2010.

[2] Goetz A., Introduction to Differential Geometry, Addison Wesley, Publ. Co., 1970.

[3] Manfredo P. do Carmo, Differential Geometry of Curves and Surfaces, Prentice-Hall, Inc., Englewood Cliffs,

New Jersey, 1976.

[4] Richard S. Millman and George D. Parker, Elements of Differential Geometry, Prentice-Hall Inc., Englewood

Cliffs, New Jersey, 1977.

[5] Weatherburn, C.E., Differential Geometry in Three Dimensions, Cambridge University Press, 1964.

Excerpt out of 77 pages

Details

Title: Lecture Notes on Differential Geometry
Course: Differential Geometry
Author: Dr./Ph.D. Prakash Dabhi (Author)
Year: 2015
Pages: 77
Catalog Number: V303166
ISBN (eBook): 9783668015722
ISBN (Book): 9783668015739
File size: 1144 KB
Language: English
Keywords: lecture, notes, differential, geometry

Quote paper: Dr./Ph.D. Prakash Dabhi (Author), 2015, Lecture Notes on Differential Geometry, Munich, GRIN Verlag, https://www.grin.com/document/303166

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