# Lecture Notes on Differential Geometry

## 77 Pages

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
1. Curves
1
2. Regular curves
3
3. Curvature and Torsion
7
4. Fundamental Theorem of Space Curves
17
5. Isoperimetric Inequality and Four Vertex Theorem
19
Chapter 2. Surfaces in R
3
27
6. Surfaces
27
7. Calculus on Surface
30
8. First Fundamental Form
35
9. Local isometries and conformal maps
36
Chapter 3. Curvature of a surface
45
10. Second Fundamental Form
45
11. Gaussian, Mean and Principal Curvatures
47
Chapter 4. Geodesics and some fundamental results
59
12. Christoffel's Symbols and tangent vector fields
59
13. Geodesics
61
14. Codazzi - Mainardi and Gauss equations
65
15. The Gauss-Bonnet Theorem
69
Bibliography
71
v

CHAPTER 1
Curve Theory
1. Curves
Definition 1.1.1. Let I = (a, b) be an interval, - a < b . A continuous map : I R
n
is called a parametrized curve in R
n
.
Example 1.1.2. Parametrization of Cartesian curves.
(i) A parametrization of a parabola y = x
2
is (t) = (t, t
2
), t R. The curve
1
(t) = (t
2
, t
4
), t
R is not a parametrization of y = x
2
. The curve (t) = (t
3
, t
6
), t R is a parametrization of
y = x
2
.
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
x
2
a
2
+
y
2
b
2
= 1.
(iii) The curve (t) = (a sec t, b tan t), t (-
2
,
2
) is a parametrization of the hyperbola
x
2
a
2
-
y
2
b
2
=
1, x > 0.
(iv) The curve (t) = (a cos
3
t, a sin
3
t), t R is a parametrization of the astroid x
2
3
+ y
2
3
= a
2
3
Example 1.1.3.
(i) Let : R R
2
be defined by (t) = (a cos t, b sin t), where a, b R\{0}. Then the image
of is an ellipse in R
2
. Its Cartesian equation is
x
2
a
2
+
y
2
b
2
= 1. In particular when a = b, then
the image of is a circle in R
2
.
(ii) Let : R R
2
be defined by (t) = (t, t
2
). Then the image of is a parabola in R
2
. Its
Cartesian equation is y = x
2
.
(iii) Let : R R
2
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
x
2
a
2
-
y
2
b
2
= 1 such
that x > 0.
(iv) Let : R R
2
be defined by (t) = (e
t
cos t, e
t
sin t). Then the image of is a logarithmic
spiral in R
2
.
(v) Let : R R
3
be defined by (t) = (a + lt, b + mt, c + nt), where a, b, c, l, m, n R and
l
2
+ m
2
+ n
2
= 0. Then the image of is a line in R
3
passing through (a, b, c) having direction
(l, m, n). Its Cartesian equation is
x-a
l
=
y-b
m
=
z-c
n
.
1

2
1. CURVE THEORY
(vi) Let : R R
3
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
3
. It is a helix with the base ellipse
x
2
a
2
+
y
2
b
2
= 1. When a = b, it is
called circular helix.
(vii) Let (t) = (cos
2
t, sin
2
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
, t
2
). Then the Cartesian form of the curve is y = (ln x)
2
.
Definition 1.1.4. A parametrized curve in R
3
is called planar if it is contained in some plane of R
3
.
It is called non-planar or twisted if it is not planar.
Let : (a, b) R
n
be a parametrized curve. Then = (
1
,
2
, . . . ,
n
), where each
i
is a
mapping from (a, b) to R. The symbol
is the derivative of and it means
= (
1
,
2
, . . . ,
n
).
Throughout it is assumed that all curves are smooth, i.e., all the derivatives of exist.
We note that a parametrized curve = (
1
,
2
,
3
) : (a, b) R
3
is planar if and only if there
are real numbers , , , R with
2
+
2
+
2
= 0 such that
1
(t) +
2
(t) +
3
(t) = for
all t (a, b).
Example 1.1.5.
(i) The curve (t) = (t, t
2
, t
3
) is not planar.
Suppose that is planar. Then the exist a, b, c, d R, a
2
+b
2
+c
2
= 0, such that at+bt
2
+ct
3
=
d for all t. It follows from Fundamental Theorem of Algebra that a = b = c = 0. This
2
+ b
2
+ c
2
= 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
1
(t) =
4
5
cos t, 1 - sin t, -
3
5
cos t) is a plane curve.
Definition 1.1.6. Let : (a, b) R
n
be a parametrized curve, and let t (a, b). The vector
(t) is
called the tangent vector to at the point (t).
If
(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
2
, t
3
), (1, 1, 1).
(c) (t) = (cos
2
t, sin
2
t), (
1
2
,
1
2
).
(d) (t) = (e
t
, t
2
), (1, 0).
(ii) Sketch the astroid (t) = (cos
3
t, sin
3
t). Calculate its tangent vector at each point. At which
point is the tangent vector zero?

2. Regular curves
3
Example 1.1.8. Consider the logarithmic spiral (t) = (e
t
cos t, e
t
sin t). We show that the angle
between (t) and
(t) is independent of t.
Here (t) = (e
t
cos t, e
t
sin t) and (t) = (e
t
cos t - e
t
sin t, e
t
sin t + e
t
cos t). Let (t) be the
angle between (t) and
(t). Then
(t) = cos
-1
(t) (t)
(t)
(t)
= cos
-1
e
2t
2e
2t
= cos
-1
1
2
.
Therefore the angle between (t) and
(t) is independent of t.
Proposition 1.1.9. Let be a parametrized curve in R
n
. If = a for some nonzero vector a R
n
,
then is part of a line.
P
. Since
= a, we have (t) = at + b. Hence is part of a line.
2. Regular curves
Definition 1.2.1. Let : (a, b) R
n
be a parametrized curve. Then the arc-length s(t) of a curve
starting at the point (t
0
) is given by
s(t) =
t
t
0
(u) du (t (a, b)).
Since
is a nonnegative continuous function, it follows that s(t
0
) = 0, s(t) 0 if t > t
0
and
s(t) < 0 if t < t
0
.
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
=
b
a
1 + f (x)
2
dx.
We may parametrize y = f (x) by (t) = (t, f (t)), t [a, b]. Then
(t) = (1, f (t)). Thus
=
b
a
(u) du =
b
a
1 + f (u)
2
du.
(ii) Compute the arc-length of the logarithmic spiral (t) = (e
kt
cos t, e
kt
sin t) starting at the point
(1, 0).
We have
(t) = (-e
kt
sin t + ke
kt
cos t, e
kt
cos t + ke
kt
sin t). Therefore
(t) = e
kt
(- sin t + k cos t)
2
+ (cos t + k sin t)
2
= e
kt
1 + k
2
.

4
1. CURVE THEORY
Since (0) = (1, 0), the arc-length of starting at the point (1, 0) is
s(t) =
t
0
e
ku
1 + k
2
du =
1 + k
2
k
(e
kt
- 1).
Definition 1.2.3. Let : (a, b) R
n
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
n
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
3
(1 + t)
3/2
,
1
3
(1 - t)
3/2
,
t
2
)
(b) (t) = (
4
5
cos t, 1 - sin t, -
3
5
cos t)
(iii) Which of the following curves are regular?
(a) (t) = (cos
2
t, sin
2
t) for t R.
(b) The same curve as in (i), but with 0 < t < /2.
(c) (t) = (t, cosh t) for t R.
Lemma 1.2.6. Let : (a, b) R
n
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.
P
. 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
n
be a regular curve. Then the arc-length of , starting at any
point of the curve, is a smooth map.
P
. We have s(t) =
t
t
0
(u) du. Then
ds
dt
=
. Let = (
1
,
2
, . . . ,
n
). Since is regular,
it follows that
2
1
+
2
2
+ · · · +
2
n
> 0. Define f : (0, ) R and g : (a, b) R by f (t) =
t
and g(t) =
2
1
(t) + · · · +
2
n
(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 =
ds
dt
.
Therefore
ds
dt
is a smooth map and hence s is a smooth map.

2. Regular curves
5
Definition 1.2.8. Let : (a, b) R
n
be a parametrized curve. A parametrized curve : (~
a, ~
b) R
n
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
n
. 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.
P
. Let : (~
a, ~
b) R
n
be a reparametrization of a regular curve : (a, b) R
n
. 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
d
d~
t
d
dt
= 1
and hence
d
d~
t
(~
t) = 0 for any ~t (~a, ~b). Since = , we have
d
d~
t
=
d
dt
d
d~
t
. Since is regular
d
dt
= 0. Since
d
d~
t
is never vanishing, it follows that
d
d~
t
is never zero (vector), i.e., is regular.
The following is a characterization of regular curves.
Proposition 1.2.11. Let : (a, b) R
n
be a parametrized curve. Then is regular if and only if
it has a unit-speed reparametrization.
P
. Let : (a, b) R
n
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
0
), i.e.,
s(t) =
t
t
0
(u) du. We note that s is a smooth map. Since is regular,
ds
dt
(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
ds
dt
(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
n
given by = s
-1
, i.e., s = . Now
d
d~
t
ds
dt
=
d
dt
. Then
d
d~
t
ds
dt
=
d
dt
=
ds
dt
, i.e.,
d
d~
t
= 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 .
P
. Since u = , we have
d
d~
t
du
dt
=
d
dt
. This implies
d
d~
t
du
dt
=
d
dt
. Therefore
du
dt
=
ds
dt
,
as is unit-speed. Since both
du
dt
and
ds
dt
are strictly monotonic continuous functions, it follows that
du
dt
(t) =
ds
dt
(t) for all t or
du
dt
(t) = -
ds
dt
(t) for all t. Hence u = ±s + c for some constant c.

6
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 -
d
d~
t
ds
dt
=
d
dt
. This implies
d
d~
t
ds
dt
=
d
dt
=
ds
dt
. Therefore
d
d~
t
= 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
3
be a regular curve which does not pass through the origin. If (t
0
) is the
point on the trace of closest to the origin and
(t
0
) = 0 for t
0
(a, b), then show that (t
0
)
is perpendicular to
(t
0
).
Define f : (a, b) R by f (t) = (t)
2
= (t)(t). Then f is a smooth map. Since (t
0
) is
point closest to the origin, f has its minimum at t
0
. Therefore
f (t
0
) = 0, i.e., (t
0
) (t
0
) = 0
(ii) Let : (a, b) R
3
be a regular curve and v R
3
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
3
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
2
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)
2
= 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
3
be a regular curve, and I is an interval. Let [a, b] I. Let (a) = p and
(b) = q.
(a) If v R
3
with v = 1, then
(q - p) · v =
b
a
(t).v dt
b
a
(t) dt.
(b) Set v =
q - p
q - p
and show that
(b) - (a)
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
b
a
(t)vdt = ((b) - (a))v. Also
b
a
(t)vdt
b
a
(t) v dt =
b
a
(t) dt. Thus
b
a
(t)vdt
b
a
(t) dt.

3. Curvature and Torsion
7
Let v =
q-p
q - p . Then it follows from above inequality that
q - p
=
(b) - (a)
b
a
(t) dt. Since
b
a
(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
3
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
3
, and let a
= 1. Then the curvature of the unit-speed curve (t) = ta + b is
zero everywhere.
(ii) Let (x
0
, y
0
) R
2
, and let R > 0. Then the curvature of the unit-speed curve (t) = (x
0
+
R cos(
t
R
), y
0
+ R sin(
t
R
)), which is a circle with center (x
0
, y
0
1
R
everywhere.
(iii) Let a, b R, and let a
2
+ b
2
= 0. Let (t) = (a cos(
t
a
2
+b
2
), a sin(
t
a
2
+b
2
),
bt
a
2
+b
2
). Then
easy computations show that (t) =
|a|
a
2
+b
2
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
3
, then its curvature is given by the formula
=
¨
×
3
.
P
. 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
ds
dt
2
=
2
= ¨
, and hence
ds
dt
d
2
s
dt
2
= ¨
. Let denote the derivative of
with respect to s. Since s = ,
ds
dt
= , i.e., =
ds
dt
. Differentiating it again, we have
=
ds
dt
¨
-
d
2
s
dt
2
ds
dt
2
dt
ds
=
ds
dt
2
¨
-
ds
dt
d
2
s
dt
2
ds
dt
4
=
( )¨
- ( ¨
)
4
=
× (¨ × )
4
.
Hence =
=
× (¨ × )
4
=
(¨ × )
4
=
¨
×
3
. The last step follows because
¨
× .
Example 1.3.4.

8
1. CURVE THEORY
(i) Let : (a, b) R
2
be a regular parametrized curve. Assume that there exists t
0
(a, b)
such that the distance |(t)| from the origin to the trace of is maximum at t
0
. We prove that
|(t
0
)| >
1
(t
0
) .
We may assume that is unit-speed. Define f : (a, b) R by f (t) =
(t)
2
= (t)(t).
Then f is a smooth map. Since (t
0
) is farthest from origin, the function f has maximum at
t
0
. Thus f (t
0
) = 0 and f (t
0
) 0. Since f (t
0
) 0, we have (t
0
(t
0
) + (t
0
)(t
0
) 0.
As is unit-speed, (t
0
(t
0
) -1. This gives |(t
0
(t
0
)| 1. Now
1 |(t
0
(t
0
)| (t
0
)
¨
(t
0
) = (t
0
) (t
0
).
Thus |(t
0
)| >
1
(t
0
) .
(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) =
a
2
+ b
2
and ¨
(t)× (t) = (-ab sin t, ab cos t, -a
2
). Therefore ¨
(t) × (t) = a
b
2
+ a
2
.
Thus (t) =
¨
×
3
=
a
a
2
+b
2
(a
2
+b
2
)
3
2
=
a
a
2
+b
2
.
Exercise 1.3.5.
(i) Compute the curvature of the following curves.
(a) (t) = (e
kt
cos t, e
kt
sin t).
(b) (t) = (t, cosh t). Also determine a point having maximum curvature.
(c) (t) = (t, t
2
, t
3
).
(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
2
be a unit-speed curve. Let t be the unit tangent of , i.e., t =
.
The unit vector n
s
obtained by rotating the tangent vector t anti-clockwise by an angle
2
is called
the signed unit normal of . Since is unit-speed, ¨
is perpendicular to t and hence ¨
and n
s
are
linearly dependent. Therefore there exists a map
s
: (a, b) R such that ¨
=
s
n
s
. The map
s
is called the signed curvature of .
Since =
¨
, it follows that = |
s
|, i.e., the absolute value of the signed curvature is the
curvature
Let R
: R
2
R
2
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
2
.
Then
R
(x, y) = R
(x(1, 0) + y(0, 1))

3. Curvature and Torsion
9
= xR
(1, 0) + yR
(0, 1)
= x(cos , sin ) + y(- sin , cos ).
Hence
R
(x, y) = (x cos - y sin , x sin + y cos ).
In fact, R
is an isometry of R
2
, i.e., it satisfies
R
(x
1
, y
1
) - R
(x
2
, y
2
) = (x
1
, y
1
) - (x
2
, y
2
)
((x
1
, y
1
), (x
2
, y
2
) R
2
).
In particular, when =
2
, R
2
(x, y) = (-y, x).
Thus if : (a, b) R
2
is a unit-speed curve, then the signed unit normal of can be computed
by the formula n
s
(t) = R
2
(t(t)) for all t.
Fix a vector a R
2
. Define T
a
: R
2
R
2
by T
a
(x) = x + a. Then T
a
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
2
of the form T
a
R
is called a direct isometry, where a R
2
and [0, 2).
Example 1.3.8. Compute the signed unit normal of the curve (t) = (e
kt
cos t, e
kt
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
kt
sin t + ke
kt
cos t, e
kt
cos t + ke
kt
sin t) and (t) =
e
kt
1 + k
2
. Hence t(t) =
1
1+k
2
(- sin t + k cos t, cos t + k sin t). Since n
s
(t) = R
2
(t(t)), we
have
n
s
(t) =
1
1 + k
2
(- cos t - k sin t, - sin t + k cos t)
Exercise 1.3.9. If : (a, b) R
2
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
2
be a unit-speed curve, and let s
0
(a, b). Let
0
R be
such that (s
0
) = (cos
0
, sin
0
)
. Then there exists a unique smooth map : (a, b) R such
that (s) = (cos (s), sin (s)) for all s (a, b) and (s
0
) =
0
.
P
. Since
: (a, b) R
2
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
0
) = (cos
0
, sin
0
), we have f (s
0
) = cos
0
and
g(s
0
) = sin
0
. Since is unit-speed, f
2
+ g
2
= 1 and hence f
f + g g = 0. Define : (a, b) R
by
(s) =
0
+
s
s
0
(f g -
f g).

10
1. CURVE THEORY
Since f and g are smooth, is a smooth map and (s
0
) =
0
.
Let F = f cos + g sin and G = f sin - g cos . Then
F
=
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
2
) + f g g =
f f
2
+ 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
0
) for all s. Now
F (s
0
) = f (s
0
) cos (s
0
) + g(s
0
) sin (s
0
) = cos
2
0
+ sin
2
0
= 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
0
) =
0
.
For the uniqueness, let : (a, b) R be a smooth map satisfying (s
0
) =
0
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
0
) = (s
0
) =
0
, we
get c = 0. Hence (s) = (s) for all s.
Definition 1.3.11. Let : (a, b) R
2
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
0
) =
0
. The map is called the turning angle of .
Proposition 1.3.12. Let : (a, b) R
2
be a unit-speed curve, and let be the turning angle of .
Then the signed curvature of can be determined by the formula
s
=
.
P
. Since is the turning angle of , we have
(s) = (cos (s), sin (s)) for all s. Hence
¨
= (- sin , cos )
. Now n
s
= R
2
( ) = (- sin , cos ). Therefore ¨
=
n
s
and hence the
signed curvature
s
of is
, i.e.,
s
=
.
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
2
is a unit-speed curve, then its signed curvature and signed unit
normal are smooth maps.
The map n
s
: (a, b) R
2
is given by n
s
(t) = R
/2
(t(t)). The map R
/2
: R
2
R
2
defined
by R
/2
(x, y) = (-y, x) is a smooth map and the map t = is a smooth map and hence the map
R
/2
t = n
s
is a smooth map, i.e., the signed unit normal is a smooth map.

3. Curvature and Torsion
11
Now the map
s
: (a, b) R is given by ¨
(t) =
s
(t)n
s
(t). Therefore
s
(t) = ¨
(t)n
s
(t).
Since both ¨
and n
s
are smooth maps, it follows that
s
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
2
whose signed curvature is k. Moreover, if : (a, b) R
2
is a unit-speed curve whose signed
curvature is k, then there is a direct isometry M of R
2
such that = M .
P
. Let s
0
(a, b). Define : (a, b) R by (s) =
s
s
0
k(u)du. Since k is smooth, is a
smooth map. Note that (s
0
) = 0. Define : (a, b) R
2
by
(s) =
s
s
0
cos (u)du,
s
s
0
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
) = 0 and hence the signed curvature
of is
. But
= k. Hence the signed curvature of is k.
Let : (a, b) R
2
be a unit-speed curve with signed curvature k. Let be the turning angle
of . Then
= k. But then (s) =
s
s
0
k(u)du + (s
0
) = (s) + , where = (s
0
) is a constant.
So we have = + . Since is the turning angle of ,
(s) = (cos (s), sin (s)) for all s. Let
a = (s
0
). Integrating the above equation, we have
(s) =
s
s
0
cos du,
s
s
0
sin du
+ a = T
a
s
s
0
cos du,
s
s
0
sin du
= T
a
s
s
0
cos( + )du,
s
s
0
sin( + )du
= T
a
s
s
0
(cos cos - sin sin )du,
s
s
0
(sin sin + cos cos )du
= T
a
cos
s
s
0
cos du - sin
s
s
0
sin du, cos
s
s
0
cos du + sin
s
s
0
sin du
= T
a
R
s
s
0
cos du,
s
s
0
sin du
= T
a
R
((s)).

12
1. CURVE THEORY
Therefore (s) = M ((s)) for all s, where M = T
a
R
is a direct isometry. Hence = M .
Exercise 1.3.15.
(i) If is a unit-speed plane curve, then show that
n
s
= -
s
t.
(ii) Let and be two plane curves. Show that, if is obtained from by applying an isometry M
of R
2
, the signed curvatures
s
and
s
of and are equal if M is direct but that
s
= -
s
if
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
2
.
Example 1.3.16.
(i) Let be a regular plane curve and let be a constant. The parallel curve
of is defined
by
(t) = (t) + n
s
(t). If
s
(t) = 1 for all values of t, then we prove that
is a regular
curve and that its signed curvature is
s
/|1 -
s
|.
Here
s
- 1 = 0. Since
s
- 1 is a continuous, it follows that
s
(t) - 1 > 0 for all t or
s
(t) - 1 < 0 for all t.
Assume that
s
(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
s
(t). Differentiating it
with respect to s we get = (t +
n
s
)
dt
ds
= (t -
s
t)
dt
ds
= (1 -
s
)t
dt
ds
. As
= 1, we
have
dt
ds
=
1
|1-
s
|
=
1
1-
s
because 1 -
s
> 0. Thus = t. Now =
t
dt
ds
= ¨
1
1-
s
=
s
1-
s
n
s
. Let n
s
be the signed unit normal of . As unit-tangents to and are same, they have
the same signed unit normal, i.e., n
s
= n
s
. Hence =
s
1-
s
n
s
. This shows that the signed
curvature of is
s
1-
s
.
Similarly, if
s
(t) - 1 > 0 for all t, then the signed curvature of will be
s
s
-1
. Hence the
result.
(ii) Let : (a, b) R
2
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
2
be defined by (t) = (
1
cos( t), -
1
sin( t)). Then
is a unit-speed and t(t) = (t) = (- sin( t), - cos( t)), ¨
(t) = (- cos( t), sin( t)).
Hence n
s
(t) = (cos( t), - sin( t)). Therefore ¨
(t) = - n
s
(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
2
such that = M . Since N is (part of) a circle for any direct isometry
N of R
2
, 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
13
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
2
t
=
1
cosh t
.
Therefore sin (t) = tanh t and tan (t) = sinh t. Let s be the arc-length of , i.e., s(t) =
t
0
(u) du = sinh t. Differentiating tan (t) = sinh t = s with respect to s we have
sec
2
= 1, i.e, (1 + tan
2
)
= 1. Therefore
=
1
1+tan
2
=
1
1+s
2
=
1
1+sinh
2
t
=
1
cosh
2
t
.
Hence
s
(t) =
1
cosh
2
t
.
(ii) Give an example of a plane curve having signed curvature
1
1+s
2
.
It is
(s) =
s
0
cos(tan
-1
u)du,
s
0
sin(tan
-1
u)du
.
See the proof of Theorem 1.3.14.
Definition 1.3.18. Let : (a, b) R
3
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
3
be a regular curve with nowhere vanishing curvature.
Then the torsion of is given by
=
( × ¨
)
...
× ¨
2
.
P
. 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)) =
2
,
and
× ¨
2
= b
2
=
2
. Hence
(
×¨
)
...
× ¨
2
=
2
2
= .
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

14
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
ds
dt
= , i.e,. =
ds
dt
. Therefore (
ds
dt
)
2
+
d
2
s
dt
2
= ¨
and
(
ds
dt
)
3
+ 2
ds
dt
d
2
s
dt
2
+
ds
dt
d
2
s
dt
2
+
d
3
s
dt
3
=
...
. It follows that × ¨
= ( × )(
ds
dt
)
3
and (
× ¨
)
...
=
( × ) (
ds
dt
)
6
. Also
× ¨
2
= ×
2
(
ds
dt
)
6
. Hence
(
×¨
)
...
× ¨
2
=
( × )
×
2
= , 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
2
)
3
2
.
(ii) Compute curvature and torsion of the following curves.
(a) (t) = (t, t
2
, t
3
).
(b) (t) = (a cos t, a sin t, bt)
(c) (t) = (e
t
cos t, e
t
sin t, e
t
)
(d) (t) = (
4
5
cos t, 1 - sin t,
3
5
cos t)
(e) (t) = (
1
3
(1 + t)
3/2
,
1
3
(1 - t)
3/2
,
t
2
)
(f) (t) = (
1 + t
2
, ln(t +
1 + t
2
))
Example 1.3.21.
(i) If : (a, b) R
3
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
3
and , : (a, b) R. By definition (t) =
¨
(t) . Let
= (
1
,
2
,
3
). Since is nowhere vanishing,
2
1
(t) +
2
2
(t) +
2
3
(t) > 0 for all t. Define
f : (0, ) R by f (t) =
t and g : (a, b) R by g(t) =
2
1
(t) +
2
2
(t) +
2
3
(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,
1
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
3
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
3
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
15
Conversely, assume that = 0. Since
b = - n, we have
b = 0, i.e, b is a constant unit vector.
Now
d
dt
( 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
3
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
d
dt
( +
1
n) = t +
1
n = t +
1
( b - t) = 0.
Therefore +
1
n = a for some constant vector a R
3
. But then - a =
1
k
, i.e., is lying
on the sphere R - a =
1
. 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
3
with nowhere vanishing curvature. Show that is spherical if
and only if
=
d
dt
2
.
We may assume that is unit-speed. Assume that is spherical. Then there exist a constant
vector a R
3
and a constant R R such that ( -a)( -a) = - a
2
= R
2
. Differentiating
it we get ( - a)t = 0. Differentiating it again we have ( - a)n = -
1
. Again differentiation
gives ( - a)b =
2
. Since {t, n, b} is an orthonormal basis of R
3
, - a = (( - a)t)t +
(( - a)n)n + (( - a)b)b. Therefore
- a = -
1
n +
2
b.
But then R
2
= - a
2
=
1
2
+
2
2
4
. Differentiating it we obtain
2
4
2
¨
-
2
(2
4
+
2
4
2
)
4
8
-
2
3
= 0, i.e.,
2
¨
-2
2
-
2
4
2
-
= 0, i.e.,
=
d
dt
2
.
Conversely, assume that
=
d
dt
2
. Let a = +
1
n -
2
b. Then
a = t -
2
n +
1
n -
d
dt
2
b -
2
b
= t -
2
n +
1
( b - t) -
d
dt
2
b -
2
(- )n
=
-
d
dt
2
b = 0.
Hence a is a constant vector. Now
- a
2
=
1
2
+
2
2
4
. Again using
=
d
dt
2
, it
follows that
1
2
+
2
2
4
is a positive constant, say, R
2
. Therefore - a
2
= R
2
, and hence
is spherical.
OR
Let =
1
and =
1
. Assume that is spherical. Then there exists a constant vector a R
3
and a constant R > 0 such that ( - a)( - a) =
- a
2
= R
2
. Differentiating it we get
( - a)t = 0. Differentiating it again we have ( - a)n = -
1
= -. One more differentiation

16
1. CURVE THEORY
gives ( - a)b = -
. Now
- a = (( - a)t)t + (( - a)n)n + (( - a)b)b
= -n -
b.
Therefore R
2
=
- a
2
=
- n - b
2
=
2
+ (
)
2
. Differentiation of the last
equation gives
+ (
)
·
= 0, which is required.
Conversely, assume that
+ (
)
·
= 0. Then
2
+ (
)
2
= 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
2
=
- n - b
2
=
2
+ (
)
2
= d. Hence
is spherical.
(v) Let be a unit-speed curve in R
3
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 -
2
n - 2 t -
3
n. Now × ¨
=
2
t +
3
b, ( × ¨
)
...
= -
3
+
4
, =
and
× ¨
=
2
2
+
2
. Let
1
be the curvature of and
1
be the torsion of . Then
1
=
× ¨
3
=
2
2
+
2
3
=
2
+
2
and
1
=
(
× ¨
)
...
× ¨
2
=
( - )
2
+
2
.
(vi) Let be a unit-speed curve in R
3
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
3
. Since is regular,
a = 0. Thus is part of a line.
(vii) Let be a unit-speed curve in R
3
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
17
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 =
1
a
. 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
3
. 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
n = -t
b
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
a
a
2
+ b
2
, a sin
a
a
2
+ b
2
,
b
a
2
+ b
2
.
4. Fundamental Theorem of Space Curves
We have seen that if : (a, b) R
3
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
3
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
3
whose curvature is k and torsion is t.
P
. Fix s
0
(, ). 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
0
), N(s
0
), B(s
0
)} is a standard basis of R
3
. Since the matrix
0
k
0
-k
0
t
0
-t 0
expressing
T,
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

18
1. CURVE THEORY
(s) = 1 for all s or (s) = -1 for all s. Since B(s
0
) = T(s
0
) × N(s
0
), we have (s) = 1 for all
s, i.e., B = T × N. Define : (, ) R
3
by
(s) =
s
s
0
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
3
are unit-speed curves with same curvature and
same torsion, then there is an isometry M of R
3
such that = M .
P
. 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
0
(, ). Since {t(s
0
), n(s
0
), b(s
0
)} and {t(s
0
), n(s
0
), b(s
0
)} are both right
handed orthonormal basis of R
3
, there is a rotation about origin that maps t(s
0
), n(s
0
) and b(s
0
)
to t(s
0
), n(s
0
) and b(s
0
) respectively. Further, there is a translation which takes (s
0
) to (s
0
)
(and this has no effect on t, n and b). By applying rotation followed by translation, we can therefore
assume that (s
0
) = (s
0
), t(s
0
) = t(s
0
), n(s
0
) = n(s
0
) and b(s
0
) = b(s
0
). 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
0
) = 3. Also observe that
A(s) = 0
for all s (you need to verify this). Therefore A is a constant map. Since A(s
0
) = 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
0
) = (s
0
), we get c = 0. Hence = . This proves the theorem.
Example 1.4.3. Describe all curves in R
3
which have constant curvature > 0 and constant torsion
.
Let : (a, b) R
3
be a unit-speed curve with constant curvature > 0 and constant torsion
. Let a =
2
+
2
and b =
2
+
2
. Define : (a, b) R
3
by
(t) =
a cos
t
a
2
+ b
2
, a sin
t
a
2
+ b
2
, b
t
a
2
+ b
2
.
Then is a unit-speed curve. The curvature of is
a
a
2
+b
2
= and its torsion is
b
a
2
+b
2
= . Therefore
, : (a, b) R
3
are unit-speed curves with same curvature and same torsion. So, there is an
isometry M of R
3
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
19
Exercise 1.4.4.
(i) A regular curve in R
3
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
n
, so that M (v) = P v + a is an isometry of
R
3
. Show that, if is a unit-speed curve in R
n
, 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
2
is a (regular) curve
: R R
2
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
2
has an `interior' and `exterior'. The
following defines interior and exterior of a simple closed curve. The set of points in R
2
that are not
on the curve is the disjoint union of two subsets of R
2
, 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 .
P
. Since is a- periodic, the length, (), of is
() =
a
0
(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) =
t
0
(u) du -
t
0
(u) du =
t
t
(u) du

20
1. CURVE THEORY
=
t+a
t
(u) du +
t+2a
t+a
(u) du + · · · +
t+ka
t+(k-1)a
(u) du +
t+ka+
t+ka
(u) du
= k () +
t+
t
(u) du.
Hence s(t ) - s(t) = k () if and only if
t+
t
(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
2
. 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
2
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
2
, and let be positively
oriented. If f and g are continuous on the closure of int() and smooth on int(), then
int()
g
x
-
f
y
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
2
with period a. Let be
positively oriented. Then
A(int()) =
1
2
a
0
(x
y - y x)dt.
P
. Take f (x, y) = -
1
2
y and g(x, y) =
1
2
x. Then both f and g are smooth functions on R
2
. It
follows from Green's theorem that
int()
(
1
2
- (-
1
2
)dxdy =
[
1
2
xdy -
1
2
ydx]. This implies
A(int()) =
[
1
2
xdy -
1
2
ydx] =
1
2
a
0
(x
y - y x)dt.
Example 1.5.6. If f : [a, b] R is a nonnegative continuous function with
b
a
f (x)dx = 0, then
f 0.
If f is not identically zero, then there is x
0
[a, b] such that f(x
0
) > 0. Take =
f (x
0
)
2
.
Assume that x
0
= a. Then there is > 0 such that |f (x) - f (x
0
)| <
for every x [a, a + ).
This implies that f (x)
f (x
0
)
2
for all x [a, a + ). Now 0 =
b
a
f (x)dx =
a+
a
f (x)dx +
b
a+
f (x)dx
a+
a
f (x
0
)
2
dx =
f (x
0
)
2
> 0. This is a contradiction.

5. Isoperimetric Inequality and Four Vertex Theorem
21
Assume that a < x
0
< b. By continuity of f , there is > 0 such that |f (x) - f (x
0
)| <
for every x [x
0
- , x
0
+ ). It follows that f (x)
f (x
0
)
2
for all x (x
0
- , x
0
+ ). Again
0 =
b
a
f (x)dx
x
0
+
x
0
-
f (x)dx
x
0
+
x
0
-
f (x
0
)
2
dx = f (x
0
) > 0, which is a contradiction.
Similar arguments shows that x
0
= 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
0
F
2
dt
0
F
2
dt
and the equality holds if and only if there is a constant A such that F (t) = A sin t for all t.
P
. Define G : [0, ] R by G(t) =
F (t)
sin t
. Note that G is a smooth map. Then
0
F
2
dt =
0
(
G sin t + G cos t)
2
dt
=
0
G
2
sin
2
tdt + 2
0
G
G sin t cos tdt +
0
G
2
cos
2
tdt
=
0
G
2
sin
2
tdt -
0
G
2
(cos
2
t - sin
2
t)dt +
0
G
2
cos
2
tdt
=
0
G
2
sin
2
tdt +
0
G
2
sin
2
tdt =
0
F
2
dt +
0
G
2
sin
2
tdt.
Hence
0
F
2
dt -
0
F
2
dt =
0
G
2
sin
2
tdt 0.
(1.5.7.1)
By equation (1.5.7.1) it follows that
0
F
2
dt =
0
F
2
dt if and only if
0
G
2
sin
2
tdt = 0 if and only
if
G
2
sin
2
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.

22
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
2
be a simple closed curve with period , and let M : R
2
R
2
be an isometry
1
.
Then there is an orthogonal matrix A and a R
2
such that M (x) = A(x) + a (x R
2
). Also note
that DM (x) = DA(x) = A (x R
2
) Let : R R
2
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
() =
0
(u) du =
0
DM ((u)) du
=
0
DA((u)) (u) du
=
0
A( (u)) du =
0
(u) du = ().
Since M is an isometry, there are a, b, c, d, e, f R
2
such that a
2
+ b
2
= c
2
+ d
2
= a
2
+ c
2
=
b
2
+d
2
= 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()) =
1
2
0
((ax + by + e)(cx + dy + f )
·
- (ax + by + e)
·
(cx + dy + f ))dt
=
1
2
0
((ax + by + e)(c x + d
y) - (a x + b
y)(cx + dy + f ))dt
=
1
2
0
(x
y - xy)dt + (ec - af )(x( ) - x(0)) + (ed - bf )(y( ) - y(0))
=
1
2
0
(x
y - xy)dt = A(int()),
as ad - bc = ±1, x(0) = x( ) and y(0) = y( ) (???).
1
We know that N : R
2
R
2
is an isometry if and only if there are a, b, c, d, e, f R
2
such that a
2
+ b
2
=
c
2
+ d
2
= a
2
+ c
2
= b
2
+ d
2
= 1
, ac + bd = ab + cd = 0 and N (x, y) = (ax + by, cx + dy) + (e, f ) OR N is an
isometry of R
2
if and only if there is a 2 × 2 orthogonal matrix A and a R
2
such that N (x) = Ax + a

5. Isoperimetric Inequality and Four Vertex Theorem
23
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
2
4A. Moreover,
2
= 4A
if and only if is a
circle.
P
. 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) =
t
, i.e., = s or s
-1
= ,
where s(t) =
t
. [Note that is a unit-speed reparametrization of with the parametrization map
s
-1
(t) =
t
, i.e., s(t) =
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
x
2
+
y
2
= r
2
+ r
2
2
and x
y - y x = r
2
. Since
= s, we have =
ds
dt
. But then
r
2
+ r
2
2
= x
2
+
y
2
=
2
=
2
(
ds
dt
)
2
=
2
2
. We also
note that A = A(int()) =
1
2
0
(x
y - y x)dt =
1
2
0
r
2
. Now,
2
4
- A =
1
4
0
( r
2
+ r
2
2
)dt -
1
2
0
r
2
dt
=
1
4
0
( r
2
+ r
2
2
- 2r
2
)dt
=
1
4
0
r
2
(
- 1)
2
dt +
1
4
0
( r
2
- r
2
)dt 0 [
Wirtinger's Inequality].
It follows from the above equation that
2
4
-A = 0 if and only if
1
4
0
r
2
(
-1)
2
dt+
0
( r
2
-r
2
)dt = 0
if and only if
0
r
2
(
- 1)
2
dt =
0
( r
2
- r
2
)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
2
- 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
x
2
a
2
+
y
2
b
2
= 1, we prove that
2
0
a
2
sin
2
t + b
2
cos
2
t dt 2
ab.
We also show that equality holds if and only if a = b.

24
1. CURVE THEORY
Consider (t) = (a cos t, b sin t). Then the trace of is the ellipse
x
2
a
2
+
y
2
b
2
= 1. Note the is
2-periodic. Therefore length of is
=
2
0
(t) dt =
2
0
a
2
sin
2
t + b
2
cos
2
t dt.
Also, the area A of the interior of is
A =
1
2
2
0
(x
y - y x)dt =
1
2
2
0
ab(cos t cos t + sin t sin t)dt =
ab
2
2 = ab.
It follows from the Isoperimetric inequality that
2
0
a
2
sin
2
t + b
2
cos
2
t dt
2
4ab = 4
2
ab,
i.e.,
2
0
a
2
sin
2
t + b
2
cos
2
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
2
is called convex if its interior is a convex subset of
R
2
.
Definition 1.5.12. Let be a regular plane curve. A point (t) on is called a vertexif
s
(t) = 0.
Theorem 1.5.13 (Four Vertex Theorem). Every convex, simple closed curve in R
2
has at least four
vertices.
Example 1.5.14.
(i) The ellipse
x
2
a
2
+
y
2
b
2
= 1 (or the curve (t) = (a cos t, b sin t)) is a convex curve.
Let (x, y), (z, w) int(), i.e.,
x
2
a
2
+
y
2
b
2
< 1 and
z
2
a
2
+
w
2
b
2
< 1. Let X =
x
a
,
y
b
and Y =
z
a
,
w
b
.
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)
2
a
2
+
(ty + (1 - t)w)
2
b
2
= tX + (1 - t)Y
2
< 1.
Therefore t(x, y) + (1 - t)(z, w) int() and hence is convex.

5. Isoperimetric Inequality and Four Vertex Theorem
25
(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)
dt
ds
. Since
= 1,
dt
ds
=
1
a
2
sin
2
t+b
2
cos
2
t
.
Therefore
t(t) = (t) =
1
a
2
sin
2
t + b
2
cos
2
t
(-a sin t, b cos t).
Differentiating with respect to s, we get
(t) =
ab
(a
2
sin
2
t + b
2
cos
2
t)
2
(-b cos t, -a sin t).
Now n
s
(t) = R
2
(t(t)) =
1
(a
2
sin
2
t+b
2
cos
2
t)
1
2
(-b cos t, -a sin t). Hence
=
ab
(a
2
sin
2
t + b
2
cos
2
t)
3
2
n
s
.
This implies that
s
(t) =
ab
(a
2
sin
2
t+b
2
cos
2
t)
3
2
. Now
s
(t) =
-3ab(a
2
-b
2
) sin t cos t
(a
2
sin
2
t+b
2
cos
2
t)
5
2
. Since a, b > 0
and a = b, we have
s
(t) = 0 if and only if sin t cos t = 0 if and only if sin 2t = 0 if and
only if t =
n
2
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
3
6. Surfaces
Definition 2.6.1. A subset S of R
3
is called a surface if for every p S there is an open subset W
p
of R
3
containing p and an open subset U
p
of R
2
such that S W
p
and U
p
are homeomorphic.
Thus if S is a surface, then for every p S there is homeomorphism
p
: U
p
S W
p
. 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
2
: z = x
2
+ y
2
} is a surface.
Let W = R
3
and U = R
2
. Then W is open in R
3
and U is open in R
2
. Consider the map
: U S W defined by (u, v) = (u, v, u
2
+ v
2
). Then is a bijection and continuous.
Since
-1
is the restriction on S of the continuous map : R
3
R
2
, (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
3
: z = y
2
- x
2
} is a surface.
The proof is same as above by considering (u, v) = (u, v, v
2
- u
2
).
(iii) The set S = {(x, y, z) R
3
: z
2
= x
2
+ y
2
, z 0} is a surface.
Consider (u, v) = (u, v,
u
2
+ v
2
).
(iv) The set S = {(x, y, z) R
3
: x
2
+ y
2
+ z
2
= 1} is a surface.
Let W
1
= {(x, y, z) R
3
: x < 0} {(x, y, z) R
3
: y = 0}, W
2
= {(x, y, z) R
3
: x >
0} {(x, y, z) R
3
: z = 0} and U = (0, 2) × (-
2
,
2
). Then both W
1
and W
2
are open
in R
3
and U is open in R
2
. For i = 1, 2, define
i
: U S W
i
by
1
(u, v) = (cos u cos v, cos u sin v, sin v)
and
2
(u, v) = (- cos u cos v, - sin v, - cos u sin v).
Then both
1
and
2
are homeomorphisms. Since
1
(U )
2
(U ) = S, it follows that S is a
surface.
27

28
2. SURFACES IN R
3
We now see another parametrization of S. Let U = {(u, v) R
2
: x
2
+ y
2
< 1}. Let
W
1
= {(x, y, z) R
3
: z > 0}, W
2
= {(x, y, z) R
3
: z < 0}, W
3
= {(x, y, z)
R
3
: y > 0}, W
4
= {(x, y, z) R
3
: y < 0}, W
5
= {(x, y, z) R
3
: x > 0} and
W
6
= {(x, y, z) R
3
: x < 0}. Then U is open in R
2
and each W
i
is open in R
3
. For
i = 1, . . . , 6, define
i
: U S W
i
by
1
(u, v) = (u, v,
1 - u
2
- v
2
),
2
(u, v) = (u, v, -
1 - u
2
- v
2
),
3
(u, v) = (u,
1 - u
2
- v
2
, v),
4
(u, v) = (u, -
1 - u
2
- v
2
, v),
5
(u, v) = (
1 - u
2
- v
2
, u, v) and
6
(u, v) = (-
1 - u
2
- v
2
, u, v).
Then each of
i
is a homeomorphism. Since
6
i=1
i
(U ) = S, the set S is a surface.
We now give another parametrization of the sphere S called the stereographic projection. Let
U
1
= R
2
and W
1
= R
3
\ {(0, 0, z) : z 0}. Then W
1
is open in R
3
. Define : U
1
S W
1
by
(u, v) =
2u
u
2
+ v
2
+ 1
,
2v
u
2
+ v
2
+ 1
,
u
2
+ v
2
- 1
u
2
+ v
2
+ 1
.
Then is a homeomorphism. We note that (U
1
) = S \ {(0, 0, 1)}. Now take U
2
= R
2
and
W
2
= R
3
\ {(0, 0, z) : z 0}. Then W
2
is open in R
3
. Define : U
2
S W
2
by
(u, v) =
2u
u
2
+ v
2
+ 1
,
2v
u
2
+ v
2
+ 1
,
1 - u
2
- v
2
u
2
+ v
2
+ 1
.
Then is a homeomorphism and that (U
2
) = S \ {(0, 0, -1)}. Thus (U
1
) (U
2
) = S.
Hence S is a surface.
(v) The set S = {(x, y, z) R
3
: x
2
+ y
2
= 1} is a surface.
Let W
1
= {(x, y, z) R
3
: x > 0} {(x, y, z) R
3
: y = 0}, W
2
= {(x, y, z) R
3
: x <
0} {(x, y, z) R
3
: z = 0}, U
1
= (0, 2) × R and U
2
= (-, ) × R. Then both W
1
and
W
2
are open in R
3
and both U
1
and U
2
are open in R
2
. For i = 1, 2, define
i
: U
i
S W
i
by
i
(u, v) = (cos u, sin u, v).
Then both
1
and
2
are homeomorphisms. Since
1
(U
1
)
2
(U
2
) = S, the set S is a surface.
Let U = {(u, v) :
1
2
<
u
2
+ v
2
<
3
2
} and W = R
3
. Then U is open in R
2
and W is open in
R
3
. Define : U S W by
(u, v) =
u
u
2
+ v
2
,
v
u
2
+ v
2
, tan(
u
2
+ v
2
- ) .

6. Surfaces
29
Then is a homeomorphism. Hence S is a surface.
(vi) Let a, b, c, d R, and let a
2
+b
2
+c
2
= 0. Then the set S = {(x, y, z) R
3
: ax+by+cz = d}
is a surface.
Here S is a plane. We may assume that c = 0. Let U = R
2
and W = R
3
. Then U is open in
R
2
and W is open in R
3
. Define : U S W by
(u, v) = (u, v,
1
c
(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
2
+ y
2
- z
2
= 1} is a surface.
Let U = {(u, v) :
1
2
<
u
2
+ v
2
<
3
2
} and W = R
3
. Then U is open in R
2
and W is open in
R
3
. Define : U S W by
(u, v) =
tan
2
(r - ) + 1
u
r
,
tan
2
(r - ) + 1
v
r
, tan(r - )
,
where r =
u
2
+ v
2
. Then is a homeomorphism. Hence S is a surface.
(viii) The set S = {(x, y, z) R
3
: z
2
= x
2
+ y
2
} is not a surface.
Suppose that S is a surface. Then there exists an open subset W of R
3
containing (0, 0, 0)
such that S W is homeomorphic to some open subset U of R
2
. 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
3
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
U
1
: (0, 2) × (a, b) and U
2
= (-, ) × (a, b), W
1
= {(x, y, z) R
3
: x > 0} {(x, y, z)
R
3
: y = 0} and W
2
= {(x, y, z) R
3
: x < 0} {(x, y, z) R
3
: z = 0}. Then U
1
,
U
2
are open in R
2
and W
1
and W
2
are open in R
3
. For i = 1, 2, define
i
: U
i
S W
i
by
i
(u, v) = (f (v) cos u, f (v) sin u, g(v)). Then both
1
and
2
are homeomorphisms and
1
(U
1
)
2
(U
2
) = 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.

30
2. SURFACES IN R
3
Let S be the torus. Let U
1
= (0, 2)×(0, 2), U
2
= (0, 2)×(-, ), U
3
= (-, )×(0, 2)
and U
4
= (-, ) × (-, ). The each U
i
is open in R
2
. Let
W
1
= R
3
\ ({(x, y, z) : x
2
+ y
2
= (a + b)
2
} {(x, y, 0) : 0 x a + b}).
W
2
= R
3
\ ({(x, y, z) : x
2
+ y
2
= (a - b)
2
} {(x, y, 0) : 0 x a + b}).
W
3
= R
3
\ ({(x, y, z) : x
2
+ y
2
= (a + b)
2
} {(x, y, 0) : -(a + b) x 0}), and
W
4
= R
3
\ ({(x, y, z) : x
2
+ y
2
= (a - b)
2
} {(x, y, 0) : -(a + b) x 0}). Then each
W
i
is open R
3
. For i = 1, 2, 3, 4, define
i
: U
i
S W
i
by
i
(u, v) = ((b + a cos v) cos u, (b + a cos v) sin u, a sin v).
Then each
i
is a homeomorphism and
4
i=1
i
(U
i
) = S. Therefore S is a surface.
Example 2.6.3. The surface S = {(x, y, z) R
3
: x
2
+ y
2
+ z
2
= 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
3
and an open
subset U of R
2
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
R
2
. This is a
contradiction as nonempty clopen subset of R
2
is R
2
only.
In fact, the same proof works for any compact surface.
7. Calculus on Surface
If : U S W R
3
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
u
and
v
respectively.
Definition 2.7.1. A surface patch : U R
3
is called a regular if is a smooth map and
u
and
v
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
i
: U
S W
i
, i = 1, 2,
1
(u, v) = (cos u cos v, cos u sin v, sin v)
and
2
(u, v) = (- cos u cos v, - sin v, - cos u sin v),
where W
1
= {(x, y, z) R
3
: x < 0} {(x, y, z) R
3
: y = 0}, W
2
= {(x, y, z)
R
3
: x > 0} {(x, y, z) R
3
: z = 0} and U = (0, 2) × (-
2
,
2
). Clearly, both the

7. Calculus on Surface
31
maps
1
and
2
are smooth maps. We show that (
1
)
u
and (
1
)
v
are linearly independent. We
have (
1
)
u
= (- sin u cos v, cos u cos v, 0) and (
1
)
v
= (- cos u sin v, - sin u sin v, cos v).
Therefore (
1
)
u
× (
1
)
v
= (cos u cos
2
v, sin u cos
2
v, cos v sin v) gives
(
1
)
u
× (
1
)
v
2
= cos
2
v > 0 as v (-
2
,
2
).
Therefore
1
is a regular surface patch. Similarly, it follows that
2
is a regular surface patch.
(ii) Show that translations and invertible linear transformations of R
3
take smooth surfaces to
smooth surfaces.
Let S be a smooth surface and let { : U
R
3
: } be an atlas of S containing regular
surface patches. Fix a R
3
. We want to show that S + a = {s + a : s S} is a smooth
surface. Let T
a
: R
3
R
3
be T
a
(x) = x + a (x R
3
). Then T
a
is a homeomorphism. Let
A = {
= T
a
: U
R
3
: }. Then
(U
) =
(T
a
)(U
) = T
a
(U
)
= S + a.
Since U
-
(U
) and
(U
)
T
a
- T
a
(
(U
)) are homeomorphism, U
T
a
- T
a
(
(U
))
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 (
)
u
=
u
and (
)
v
=
v
. Since
u
and
v
are linearly independent, it follows that (
)
u
and (
)
v
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
3
R
3
be an invertible linear map. Then A = {T
: U
R
3
: } 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
R
3
: } be an
atlas of S such that each
is a regular patch. Let = { :
(U
) W = }.
Let A = {
=
|
U
-1
(W )
: U
-1
(W ) R
3
: }. We show that A is an
atlas of W and each
in A is a regular patch. Clearly,
(U
-1
(W )) W . Let
q W S =
(U
). Then q
(U
) for some . But then q W
(U
), i.e.,
. It also follows that q
(U
-1
(W )). Therefore W
(U
-1
(W )).
Since
is a restriction of
on an open subset of U
, the map
is a smooth map and
(
)
u
= (
)
u
and (
)
v
= (
)
v
. Therefore (
)
u
and (
)
v
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
3
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
3
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.

32
2. SURFACES IN R
3
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
p
S.
Proposition 2.7.5. Let : U R
3
be a patch of a surface S containing a point p of S, i.e.,
p = (u
0
, v
0
)
for some (u
0
, v
0
) U
. Then the tangent space to S at p is a vector subspace of R
3
spanned by the vectors
u
(u
0
, v
0
)
and
v
(u
0
, v
0
)
.
P
. Let be a curve in S passing through p. Then (t) = (u(t), v(t)) and (t
0
) = (u(t
0
), v(t
0
)) =
(u
0
, v
0
) = p for some t
0
. Then the tangent vector to at p = (t
0
) will be
(t
0
) =
u
(u(t
0
), v(t
0
)) u(t
0
) +
v
(u(t
0
), v(t
0
)) v(t
0
)
=
u
(u
0
, v
0
) u(t
0
) +
v
(u
0
, v
0
) v(t
0
).
Hence any tangent vector to S at p is a linear combination of
u
(u
0
, v
0
) and
v
(u
0
, v
0
), i.e., T
p
S
span{
u
(u
0
, v
0
),
v
(u
0
, v
0
)}.
Conversely, let a span{
u
(u
0
, v
0
),
v
(u
0
, v
0
)}. Then a =
u
(u
0
, v
0
)+
v
(u
0
, v
0
) for some
, R. Consider the curve (t) = (u
0
+ t, v
0
+ t). Then is lying on the patch and and
(0) = (u
0
, v
0
) = p. Now the tangent vector to at 0 will be (0) =
u
(u
0
, v
0
)+
v
(u
0
, v
0
) = a.
This shows that a is tangent vector to some curve in S passing through p, i.e., a T
p
S. Therefore
span{
u
(u
0
, v
0
),
v
(u
0
, v
0
)} T
p
S.
Definition 2.7.6. Let S be a surface, and let p be a point in the patch : U R
3
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
(u, v) ×
v
(u, v)
u
(u, v) ×
v
(u, v)
is called the unit normal to S at p.
Definition 2.7.7. Let S
1
and S
2
be smooth surfaces with surface patches
1
: U
1
R
3
and
2
:
U
2
R
3
.
(i) A map f : S
1
S
2
is called a smooth map if the map
-1
2
f
1
: U
1
U
2
is a smooth
map.
(ii) A map f : S
1
S
2
is called a diffeomorphism if f is bijective and both f and f
-1
are smooth
maps.
(iii) A map f : S
1
S
2
is called a local diffeomorphism if for every p S
1
there exists an open
subset O of S
1
containing p such that f
|O
: O f (O) is a diffeomorphism.
Example 2.7.8. Let a, b, c > 0. We construct a diffeomorphism between the ellipsoid
x
2
a
2
+
y
2
b
2
+
z
2
c
2
= 1
and the unit sphere x
2
+ y
2
+ z
2
= 1.
Let S
1
be the ellipsoid and S
2
be the unit sphere. Define f : S
1
S
2
by f (x, y, z) = (
x
a
,
y
b
,
z
c
).
Clearly, f is bijective. Check that f and f
-1
are smooth maps.

7. Calculus on Surface
33
Exercise 2.7.9. Let C be a collection of smooth surfaces. Two smooth surfaces in C are said to be
related, S
1
S
2
, if S
1
and S
2
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
p
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
0
) = p and (t
0
) = w
for some t
0
. Then = f is a curve in S passing through f (p) ((t
0
) = f ((t
0
)) = f (p)). Let
w be the tangent vector to at the point f (p), i.e., w = (t
0
). (Hence we have a map which maps
w T
p
S to w T
f (p)
S.) The map D
p
f : T
p
S T
f (p)
S given by D
p
f (w) = w is called the
derivative of f , D
p
f , at p.
We now show that the definition of D
p
f (w) depends on f , p and w.
Let : U R
3
be a surface patch containing p, say, p = (u
0
, v
0
), and let , be smooth
functions such that f ((u, v)) = ((u, v), (u, v)). Let w =
u
+ µ
v
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
0
) = and
v(t
0
) = µ. Since the corresponding curve on S is (t) = ((u(t), v(t)), (u(t), v(t))), we have
D
p
f (w) =
u
( u
u
+ v
v
) +
v
( u
u
+ v
v
),
the derivatives of u and v are evaluated at t
0
. Thus
D
p
f (w) =
u
(
u
+ µ
v
) +
v
(
u
+ µ
v
).
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
p
f : T
p
S T
f (p)
S is a linear map.
P
. Let w
1
=
1
u
+ µ
1
v
and w
2
=
2
u
+ µ
2
v
be in T
p
S, and let R. Then
D
p
f (w
1
+ w
2
) = D
p
f ((
1
+
2
)
u
+ (µ
1
+ µ
2
)
v
)
= [(
1
+
2
)
u
+ (µ
1
+ µ
2
)
v
]
u
+ [(
1
+
2
)
u
+ (µ
1
+ µ
2
)
v
]
v
= [(
1
u
+ µ
1
v
)
u
+ (
1
u
+ µ
1
v
)
v
]
+(
2
u
+ µ
2
v
)
u
+ (
2
u
+ µ
2
v
)
v
= D
p
f (w
1
) + D
p
f (w
2
).
Hence the map D
p
f is linear.
Note that the matrix of the linear map D
p
f with respect to the basis {
u
,
v
} of T
p
S and
{
u
,
u
} of T
f (p)
S is
u
v
u
v
.
Proposition 2.7.11. Let S
1
, S
2
and S
3
be smooth surfaces.
(i) If p S
1
, then the derivative at p of the identity map from S
1
to itself is the identity map
from T
p
S
1
to itself.

34
2. SURFACES IN R
3
(ii) (Chain rule) If f : S
1
S
2
and g : S
2
S
3
are smooth maps, then for all p S
1
,
D
p
(g f ) = D
f (p)
g D
p
f
.
(iii) If f : S
1
S
2
is a diffeomorphism, then for all p S
1
, the linear map D
p
f : T
p
S
1
T
f (p)
S
2
is invertible.
P
. (i) Here I : S
1
S
1
is the identity map, i.e., I(x) = x for all x S
1
. Let p be a point
on S
1
, and let w T
p
S. Let be a curve in S
1
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
p
I(w) = w. Since w is an arbitrary point of T
S
, then map D
p
I is the identity map, i.e.,
D
p
I(w) = w for all w T
p
S
1
.
(ii) Let p be a point on S
1
, and let w T
p
S
1
. Let be a curve in S
1
passing through p whose
tangent vector at p is w. Let
1
= f . Then
1
is a curve in S
2
passing through f (p). Let w
1
be the
tangent vector to
1
at f (p). Then D
p
f (w) = w
1
. Let
2
= g
1
= g f . Then
2
is curve in
S
3
passing through g(f (p)). Let w
2
be the tangent vector to
2
at g(f (p)). Then D
f (p)
g(w
1
) = w
2
,
i.e, D
f (p)
g(D
p
f (w)) = w
2
, i.e., (D
f (p)
g D
p
f )(w) = w
2
. Also, w
2
is the tangent vector to (g f )
at g f (p). Therefore D
p
(g f )(w) = w
2
. This shows that D
p
(g f )(w) = (D
f (p)
g D
p
f )(w).
Since w is an arbitrary point of T
p
S
1
, we get D
p
(g f ) = D
f (p)
g D
p
f .
(iii) Since f is a diffeomorphism, the map f
-1
is smooth and the map f
-1
f : S
1
S
1
is the
identity map. Let p S. Then I = D
p
(f
-1
f )
Proposition 2.7.12. Let S
1
and S
2
be smooth surfaces and let f : S
1
S
2
be a smooth map. Then
f
is a local diffeomorphism if and only if the linear map D
p
f : T
p
S
1
T
f (p)
S
2
is invertible for
all p S
1
.
P
. Assume that f is a local diffeomorphism. Let p S
1
. Since f is a local diffeomorphism, there
exists an open subset O of S
1
containing p such that f
|O
: O f (O) is a diffeomorphism. It follows
from Proposition 2.7.11 (ii) that D
p
f is invertible.
The converse follows from the Inverse Function Theorem.
Example 2.7.13.
(i) Let f : S
1
S
2
be a local diffeomorphism, and let be a regular curve in
S
1
. We show that f is a regular curve in S
2
.
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
p
f
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
2
+ y
2
= z
2
, 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
S
1
= {(0, y, z)|y > 0} and S
2
= {(x, y, z) : x
2
+ y
2
= z
2
, z > 0}. Then S
1
and S
2

8. First Fundamental Form
35
are smooth surfaces. Let U
1
= {(u, v) : u > 0}, U
2
= R
2
\{0} and W = R
3
. Consider
: U
1
W S
1
defined by
1
(u, v) = (0, u, v) and : U
2
W S
2
defined by
(u, v) = (u, v,
u
2
+ v
2
). Then
1
is a surface patch of S
1
and is a surface patch of S
2
.
Also note that (U
1
) = S
1
and (U
2
) = S
2
. Since f (U
1
) is a subset of S
2
= (U
2
), there ex-
ists smooth maps , : U
1
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)
2
+ (u, v)
2
). This gives (u, v) = u cos v
and (u, v) = u sin v. Fix p = ((u
0
, v
0
)) S
1
. Then the matrix of D
p
f with respect to the
basis {
u
(u
0
, v
0
),
v
(u
0
, v
0
)} of T
p
S
1
and the basis {
u
((u
0
, v
0
), (u
0
, v
0
)),
u
((u
0
, v
0
), (u
0
, v
0
))}
is
u
(u
0
, v
0
)
v
(u
0
, v
0
)
u
(u
0
, v
0
)
v
(u
0
, v
0
)
=
cos v
0
-u
0
sin v
0
sin v
0
u
0
cos v
0
. Now D
p
f is invertible if and only if
the above matrix is invertible if and only if u
0
= 0. But u
0
> 0. Hence D
p
f is invertible. Since
p is an arbitrary point of S
1
, it follows that D
p
f is invertible for every p S
1
. 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
p
: T
p
S × T
p
S R given by
I
p
(v, w) = v, w
p
= vw
(v, w T
p
S).
Suppose that : U R
3
be a surface patch of a surface S. Fix p S. Then any tangent
vector to S at p is a linear combination of
u
and
v
. Define du, dv : T
p
S R by du(w) = and
dv(w) = µ if w =
u
+ µ
v
. Then both du and dv are linear maps. Now
I(w, w) = w, w =
2
u
,
u
+ 2µ
u
,
v
+ µ
2
v
,
v
.
Let E =
u
2
=
u
u
, F =
u
v
and G =
v
2
=
v
v
. 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)
2
+
2F du(w)dv(w) + Gdv(w)
2
or
·, · = Edu
2
+ 2F dudv + Gdv
2
.
(2.8.1.1)
The expression Edu
2
+ 2F dudv + Gdv
2
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
u
+ v
v
and hence
= ( , )
1/2
=
F u
2
+ 2F u v + G v
2
. If
p = (t
0
) and q = (t
1
), then the length of the curve (on the surface patch ) is
t
1
t
0
F u
2
+ 2F u v + G v
2
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
2
+ y
2
= 1.

36
2. SURFACES IN R
3
We know that the surface patch of the above surface is given by (u, v) = (cos u, sin u, v).
Now
u
= (- sin u, cos u, 0) and
v
= (0, 0, 1). Therefore E =
u
u
= 1, F =
u
v
= 0 and
G =
v
v
= 1. Hence the first fundamental form on the surface is du
2
+ dv
2
.
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
2
+ v
2
)
(v) (u, v) = (cosh u, sinh u, v)
(vi) (u, v) = (u, v, u
2
+ v
2
)
Example 2.8.4. When applying an isometry of R
3
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
3
to R
3
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
3
be a surface
patch of a surface S, and let T : R
3
R
3
be a linear isometry. Then : U R
3
is a surface
patch of the surface T (S). It follows that
u
= T (
u
) and
v
= T (
v
) (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 =
u
,
u
=
T (
u
), T (
u
) =
u
,
u
= 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
3
be a surface patch of S. Let T (v) = av be a dilation, where a is a nonzero
real. Then T = : U R
3
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 =
u
u
= a
2
u
u
= a
2
E. Similarly, we get
F = a
2
F and G = a
2
G. Thus the first fundamental form on T (S) is a
2
times the first fundamental
form on S.
9. Local isometries and conformal maps
Definition 2.9.1. Let S
1
and S
2
be smooth surfaces. A smooth map f : S
1
S
2
is called a local
isometry if it takes any curve in S
1
to a curve of same length in S
2
.
A local isometry that is a diffeomorphism is called an isometry.
Thus if f : S
1
S
2
is a local isometry, then for any curve on S
1
, the curves and f
have the same length.
Example 2.9.2.
(i) Let S
1
, S
2
and S
3
be surfaces. Let f : S
1
S
2
and g : S
2
S
3
be local isometry. Then
g f : S
1
S
3
is a local isometry (i.e., composition of local isometries is a local isometry).

9. Local isometries and conformal maps
37
Let be a curve in S
1
. Since f is a local isometry f is a curve in S
2
of the same length.
Since g is a local isometry, g f is a curve in S
3
of the same length. Hence if is a curve
S
1
, 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
1
S
2
be an isometry. Then by definition f
-1
: S
2
S
1
is a diffeomorphism. Let
be a curve in S
2
. 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
2
+y
2
= z
2
, 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
1
0
(u) du =
2. Now f : (1, 2) R
3
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
1
S
2
be a smooth map, and let p S
1
. Define f
: T
p
S
1
× T
p
S
1
R
by
f
v, w
p
= D
p
f (v), D
p
f (w)
f (p)
(v, w T
p
S
1
).
Example 2.9.4. The map f
is a symmetric bilinear map.
Let u, v, w T
p
S
1
and , µ R. Then
f
u, v
p
=
D
p
f (u), D
p
f (v)
f (p)
= D
p
f (u)D
p
f (v) = D
p
f (v)D
p
f (u)
=
D
p
f (v), D
p
f (u)
f (p)
= f
v, u
p
.
Therefore f
is symmetric. Now
f
u + µv, w
p
=
D
p
f (u + µv), D
p
f (w)
f (p)
= D
p
f (u) + µD
p
f (v), D
p
f (w)
f (p)
= D
p
f (u), D
p
f (w)
f (p)
+ µ D
p
f (v), D
p
f (w)
f (p)
= f
u, w
p
+ µf
v, w
p
.
Hence f
is bilinear.
The following is an important fact. Let f : S
1
S
2
be a smooth map. Let be a curve in
S
1
. Then
(t
0
) is a tangent to S
1
at the point (t
0
) = p. Now = f is a curve in S
2
passing
through f ((t
0
)) = f (p). Therefore by definition D
p
f ( (t
0
)) = (t
0
). Since t
0
arbitrary, we have
D
p
f ( ) =
Example 2.9.5. Let T : R
n
× R
n
R be a symmetric bilinear map. If T (x, x) = 0 for all x R
n
,
then T = 0.
Let x, y R
n
. Then
0 = T (x + y, x + y) = T (x, x) + 2T (x, y) + T (y, y) = 2T (x, y).

38
2. SURFACES IN R
3
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
n
and S(x, x) =
T (x, x) for all x, then S = T .
Theorem 2.9.6. Let S
1
and S
2
be smooth surfaces, and let f : S
1
S
2
be a smooth map. Then
f
is a local isometry if and only if the symmetric bilinear maps ·, ·
p
and f
·, ·
p
on T
p
S
1
are
equal for every p S
1
.
P
. If is curve in S
1
, then the length of the part of with endpoints (t
0
) and (t
1
) is
t
1
t
0
,
1/2
dt.
Then = f is a curve in S
2
. The length of between the points (t
0
) and (t
1
) is
t
1
t
0
,
1/2
dt =
t
1
t
0
D
f ( ), D
f ( )
1/2
dt =
t
1
t
0
f
,
1/2
dt
(2.9.6.1)
Assume that the symmetric bilinear maps ·, ·
p
and f
·, ·
p
are equal for every p S
1
. Then it
follows from the equation (2.9.6.1) that any curve on S
1
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
2
have the same length, i.e.,
t
1
t
0
,
1/2
dt =
t
1
t
0
f
,
1/2
dt.
Suppose that f
·, ·
p
= ·, ·
p
for some p S
1
. Then there is v T
p
S
1
such that f
v, v
p
=
v, v
p
. Let : (a, b S
1
R
3
such that (s) = p and
(s) = v for some s (a, b). Then
(s), (s)
1
2
(s)
= f
(s), (s)
1
2
(s)
. We may assume that
(s), (s)
1
2
(s)
-f
(s), (s)
1
2
(s)
> 0.
By continuity of the map g(u) =
(u), (u)
1
2
- f
(u), (u)
1
2
(u (a, b)) there exists > 0
and c > 0 such that
(u), (u)
1
2
(u)
- f
(u), (u)
1
2
(u)
> c for all u (s - , s + ). But
then
s+
s-
( (u), (u)
1
2
(u)
- f
(u), (u)
1
2
(u)
)du 2c > 0. i.e.,
s+
s-
(u), (u)
1
2
(u)
du >
s+
s-
f
(u), (u)
1
2
(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 ·, ·
p
= f
·, ·
p
for every p S
1
.
Thus a smooth map f : S
1
S
2
is a local isometry if and only if for every p S
1
,
D
p
f (v), D
p
f (w)
f (p)
= v, w
p
(v, w T
p
S
1
).
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
39
(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
1
and S
2
be smooth surfaces, and let f : S
1
S
2
be a smooth map. Then
f
is a local isometry if and only if for every p S
1
, the linear map D
p
f : T
p
S
1
T
f (p)
S
2
is an
isometry.
P
. Assume that f is a local isometry. Fix p S
1
. Since f is a local isometry, f
u, v
p
= u, v
p
for all u, v T
p
S
1
, i.e., D
p
f (u), D
p
f (v)
p
= u, v
p
for all u, v T
p
S
1
. This shows that D
p
f is
an isometry. Since p S
1
is arbitrary, we are done.
Conversely, assume that D
p
f is isometry for all p S
1
. Fix p S
1
. Since D
p
f is an isometry,
it follows that D
p
f (u), D
p
f (v)
p
= u, v
p
for all u, v T
p
S
1
, i.e., f
u, v
p
= u, v
p
for all
u, v T
p
S
1
. This shows that f
·, ·
p
= ·, ·
p
. Since p S
1
is arbitrary, it follows that f is a local
isometry.
Let : U R
3
be a surface patch. Fix (u
0
, v
0
) in U . Define
u
0
and
v
0
by
u
0
(v) = (u
0
, v)
and
v
0
(u) = (u, v
0
). Then both
u
0
and
v
0
are curves on passing through (u
0
, v
0
) (
u
0
(v
0
) =
(u
0
, v
0
) =
v
0
(u
0
)). Now
u
0
(v
0
) =
v
(u
0
, v
0
) and
v
0
(u
0
) =
u
(u
0
, v
0
). Therefore
u
(u
0
, v
0
)
and
v
(u
0
, v
0
) are tangent vectors at (u
0
, v
0
). The curves
u
and
v
are called parameter curves
to the surface patch .
Let f : S
1
S
2
be a smooth map. Let : U R
3
be a surface patch of S
1
. Let = f :
U R
3
. Let (u, v) U . Then
u
(u, v) =
lim
h0
f (u + h, v) - f (u, v)
h
=
lim
h0
f
v
(u + h) - f
v
(u)
h
= (f
v
)
·
(u)
Since
v
(u) =
u
(u, v), by definition of derivative D
p
f of f at p = (u, v) we have D
p
f (
v
(u)) =
(f
v
)
·
(u), i.e., D
p
f (
u
(u, v)) =
u
(u, v) or D
p
f (
u
) =
u
. Similarly, we get D
p
f (
v
) =
v
.
Corollary 2.9.9. Let S
1
and S
2
be smooth surfaces, and let f : S
1
S
2
be a local diffeomorphism.
Then f is a local isometry if and only if for any surface patch of S
1
, the patches and f of
S
1
and S
2
, respectively, have the same first fundamental form.
P
. Let E, F and G be the first order magnitudes of and E
1
, F
1
and G
1
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
1
, F = F
1
and G = G
1
. Since f is a local isometry f
v, w
p
= v, w
p
for all p S
1
.
In particular, f
u
,
u p
=
u
,
u p
, f
u
,
v p
=
u
,
v p
and f
v
,
v p
=
v
,
v p
. Now,
E =
u
,
u p
= f
u
,
u p
= D
p
f (
u
), D
p
f (
v
)
f (p)
=
u
,
u f (p)
= E
1
.

40
2. SURFACES IN R
3
Similar arguments shows that F = F
1
and G = G
1
. 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
1
,
F = F
1
and G = G
1
. Fix p = (u, v). Let w =
1
u
+ µ
1
v
, z =
2
u
+ µ
2
v
T
p
S. Then
f
w, z
p
=
D
p
f (w), D
p
f (z)
f (p)
= D
p
f (
1
u
+ µ
1
v
), D
p
f (
2
u
+ µ
2
v
)
f (p)
=
1
2
D
p
f (
u
), D
p
f (
u
)
f (p)
+ (
1
µ
2
+
2
µ
1
) D
p
f (
u
), D
p
f (
v
)
f (p)
1
µ
2
D
p
f (
v
), D
p
f (
v
)
f (p)
=
1
2
u
,
u p
+ (
1
µ
2
+
2
µ
1
)
u
,
v p
+ µ
1
µ
2
v
,
v p
=
1
u
+ µ
1
v
,
2
u
+ µ
2
v p
= w, z
p
.
Hence f
·, ·
p
= ·, ·
p
. Since p S
1
is arbitrary, f
·, ·
p
= ·, ·
p
for every p S
1
, i.e., f is a
local isometry.
Definition 2.9.10. Let S
1
and S
2
be smooth surfaces. A local diffeomorphism f : S
1
S
2
is called
a conformal map if it preserves angle between curves, i.e., if and
1
are any two curves on S
1
intersecting at a point p S
1
and if
2
and
2
are their images under f , then the angle between
1
and
1
at p is equal to the angle between
2
and
2
at f (p).
If : U R
3
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
1
and S
2
be smooth surfaces, and let f : S
1
S
2
be a local diffeomorphism.
Then f is conformal if and only if there is map : S
1
R such that f
v, w
p
= (p) v, w
p
for
all p S
1
and for all v, w T
p
S
1
.
P
. Let and be two curves on S
1
that intersects at p S
1
. If is an angle between these two
curves, then
cos =
,
p
=
,
p
,
1/2
p
,
1/2
p
.
Let be the corresponding angle of intersection between the curves f and f on S
2
at f (p).
Then
cos =
D
p
f ( ), D
p
f ( )
f (p)
D
p
f ( )
D
p
f ( )
=
D
p
f ( ), D
p
f ( )
f (p)
D
p
f ( , D
p
f ( )
1/2
f (p)
D
p
f ( ), D
p
f ( )
1/2
f (p)

9. Local isometries and conformal maps
41
=
f
,
p
f
,
1/2
p
f
,
1/2
p
.
Suppose that there is a map : T
p
S
1
R such that f
v, w
p
= (p) v, w
p
for all p S
1
and for
all v, w T
p
S
1
. Then
cos =
f
,
p
f
,
1/2
p
f
,
1/2
p
=
(p) ,
p
(p)
1/2
,
1/2
p
(p)
1/2
,
1/2
p
=
,
p
,
1/2
p
,
1/2
p
= cos ,
which proves that the map f is conformal.
Conversely, assume that f is conformal. Then
f
,
p
f
,
1/2
p
f
,
1/2
p
=
,
p
,
1/2
p
,
1/2
p
(2.9.12.1)
Since every tangent vector to S
1
is a tangent vector to a curve in S
1
, the equation (2.9.12.1) implies
that
f
v, w
p
f
v, v
1/2
p
f
w, w
1/2
p
=
v, w
p
v, v
1/2
p
w, w
1/2
p
for all p S
1
and v, w T
p
S
1
. Fix p S
1
. Let {x, y} be an orthonormal basis of T
p
S
1
. Let
= f
x, x , µ = f
x, y and = f
y, y . Let w = cos x + sin y T
p
S
1
. Then w makes an
angle with x. Therefore
cos =
x, w
p
x, x
1/2
p
w, w
1/2
p
=
cos + µ sin
( cos
2
+ 2µ sin cos + sin
2
)
.
Take =
2
. Then µ = 0. Hence we get = cos
2
+ sin
2
, which gives = µ, say, (p). Let
v =
1
x +
1
y, w =
2
x +
2
y T
p
S
1
. Then
f
v, w
p
= f
1
x +
1
y,
2
x +
2
y
p
=
1
2
f
x, x
p
+ (
1
2
+
2
1
)f
x, y
p
+
1
2
f
y, y
p
= (p)(
1
2
+
1
2
) = (p)(v · w)
= (p) v, w
p
.

42
2. SURFACES IN R
3
Therefore f
v, w
p
= (p) v, w
p
for all v, w T
p
S
1
. Since p S
1
arbitrary, for every p S
1
there exists (p) R (hence there is a map : S
1
R) such that f
v, w
p
= (p) v, w
p
for all
v, w T
p
S
1
. Hence the theorem.
Corollary 2.9.13. A local diffeomorphism f : S
1
S
2
is conformal if and only if for any surface
patch of S
1
, the first fundamental forms of patches of S
1
and f of S
2
are proportional.
P
. Let E, F and G be the first order magnitudes of and E
1
, F
1
and G
1
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
1
, F = F
1
and G = G
1
. Since f is conformal, there is a map : S
1
R
such that f
v, w
p
= (p) v, w
p
for all p S
1
. Fix p S
1
. Then f
u
,
u p
= (p)
u
,
u p
,
f
u
,
v p
= (p)
u
,
v p
and f
v
,
v p
= (p)
v
,
v p
. Now,
(p)E = (p)
u
,
u p
= f
u
,
u p
= D
p
f (
u
), D
p
f (
v
)
f (p)
=
u
,
u f (p)
= E
1
.
Similar arguments shows that (p)F = F
1
and (p)G = G
1
. 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
1
.
Then (p)E = E
1
, (p)F = F
1
and (p)G = G
1
. It means that (p)
u
,
u p
= (f )
u
, (f
)
u p
= f
u
,
u p
, (p)
u
,
v p
= (f )
u
, (f )
v p
= f
u
,
v p
and (p)
v
,
v p
=
(f )
v
, (f )
v p
= f
v
,
v p
. Let w =
1
u
+ µ
1
v
, z =
2
u
+ µ
2
v
T
p
S. Then
f
w, z
p
=
D
p
f (w), D
p
f (z)
f (p)
= D
p
f (
1
u
+ µ
1
v
), D
p
f (
2
u
+ µ
2
v
)
f (p)
=
1
2
D
p
f (
u
), D
p
f (
u
)
f (p)
+ (
1
µ
2
+
2
µ
1
) D
p
f (
u
), D
p
f (
v
)
f (p)
1
µ
2
D
p
f (
v
), D
p
f (
v
)
f (p)
=
1
2
f
u
,
u p
+ (
1
µ
2
+
2
µ
1
)f
u
,
v p
+ µ
1
µ
2
f
v
,
v p
= (p)(
1
2
u
,
u p
+ (
1
µ
2
+
2
µ
1
)
u
,
v p
+ µ
1
µ
2
v
,
v p
)
= (p)
1
u
+ µ
1
v
,
2
u
+ µ
2
v p
= (p) w, z
p
.
Hence f
·, ·
p
= (p) ·, ·
p
. Since p S
1
is arbitrary, f
·, ·
p
= (p) ·, ·
p
for every p S
1
, 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
2
+ dv
2
) for some smooth map .
Example 2.9.14. Let : U V be a diffeomorphism between open subsets of R
2
. 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
u
= g
v
and f
v
= -g
u
) or (f
u
= -g
v
and f
v
= g
u
).
We have (u, v) = (f (u, v), g(u, v)). Therefore
u
= (f
u
, g
u
) and
v
= (f
v
, g
v
). This gives
E = f
2
u
+ g
2
u
, F = f
u
f
v
+ g
u
g
v
and F = f
2
v
+ g
2
v
. Assume that is conformally parametrized.
Then its first fundamental form is of the form (du
2
+ dv
2
), where is a smooth function of u and

9. Local isometries and conformal maps
43
v. This gives F = f
u
f
v
+ g
u
g
v
= 0 and f
2
u
+ g
2
u
= E = G = f
2
v
+ g
2
v
. It means the vectors (f
u
, g
u
)
and (f
v
, g
v
) are orthogonal and have the same norm. Therefore (f
v
, g
v
) can be obtained by rotating
(f
u
, g
u
) by an angle /2 or 3/2. Hence (f
v
, g
v
) = (g
u
, -f
u
) or (f
v
, g
v
) = (-g
u
, f
u
). Therefore
(f
u
= -g
v
and f
v
= g
u
) or (f
u
= g
v
and f
v
= -g
u
).
Conversely, if (f
u
= -g
v
and f
v
= g
u
) or (f
u
= g
v
and f
v
= -g
u
), then F = 0 and E =
G = f
2
u
+ g
2
u
. Therefore the first fundamental form of is (f
2
u
+ g
2
u
)(du
2
+ dv
2
) and hence is
conformally parametrized.
Definition 2.9.15. Let : U R
3
be a surface patch, and let R U . Then the surface area, A
(R),
of the part (R) of (U ) is
A
(R) =
R
u
×
v
dudv.
Proposition 2.9.16. Let : U R
3
be a surface patch. Then
u
×
v
=
EG - F
2
.
P
. We know that if a , b, c, d R
3
, then (a × b)(c × d) = (a c)(b d) - (a d)(bc). Now
u
×
v
2
= (
u
u
)(
v
v
) - (
u
v
)
2
= EG - F
2
. Therefore
u
×
v
=
EG - F
2
.
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 (-
2
,
2
)).
Now
u
= (-r cos v sin u, r cos v cos u, 0),
v
= (-r sin v cos u, -r sin v sin u, r cos v). There-
fore E =
u
u
= r
2
cos
2
v, F =
u
v
= 0 and G = r
2
, and EG - F
2
= r
4
cos
2
v. Now the
surface area, A, of the sphere is
A =
(0,2)×(-
2
,
2
)
r
2
cos v dudv
= r
2
/2
-/2
2
0
cos v dudv
= r
2
/2
-/2
cos vdv
2
0
du
= 4r
2
.
(ii) Compute the area of {(x, y, z) : x
2
+ y
2
= 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),

44
2. SURFACES IN R
3
where R = {(u, v) : u (0, 2), |v| 1}. Therefore the surface area A of (R) is
A =
(0,2)×[-1,1]
1dv =
2
0
du
1
-1
dv
= 4.
(iii) Let a > 0 be fixed. Let : (, ) R
3
be a unit-speed curve in R
3
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
= 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 =
t
t
= (1 - a cos )
2
+ a
2
, F =
t
= a
2
and
G =
= a
2
. Now EG - F
2
= ((1 - a cos )
2
+ a
2
)a
2
- a
4
2
= a
2
(1 - a cos )
2
.
Now
A
=
(,)×(0×2)
EG - F
2
dtd
=
2
0
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|
1
a
.
(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
3
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
u +
v
v +
1
2
[
uu
(u)
2
+ 2
uv
uv +
vv
(v)
2
] +
u
u +
v
v +
1
2
[
uu
(u)
2
+ 2
uv
uv +
vv
(v)
2
].
Hence ((u + u, v + v) - (u, v))N =
1
2
[L(u)
2
+ 2M (u)(v) + N (v)
2
], where L =
uu
N,
M =
uv
N and N =
vv
N. The expression Ldu
2
+ 2M dudv + N dv
2
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
2
denote the unit sphere in R
3
, i.e., S
2
= {(x, y, z) : x
2
+ y
2
+ z
2
= 1}.
Definition 3.10.2. Let S be a smooth surface. The map G : S S
2
defined by G(p) = N
p
, where
N
p
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
3
(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
p
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
p
G : T
p
S
T
G(p)
S
2
is a linear map. Note that T
p
S = T
G(p)
S
2
. Hence D
p
G : T
p
S T
p
S is a linear map.
Example 3.10.3. Find the image of the Gauss map for (u, v) = (u, v, u
2
+ v
2
), u, v R.
We have
u
= (1, 0, 2u) and
u
= (0, 1, 2v). Therefore
u
×
v
= (-2u, -2v, 1). This gives
N =
1
1+4u
2
+4v
2
(-2u, -2v, 1). Hence the image of the Gauss map is {
1
1+4u
2
+4v
2
(-2u, -2v, 1) :
u, v R}.
45

46
3. CURVATURE OF A SURFACE
Definition 3.10.4. Let p be a point on an oriented surface S.
(i) The map W
p
: T
p
S T
p
S defined by W
p
= -D
p
G is called the Weingarten map.
(ii) The second fundamental form on a surface S at p S is a bilinear map
II
p
: T
p
S × T
p
S R defined by
II
p
v, w
p
= W
p
(v), w
p
(v, w T
p
S).
Since W
p
= -D
p
G and D
p
G is a linear map, the map W
p
: T
p
S T
p
S is a linear map.
Example 3.10.5. Let : U R
3
be a regular patch of an oriented surface. Then D
p
G(
u
) = N
u
and D
p
G(
v
) = N
v
.
Fix a point p = (u, v) on the surface. Consider the parametric curves
v
(t) = (t, v) and
u
(t) = (u, t). Then
v
(u) = (u, v) = p =
u
(v). Now
v
(u) =
u
(u, v) and
u
(v) =
v
(u, v).
Now G(
v
)(t) = N
v
(t) = N(t, v). Hence tangent to the curve G
v
at G (u, v) (or the tangent
to G
v
at G(p)) is (G
v
)
·
(u) = N
u
(u, v). Hence D
p
G(
u
(u, v)) = N
u
(u, v) or D
p
G(
u
) = N
u
.
One may similarly show that D
p
G(
v
) = N
v
.
Lemma 3.10.6. Let be a surface patch of an oriented surface with the unit normal N. Then
N
u
u
= -L
, N
u
v
= -M = N
v
u
and N
v
v
= -N
.
P
. Since
u
and
v
generates the tangent space and N is orthogonal to the tangent space, we have
N
u
= N
v
= 0. Differentiating N
u
= 0 partially with respect to u, we obtain N
u
u
+ N
uu
= 0.
But N
uu
= L. Hence N
u
u
= -L. By differentiating N
u
= 0 partially with respect to v, we get
N
v
u
= -M . Similarly, by differentiating N
v
= 0 partially with respect to u and v, we get the
relations N
v
u
= -M and N
v
v
= -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
II
p
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
p
S,
where du, dv : T
p
S R are defined by du(w) = and dv(w) = µ when w =
u
+ µ
v
T
p
S.
P
. Since D
p
G(
u
) = N
u
and D
p
G(
v
) = N
v
, we have W
p
(
u
) = -N
u
and W
p
(
v
) = -N
v
.
Let w =
1
u
+ µ
1
v
and x =
2
u
+ µ
2
v
be in T
p
S. Then
II
p
w, x
p
=
W
p
(w), x
p
=
W
p
(
1
u
+ µ
1
v
),
2
u
+ µ
2
v p
=
1
W
p
(
u
) + µ
1
W
p
(
v
),
2
u
+ µ
2
v p
=
-
1
N
u
- µ
1
N
v
,
2
u
+ µ
2
v p
= -
1
2
N
u
u
-
1
µ
2
N
u
v
-
2
µ
1
N
v
u
- µ
1
µ
2
N
v
v
= 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
47
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.
P
. One can check (means you) that II
p
is linear in both the variable (by fixing one), i.e., II
p
is
bilinear. It follows from the Proposition 3.10.7 that II
p
w, x
p
= II
p
x, w
p
for all x, w T
p
S,i.e.,
II
p
is symmetric. Hence the second fundamental form II
p
is a symmetric bilinear map.
Let x, w T
p
S. Since II
p
w, x
p
= II
p
x, w
p
, we have W
p
(w), x
p
= W
p
(x), w
p
=
w, W
p
(x)
p
, i.e, W
p
(w), x
p
= w, W
p
(x)
p
. Hence W
p
is symmetric.
11. Gaussian, Mean and Principal Curvatures
Definition 3.11.1. Let W
p
be the Weingarten map on an oriented surface S at p S. Then the
Gaussian curvature, K
p
, and the mean curvature, H
p
, at p are defined by
K
p
= det(W
p
) and H
p
=
1
2
trace(W
p
).
Proposition 3.11.2. Let : U R
3
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
p
with respect to the
basis {
u
,
v
} of T
p
S is
E
F
F
G
-1
L
M
M
N
.
P
. Since N is a unit vector, N N = 1. This will imply that N
u
N = 0 and N
v
N = 0. Hence N
u
and N
v
(and so -N
u
and -N
v
) are in the tangent space T
p
S. Therefore there exist , , , R
such that -N
u
=
u
+
v
and -N
V
=
u
+
v
. Since W
p
(
u
) = -N
u
and W
p
(
v
) = -N
v
,
we have W
p
(
u
) =
u
+
v
and W
p
(
v
) =
u
+
v
. Therefore the matrix of W
p
with respect
to basis {
u
,
v
} of T
p
S is
. Taking dot product with
u
and
v
on both the sides of the
equations -N
u
=
u
+
v
and -N
V
=
u
+
v
, we obtain L = E + F , M = F + G,
M = E + F and N = F + G. The last four equations can be written as
L
M
M
N
=
E
F
F
G
.
This gives
=
E
F
F
G
-1
L
M
M
N
,
which proves the result.

48
3. CURVATURE OF A SURFACE
We have
E
F
F
G
-1
=
1
EG-F
2
G
-F
-F
E
.
Corollary 3.11.3. Let be a surface patch of an oriented surface S containing point p = (u, v).
Then the Gaussian curvature K
p
and the mean curvature H
p
are
K
p
=
LN - M
2
EG - F
2
and H
p
=
LG - 2M F + N E
2(EG - F
2
)
.
P
. Since K
p
= det(W
p
), we have
K
p
= det
E
F
F
G
-1
L
M
M
N
= det
E
F
F
G
-1
det
L
M
M
N
= det
E
F
F
G
-1
det
L
M
M
N
=
LN - M
2
EG - F
2
.
Now
E
F
F
G
-1
L
M
M
N
=
1
EG-F
2
LG - M F
M G - N F
M E - LF
N E - M F
, and hence H
p
=
1
2
trace(W
p
) =
LG-2M F +N E
2(EG-F
2
)
.
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
u
=
(1, 0, f
u
),
v
= (0, 1, f
v
),
uu
= (0, 0, f
uu
),
uv
= (0, 0, f
uv
) and
vv
= (0, 0, f
vv
). Therefore
E = 1 + f
2
u
, F = f
u
f
v
and G = 1 + f
2
v
. Now N =
u
×
v
u
×
v
=
1
1+f
2
u
+f
2
v
(-f
u
, -f
v
, 1).
So, L =
uu
N =
f
uu
1+f
2
u
+f
2
v
, M =
uv
N =
f
uv
1+f
2
u
+f
2
v
and N =
vv
N =
f
vv
1+f
2
u
+f
2
v
. Now
EG - F
2
= 1 + f
2
u
+ f
2
v
and LN - M
2
=
f
uu
f
vv
-f
2
uv
1+f
2
u
+f
2
v
. Thus
K =
LN - M
2
EG - M
2
=
f
uu
f
vv
- f
2
uv
(1 + f
2
u
+ f
2
v
)
2
,
also
H =
LG - 2M F + N E
2(EG - F
2
)
=
(1 + f
2
v
)f
uu
- 2f
u
f
v
f
uv
+ (1 + f
2
u
)
(1 + f
2
u
+ f
2
v
)
3
2
.

11. Gaussian, Mean and Principal Curvatures
49
(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
2
+ g
2
= 1.
We have
u
= (f cos v, f sin v, g ),
v
= (-f sin v, f cos v, 0),
uu
= (f cos v, f sin v, g ),
uv
= (-f sin v, f cos v, 0) and
vv
= (-f cos v, -f sin v, 0). Therefore E = 1, F = 0
and G = f
2
. 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 )
2
f
2
= -
f (g )
2
f
and H =
-f
2
f g +f g
2f
2
=
g (1-f f )
f
.
(iii) Let : (a, b) R
3
be a unit-speed curve with nowhere vanishing curvature. Define :
(a, b) × (0, ) R
3
by (u, v) = (u) + vt(u). We shall compute the Gaussian curvature
and the mean curvature of this surface.
Here
u
= t + v t = t + vn,
v
= t,
uu
= t + v n + v
n = n + v n + v( b - t),
uv
= n and
vv
= 0. Thus E = 1 + v
2
2
, F = 1 and G = 1. Also,
u
×
v
= -vb gives
N = -b. Now L = -v , M = N = 0. Thus LN - M
2
= 0 implies that the Gaussian
curvature is zero everywhere. Also, H =
-v
2v
2
2
= -
2v
.
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
3
be a regular surface patch, and let (u
0
, v
0
) U
. For > 0, let
B
= {(u, v) : (u - u
0
)
2
+ (v - v
0
)
2
<
2
}. Let K be the Gaussian curvature of the surface at p.
Then lim
0
A
N
(B
)
A
(B
)
= |K|
.
P
. Since (u
0
, v
0
) U and U is open, there is
0
> 0 such that B
0
U . We have N
u
× N
v
=
LN -M
2
EG-G
2
(
u
×
v
) = K(
u
×
v
). If <
0
, then
A
N
(B
)
A
(B
)
=
B
N
u
× N
v
dudv
B
u
×
v
dudv
=
B
|K|
u
×
v
dudv
B
u
×
v
dudv
.
Let
> 0. Since K is continuous, there exists > 0 ( <
0
) such that |K(u
0
, v
0
)| - < |K(u, v)| <
|K(u
0
, v
0
)| + for all (u, v) B
. Let 0 < < . Then
(|K(u
0
, v
0
)| - )
B
u
×
v
dudv
B
|K|
u
×
v
dudv

50
3. CURVATURE OF A SURFACE
(|K(u
0
, v
0
)| + )
B
u
×
v
dudv.
This gives
|K(u
0
, v
0
)| -
B
|K|
u
×
v
dudv
B
u
×
v
dudv
|K(u
0
, v
0
)| + ,
i.e.,
|K(u
0
, v
0
)| -
A
N
(B
)
A
(B
)
|K(u
0
, v
0
)| + .
Hence
A
N
(B
)
A
(B
)
- |K(u
0
, v
0
)| <
for every 0 < < ,
i.e., lim
0
A
N
(B
)
A
(B
)
= |K(u
0
, v
0
)|.
Example 3.11.7. For a plane, the image of the unit normal N is a point only. Hence A
N
(R
) = 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
N
(R
) = 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
N
(R
) = A
(R
) 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