GLOSSARY
Notations :
¡3
:
3 - dimensional Euclidean space;
p, q
:
Point in ¡3 ⇒ p = ( p1, p2 , p3 ) , q = ( q1, q2 , q3 ) ;
f, g
:
Real valued functions on ¡ 3 ;
v, w :
Tangent vectors in ¡ 3 ;
vp
Tangent vector in ¡ 3 at point p;
:
V,W :
Vector fields in ¡ 3 ;
α,β
Curves in ¡ 3
:
( )
T p ¡3 :
U T p ( ¡3 )
p∈¡3
Tangent space of ¡ 3 at p.
: The set of all tangent vectors in ¡ 3 .
Vector Space : Let V be a non empty set whose elements are called vectors and ( F , +, ⋅) a field.
Then (V , +, ) is called a vector space over the field F if
(i)
( V , +) is a abelian group.
(ii)
a ⋅ ( v + w ) = av + aw,
∀v , w ∈ V
(iii)
( a + b) ⋅ v = av + bv ,
∀a, b ∈ F
(iv)
( ab ) ⋅ v = a ( bv ) = b ( av )
(v)
1⋅ v = v
(vi)
0⋅v = 0
V is also called linear space.
Lines, Planes and Spheres :
We recall the vector equation of lines, planes and spheres in ¡ 3 .
(i)
Geometrically, a straight line is determined by a point on the line and a vector parallel to it.
(ii)
A plane is determined by a point on the plane and the vector perpendicular to the plane.
(iii)
A sphere is determined by its center and its radius.
Differential Geometry
(1)
Definition (Line) :
z
−p=
α (t)
The line through the point p ∈ ¡ 3 and parallel to the vector
tv
α (t)
v ≠ 0 has equation
p = α (0)
α ( t ) = p + tv
v
... (1)
where p = α (0)
y
x
Note : If p = α (0) = ( x0 , y0 , z 0 ) , v = ( v1, v2 , v3 )
and α (t ) = ( x (t) , y (t ), z( t) ) , then equation (1) gives
x (t ) − x(0) = tv1 , y (t ) − y (0) = tv2 , z (t ) − z (0) = tv3
Eliminating t, we get
x − x0 y − y0 z − z0
=
=
v1
v2
v1
... (2)
This required equation of straight line in a space passing through the point ( x0 , y0 , z0 ) and having
direction cosines ( v1, v2 , v3 ) .
α (t) −
)
(x2− x1
x1 = t
Similarly, the line through the point x1 and x2 in ¡ 3 has equation
x2
α (t ) = x1 + t ( x2 − x1 )
x1
α (t)
Definition (Plane) :
z
The plane through x0 is given by
( x − x0 ) ⋅ nˆ = 0
where n̂ is the unit normal to the plane and x is any
other point on the plane.
x
x
−
... (3)
x
0
n
x0
o
y
We can also write equation of the plane as
( r − α ) ⋅ nˆ = 0
... (4)
x
where α is the fixed point and r is any other point on the plane.
Differential Geometry
(2)
Definition (Sphere) :
The equation of the sphere in ¡ 3 with center at c and radius r > 0 is given by
( r − c ) ( r − c ) = r2
... (5)
where r = ( x , y, z) any point on the sphere.
c = ( a , b, c) center of the sphere.
r = radius of the sphere.
Note : We write equation (5) as
(r − c ) 2 = r2
⇒ ( x − a) + ( y − b) + ( z − c) = r 2
2
2
2
This is in standard form.
Differential Geometry
(3)
CHAPTER - I
EUCLIDEAN SPACE
Unit I
Introduction :
In this unit we establish some basic definitions and some part of elementary calculus which
deals with differentiation of a function of more variables. Initial language used throughout the book is
formulated in this unit.
Euclidean Space of 3-Dimension :
The set of all ordered triplets of real numbers is called an Euclidean space of 3-dimension.
Thus ¡ 3 = {( p1 , p2 , p3 ) / pi ∈ ¡ , i = 1,2,3} . The triplet ( p1, p2 , p3 ) is called a point of ¡ 3 and is
denoted by p.
If p = ( p1, p2 , p3 ) and q = ( q1, q2 , q3 ) are two points of then we say that,
(i)
p = q iff p1 = q1 , p2 = q2 , p3 = q3
(ii)
p + q = ( p1 + q1, p2 + q2 , p3 + q3 )
(iii)
a ⋅ p = ( ap1, ap 2 , ap3 ) for a scalar .
Note :
(i)
The point 0 = ( 0,0,0) is called the the origin of ¡ 3 .
(ii)
The above two operations addition and scalar multiplication on ¡ 3 satisfy the axioms of
vector space. This shows that ¡ 3 is a vector space.
Definition : A real valued function f on ¡ 3 is differentiable (or infinitely differentiable or of class C ∞ )
provided all partial derivatives of f of all orders exist and are continuous.
Differentiable real valued function f and g may be added and multiplied in a similar way to get
differentiable real valued function.
(f + g) (p)
=
f(p) + g(p)
(fg) (p)
=
f(p) . g(p)
Differential Geometry
1
at each point p of ¡ 3 .
Definition : Let x, y, z be the real valued functions on ¡ 3 .
⇒ x : ¡ 3 → ¡ , y : ¡3 → ¡ and z : ¡3 → ¡
such that at each point p of ¡ 3 ,
x ( p ) = p1, y( p ) = p2 , z ( p ) = p3
The values of the functions x, y, z at a point p are the numbers p1, p2 , p3 respectively. These
functions x, y, z are called the natural coordinate functions. Hence a point p = ( p1, p2 , p3 ) of ¡ 3 can
also be written as (x(p), y(p), z(p)) for each point p of ¡ 3 .
Note : Eliminatory calculus does not make a sharp distinction between numbers p1, p2 , p3 and the
functions x, y, z.
Tangent Vector in ¡ 3 :
Definition : A tangent vector v p at point p in ¡ 3 consist of two points of ¡ 3
(i) its vector part v and (ii) its point of application p (initial point).
p+v
We shall always picture a vector v p as an arrow from the point p to
the point p + v . Thus a tangent vector to ¡ 3 at a point p is a directed line
in the direction of a given vector v which emerges at p and terminates at
p+ v .
v
vp
p
z
p + v = (3, 4, 5)
vp
e.g. If p = (1, 1, 3) and v = (2, 3, 2) then the vector at
point p, v p is an arrow starting from (1, 1, 3) and ending
at (1 + 2, 1 + 3, 3+2). Thus a point ( p1, p2 , p3 ) in ¡ 3
can also represent the vector from origin.
p = (1, 1, 3)
y
Definition : Two tangent vectors v p and wq are said to be equal ifff
(i)
they have the same vector part v = w and
(ii)
the same point of application. p = q.
Differential Geometry
2
vp
Parallel Vectors : Two tangent vectors v p and vq with same vector part
vq
but different points of application are said to be parallel.
p
q
Note : It is essential to recognize that v p and vq are different tangent vectors if p ≠ q .
( )
Tangent space T p ¡3 :
Let p be a point of ¡ 3 . The set consisting of all tangent vectors that have p as its point of
( )
application is called tangent space of at p and it is denoted by T p ¡3 .
z
Note :
33
E
¡
3 3
TTpp( E¡
)
(i) Elements of T p ¡3 are called tangent vectors.
( )
( )
p
y
(ii) All the tangent vectors in a given tangent space have the same
point of application.
x
Worked examples........................................................................................................................•
Example : Draw the following vectors
(i) (1,0,0) p
(ii) (0,1,0) p
(iii) (0,0,1) p
where p = ( p1, p2 , p3 ) is a point of ¡ 3 .
Solution : (1,0,0) p is the vector starting at ( p1, p2 , p3 ) and ending at (1 + p1 , p2 , p3 ) .
Similarly for other vectors.
z
(p1, p2, 1 + p3)
( p1, 1 + p2, p3 )
( 1 + p1 , p2 , p3 )
y
x
Note : Several tangent vectors can be drawn at a point of ¡ 3 .
Differential Geometry
3
p
wp
u
vp + wp
w
v
( )
a tangent space T p ¡3 of ¡ 3 at p.
v+w
p
We define vector addition and scalar multiplication of ¡ 3 to form
vp
Definition : Let v p and wp be tangent vectors at the same point p. Then we define their addition at
( v + w) p
= v p + w p . This is just usual "Parallelogram law for addition of vectors. And
( av ) p = a ⋅ v p ,
this merely stretches a tangent vector by a if a > 0 and reversing its direction if
p as
a < 0, a ∈¡ .
( )
Note : These operations on each tangent space T p ¡3 shows that it is a vector space.
Note : The set of all tangent vectors in ¡ 3 is denoted by
U Tp ( ¡ 3 ) .
p∈¡3
Tangent Vector Field (Vector Field) :
Definition : A vector field V on ¡ 3 is a function that assigns to each point p of ¡ 3 a tangent vector
V ( p) to ¡ 3 at p.
Thus a vector field is just a big collection of tangent vectors one at each point of R3 .
i.e.
V : R3 →
U T p ( ¡3 )
p∈¡ 3
e.g. The gravitational force field of the Earth assigns to each point of a space, and is directed at the
center of the Earth.
Magnetic field force is another example of vector field.
Definition : Let V and W be two vector fields, define V + W at point p of ¡ 3 by
and
(V + W ) ( p ) = V ( p ) + W ( p )
∀p ∈ ¡ 3
( f V ) ( p ) = f ( p ) ⋅V ( p )
∀p .
where f : ¡ 3 → ¡ is a real valued function on ¡ 3 .
Differential Geometry
4
Natural Frame Field :
To know how an arbitrary vector field looks like in ¡ 3 , we define special vector fields
Ui , i = 1, 2, 3 on ¡ 3 .
Ui : ¡3 →
U Tp ( ¡ 3 )
p∈¡3
such that U1 ( P) = (1,0,0) p , U 2 ( P) = (0,1,0) p , U3 ( P ) = (0,0,1) p for each
p ∈ ¡ 3 . Then
U1 , U2 , U3 are collectively called the natural frame fields on ¡ 3 . Thus Ui , i = 1, 2, 3 are the unit
vector fields in the positive x, y, z directions, which will serve as a 'basis' for all vector fields.
Theorem : Any vector field V of ¡ 3 is uniquely expressed as
V = v1U1 +v2 U2 + v3 U3
where v1 , v2 , v3 are three real valued functions on ¡ 3 evaluated at p.
(The functions v1 , v2 , v3 are called the Euclidean coordinate functions of V .)
Proof : A vector field V assigns to each point p a tangent vector V(p) at p. Thus,
V : ¡3 →
U T p ( ¡3 )
p∈¡ 3
Such that V ( p) is a tangent vector at a point p. The tangent vector can be put as a unique triple
( v1( p ), v2 ( p), v3( p) ) . Each vi ( p) , depends on p, is a real number. Thus,
V ( p) = ( v1 ( p), v2 ( p), v3 ( p) ) ,
= ( v1 ( p),0,0 ) p + ( 0, v 2 ( p),0 ) p + ( 0,0, v3 ( p ) ) p ,
= v1 ( p) (1,0,0 ) p + v2 ( p ) ( 0,1,0 ) p + v3 ( p ) ( 0,0,1) p ,
= v1 ( p)U1 p + v 2( p)U 2 p + v 3( p)U 3( p) ,
(
)
(
)
(
)
= v1U1 ( p ) + v2U 2 ( p) + v3U3 ( p) ,
(
)
V ( p) = v1U1 + v2U2 + v3U3 ( p ) .
Since this is true for every point p of ¡ 3 , it follows that
Differential Geometry
5
V = v1U1 +v2 U2 + v3 U3 ,
or
V = ∑ vi Ui ,
where Ui are natural frame fields.
Note :
(i)
A vector field may always be expressed in terms of their Euclidean coordinate functions.
(ii)
A vector field V is differentiable provided its Euclidean coordinate functions are differentiable
Worked examples ........................................................................................................................•
Example 1 : Let v = (-2, 1, -1) and w = (0, 1, 3). If p is any point of ¡ 3 then express the tangent
vector 3v p − 2 wp as a linear combination of U1 ( p), U 2 ( p),U3 ( p ) .
Solution : Given that v p = ( −2,1, −1) p , wp = (0,1,3) p be two tangent vectors defined at point p of
¡ 3 . By definition of addition of two tangent vectors and scalar multiplication, we have
3v p − 2 wp = 3 ( −2,1 −1) p − 2 ( 0,1,3) p
= ( −6,3, −3) p + 2 ( 0, −2, −6 ) p
= ( −6,1, −9 ) p
= ( −6,0,0 ) p + ( 0,1,0 ) p + ( 0,0, −9 ) p
= −6 (1,0,0 ) p + 1( 0,1,0) p − 9 ( 0,0,1) p
or
3v p − 2 wp = −6U1( p) + U 2 ( p ) − 9U3 ( p) .
Example 2 : Let V = xU1 + yU2 , W = 2 x2U 2 − U3 be two vector fields. Compute the vector field
W − xV and find its value at P (-1, 0, 2).
Solution : Let V and W be two vector fields given by
V = xU1 + yU2 , W = 2 x2U 2 − U3
Differential Geometry
6
W − xV = 2 x 2U2 − U3 − x 2U1 − xyU 2
Then,
(
)
= − x 2U 1 + 2 x2 − xy U 2 − U 3
(W − xV ) p = ( − x2U 1 + ( 2 x 2 − xy )U 2 − U 3 ) ( p)
Therefore
(
)
= − x 2 ( p)U1 ( p) + 2 x 2 − xy ( p )U 2 ( p ) −U 3 ( p )
Where the point p = (-1, 0, 2)
Thus
(W − xV ) p = −U 1( p ) + 2U 2 ( p ) −U 3 ( p )
= −1(1,0,0) p + 2(0,1,0) p − (0,0,1) p
= (−1,0,0) p + (0,2,0) p − (0,0, −1) p
(W − xV ) p = (−1,2, −1) p
Example 3 : If V = y 2 U 1 − x 2U 3 , W = x 2U 1 − zU 2 . Find the function f and g such that the vector
field f V + gW can be expressed in terms of U 2 and U 3 only..
Solution : Two vector fields V and W are given by
V = y 2 U 1 − x 2U 3 , W = x 2U 1 − zU 2
We define,
(
) (
f V + gW = f y 2U 1 − x 2U 3 + g x 2U 1 − zU 2
(
)
)
f V + gW = fy 2 + gx 2 U 1 − gzU 2 − fx 2U 3
Given that
f V + gW = − gzU 2 − fx 2U 3
This emplies that the coefficient of U 1 is zero.
⇒ fy 2 + gx 2 = 0
or
fy 2 = − gx 2
Let f be the function given by f = − x 2 h , where h is real valued function on ¡ 3 .
Consequently, we have g = y 2 h . Thus
f V + gW = − gzU 2 − fx 2U 3 ,
Differential Geometry
7
where f = − x 2 h and g = y 2 h
Example 4 : Let V 1 = U 1 − xU 3 , V 2 = U 2 , V 3 = xU 1 + U 3 .
(a)
Prove that the vectors V 1 ( p) , V 2 ( p ) , V 3 ( p) are linearly independent at each point of ¡ 3 .
(b)
Express the vector field xU 1 + yU 2 + zU 3 as a linear combination of V 1 , V 2 and V 3 .
Solution :
(a)
to show V 1 = U 1 − xU 3 , V 2 = U 2 , V 3 = xU 1 + U 3 are linearly independent at each point of
¡ 3 , consider the sum
α , β ,γ ∈ ¡
α V 1 ( p) + β V 2 ( p) + γ V 3 ( p ) = 0
(
)
(
)
⇒ α U 1 − xU 3 ( p ) + β U 2 ( p) + γ xU 1 + U3 ( p ) = 0
(
)
(
)
⇒ α U 1( p) − x( p )U 3 ( p ) + βU 2 ( p) + γ x( p )U1 ( p) + U 3 ( p) = 0
⇒ α  (1,0,0 ) p − p1 ( 0,0,1) p  + β ( 0,1,0) p + γ  p1 (1,0,0 ) p + ( 0,0,1) p  = 0




⇒ α (1,0, − p1 ) p + β ( 0,1,0 ) p + γ ( p1 ,0,0 ) p = 0
⇒ (α + γ p1, β , −α p1 + γ ) = 0
This shows that,
α + γ p1 = 0 , β = 0 , −α p1 + γ = 0
This gives
α = 0, β = 0, γ = 0 .
Hence V 1 , V 2 , V 3 are linearly independent at each point of .
(b)
Consider V = xU 1 + yU 2 + zU 3 be a vector field.
Define V = α V1 + β V2 + γ V3
where α , β , γ are to be determined. Thus we have
xU 1 + yU 2 + zU 3 = αV 1 + βV 2 + γV 3
(
)
(
= α U 1− xU 3 + β U 2 + γ xU 1 + U 3
xU 1 + yU 2 + zU 3 = ( α + γ x ) U 1 + β U 2 + ( −α x + γ ) U 3
Differential Geometry
8
)
⇒ α + xγ = x
β = y and
−α x + γ = z
Solving these equation for α , β we γ obtain
x (1 − z )
x2 + z
α= 2
, β = y, γ = 2
x +1
x +1
Thus,
 x2 + z 
 x (1 − z ) 
V = 2
 V 3
 V 1 + yV 2 +  2
 x +1 
 x +1 
•....................................................................................................................................................•
Directional Derivative :
Introduction :
Besides the knowledge of vector fields and their differentiability a concept which plays an
important role in curve theory is the directional derivative.
Partial derivative is a some what unsatisfactory generalization of the usual derivative because
existence of all the partial derivatives at a point does not necessarily imply continuity of the function at
that point. The trouble with partial derivative is that they treat a function of several variables as a
function of one variable at a time. The partial derivative describes the rate of change of a function in the
direction of each coordinate axis.
There is a straight generalization of usual derivative called the directional derivative which
studies the rate of change of a function in an arbitrary direction.
Definition : Let f : ¡ 3 → ¡ be a differentiable real valued function on ¡ 3 and v p a tangent vector
to R3 then the number
vp [ f ] =
d
 f ( p + tv ) 
t =0
dt 
is called the directional derivative of f w.r.t v p .
Note : This definition appears in elementary calculus with additional restriction that v p be a unit
vector.
Differential Geometry
9
Note : The directional derivative v p [ f ] represents initial rate of change in f as p moves in the direction
of v .
Worked examples..........................................................................................................................•
Example 1 : Calculate v p [ f ] for the function f = x2 yz with p = (1, 1, 0) and v = (1, 0, -3).
Solution : By definition,
vp [ f ] =
d
 f ( p + tv ) 
t =0
dt 
... (1)
where p + tv = (1, 1, 0) + t (1, 0, -3)
= (1 + t, 1, -3t)
Thus
f ( p + tv ) = f (1 + t, 1, – 3t)
= −3 (1 + t ) t
2
as f (x , y , z) = x 2 yz
f ( p + tv ) = −3t − 6t 2 − 3t 3
Substituting this in (1) we get
vp[ f ] =
d
 −3t − 6t 2 − 3t 3 
 t =0
dt 
v p [ f ] = − 3 −12t − 9t 3 
t =0
v p [ f ] = −3
Example 2 : Compute v p [ f ] for f = x12 + 2 x2 x3 , v = (2, 1, -3), p = (0, -1, 3)
Answer : v p [ f ] = 12
•...................................................................................................................................................•
Directional derivative of any function f can also be expressed in terms of partial derivative of f.
In this connection we prove the following theorem.
Differential Geometry
10
Theorem : If v p = ( v1, v2 , v3 ) is a tangent vector to ¡ 3 at a point p then prove that
v p [ f ] = ∑ vi
i
∂f
( p)
∂xi
Proof : Let v = ( v1, v2 , v3 ) and p = ( p1, p2 , p3 ) .
Thus
p + tv = ( p1 + tv1, p2 + tv2 , p3 + tv3 )
f ( p + tv ) = f ( pi + tvi ) ,
for i = 1, 2, 3
or
f ( p + tv ) = f ( xi )
for xi = pi + tvi
Thus
df ∂f dx1 ∂f dx2 ∂f dx3
=
+
+
dt ∂x1 dt ∂x 2 dt ∂x 3 dt
∀i
∂f dxi
i =1 ∂xi dt
3
=∑
3
df
∂f
=∑
vi
dt i =1 ∂xi
Thus the directional derivative of f w. r. t. v p becomes
d
∂f
 f ( p + tv )  = ∑ vi
( p)
t =0
dt
∂xi
i
vp [ f ] =
... (2)
We can solve above example 1 by applying the above theorem.
Worked examples..........................................................................................................................•
Example 3 : Find v p [ f ] for f = x2 yz with p = (1, 1, 0) and v = (1, 0,-3).
Solution : Given that v = (1, 0, -3); p = (1, 1, 0)
By definition (2) we have
v p [ f ] = ∑ vi
i
= v1
∂f
( p)
∂xi
∂f
∂f
∂f
( p ) +v 2
( p) + v3
( p)
∂x1
∂ x2
∂ x3
( )
= 1( 2 xyz ) p + 0 + ( −3) x2 y
Differential Geometry
11
p
v p [ f ] = −3
Example 4 : If v p is a tangent vector to R3 , where v = (2, -1, 3) and p = (2, 0, -1). Compute
v p [ f ] when
(i) f = y 2 z
(ii) f = e x cos y
Solution : The directional derivative of f w.r.t. v is given by
v p [ f ] = ∑ vi
i
∂f
( p)
∂xi
⇒ v p [ f ] = v1
... (1)
∂f
∂f
∂f
( p) + v 2
( p) + v3
( p)
∂x1
∂x2
∂x3
where v = (2, -1, 3) and p = (2, 0, -1)
(i)
For
f = y2 z
(
v p [ f ] = −2 yz + 3 y 2
)
p
vp [ f ] = 0
(ii)
If f = e x cos y then
(
v p [ f ] = 2 e x cos y
)
p
(
+ ( −1) −e x sin y
(
= 2e x cos y + e x sin y
)
)
p
p
v p [ f ] = 2e 2
•....................................................................................................................................................•
Properties of Directional Derivative :
Theorem : Let f and g be real valued functions on E3 . v p and wp are tangent vectors on E3 and a,
b are real numbers, show that
(i)
( av p + bwp )[ f ] = av p [ f ] + bwp [ f ]
linear in v p
(ii)
v p [ af + bg ] = av p [ f ] + bv p [ g ]
linear in f
Differential Geometry
12
(iii)
v p [ fg ] = v p [ f ] g ( p) + f ( p )v p [ g ]
Leibnitz Rule
Proof : The directional derivative of f w. r. t. v is defined by
v p [ f ] = ∑ vi
i
(i)
∂f
( p)
∂xi
... (1)
Using (1) we write
( av p + bw p )[ f ] = ∑ ( avi + bwi ) ∂∂xf
i
= ∑ avi
∂f
∂f
( p ) + b ∑ bwi
(p)
∂xi
∂xi
i
= a ∑ vi
∂f
∂f
( p) + b ∑ wi
(p)
∂xi
∂xi
i
i
i
(
⇒ av p + bw p
(ii)
( p)
i
)[ f ] = av p [ f ] + bwp [ f ]
Now consider
v p [ af + bg ] = ∑ vi
i
∂
( af + bg ) ( p )
∂xi
= ∑ vi
i
∂
∂
( af ) ( p ) + ∑v i ( bg ) ( p )
∂xi
∂xi
i
= a ∑ vi
i
∂f
∂g
( p) + b ∑ vi
( p)
∂xi
∂xi
i
⇒ v p [ af + bg ] = av p [ f ] + bv p [ g ]
(iii)
Next consider
v p [ fg ] = ∑ v i
i
∂
( fg ) ( p )
∂xi
 ∂
∂ 
= ∑ vi 
g+ f
 ( p)
∂xi 
i
 ∂xi
 ∂f

∂g
= ∑ vi  ( p) g( p) + f ( p )
( p) 
∂xi
i
 ∂xi

Differential Geometry
13
⇒ v p [ fg ] = g ( p) ∑ vi
i
∂f
∂g
( p ) + f ( p )∑ vi
( p)
∂xi
∂xi
i
⇒ v p [ fg ] = v p [ f ]g ( p ) + f ( p )v p [ g ]
Directional Derivative of f in the Direction of the Vector Field :
The process of directional derivative of a function f with respect to a tangent vector can also be
repeated with respect to a vector field (by applying the point wise principle).
Since a vector field V on ¡ 3 is a function that assigns to each point p of ¡ 3 a tangent vector
3
V (p) to ¡ at p. Thus we define
Definition : The directional derivative of a function f w. r. t. the vector field V is the directional
derivative of f w. r. t. the tangent vector V (p) at p. Thus we have
V [ f ] ( p) = V ( p) [ f ] = ∑ vi
i
∂f
( p)
∂xi
We know that every vector field V is expressible in terms of vector fields U i . We will see in the
following result that what does the directional derivative of a function look like in the direction of U i .
Result : If U 1 , U 2 , U 3 are natural frame fields at p, then show that
U1[ f ] =
∂f
∂f
∂f
, U2[ f ]=
, U3[ f ]=
∂y
∂x
∂z
Proof : By definition, we have
U 1 [ f ] ( p ) =U 1 (p ) [ f
= ∑ vi
i
where
]
∂f
( p)
∂xi
U 1 = ( v1, v2 , v3 ) = (1,0,0)
⇒ U 1 [ f ]( p ) = 1
∂f
( p)
∂x1
since p is an arbitrary point of E3 , we have
Differential Geometry
14
U1[ f ] =
∂f
∂x1
∂f
∂f
Similarly we have U 2 [ f ] = ∂x , U 3 [ f ] = ∂x
2
3
In general,
U i [ f ]( p ) =
∂f
( p)
∂xi
⇒Ui[ f ]=
∀i = 1,2,3.
∂f
∂xi
∀i = 1,2,3.
where ( x1, x2 , x3 ) = ( x, y , z )
Theorem : If V and W are vector fields on ¡ 3 and f, g, h are real valued functions, then show that,
(i)
( f V + gW ) [h] = f V [h ] + gW [ h]
(ii)
V [ af + bg ] = aV [ f ] + bV [ g ]
(iii)
V [ fg ] = V [ f ] g + f V [ g ]
a, b ∈ ¡
Proof : Let p ∈ ¡ 3 be an arbitrary point. Since for any function f and a vector field V , f V is a new
vector field such that
( f V ) ( p ) = f ( p )V ( p )
(i)
Consider
( f V + gW ) [h] ( p) = ( f V + gW ) ( p) [h]
( )
( )
=  f V ( p) + gW ( p)  [ h]
=  f ( p)V ( p) + g ( p )W( p )  [ h ]
= f ( p)V ( p) [ h ] + g ( p )W( p) [ h]
= f ( p)V [ h ] ( p) + g ( p)W [ h ] ( p )
{
}
= f V [ h] + gW [ h] ( p )
Since the point p is an arbitrary, we have
( f V + gW ) = f V [h ] + gW [ h ]
Differential Geometry
15
(ii)
Now consider
V [ af + bg ] ( p ) = V ( p) [ af + bg ]
= V ( p) [ af ] + V ( p) [ bg ]
= aV ( p ) [ f ] + bV ( p) [ g]
{
}
{
}
= aV [ f ] + bV [ g ] ( p )
⇒ V [ af + bg ] ( p ) = aV [ f ] + bV [ g ] ( p)
Aliter
V [ af + bg ] ( p ) = V ( p) [ af + bg ]
 ∂

= ∑ vi ( p )  ( af + bg )  p
i
 ∂xi

 ∂f

∂g
= ∑ vi ( p) a
( p) + b ( p)
∂xi
i
 ∂xi

= a ∑ vi ( p )
i
∂f
∂g
( p ) + b ∑ vi ( p) ( p )
∂xi
∂xi
i
{
}
{
}
V [ af + bg ] ( p ) = aV [ f ] + bV [ g ] ( p )
⇒ V [ af + bg ] = aV [ f ] + bV [ g ]
(iii)
Consider
V [ fg ] ( p) = V ( p) [ fg ]
 ∂

= ∑ vi ( p ) 
( fg ) ( p )
i
 ∂xi

 ∂g ∂f
= ∑ vi ( p )  f
+
∂
x
∂xi
i
i

= ∑ vi ( p ) f ( p)
i
Differential Geometry

g  ( p)

∂g
∂f
( p) + ∑ vi ( p ) f ( p)
( p) g( p)
∂xi
∂xi
i
16

 

∂g
∂f
= f ( p)  ∑ vi ( p )
( p)  +  ∑ vi ( p ) f ( p)
( p ) g ( p )
∂xi
∂xi
 i
  i

= f ( p)V [ g ] ( p) + V [ f ] ( p) g( p)
{
}
= f V [ g] + V [ f ] g ( p)
Since the point p is an arbitrary, we have
V [ fg ] = V [ f ] g + f V [ g ]
Worked examples ..........................................................................................................................•
Example 5 : If V = xU 1 − y 2U 3 , f = x 2 y + z 3 , find V [ f ] and V V [ f ] .
Solution :
(i)
By definition
V [ f ] ( p) = V ( p) [ f ]
(
)
= xU 1 − y 2U 3 ( p) [ f ]
=
(( xU ) ( p) − ( y U ) (p ) )[ f ]
2
1
3
( )
p
U 3 ( p )[ f
( )
p
U 3 [ f ]( p )
= ( x ) p U 1 ( p) [ f ] − y 2
= ( x ) pU 1 [ f ] ( p) − y 2
= (x ) p
∂f
( p) − y2
∂x
( )
∂f
( p)
p ∂x
( ) (3 z )
= ( x ) p ( 2 xy ) p − y2
(
)
(
)
V [ f ]( p ) = 2 x2 y − 3 y2 z2
⇒ V [ f ] = 2x2 y − 3y2 z 2
Differential Geometry
2
p
p
17
p
]
Aliter
We have
V [ f ] ( p) = V ( p) [ f ]
= ∑ vi ( p)
i
= v1 ( p)
∂f
( p)
∂xi
∂f
∂f
∂f
( p ) + v 2 ( p)
( p) + v3 ( p)
( p)
∂x1
∂x2
∂x3
( )
= x( p)(2 xy)( p ) + 0 − y 2 ( p) 3 z 2 ( p)
(
)
= 2 x 2 y − 3 y 2 z 2 ( p)
⇒ V [ f ] = 2x 2 y − 3 y 2 z 2
(ii)
Consider
V  V [ f ] ( p ) = V  2x 2 y − 3y 2 z 2  ( p)
= V ( p)  2 x 2 y − 3y 2 z 2  ( p)
= ∑ vi ( p )
i
∂f
2 x 2 y − 3 y 2 z 2 ( p)
∂xi
(
)
(
)
= x( p)(4 xy )( p ) + 0 − y 2 ( p) −6 y 2 z ( p )
(
)
= 4 x 2 y + 6 y 4 z ( p)
⇒ V  V [ f ] = 4 x 2 y + 6 y 4 z
Example 6 : Let V = y 2 U 1 − xU 3 , f = xy , g = z 3 . Compute the following functions.
(i) V [ f ]
(ii) V [ g ]
(iii) V [ fg ]
(iv) V  f 2 + g 2 
(v) V V [ f ]
(vi) f V [ g ] − gV [ f ]
Differential Geometry
18
Solution : Let p ∈ ¡ 3 be any arbitrary point.
(i)
Consider
V [ f ] ( p) = V ( p) [ f ]
= ∑ vi ( p)
i
= y 2 ( p)
∂f
( p)
∂xi
∂f
∂f
( p) − x( p) ( p)
∂x
∂z
= y 2 ( p) y ( p ) − x( p )(0)
= y 3 ( p)
⇒ V [ f ] = y3
... (1)
Similarly we prove
(ii)
V [ g ] = −3 xz 2
(iii)
We know the formula
... (2)
V [ fg ] = f V [ g ] + gV [ f ]
Using (1) and (2) we get
V [ fg ] = y3 z 3 − 3x 2 yz 2
(iv)
... (3)
Consider
(
)
V  f 2 + g 2  ( p) = V  f 2  +V  g 2  ( p)
... linearity
= V  f 2  ( p ) + V  g 2  ( p)
= V ( p)  f 2  + V ( p)  g 2 
(
)
(
)
= y 2U1 − xU 3 ( p )  f 2  + y 2U 1 − xU 3 ( p)  g 2 
= y 2 ( p)U 1 ( p)  f 2  − x ( p)U 3 ( p)  f 2  + y 2 ( p)U1 ( p)  g 2  − x ( p)U 3 ( p )  g 2 
= y 2 ( p)
Differential Geometry
∂f 2
∂f 2
∂g 2
∂g 2
( p) − x( p)
( p) + y2 ( p)
( p) − x ( p)
( p)
∂x
∂z
∂x
∂x
19
(
)
= y 2 ( p) ( 2 f ⋅ y) ( p) − x( p) ( 2 f ⋅ 0 ) ( p) + 0 − x( p) 6 gz 2 ( p)
(
)
V  f 2 + g 2  ( p) = 2 xy 3 − 6 xz 5 ( p )
⇒ V  f 2 + g 2  = 2 xy3 − 6 xz 5
(v)
... (4)
Using equation (1) we have
V  V [ f ] = V  y 3 
V  V [ f ] ( p ) = V  y 3  ( p)
= V ( p)  y 3 
⇒
(
)
= y 2U 1 − xU 3 ( p )  y 3 
= y 2 ( p)U 1 ( p)  y 3  − x( p )U 3 ( p )  y 3 
= y2 ( p)
∂y 3
∂y3
− x( p )
( p)
∂x
∂z
V V [ f ] = 0
... (5)
Example 7 : If X , Y and V are vector fields in ¡ 3 , prove that
V  X ⋅ Y  = ∑ xi V [ yi ] + ∑V [ xi ] yi
i
i
Solution : Any vector field of ¡ 3 is uniquely expressed as a linear combination of Natural frame
fields. Thus we have
X = ∑ xi U i and Y = ∑ yi U i
i
i
where xi are yi the Euclidean coordinate functions.
Therefore X ⋅ Y = ∑ xi yi is a scalar function.
i
Differential Geometry
20
Now consider


V  X ⋅ Y  ( p) = V ( p)  X ⋅ Y  = V ( p ) ∑ xi yi 
 i

= ∑ vk ( p )
k

∂ 
 ∑ xi yi  ( p )
∂xk  i



∂y
∂x
= ∑ vk ( p )∑  xi ( p) i ( p ) + yi ( p) i ( p) 
∂xk
∂xk
k
i 





∂y
∂x
= ∑ xi ( p )  ∑ vk ( p ) i ( p)  + yi ( p)  ∑ vk ( p ) i ( p ) 
∂xk
∂xk
i
 k

 k

= ∑ xi ( p )V ( p) [ yi ] + ∑V ( p) [ xi ] yi ( p )
i
i


=  ∑ xi V [ yi ] + ∑ V [ xi ] yi  ( p)
 i
i

As p is any arbitrary point in ¡ 3 , hence we have
V  X ⋅ Y  = ∑ xi V [ yi ] + ∑V [ xi ] yi
i
i
•.....................................................................................................................................................•
Curves in ¡ 3 :
We now define the curves in ¡ 3 and discuss their geometric properties.
Definition : Let I be an open interval in the real line ¡ . A curve in ¡ 3 is differentiable function
α : I → ¡3
From an open interval I into ¡ 3 .
z
Note :
(i)
I may be ¡ .
c
α (t)
If we regard the parameter t as the time variable,
then one can picture a curve in ¡ as a path of moving point
whose position vector at any time t is α (t ) where
3
α (t ) = (α1, α2 , α3 ) . At each time t in some open interval,
Differential Geometry
21
y
(
t
)
I
x
Imα is located at the point α (t ) in ¡ 3 . Here α1, α2 , α 3 are real valued functions of real variable t
and are called the Euclidean coordinate functions.
The expression (α1 (t) , α2 (t ), α3 ( t ) ) of α (t ) is called a parametric representation of the curve.
(ii)
(iii)
To make the differentiability meaningful we have open interval I of ¡ . We say that α is
differentiable if each α i is differentiable and we write
(
α '(t ) = α1' ( t ), α2' (t ), α 3' (t )
)
e.g. A line is a simplest type of curve in Euclidean space.
Definition (Straight line) :
The curve α : ¡ → ¡3 such that
α (t ) = p + tq = ( p1 + tq1, p2 + tq2 , p3 + tq3 ) , q ≠ 0
is a straight line passing through the point p = α (0) in the direction of q.
Ex. (i) Circle : The curve
α (t ) = ( a cos t , a sin t ,0 )
z
represents a circle in the xy plane with radius a and centered at origin.
(ii) Helix : The curve α : ¡ → ¡3 defined by
α
y
x
α (t ) = ( a cos t , a sin t , bt ) , where a > 0 , b ≠ 0
is called a helix. If is a curve on a cylinder making a constant angle with a fixed
line.
Remark : We could picture a curve for 0 ≤ t ≤ 1 by computing α (t ) . In this way we visualize a
curve α in ¡ 3 as a moving point.
Velocity Vector of a curve :
If we visualize a curve α in ¡ 3 as a moving point, then at every time t there is a tangent vector at the
point α (t ) which gives instantaneous velocity of α at that time.
Differential Geometry
22
Definition :
α (t)
c
α(t)
(
I
Let α : I → ¡3 be a curve in ¡ 3 such that
(
)
α (t ) = α1 (t ), α2 (t ), α3 ( t ) ∀t ∈ I , then the velocity vector of
α at time t is defined to be the tangent vector
 dα dα dα 
α '(t ) =  1 , 2 , 3 
 dt dt dt α (t )
)
at the point α (t ) in R3 .
Geometrical interpretation :
We know the derivative at t of a real valued function f on R is given by
f ( t + δ t ) − f (t)
df
(t ) = lim
δ t →0
dt
δt
(
Now if f is replaced by a curve α (t ) = α1 ,α2 , α 3
) we get
 α ( t + δ t ) − α1(t ) α2 ( t + δ t ) − α 2 (t ) α3 ( t + δ t ) − α 3 (t ) 
1
α ( t +δ t ) −α ( t )  =  1
,
,

δt
δt
δt
δt


This is a vector from α (t ) to α ( t + δ t ) , scalar multiplied by
]
α( t)
)−
t
δ
+
1 [ α(t
α (t + δt)
δt
α(t)
1
.
δt
Now as δ t gets smaller, α ( t + δ t ) approaches α (t ) and in the limit
as ∂t → 0 we get a vector tangent to the curve α at the point α (t ) ,
viz;
 dα (t ) d α (t ) dα (t ) 
α '(t ) =  1 , 2 , 3 
dt
dt 
 dt
This shows that the standard limit operation for derivative gives rise to our definition of the velocity of
a curve.
Reparametrization of Curves :
Parametric equations of a curve are not unique. We can reparametrize the given curve by another
parameters. This is given in the following definition.
Differential Geometry
23
Definition :
Let I and J be open intervals in ¡ . Let α : I → ¡3 be a curve and h : J → I be a differentiable real
valued function. Then the composite function β = α ( h) : J → ¡ 3 is a curve called the reparametrization
of α by h.
z
( s)
ββ(S)
S
h
E3
¡
3
α (t) = β(s) = α(h(s))
α
(s)
tt == hh(S)
J
y
I
x
At each time s in the interval J, the curve β is at the point β ( s ) = α ( h( s )) reached by the curve α at
time h(s) in the interval I. Thus β does follow the route of α , but β generally reaches a given point
on the route at a different time than α does.
Relation between the Velocities of a Curve α and a reparametrization :
Lemma : If β is a reparametrization of α by h then prove that β '( s) = h '( s )α '( h( s )) .
Proof : Let α : I → ¡3 be a curve such that
α (t ) = (α1( t ), α2 (t ), α3 (t ) )
then the composite function
β ( s) = α (h ) : J → ¡ 3
is the reparametrization of α by h such that
β ( s) = α ( h( s ) ) = (α1 ( h( s ) ) , α2 ( h( s ) ) , α3 ( h( s ) ) )
... (1)
We know the chain rule for the derivative of a composition of real valued functions f and g.
( g ( f ) ) ' = g '( f ) ⋅ f '
Using this formula we write
 ai ( h ( s) )  ' = ai ( h ( s) ) ⋅ h '( s )
'
⇒ β '( s) =  a (h (s) )  '
Differential Geometry
24
(
= a1' ( h( s ) ) ⋅ h '( s), a2' ( h( s ) ) ⋅ h '( s), a3' ( h ( s) ) ⋅ h '( s )
(
= h '( s ) a1' ( h( s ) ) , a2' ( h ( s ) ) , a3' ( h ( s) )
)
)
⇒ β '( s) = h '( s )α '( h( s ) )
This gives the relation between the velocities of curve α and the reparametrization of α by h.
Lemma : Let α be a curve in ¡ 3 and f be a differentiable function on ¡ 3 then show that
α '(t ) [ f ] =
df
( α (t) )
dt
Proof : Let α : I → ¡3 be a curve in ¡ 3 such that
α (t ) = (α1( t ), α2 (t ), α 3 (t ) ) ∈¡3
The tangent vector to the curve α at point α (t ) is given by
(
)
α '(t ) = α1' ( t ), α2' (t ), α 3' (t ) . Also f ( α ) = f (α1, α2 , α3 )
Thus the rate of change of f in the direction of the curve is given by
α '(t ) [ f ] = ∑ α1' (t )
i
⇒ α '(t ) [ f ] = α 1' (t )
∂f
(α (t ) )
∂α i
∂f
∂f
∂f
(α ( t) ) +α 2' (t )
( α (t) ) +α 3' (t)
( α (t) )
∂α1
∂α2
∂α3
⇒ α '(t ) [ f ] =
∂f d α1 ∂ f dα 2
∂ f dα 3
+
+
∂α1 dt ∂α 2 dt ∂α 3 dt
⇒ α '( t ) [ f ] =
df
( α (t ) )
dt
This gives the directional derivative of f along the curve .
Worked examples..........................................................................................................................•
Example 1 : Find the coordinate function of the curve β = α ( h( s ) ) where
(
)
where 0 < t <
α (t ) = 2cos 2 t ,sin2,2sin
t
t ,
Differential Geometry
25
π
4
and h is a function on J : 0 < s < 1, such that h ( s) = sin −1 s .
Solution : Let α : I → ¡3 be a curve defined by
(
α (t ) = 2cos 2 t ,sin2,2sin
t
t
)
... (1)
We reparametrize α by h, such that
β ( s) : J → ¡ 3 such that
β ( s) = α ( h( s ) ) ,
where h ( s) = sin −1 s .
To reparametrize α by h(s) we first write equation (1) as
(
α (t ) = 2 − 2sin 2 t ,2sin t 1 − sin 2 t ,2sin t
)
Thus reparametrization of α (t ) gives
(
α ( h( s )) = β ( s ) = 2 − 2 s2 , 2 s 1− s2 t ,2s
)
where 2 − 2 s 2 , 2 s 1 − s 2 , 2 s are called coordinate functions.
Example 2 : Show that the curves given by
(
)
α = t ,1 + t 2 , t , β = ( sin t ,cos t , t ) , γ = ( sinh t ,cosh t , t )
have the same initial velocity v p . If f = x2 − y2 + z2 . Compute v p [ f ] by evaluating f on each of
the curves.
(
)
Solution : We are given that α : I → ¡3 is a curve defined by α = t ,1 + t 2 , t .
The velocity of α is the tangent vector α ' , thus we have
α '( t ) = (1, 2t, 1 )
Hence the initial velocity of is given by
α '( t ) t =0 = (1,0,1)
... (1)
Similarly, for the curve β = ( sin t ,cos t , t ) , the velocity is the tangent vector β ' and is given by
β ' = ( cos t , − sin t ,1)
Differential Geometry
26
⇒ β '(t )t = 0 = (1,0,1)
... (2)
Similarly, for the velocity of the curve γ is given by
γ '(t )t =0 = ( cosh t ,sinh t ,1) t =0
⇒ γ '(t )t =0 = (1,0,1)
... (3)
Equation (1), (2) and (3) shows that the curves α , β , γ have the same initial velocity..
To find v p [ f ] along the curve α is equivalent to find α '(t ) [ f ] . Thus along we have
α '(t ) [ f ] =
(i)
df ( a (t ) )
dt
(
=
d 2
x − y2 + z 2
dt
=
d 2
2
 t − 1+ t
dt 
(
)
)
2
(
a (t )=t ,1+ t 2 , t 


+ t 2 

)
⇒ α '(t ) [ f ] = 2t − 4t + 4t 3 + 2t
α '( t ) [ f ] = −4t 3
or
(ii)
Similarly, the rate of change of f along a curve β is given by
β '(t ) [ f ] =
(
d 2
x − y2 + z 2
dt
)
β (t )=( sin t ,cos t , t )
⇒ β '(t ) [ f ] = 4sin t cost + 2t
and
(iii)
γ '(t ) [ f ] =
(
d 2
x − y2 + z2
dt
)
γ (t )= ( sinh t ,cosh t ,t )
This gives,
γ '( t ) [ f ] = 2t
Example 3 : Find a straight line passing through the points (1, -3, -1) and (6, 2, 1). Does this line
meet the line through the points (-1, 1, 0) and (-5, -1, -1) ?
Differential Geometry
27
Solution : We know the vector part of the line starting at p (1, -3, -1) and ending at point
p + v = (6, 2, 1) is given by
v = (6, 2, 1) - (1, -3, -1)
⇒ v = (5, 5, 2)
... (1)
Thus the equation of the line through the point (1, -3, -1) and (6, 2, 1) is
α (t ) = p + tv = (1, −3, −1) + t (5,5,2)
α (t ) = p + tv = (1 +5t , − 3 + 5t , − 1+ 2t )
... (2)
Similarly, the vector part of the line through the points (-1, 1, 0) and (-5, -1, -1) is
v = (-4, -2, -1)
... (3)
Therefore the line through the points (-1, 1, 0) and (-5, -1, -1) is
α ( s ) = p + sv = ( −1,1,0 ) + s(− 4,− 2, − 1)
α ( s ) = ( −1 − 4s ,1 − 2s , −4s )
... (4)
We see from the equations (1) and (3) that the vector parts of two lines are not same. This implies that
the two lines are not parallel and hence they may intersect at a point. The point of intersection of two
lines (2) and (4) is obtained by solving any two of the equations
1 + 5t = -1 - 4s,
-3 + 5t = 1 - 2s,
-1 + 2t = -s
This gives t = 2, s = -3.
Thus the point of intersection is (11, 7, 3).
•...................................................................................................................................................•
Regular Curves :
Introduction : Let α : I → ¡3 be a curve. If
dα
= 0 on an interval I, then α (t ) is constant in that
dt
dα
= 0 at some point, then graph of α can
dt
have sharp corner which is geometrically unappealing. Hence we will only work with regular curves.
interval which is geometrically uninteresting. If further,
Definition : A curve α : I → ¡3 is called regular curve if α '(t ) ≠ 0 , ∀t ∈ I .
Differential Geometry
28
Note :
(i)
A regularity condition says that the speed is always non-zero and hence particle never stops
moving even instantaneously.
(ii)
A regular curve can have no corners or cusps because at cusps the function is not differentiable.
(iii)
Speed is not an absolute measure. A curve can have different speed through reparametrization.
(iv)
The regularity condition
e.g.
(1)
dα
≠ 0 , ∀t ∈ I implies that the set
dt
The curve α : ¡ → ¡3 defined by
(
α ( t ) = t 3,0,0
is not a regular curve because
(2)
)
dα
dα
= 3t 2 ,0,0 and
= 0 at t = 0.
dt
dt
(
)
Let α : ¡ → ¡3 be a curve given by
(
α (t ) = t 3 + t ,0,0
We see that
 dα 

 is linearly independent.
 dt 
)
dα
≠ 0 at every t ∈¡
dt
⇒ the curve is a regular..
EXERCISE ........................................................................................................•
Which of the following are regular curves.
(1)
α (θ ) = ( cosθ ,1 − cos θ − sin θ , −sin θ )
(2)
β (θ ) = 2sin 2 θ ,2sin θ tan θ , 0
(3)
γ (θ ) = cos θ ,cos 2 θ ,sin θ
(
(
)
)
•........................................................................................................................•
Differential Geometry
29
Unit II
Differential forms :
Introduction :
The calculus of differential forms developed in the early part of the last century by Eli Cartan
(1869-1951). It is one of the most useful and powerful analytical tool of modern mathematics for the
use of engineers and scientists. There are certain problems normally very difficult to solve by using
tensor only for which the results are more quickly and directly obtained with differential forms.
Definition :
(i)
1- form :
( )
A 1-form φ is a linear real valued function defined on the tangent space T p ¡3
( )
φ : T p ¡3 → ¡
i.e.
such that
(ii)
( )
φ v p ∈¡
Linearity :
( )
If v p and wp are tangent vectors of T p ¡3 at a point p and a, b are real scalars, then
(
)
( )
( )
φ av p + bwp = aφ v p + bφ wp
(iii)
∀p ∈ ¡ 3 , a , b ∈ ¡ 3
Addition of two 1-forms :
Two 1-forms can be added pointwise. If φ and ψ are two 1-forms, then
(φ +ψ ) ( v p ) = φ ( vp ) +ψ ( vp )
(iv)
... (1)
Scalar Multiplication : If f is a real valued function on ¡ 3 and φ is a 1-form then f φ is 1form satisfying
( )
( f φ ) (vp ) = f ( p)φ ( vp )
Differential Geometry
∀v p ∈ T p ¡3
30
... (2)
( )
( )
If *Tp ¡3 is a set of all linear maps from T p ¡3 onto ¡ with binary operation addition
( )
defined in (1) and an operation for scalar multiplication (2), then it is easily established that *Tp ¡3
is a vector space. This vector space is called co-tangent space or dual space.
1-form φ on a vector field :
Since vector field gives tangent vector at each point of ¡ 3 , hence 1-form φ on a vector field V is a
( )
( )
real valued function φ V at each point p of ¡ 3 . The value of φ V is the real number denoted by
(
)
φ V ( p) .
Note :
(i)
One can view 1-form as a machine which converts vector field into a real valued function and
( )
at any point p of ¡ 3 the value of φ V (p) is a number..
(ii)
A 1-form φ is differentiable whenever the vector field V is differentiable.
(iii)
A 1-form φ on a vector field is linear in both φ and V .
i.e.
φ ( f V + gW ) = f φ (V ) + gφ (W )
( f φ + gψ ) ( V ) =
and
f φ (V ) + gψ (V )
where f and g are functions.
Now given a real valued function f on ¡ 3 , we see how to convert it into a 1-form.
Definition : For every real valued differentiable function f on ¡ 3 , the differential df is a 1-form
whose value on a tangent vector v p is v p [ f ] .
Hence
( )
df : Tp ¡3 → ¡
is a 1-form defined by
( )
df v p = v p [ f ]
This shows that the value of df at v p is equal to the directional derivative of f w.r.t. v p .
Differential Geometry
31
Worked examples .........................................................................................................................•
Example 1 : In view of the definition, for coordinate function xi , i = 1, 2, 3; the differentials dxi are
1-forms.
By definition we have
( )
dxi : T p ¡3 → ¡ , such that
( )
dxi v p = v p [ xi ]
= ∑ v j ( p)
j
∂xi
∂x j
= ∑ v jδ ij
j
( )
dxi v p = vi
This shows that the action of the differentials dxi (1-forms) on an arbitrary tangent vector v p gives its
ith component vi - a scalar and does not depend on the point of application p.
Result : If φ is a 1-form on ¡ 3 , then prove that φ = ∑ f i dxi , where f i = φ (U i )
i
Proof : Let v p = ∑ vi U i ( p) be a tangent vector. Since 1-form is a real valued function on vectors.
i
Now to prove
φ = ∑ f i dxi , where f i = φ (U i ) is a 1-form.
i
We have to prove that φ and
∑ fi dxi
i
have the same value when operated on an arbitrary vector..
Hence consider
φ ( vp ) = φ
(∑ v U ( p ))
i
i
i
= ∑ viφ ( U i ( p ) ) ........ linearity of 1-form φ ( f V + gW ) = f φ (V ) + gφ (W )
i
Differential Geometry
32
φ ( v p ) = ∑ vi f i ( p )
... (1)
i
Now consider
(∑ f dx ) ( v ) = ∑ ( f dx )( v )
i
i
p
i
i
i
p
i
(Q ( f φ + gψ ) ( v p ) = f ( p )φ ( v p ) + g ( p)ψ ( v p ) )
= ∑ f i ( p) dxi ( v p )
i
= ∑ f i ( p )vi
(Q dxi ( v p ) = vi )
i
⇒
(∑ f dx ) (v ) = ∑ v f ( p)
i
i
p
i i
i
... (2)
i
From equation (1) and (2), we see that φ and
∑ fi dxi
have the same effect when operated on a
i
tangent vector v p . Since v p is an arbitrary vector, we have therefore
φ = ∑ f i dxi
... (3)
i
Note : In elementary calculus for a real valued function f we define
df =
∂f
∂f
∂f
dx + dy + dz
∂x
∂y
∂z
But it is not made clear what this expression means. Now we give rigorous treatment to this expression.
Result : If f is a differentiable function on ¡ 3 , then show that df = ∑
i
∂f
dxi is a 1-form.
∂xi
Proof : Since f is differentiable function on ¡ 3 , then by definition the differential df is 1-form.
But 1-form by definition is a function on vectors, such that
df ( v p ) = v p [ f ]
df ( v p ) = ∑ vi
i
Similarly, the value of
∂f
∑ ∂x
i
i
∂f
( p)
∂xi
... (1)
dxi on an arbitrary vector v is given by
p
 ∂f

 ∂f

 ∑ ∂x dxi  ( v p ) = ∑  ∂x dxi  ( v p )
i
i
 i

i 

Differential Geometry
33
=∑
∂f
( p ) dxi ( v p )
∂xi
=∑
∂f
( p ) ⋅ vi
∂xi
i
i
∂f
 ∂f

⇒ ∑
dxi  ( v p ) = ∑ vi
( p)
∂xi
 i ∂xi

i
From equations (1) and (2), we see that the 1-form df and
... (2)
∂f
∑ ∂x
i
dxi have the same value on every
i
∂f
tangent vector. Hence df = ∑ ∂x dxi represents 1-form.
i
i
Note : If f and g are differentiable functions on ¡ 3 then d ( f + g) = df + dg.
Result : Let fg be the product of differentiable functions f and g on ¡ 3 then prove that
d(fg) = g df + f dg
Proof : If f is differentiable functions on ¡ 3 then
df = ∑
i
∂f
dxi
∂xi
Using this we write
d ( fg ) = ∑
i
∂( fg )
dxi
∂xi
∂g 
 ∂f
= ∑
g+ f
dx
∂xi  i
i  ∂xi
∂f


=∑
dxi  g +
 i ∂xi

 ∂g

f ∑
dxi 
 i ∂xi

d ( fg ) = gdf + fdg
Result : If f : ¡ 3 → ¡ and h : ¡ → ¡ are differentiable functions, so that the composite function
h ( f ) : ¡ 3 → ¡ is also differentiable. Then show that
d ( h ( f ) ) = h ' ( f ) df
where dash is just the ordinary derivative.
Differential Geometry
34
Proof : Given that f and h are differentiable functions and the composite function
h ( f ) : ¡3 → ¡
is also differentiable on ¡ 3 . By using above result we have,
d ( h ( f )) = ∑
i
∂
( h ( f ) ) ⋅ dxi
∂xi
= ∑ h '( f
)
i
= h '( f ) ∑
i
∂f
⋅ dxi
∂xi
∂f
dxi
∂xi
d ( h ( f ) ) = h ' ( f ) df
Worked examples .........................................................................................................................•
Example 2 : If f = ( x 2 − 1) y + ( y 2 + 2 ) z . Compute v p [ f ] and hence the 1-form df.
Solution : Let v p be tangent vector at point p. By definition
df ( v p ) = v p [ f ]
... (1)
∂f
where v p [ f ] = ∑ vi ∂x ( p)
i
= v1
i
∂f
∂f
∂f
( p ) +v 2
( p) + v3
( p)
∂x1
∂ x2
∂ x3
We know the action of the differentials dxi on an arbitrary tangent vector v p gives its ith component.
Hence we write
vp [ f ] =
∂f
∂f
∂f
( p ) dx1 ( v p ) +
( p) dx2 ( vp ) +
( p )dx3 ( v p )
∂x1
∂x2
∂x3
Substituting this in equation (1) we get
∂f
∂f
 ∂f

df ( v p ) = 
dx1 +
dx2 +
dx3  ( v p )
∂x 2
∂x 3
 ∂x1

Differential Geometry
35
... (2)
Since v p is an arbitrary vector, hence we have
df =
∂f
∂f
∂f
dx + dy + dz
∂x
∂y
∂z
... (3)
⇒ df = 2 xydx + ( x 2 + 2 yz − 1) dy + ( y 2 + 2 ) dz
... (4)
Aliter :
We also have
vp[ f ] =
d
f ( p + tv ) |t =0
dt
where p = ( p1, p2 , p3 ) and v = ( v1, v2 , v3 )
Thus
vp [ f ] =
d
{ f ( p1 + tv1, p2 + tv2 , p3 + tv 3)} |t= 0
dt
As f (x , y , z ) = ( x 2 − 1) y + ( y 2 + 2 ) z , hence we have
vp [ f ] =
{
}
d 
2
2



( p1 + tv1 ) −1 ( p2 + tv2 ) + ( p2 + tv2 ) + 2  ( p3 + tv 3) |t = 0
dt
 ( p + tv )2 − 1 v + 2v ( p +tv )( p + tv ) + 
1
1
1
1
2
2
 1
 2

=

2
+2 ( p3 + tv3 ) ( p 2 + tv2 ) v2 + ( p2 + tv2 ) + 2  v3 
|t =0
⇒ v p [ f ] = ( p12 −1)v 2 + 2 p1 p2 v1 + 2 p2 p 3 v1 + ( p22 + 2 ) v3
... (5)
Substituting this in equation (1) we obtain
df ( v p ) = 2 p1 p2v1 + ( p12 + 2 p2 p3 − 1)v2 + ( p22 + 2 ) v3
But we know
Hence
dxi ( v p ) = vi
and also
xi ( p ) = pi
∀, i = 1,2,3
df ( v p ) = 2 p1 p 2dx1 ( v p ) + ( p 12 + 2 p2 p3 − 1) dx2 ( v p ) + ( p22 + 2) dx3 ( v p )
(
)
df ( v p ) = 2 xydx + ( x 2 + 2 yz − 1) dy + ( y 2 + 2 ) dz ( v p )
⇒ df = 2 xydx + ( x 2 + 2 yz − 1) dy + ( y 2 + 2 ) dz
Differential Geometry
36
EXERCISE .........................................................................................................•
1.
Let f be a differentiable function defined as f = ( x3 − 2 ) z + ( yz − 1) x . Find v p [ f ] and
hence or otherwise determine the 1-form df.
2.
In each case find v p [ f ] and hence or otherwise determine the 1-form df, for v = (1, 2, -3),
p = (0, -2, 1).
(i) f = xy 2 − yz 2
(iii) f = sin( xy)cos( xz )
(ii) f = xe yz
•.....................................................................................................................................................•
Result : If φ = f1dx1 + f 2dx2 + f 3dx3 is a 1-form and V = ∑ vi U i is a vector field, show that the
i
1-form φ evaluated on the vector field V is the function φ (V ) = ∑ f ivi .
i
Proof : Consider the value of 1-form φ on an arbitrary vector field V as φ (V ) . Thus
φ (V ) =
( ∑ f dx ) ∑ v U
i
i
i
j
j
j



= ∑ f i dxi ( v1U1 + v 2U 2 + v 3U 3 )
i
= ( f1dx1 + f 2dx2 + f3dx3 ) v1U1 + ( f1dx1 + f 2dx2 + f 3dx3 ) v 2U 2 +
+ ( f1dx1 + f 2 dx2 + f 3dx3 ) v3U 3
= f1v1dx1 (U 1 ) + f 2v1dx2 (U1 ) + f3v1dx3 (U 1 ) +
+ f1v2dx1 (U 2 ) + f 2v2dx2 (U 2 ) + f 3v2dx3 ( U 2 ) +
+ f1v3dx1 (U 3 ) + f 2 v2 dx2 (U 3 ) + f 3v 3dx3 (U 3 )
Since dxi (U j ) = δ ij ∀i , j hence we have
φ (V ) = f1v1 + f 2v2 + f 3v3
or
φ (V ) = ∑ f ivi
Differential Geometry
i
37
Worked examples ..........................................................................................................................•
Example 3 : Evaluate the 1-form φ = x2 dx − y 2 dz on the vector fields
(ii) W = xy (U 1 − U 3 ) + yz (U 1 − U 2 )
(i) V = xU 1 + yU 2 + xU 3
(iii)
1
1
V+ W
x
y
Solution : Since φ = x2 dx − y 2 dz
Thus the value of φ on the vector field V is given by
φ ( V ) = ( x 2dx − y 2dz ) (V )
= x 2dx (V ) − y 2dz (V )
= x2v1 − y 2v3
Since V = xU 1 + yU 2 + zU 3 ⇒ v1 = x , v 2 = y , v3 = z
Therefore, φ ( v ) = x 3 − y 2 z
... (1)
Similarly, we find
φ (W ) = x3 y + x2 yz + xy 3
Let
1
1
z= V+ W
x
y
Thus
1 
1
φ ( z ) = ( x 2dx − y 2dz )  V + W 
y 
x
2
... (2)
2
y
x
= xdx (V ) − dz (V ) + dx (W ) − ydz (W )
x
y
y2
x2
= xv1 −
v + w − yw3
x 3 y 1
1
1
φ V + W
y
x
y2 z

2
3
2
2
 = x − x + x + x z + xy

Differential Geometry
38
... (3)
EXERCISE .........................................................................................................•
Evaluate the 1-form φ = yzdx − x 2 dz on the following vector fields.
(i)
V = yzU 1 + xzU 2 + xyU 3
(ii)
W = xz (U 2 − U 3 ) + yz 2 (U 1 − U 2 )
(iii)
X=
1
1
V+ W
xyz
z
•....................................................................................................................................................•
Exterior Product (Wedge Product) :
Introduction :
We have seen that two 1-forms can be added to obtain another 1-form. Similarly 2-forms and
3-forms can be added to get another form. Now these arises a natural question about what happens if
two 1-forms are multiplied together. Is multiplication permissible ? Answer is yes. We denote the
multiplication operation by ∧ and call it as exterior multiplication or wedge product.
These two operations addition and multiplication of 1-forms obey the usual associative and
distributive laws. However the multiplication is not commutative.
(i)
(φ ∧ψ ) ∧ θ = φ ∧ (ψ ∧θ )
(ii)
φ ∧ (ψ + θ ) = φ ∧ ψ + φ ∧ θ
(iii)
φ ∧ ψ ≠ ψ ∧ φ but φ ∧ ψ = −ψ ∧ φ
(iv)
φ ∧φ = 0
In particular,
dxi ∧ dx j = −dx j ∧ dxi
∀i , j
For i = j we have
dxi ∧ dxi = 0
∀i
Remarks :
(i)
On three dimensional Euclidean space, all p-forms with p > 3 are zero. This is a consequence
of the anti-commutativity of wedge product. Because a product of more than three d xi' s in
¡ n must contain some dxi twice.
Differential Geometry
39
(ii)
A o-form is just a differentiable function f.
A 1-form is an expression φ = fdx + gdy + hdz
A 2-form is an expression of the form ψ = fdx ∧ dy + gdx ∧ dz + hdy ∧ dz
A 3-form is an expression of the form φ = fdx ∧ dy ∧ dz
In general , any p-form on ¡ n is of the form
φ = fdx1 ∧ dx 2 ∧ dx3 ∧ .... ∧ dx n
Worked examples .........................................................................................................................•
Example 4 : Let φ = xdx − ydy and ψ = zdz + xdz , then find φ ∧ ψ .
Solution : Consider
φ ∧ ψ = ( xdx − ydy )∧ ( zdz + xdz )
= xzdx ∧ dx + x 2 dx ∧ dz − yzdy ∧ dx − yxdy ∧ dz
Since
dx ∧ dx = 0 , dx ∧ dy = −dy ∧ dx
Thus
φ ∧ ψ = x 2 dx ∧ dz + yzdx ∧ dy − xydy ∧ dz
This shows that, the product of two 1-forms is a 2-form.
EXERCISE .........................................................................................................•
Let φ = xydx + dz , ψ = sin zdx + cos zdy , 0 = dy + zdz . Find φ ∧ ψ ,ψ ∧ θ , φ ∧ ψ ∧ θ .
•...........................................................................................................................•
Result : If φ and ψ are 1-forms, then φ ∧ ψ = −ψ ∧ φ .
Proof : Let φ = ∑ f i dxi , ψ = ∑ gdx
i i be two 1-forms.
i
i
Consider the wedge product of φ and ψ ,
φ ∧ψ =
( ∑ f dx ) ∑ g dx 
i
i
Differential Geometry
i
j
j
j
40
= ∑∑ f i g jdxi ∧ dx j
i
j
= −∑∑ fi g j dx j ∧ dxi
i
j
=  ∑ g j dx j  ∧
 j



( ∑ f dx )
i
i
i
φ ∧ ψ = −ψ ∧ φ
Exterior Derivative :
Definition : If φ = ∑ f i dxi is a 1-form on ¡ 3 , then the exterior derivative of φ is 2-form and is
i
given by
d φ = ∑ df i ∧ dxi
i
If φ = f1dx1 + f 2dx2 + f 3dx3 then the exterior derivative is given by
d φ = df1 ∧ dx1 + df 2 ∧ dx2 + df 3 ∧ dx3
(... a 2-form)
In particular if φ = fdx then d φ = df ∧ dx
Note :
(i)
Exterior derivative d is linear. i.e. d ( aφ + bψ ) = ad φ + bdψ .
(ii)
Exterior derivative of any p-form is p+1 form.
Worked examples ..........................................................................................................................•
Example 5 : If φ = f1dx1 + f 2dx2 + f 3dx3 is 1-form, then find its exterior derivative dφ .
Show that d 2φ = 0 .
Solution : Let the 1-form φ be
φ = f1dx1 + f 2dx2 + f 3dx3
then
d φ = df1 ∧ dx1 + df 2 ∧ dx2 + df 3 ∧ dx3
Differential Geometry
41
∂f
∂f
∂f
∂f
 ∂f

 ∂f

=  1 dx1 + 1 dx2 + 1 dx3  ∧ dx1 +  2 dx1 + 2 dx2 + 2 dx3  ∧ dx2 +
∂
x
∂
x
∂
x
∂
x
∂
x
∂
x
2
3
2
3
 1

 1

∂f
∂f
 ∂f

+  3 dx1 + 3 dx2 + 3 dx3  ∧ dx3
∂
x
∂
x
∂
x
2
3
 1

=−
∂f1
∂f
∂f
dx1 ∧ dx2 − 1 dx1 ∧ dx3 + 2 dx1 ∧ dx2 −
∂x2
∂ x3
∂ x1
−
∂f 2
∂f
∂f
dx2 ∧ dx3 + 3 dx1 ∧ dx3 + 3 dx2 ∧ dx3
∂x3
∂x1
∂x2
∂f 
∂f 
∂f 
 ∂f
 ∂f
 ∂f
=  2 − 1  dx1 ∧ dx2 +  3 − 1  dx1 ∧ dx3 +  3 − 2  dx2 ∧ dx3
 ∂x1 ∂x2 
 ∂x1 ∂x3 
 ∂x2 ∂x3 
... (1)
Taking the exterior derivative of (1) we get
 ∂ 2 f2
∂ 2 f1 
d (d φ ) = 
−
 dx1 ∧ dx2 ∧ dx3 +
 ∂x1∂x3 ∂x2∂x3 
 ∂ 2 f3
∂ 2 f1 
+
−
 dx2 ∧ dx1 ∧ dx3 +
 ∂x1∂x2 ∂x3∂x2 
 ∂ 2 f3
∂ 2 f2 
+
−
 dx1 ∧ dx 2 ∧ dx3
 ∂x2∂x1 ∂x3∂x1 
 ∂ 2 f2
∂ 2 f3
∂ 2 f3
∂ 2 f1
∂ 2 f1
∂2 f2 
d φ =
−
−
+
+
−
 dx1 ∧ dx2 ∧ dx3
 ∂x1∂x2 ∂x2∂x3 ∂x1∂x2 ∂x3∂x2 ∂x2∂x1 ∂x3∂x1 
2
⇒ d 2φ = 0
Example 6 : For any 2-form w, prove that d 2w = 0
Solution : Let w = Pdx ∧ dy + Qdy ∧ dz + Rdz ∧ dx be any 2-form. Take exterior derivative of w,,
we get
dw =
∂P
∂Q
∂R
dz ∧ dx ∧ dy +
dx ∧ dy ∧ dz +
dy ∧ dz ∧ dx
∂z
∂x
∂y
 ∂P ∂Q ∂R 
⇒ dw = 
+
+  dx ∧ dy ∧ dz
 ∂z ∂x ∂y 
Differential Geometry
42
... (1)
Now taking exterior derivative of (1) we get d 2w = 0 , because the product of more than
three d xi' s in ¡ 3 must contain some dxi twice.
•....................................................................................................................................................•
Theorem : Let f and g be functions, φ and ψ are 1-forms. Then show that
(i)
d (φ ∧ ψ ) = dφ ∧ ψ − φ ∧ dψ
(ii)
d ( f φ ) = df ∧ φ + fd φ
Proof : Case (i) : Let φ = fdx and ψ = gdx be 1-forms.
⇒ d φ = df ∧ dx and dψ = dg ∧ dx
... (1)
Consider
φ ∧ ψ = fgdx ∧ dx
⇒ φ ∧ψ = 0
⇒ d (φ ∧ ψ ) = 0
... (2)
Now consider
d φ ∧ ψ − φ ∧ dψ = df ∧ dx ∧ gdx − fdx ∧ dg ∧ dx
d φ ∧ ψ − φ ∧ dψ = 0
... (3)
From (2) and (3) we have
d (φ ∧ ψ ) = dφ ∧ ψ − φ ∧ dψ
Case (ii) : Let φ = fdx and ψ = gdy be 1-forms.
∂f
∂f 
 ∂f
d φ =  dx + dy + dz 
∂y
∂z 
 ∂x
⇒ dφ =
∂f
∂f
dy ∧ dx + dz ∧ dx
∂y
∂z
Consider now
∂f
 ∂f

d φ ∧ψ =  dy ∧ dx + dz ∧ dx  ∧ gdy
∂z
 ∂y

⇒ d φ ∧ψ =
∂f
⋅ gdx ∧ dy ∧ dz
∂z
... (4)
Similarly we find
dψ =
Differential Geometry
∂g
∂g
dx ∧ dy +
dz ∧ dy
∂x
∂z
43
Hence
φ ∧ dψ = − f
∂g
dx ∧ dy ∧ dz
∂x
... (5)
From (4) and (5) we have
∂g 
 ∂f
d φ ∧ ψ − φ ∧ dψ =  g
+f
 dx ∧ dy ∧ dz
 ∂z
∂z 
... (6)
Also we have
φ ∧ ψ = fdx ∧ gdy
φ ∧ ψ = fgdx ∧ dy
Taking exterior derivative we get
d (φ ∧ ψ ) = d ( fg ) ∧ dx ∧ dy
= ( gdf + fdg ) ∧ dx ∧ dy
∂g 
 ∂f
d (φ ∧ ψ ) =  g
+f
 dx ∧ dy ∧ dz
 ∂z
∂z 
... (7)
From equation (6) and (7) we have
d (φ ∧ ψ ) = dφ ∧ ψ − φ ∧ dψ
(ii)
... (8)
Claim :
Now we prove d ( f φ ) = df ∧ φ + fd φ
In general the result (8) is true given by
d (φ ∧ψ ) = d φ ∧ψ + ( −1) deg of φ φ ∧ dψ
... (9)
If φ is a zero-form, then from equation (9) we have
d ( f φ ) = df ∧ φ + ( −1) 0 fdφ
⇒ d ( f φ ) = df ∧ φ + fdφ
Worked examples .......................................................................................................................•
Example 7 : For any function f show that d(d(f)) = 0. Deduce that d ( fdg ) = df ∧ dg .
Solution : Let f be an arbitrary function in ¡ 3 .
⇒ f = f (x , y , z )
Differential Geometry
44
Thus
df =
∂f
∂f
∂f
dx + dy + dz is a 1-form.
∂x
∂y
∂z
... (1)
Taking exterior derivative of (1) we get
 ∂f 
 ∂f 
 ∂f 
d ( d ( f ) ) = d  dx  + d  dy  + d  dz 
 ∂x 
 ∂z 
 ∂y 
 ∂2 f
  ∂2 f

∂2 f
∂2 f
d2f =
dy ∧ dx +
dz ∧ dx  + 
dx ∧ dy +
dz ∧ dy  +
∂z∂x
∂z ∂y
 ∂y∂x
  ∂x∂y

 ∂2 f

∂2 f
+
dx ∧ dz +
dy ∧ dz 
∂y∂z
 ∂x∂z

 −∂ 2 f ∂ 2 f 
 ∂2 f ∂2 f 
d2f =
+
dx
∧
dy
+
−


 dx ∧ dz +
 ∂x∂z ∂z∂x 
 ∂y∂x ∂x∂y 
 ∂2 f ∂2 f 
+
−
 dy ∧ dz
 ∂y∂z ∂z∂y 
⇒d2f =0
(ii)
Since f and g are 0-forms. Hence df and dg are 1-forms. Hence we have
d ( fdg ) = df ∧ dg + ( −1) deg of f f ∧ d ( dg )
d ( fdg ) = df ∧ dg
Example 8 : Let f and g be real valued functions on ¡ 2 . Prove that
df ∧ dg =
∂f
dx
∂f
dy
∂g
dx
∂g
dy
dx ∧ dy
Solution : Let f and g be real valued functions on ¡ 2 .
Therefore
df =
Differential Geometry
∂f
∂f
∂g
∂g
dx + dy and dg =
dx +
dy
dx
dy
dx
dy
45
Thus
∂f
∂g 
 ∂f
  ∂g
df ∧ dg =  dx + dy  ∧  dx +
dy
dy   dx
dy 
 dx
 ∂f ∂g ∂f ∂g 
=
−
 dx ∧ dy
 dx dy dy dx 
Thus
df ∧ dg =
∂f
dx
∂f
dy
∂g
dx
∂g
dy
dx ∧ dy
Example 9 : For any three 1-forms φi = ∑ f ij dx j , prove that
j
f11
f12
f13
φ1 ∧ φ2 ∧ φ 3 = f 21
f 31
f 22
f32
f 23 dx1 ∧ dx 2 ∧ dx3
f 33
Solution : Given that
φ1 = f11dx1 + f12dx2 + f13dx3
φ2 = f 21dx1 + f 22dx2 + f 23dx3
φ3 = f 31dx1 + f 32dx2 + f 33dx3
be three 1-forms. Taking the wedge product we get
φ1 ∧ φ2 ∧ φ3 = ( f11 dx1 + f12 dx2 +f12 dx3 ) ∧ f 21dx1 ∧( f 32 dx2 + f 33 dx3 ) +
+ f 22 dx2 ∧ ( f 31dx1 + f33 dx3 ) + f 23 dx3 ∧( f 31dx1 + f32 dx2 ) 
= ( f11dx1 + f12 dx2 + f12 dx3 ) ∧  ( f 21 f 32 − f 22 f 31 ) dx1 ∧ dx2 +
+ ( f 22 f 31 − f 21 f 33 ) dx3 ∧ dx1 + ( f 22 f 33 − f 23 f 32 ) dx2 ∧ dx3 
= f13 ( f 21 f 32 − f 22 f 31 ) dx1 ∧ dx2 ∧ dx3 +
+ f12 ( f 23 f 31 − f 21 f 33 ) dx1 ∧ dx2 ∧ dx3 +
+ f11 ( f 22 f 33 − f 23 f 32 ) dx1 ∧ dx2 ∧ dx3
Differential Geometry
46
φ1 ∧ φ2 ∧ φ3 =  f13 ( f 21 f32 − f 22 f 31 ) + f12 ( f 23 f 31 − f 21 f 33 ) +
+ f11 ( f 22 f 33 − f 23 f 32 ) dx1 ∧ dx 2 ∧ dx3
f11
f12
f13
⇒ φ1 ∧ φ2 ∧ φ 3 = f 21
f 31
f22
f32
f 23 dx1 ∧ dx2 ∧ dx3
f33
Example 10 : Let w = zdx ∧ dy − 2 xf ( x )dy ∧ dz + yf (x ) dz ∧ dx be a 2-form. Find f if
(i) dw = dx ∧ dy ∧ dz and
(ii) d ( d ( w) ) = 0
Solution : Given 2-form is
w = zdx ∧ dy − 2 xf ( x )dy ∧ dz + yf (x ) dz ∧ dx
... (1)
Taking its exterior derivative we get
dw = dz ∧ dx ∧ dy − 2 ( f ( x ) + xf '( x) ) dx ∧ dy ∧ dz + ( f (x ) dy ) dy ∧ dz ∧ dx
= (1 − 2 xf '( x) − 2 f ( x) + f ( x) ) dx ∧ dy ∧ dz
⇒ dw = (1 − 2 xf '( x) − f ( x) ) dx ∧ dy ∧ dz
Given that dw = dx ∧ dy ∧ dz
... (3)
From equations (2) and (3) we have
1 − 2 xf '( x ) − f ( x) = 1
⇒ f '( x ) +
1
f (x) = 0
2x
Integrating we get
f (x) =
c1
x
where c1 is constant of integration.
Now to show that d 2 ( w) = 0 , we take exterior derivative of (2).
d 2 ( w) = ( −2 f '( x) − 2 f ''( x) − f '( x) ) dx ∧ dx ∧ dy ∧ dz
⇒ d 2 ( w) = 0
Differential Geometry
... (2)
47
1
∑ aij dxi ∧ dx j , where aij + a ji = 0 . Prove that
2
Example 11 : Let w =
1  ∂aij ∂a jk ∂aki 
+
+

 dxi ∧ dx j ∧ dxk
6  dxk
dx i dx j 
dw =
Solution : Since w =
where
1
∑ aij dxi ∧ dx j
2
... (1)
aij + a ji = 0
... (2)
Taking its exterior derivative of (1) we get
dw =
1 ∂aij
∑ dx dxk ∧ dxi ∧ dx j
2
k
⇒ dw =
1 ∂aij
∑ dx dxi ∧ dxj ∧ dxk
2
k
.
By cyclic permutation on indices i, j, k twice in turn in equation (3) we get
∂a
1
dw = ∑ jk dx j ∧ dxk ∧ dxi
2
dxi
and
dw =
1
∂aki
dx ∧ dxi ∧ dx j
∑
2
dx j k
... (3)
... (4)
... (5)
Adding equations (3), (4) and (5) we get
3dw =
1  ∂aij ∂a jk ∂aki 
∑  + dx + dx  dxi ∧ dxj ∧ dxk
2  dxk
i
j 
⇒ dw =
1  ∂aij ∂a jk ∂aki 
∑  + dx + dx  dxi ∧ dxj ∧ dxk
6  dxk
i
j 
... (6)
EXERCISE .....................................................................................................•
1.
If x = r sin θ cos φ , y = r cos θ cos φ , z = r cos θ prove that
dx ∧ dy ∧ dz = −r 2 cos φ dr ∧ dθ ∧ dφ
2.
Let v p be the tangent vector given by v = (-1, 2, 1) at p = (2, -2, 1). If f = xy 2 − yz 2 ,
g = xe y are two differentiable functions. Compute df and dg and find the directional derivative.
ANSWERS .....................................................................................................•
2.
v p [ fg ] = −6e−2
•.......................................................................................................................•
Differential Geometry
48
Unit III
Dot Product :
Definition : Let p = ( p1, p2 , p3 ) and q = ( q1, q2 , q3 ) be two points of ¡ 3 . The dot product of p
and q is denoted by p ⋅ q and is defined to be the number .
p1q1 + p2q2 + p3q3
p ⋅ q = p1q1 + p2q2 + p3q3
Thus
The dot product has following properties.
(i)
Bilinearity : ( ap + bq ) ⋅ r = ap ⋅ r + bq ⋅ r
r ⋅ ( ap + bq )⋅ = ar ⋅ p + br ⋅ q
(ii)
Symmetry : p ⋅ q = q ⋅ p
(iii)
Positive definiteness : p ⋅ p ≥ 0
The proof is obvious and left as an exercise.
Norm of p :
Definition : Let p = ( p1, p2 , p3 ) be a point is ¡ 3 . The norm of p is denoted by ||p|| and is defined
as a real valued function on ¡ 3 .
: ¡3 → ¡
i.e.
such that
p = ( p ⋅ p) = (
1
2
⇒ p
2
3
= ∑ pi
p12
+
p22
+
1
2 2
p3
)
2
i=1
Theorem : Prove that the norm has the following fundamental properties.
(i)
(ii)
p+q ≤ p ⋅ q
ap = a ⋅ p
where a is the absolute value of the number a .
Differential Geometry
49
Proof : Let p = ( p1, p2 , p3 ) and q = ( q1, q2 , q3 ) be two points of ¡ 3 . Then by definition of the
norm, we have
(i)
p+q
2
3
= ∑ pi + qi
2
i =1
3
= ∑ ( pi + qi
i =1
3
≤ ∑ pi + qi
i =1
)( pi + qi )
( pi + qi )
3
3
i =1
i =1
... by triangle inequality pi + qi ≤ pi + q i
= ∑ pi + qi pi + ∑ pi + qi qi
3
3
i =1
i=1
= ∑ ( pi + qi ) pi +∑ ( pi + qi ) qi
≤ p + q p + p +q q
(
2
p + q = p +q
... by Schwartz inequality p ⋅ q ≤ p + q
p + q
)
⇒ p+ q ≤ p + q
(ii)
Now consider
1
ap = [ ( ap ) ⋅ (ap ) ] 2
1
=  a ( p1, p2 , p3 ) ⋅ a ( p1, p2 , p3 )  2
1
=  ( ap1, ap 2 , ap3 ) ⋅ ( ap1, ap 2 , ap3 ) 2
=  a
2
p12
+a
=  a
2
(
+
Differential Geometry
p12
2
p22
p22
+a
+
2
p32
1
22
p3 
1
2
)
50
= a p
⇒ ap = a p
Euclidean Distance :
Definition : The dot product of tangent vectors v p and wp at the same point of ¡ 3 is the number
defined by
v p ⋅ wp = v ⋅ w
Note :
vp
2
= v p ⋅ wp = v ⋅ v = v
2
⇒ vp = v
( 3,4,2 ) p ⋅ (1,2,4 ) p = 3(1) +4(2) +2(4) = 19
e.g.
Worked examples...........................................................................................................................•
Example 1 : Let v = (1, 2, -1) and w = (-1, 0, 3) be tangent vectors at a point p.
v w v
,
,
, v × w, v × w .
v w v
Compute v ⋅ w,
Solution : Given that v = (1, 2, -1), w = (-1, 0, 3)
Thus
⇒ v
2
= v ⋅ v = (1,2, −1) ⋅ (−1,0,3) = 6
⇒ v
2
= 6
v  1 2
1 
=
,
,−

v  6 6
6
w  1
3 
,0,
Similarly we find w =  −

10 
 10
Next,
v
v
2
=
v v
⋅
v v
1   1 2
1 
 1 2
=
,
,−
,
,−
 ⋅

6  6 6
6
 6 6
Differential Geometry
51
1 2 1
= + + 
 6 6 6
=1
w
w
Similarly we find
2
=1
U1 U2
v ×w= 1
2
−1 0
Next
U3
−1 = (6, −2,2)
3
2
v × w = (6, −2,2) ⋅ (6, −2,2)
and
⇒ v × w = 2 11
•....................................................................................................................................................•
Remark : For any non zero vector v,
v
v
=1
Angle between Two Tangent Vectors :
wp
The angle θ between two tangent vectors v p and wp at the same
point p is defined as
θ
p
vp
v p ⋅ wp = v ⋅ w = v ⋅ w cos θ
Note : cos θ is not unique unless 0 ≤ θ ≤ π .
If θ =
π
then v p ⋅ w p = 0 .
2
In this case the two vectors v p and wp are orthogonal.
Unit vector : A vector of norm (length) one is called a unit vector.
Frame :
Definition : A set {e1, e2 , e3} of three mutually orthogonal unit vectors tangent to ¡ 3 at p is called the
frame at the point p.
Differential Geometry
52
Thus if {e1, e2 , e3} is a frame, then
e1 ⋅ e1 = e2 ⋅ e2 = e3 ⋅ e3 = 1
e1 ⋅ e2 = e1 ⋅ e3 = e2 ⋅ e3 = 0
By the symmetry of the dot product, we also have
e2 ⋅ e1 = e3 ⋅ e1 = e3 ⋅ e1 = 0
Now using the index notation, all the nine equations may be concisely expressed as
ei ⋅ e j = δ ij
∀1 ≤ i, j ≤ 3
where
1 when i = j
δ ij = 
0 when i ≠ j
is the Kronecker delta.
Note : At each point p of ¡ 3 , we see that the vectors U 1 ( p) , U 2 ( p ) , U 3 ( p ) (Natural frame
field) form a frame at the point p, since
U i ⋅U j = δ ij
∀i , j = 1,2,3
Worked examples .........................................................................................................................•
Example 2 : Prove that the tangent vectors
e1 =
(1,2,1)
1
1
( −2,0,2) , e3 =
(1, −1,1)
, e2 =
6
8
3
constitute a frame. Express v = (6, 1, -1) as a linear combination of e1, e2 , e3 .
Solution : To show that the set of vectors e1, e2 , e3 form a frame, we simply prove that they satisfy
ei ⋅ e j = δ ij
Consider
∀i , j = 1,2,3
 1 2 1   1 2 1 
e1 ⋅ e1 = 
,
,
,
,
⋅

 6 6 6  6 6 6
1 4 1
e1 ⋅ e1 =  + +  = 1
 6 6 6
⇒ e1 ⋅ e1 = 1
Differential Geometry
53
Also
2 
 1 2 1   −2
e1 ⋅ e2 = 
,
,
,0,
⋅

8
 6 6 6  8
e1 ⋅ e2 = 0
Similarly we readily show that ei ⋅ e j = δ ij
∀i , j
This proves that e1, e2 , e3 constitute a frame. The next part we will prove after the following theorem
is proved.
Theorem : Let e1, e2 , e3 be a frame at a point p of ¡ 3 . If v is any tangent vector to ¡ 3 at p then
show that v = ( v ⋅ e1 ) e1 + ( v ⋅ e2 ) e2 + ( v ⋅ e3 ) e3
Proof : Let e1, e2 , e3 be a frame at a point p of ¡ 3 .
⇒ ei ⋅e j = δ ij
... (1)
Now we prove that the vector e1, e2 , e3 are linearly independent. Therefore, consider
a1e1 + a2 e2 + a3e3 = 0
⇒ ( a1e1 + a2 e2 + a3e3 ) ⋅ e = 0 for any j = 1, 2, 3.
⇒ aj = 0
for any j = 1, 2, 3.
⇒ e1, e2 , e3 are linearly independent.
( )
The Euclidean space ¡ 3 is of dimension 3. As the tangent space T p ¡3 is isomorphic to ¡ 3 , this
( )
implies that T p ¡3 has dimension 3. Thus three independent vectors e1, e2 , e3 form as basis for
( )
( )
T p ¡3 . Thus each vector of T p ¡3 can be written uniquely as
v = c1e1 + c2e2 + c3e3
Thus
( v ⋅ e1 ) = ( c1e1 + c2e2
+c3e3 ) ⋅ e1
( v ⋅ e1 ) = c1
Similarly
( v ⋅ e2 ) = c2
and
( v ⋅ e3 ) = c3
Differential Geometry
... by (1)
54
Substituting these values in equation (2) we get,
v = ( v ⋅ e1 ) e1 + ( v ⋅ e2 ) e2 + ( v ⋅ e3 ) e3
... (3)
Note : This expression of v is called orthonormal expression of v in terms of the frame e1, e2 , e3 .
1
1
1
(1,2,1) , e2 =
( −2,0,2) and e3 =
(1, −1,1) we express
Thus for v = (6, 1, -1) and e1 =
6
8
3
on using (3)
7
7
4
e1 −
e2 +
e3
6
2
3
v=
Result : Let v and w be tangent vectors at he same point p. Prove that v × w is orthogonal to both
v and w and has length
v × w = ( v ⋅ v )( w ⋅ w) − ( v ⋅ w)
2
2
Proof : By definition we have
U 1( p ) U 2 ( p ) U 3 ( p )
v ×w =
v1
v2
v3
w1
w2
w3
= ( v2w3 − w2v3 ) U1( p ) + ( w1v3 − v1w3 )U 2( p) + ( v1w2 − w1v2 ) U 3( p)
= ( v2w3 − w2v3, v3 w1 − v1w 3, v1 w2 − w1v2 )
= ( c1 , c2 , c3 )
... (1)
Consider,
( v × w ) ⋅ v = ( c1, c2 , c3 ) ⋅ ( v1, v2 , v3 )
= c1v1 + c2v2 + c3v3
= ( v2w3 − w2v3 ) v1 + ( v3w1 − v1w3 ) v2 + ( v1w2 − w1v2 ) v3
⇒ ( v × w ) ⋅v = 0
... (2)
Similarly we prove that ( v × w ) ⋅ w = 0
... (3)
equation (2) and (3) shows that v × w is orthogonal to both v and w .
Differential Geometry
55
Now consider
2
v × w = ( v × w ) ( v × w)
= ( c1, c2, c3 ) ⋅ ( c1, c2 , c3 )
... by equation (1)
= c12 + c22 + c32
= ( v2w3 − w2v3 ) + ( v3w1 − v1w3 ) + ( v1w2 − w1v2 )
2
2
2
= v22w32 + w22v32 + v32 w12 + v12w32 + v12 w22 + w12v22 −
−2v2w 2v 3w3 − 2v1v 3w1w3 − 2v1v 2w 1w 2
We write this as
v ×w
2
= ∑ vi2w2j − 2 ∑ vi wi v j w j
i≠ j
... (4)
i< j
Now we consider
 3
 3
  3

( v ⋅ v ) ⋅ ( w ⋅ w ) − ( v ⋅ w) =  ∑ vi2  ∑ w2j  −  ∑ vi wi 
 i =1   j =1   i =1

2
2
 3

= ∑ vi2w2j −  ∑ vi2w2j + 2 ∑ vi wi v j w j 
i, j
 i= j
i< j

⇒ ( v ⋅ v ) ( w ⋅ w ) − ( v ⋅ w ) = ∑ vi w j − 2∑ vi wi v jw j
2
i≠ j
2
i< j
From equation (4) and (5) we have
v × w = ( v ⋅ v )( w ⋅ w) − ( v ⋅ w)
2
2
Note :
(i)
The cross product can also be defined as
v × w = v ⋅ w sinθ nˆ
where is a unit vector. Therefore the length of the cross product is given by
0 ≤θ ≤ π
v × w = v ⋅ w sinθ
Differential Geometry
56
... (5)
(ii)
Let u = ( u1, u2, u3 ) , v = ( v1, v2 , v3 ) , w = ( w1, w2 , w3 ) then
u1
u2
u3
u ⋅ v × w = v ⋅ w× u = w⋅ u × v = v1
w1
v2
w2
v3
w3
Result : u ⋅ v × w ≠ 0 iff u , v , w are linearly independent.
( )
Proof : Let u = ( u1, u2, u3 ) , v = ( v1, v2 , v3 ) , w = ( w1, w2 , w3 ) be vectors of T p ¡3 .
Assume u ⋅ v × w ≠ 0
Claim : We show that u , v , w are linearly independent.
Suppose that u , v , w are linearly dependent.
⇒ u = av + bw
Consider
av1 + bw1 av 2 + bw2 av 3 + bw3
u ⋅v × w =
v1
w1
v2
w2
v3
w3
av1
av2
av3
bw1
bw2
bw3
u ⋅ v × w = v1
w1
v2
w2
v3 + v1
w3
w1
v2
w2
v3
w3
v1
v2
v3
w1
w2
w3
u ⋅ v × w = a v1
w1
v2
w2
v3 + b v1
w3
w1
v2
w2
v3
w3
⇒ u ⋅ v ×w = 0
This is contradiction to the statement u ⋅ v × w ≠ 0 .
This is due to the wrong assumption that u , v , w are linearly dependent.
⇒ u , v , w are linearly independent.
Conversely, let u , v , w are linearly independent.
Claim : we prove that u ⋅ v × w ≠ 0
Let us assume that u ⋅ v × w = 0
⇒ u is orthogonal to v × w .
Differential Geometry
57
Since v and w are both orthogonal to v × w . This shows that u , v , w are all orthogonal to v × w .
⇒ u can be written as a linear combination of v and w .
⇒ u , v and w are linearly dependent. Which is wrong.
⇒ u ⋅ v ×w ≠ 0
Worked examples .........................................................................................................................•
Example 1 : Prove that v × w ≠ 0 iff v and w are linearly independent and also show that v × w
is the area of the parallelogram with sides v and w .
Solution : Let v and w be two vectors, and v × w ≠ 0 .
We claim that v and w are linearly independent.
Let us suppose that v and w are linearly dependent.
⇒ v = aw for some a
Thus v × w = aw × w = 0
This is contradiction showing that v and w are linearly independent.
Conversely,
Let v and w are linearly independent.
We prove that v × w ≠ 0
Suppose that v × w = 0
⇒ for any vector u , u ⋅ v × w = 0 .
⇒ u , v , w are linearly dependent.
Inperticular taking u = v we get u , v , w are linearly dependent, which is contradiction.
Thus we have v × w ≠ 0 .
Further let v and w be two sides of a parallelogram ABCD inclined at an angle .
By definition, we have
wB
v × w = v ⋅ w sinθ nˆ
C
Thus
w
A
v × w = v ⋅ w sinθ
1

v × w = 2  v ⋅ w sin θ 
2

v
= 2 ( area of a ∆ABD )
vD
⇒ v × w = area of parallelogram ABCD.
Differential Geometry
58
Speed of Curve :
This concept is used in the description of the curve.
Definition : Let α : I → ¡3 be a curve. The speed of is defined to be the length of the velocity
vector. It is denoted by v (t ) . Thus
speed of α = v (t ) = α '(t )
If α (t ) = (α1( t ), α2 (t ), α3 (t ) ) is a curve. Its velocity vector is
 dα (t ) d α (t ) dα (t ) 
α '(t ) =  1 , 2 , 3 
 dt
dt
dt 
Hence the speed v (t ) of α is given by
v = α '(t ) =
((
) +( ) +( )
2
α1'
2
α2'
2
α 3'
)
1/2
Note : If a '(t ) = 1 , then the curve has unit speed.
Arc Length of Curve : The arc length of the curve α : I → ¡3 from a fixed point t = a on the curve
to any point t = b on the curve is defined to be the number given by
b
s = ∫ α '(t ) dt
a
b
⇒s=∫
(α1' ) + ( α2' ) + ( α3' )
2
2
2
dt
... (1)
a
This is exactly analogues to the formula for the distance covered by the particle in physics as
b
s = ∫ vdt
... (2)
a
z
α (b)
α(t)
α(a)
E¡ 33
x
(
a
Differential Geometry
I
)
b
59
y
Remark : Given a curve α one can construct many new curves which follow the same route as α
but travel at different speeds.
e.g. Let α : ¡ → ¡3 be a curve given by α (t ) = tq and β : ¡ → ¡ 3 be defined by β (t ) = 2tq .
We see that route of these curves is the same.
However,
α '( t ) = q = ( q1, q2 , q3 ) .
Hence speed of α is
v1 ( t ) = a '(t )
⇒ v1(t ) = q12 + q22 + q32
... (1)
while for other curves β ,
β '(t ) = 2 ( q1, q2, q3 )
Hence the speed of β is
v2 ( t ) = β '(t )
⇒ v2 ( t ) = 2 q12 + q22 + q32
... (2)
From equations (1) and (2) we see that, speed of two curves is different.
Worked examples ......................................................................................................................•
Example 2 : Find the arc of the circle α (t ) = ( a cos t , a sin t ,0 ) , 0 ≤ t ≤ 2π .
Solution : Consider the curve α : I → ¡3 defined by
α (t ) = ( a cos t , a sin t ,0 )
where
... (1)
t ∈ I :0 ≤ t ≤ 2π .
The velocity vector of α at time t is given by
α '(t ) = (α1' ( t ), α2' (t ), α 3' (t ) )
⇒ α '( t ) = ( −a sin t , a cos t , 0)
Hence the speed of α = α '(t )
⇒ v (t ) = a
Differential Geometry
60
⇒ the curve has constant speed. The arc length of the curve is given by s =
2π
∫
α '(t ) dt
0
⇒s=
2π
∫ adt
0
⇒ s = 2π a
•....................................................................................................................................................•
Definition : A curve α : ¡ → ¡3 is called periodic if there is a number p > 0 such that
α ( p + t) = α (t )
The smallest number p which satisfies this property is called as a period of α .
Theorem : If α is a regular curve in ¡ 3 , then there exists a reparametrization β of α such that β
has unit speed.
Proof : Let α : I → ¡3 be a regular curve.
⇒ α '( t ) ≠ 0
∀t ∈ I
.
... (1)
The arc length of the curve α from any point t = a on the curve to any point t is given by
t
s( t ) = ∫ α '(u ) du
... (2)
a
Differentiating equation (2) w.r.t. t we get
s '( t ) = α '(t )
.
... (3)
Since α is regular, henceα '(t ) ≠ 0 , ∀t ∈ I
⇒ α '(t ) > 0
⇒ s '(t ) > 0
By standard theorem in calculus, we know that there exists an inverse function t = t(s) such that
dt
1
=
>0
ds ds / dt
Let β ( s ) be the reparametrization of α by t.
⇒ β ( s) = α ( t ( s) )
⇒ β '( s) = α ' ( t ( s ) ) ⋅ t '( s )
Differential Geometry
61
Hence the speed of β gives
β '( s ) = α ' ( t ( s ) ) t '( s )
=
dt
α ' (t( s))
ds
=
dt ds
ds dt
⇒ β '( s ) = 1
... (4)
This proves the speed of β is unity..
Example 3 : Find the unit speed reparametrization of helix α defined by
α (t ) = ( a cos t , a sin t , bt ) ,
b≠0
Solution : Let α : I → ¡3 be a helix defined by
α (t ) = ( a cos t , a sin t , bt ) ,
b≠0
... (1)
b≠0
.. (2)
The velocity of the curve α is given by
α '(t ) = ( −a sin t , a cos t , b ) ,
Hence the speed of α is given by
s( t ) = α '( t ) = a 2 + b2 = c ( say )
... (3)
This shows that the curve α has a constant speed. The arc length of the curve α from t = 0 to any
point t on the curve is given by
t
s( t ) = ∫ α '(u ) du
0
t
= c ∫ du
0
s( t ) = ct
or
t=
s
c
... (4)
This gives the relation between the parameter of α and its arc length. Substituting this in equation (1)
we get
Differential Geometry
62
s
β ( s) = α ( t ( s) ) = α  
c 

s
 s   s
⇒ β ( s) =  a cos   , a sin   ,b   
c 
 c   c

... (5)
where speed of β is
 a s a
 s  b
β '( s) =  − sin   , cos   , 
 c c
 c c 
 c
Thus its speed is given by
β '( s ) =
a2 + b 2
c2
⇒ β '( s ) = 1
(by (3))
⇒ β has unit speed.
•....................................................................................................................................................•
Remark : The importance of a curve being parametrized by arc length is that its velocity vector field
is its unit tangent vector field.
Worked examples ............................................................................................................................•
Example 4 : Find the unit parametrization of a circle of radius r and hence compute the tangent vector
field of the curve.
Solution : Let α : I → ¡3 be a circle of radius r defined by
α '(t ) = (r cos t , r sin t , 0) ,
r > 0 , 0 ≤ t ≤ 2π
The velocity vector to the circle is
α '(t ) = ( − r sin t , r cos t ,0 )
⇒ speed of the circle α '(t ) = r
⇒ the circle has constant speed r..
Thus the arc length of the curve α from t = 0 to any point t on the curve is given by
t
s( t ) = ∫ α '(u ) du
0
Differential Geometry
63
... (1)
t
= ∫ rdu
0
s( t ) = rt
or
t=
... (2)
s
r
Substituting this in equation (1) we get
s 
s
s 
β ( s) = α ( t ( s) ) = α   =  r cos   , r sin   , 0 
r 
r 
r 
... (3)
We check that

s
 s 
β '( s) =  − sin   ,cos   , 0 
r
r  

and hence .
β '( s ) = 1
Hence the curve β ( s ) is the unit speed parametrization of a circle of radius r. Now the tangent to the
curve β ( s ) is obtained by differentiating equation (3) w.r.t. s. Then

 s
s 
T ( s ) = β '( s ) =  − sin   ,cos   , 0 
r 
r 

Example 5 : Show that the curve α (t ) = ( cosh t ,sinh t , t ) has arc length function s( t ) = 2 and find
a unit speed reparametrization of α .
Solution : The curve α : I → ¡3 is defined by
α (t ) = ( cosh t ,sinh t , t )
... (1)
Velocity vector of α is given by
α '(t ) = ( sinh t ,cosh t ,1)
... (2)
Now the arc length function s (t) of the curve α from t = 0 is given by
t
s( t ) = ∫ α '( t ) du
0
= 2sinh t
Differential Geometry
... (3)
64
 s 
⇒ t (s ) = sinh −1 

 2
Thus the reparametrization of α by t is β ( s ) given by

 s 
β ( s) = α ( t ( s ) ) = α  sinh −1 

 2 



s  s
s 
=  cosh  sinh −1 
,sinh − 1 
,

 2  2
 2 



s2 s
 s  
⇒ β ( s) =  1 + ,
,sinh −1 

2 2
 2 

... (4)
s
1 1
1

β '( s) = 
,
,
2
2 2
s2
 2 1+ s
1+

2
2

... (5)
We see that





Thus the speed of β is
β '( s ) =
s2
2 (2 + s2 )
+
1
1
+
=1
2 ( 2 + s2 )
⇒ β '( s ) = 1
Showing that β has unit speed parametrization.
Example 6 : Determine the curve α : I → ¡3 such that α (0) = (1,0, −3 ) and
α '(t ) = ( t , et , t 2 )
Solution : The curve α : I → ¡3 be a curve such that
α (t ) = (α1( t ), α2 (t ), α3 (t ) )
where α '(t ) = ( t , et , t 2 )
⇒ α1' = t ⇒ α1 (t ) =
t2
+ c1
2
α 2' = et ⇒ α 2 (t ) = e t + c 2
Differential Geometry
65
α 3'
t3
= t ⇒ α3 (t ) = + c3
3
2
where c1, c2 , c3 are constants of integration and are to be determined. Given that
⇒ α1 (0) = 1, α 2(0) = 0, α 3(0) = −3
α (0) = (1,0, −3) ,
⇒ c1 = 1, c2 = −1 and c3 = −3
Hence the required curve is
 t2

t3
t
α (t ) =  +1, e −1, − 3 
2
3

b
Example 7 : If α is a plane curve given by α (t ) = ( x (t) , y (t ),0 ) , then prove that l = ∫
a
a ≤x ≤b .
Solution : The curve α is given by α (t ) = ( x (t) , y (t ),0 )
Thus its velocity vector is α '(t ) = ( x (t) , y (t ),0 ) . And hence its speed is
α '(t ) =
( x (t ) ) 2 + ( y(t ) ) 2
2
 dx   dy 
=   + 
 dt   dt 
2
  dy 2 
   
 dx 
=   + 1 +  dt 2 
 dt    dx  
   
  dt  
2


dx
=
1+ 
dt


=
Differential Geometry
dy
dt
dx
dt
2





dx
 dy 
1+  
dt
 dx 
2
66
2
 dy 
1 +   dx ,
 dx 
Consequently, the length of the curve
b
b
2
b
2
 dy  dx
 dy 
l = ∫ α '( t ) dt = ∫ 1 +  
⋅ dt = ∫ 1 +   dx
 dx  dt
 dx 
a
a
a
Example 8 : Find the arc length function for the curve α (t ) = ( a cos3 t, a sin3 t , 0 ) . Obtain its
corresponding unit speed reparametrization.
Solution : A curve α : I → ¡3 is defined by
α (t ) = ( a cos3 t, a sin3 t , 0 )
.... (1)
⇒ α '(t ) = ( −3a cos 2 t ⋅ sin t , 3a sin 2 t ⋅ cos t, 0 )
... (2)
The speed of α is
v (t ) = α '( t ) = 3a sin t ⋅ cos t
... (3)
Hence the arc length of the curve α from t = 0 to any point t on the curve is given by
t
s( t ) = ∫ α '(u ) du
... (4)
0
t
= 3a ∫ sin u ⋅ cos udu
0
s( t ) =
3a
sin 2 t
2
Hence the unit speed reparametrization of α is obtained by putting equation (4) and (1), then
  2s  3 / 2  2s  3 / 2 
β ( s) = α ( t ( s ) ) =  a  1 −  , a   , 0 
 3a 
  3a 

... (5)
To see that this curve has unit speed, we find
  2s 1 / 2  2s  1 / 2 
β '( s) =  −  1 −  ,   , 0 
  3a   3a 

⇒ β '( s ) = 1
•.....................................................................................................................................................•
Differential Geometry
67
Definition : A reparametrization α ( h) of a curve α is said to be orientation preserving if h ' ≥ 0 and
orientation reversing if h ' ≤ 0 .
Note : For a regular curve we have
ds
> 0 . This implies that a unit speed reparametrization is always
dt
orientation preserving.
A Vector Field on a Curve :
A vector field Y on a curve α : I → ¡3 is a function that assigns to each number t in I a tangent vector
Y (t) to at the point ¡ 3 .
Note :
(i)
For any curve α : I → ¡3 , the velocity vector α ' satisfies this definition. However, unlike
α ' , arbitrary vector field on need not be tangential to α but may point in any direction.
(ii)
If Y is a vector field on α : I → ¡3 , then for each t ∈ I , we write
Vector Part
Y ( t ) = ( y1 ( t), y2 (t ), y 3 (t ) )α (t )
Point of application
3
⇒ Y (t ) = ∑ yi (t )U i (α ( t ) )
i =1
where yi (t ) , i = 1, 2, 3are called Euclidean coordinate
functions of Y .
The properties of vector fields on curves are analogues to those of vector fields on ¡ 3 . The operations
of addition, scalar multiplication, dot product and cross product of vector fields on the same curve are
defined as follows.
(i)
Addition of vector fields : ( Y + Z ) (t ) = Y (t ) + Z (t )
(ii)
Scalar multiplication of vector fields : ( aY ) (t ) = aY (t )
(iii)
Dot product of vector fields :
Y ⋅ Z = ( y1, y2 , y3 )( z1, z2 , z3 ) = ( y1z1 + y2 z2 + y3 z3 )
(iv)
Cross product of vector fields :
Differential Geometry
68
U1 U2 U 3
(v)
Y × Z = y1
y2
y3
z1
z2
z3
Differentiation of vector field : If Y = ∑ yi U i
is a vector field
i
⇒ Y ' = ∑ y 'i U i
i
( aY + bZ ) ' = aY ' + bZ ' , a , b ∈ ¡
(vi)
Linearity :
(vii)
Leibnitz property :
( f Y ) = df Y + f Y '
'
dt
( Y ⋅ Z ) ' = Y ⋅' Z +Y ⋅ Z '
Note : If Y has a constant length, then we have
Y
2
= constant
⇒ Y ⋅Y = constant
⇒ Y ⋅Y '+ Y '⋅ Y = 0
⇒ Y ⋅Y ' = 0
⇒ Y and Y ' are orthogonal at each point.
Acceleration of Curve :
Definition : The derivative α " of the velocity α ' of a curve α is called acceleration of the curve.
Thus if α = (α1, α2 , α3 ) , then
α " = ( α1" , α2" , α3" )α
Acceleration is the vector field on α ; and not tangent to the curve.
Parallel Vector Field on a Curve :
We have seen that the tangent vectors are parallel if they have the same vector parts. Now a vector
field Y on a curve is said to be parallel provided all its tangent vectors are parallel.
Differential Geometry
69
⇒ If a vector field Y is parallel, then its Euclidean coordinate functions are constants. Thus a parallel
vector field is given by
Y = ∑ ci U i = ( c1, c2, c3 )α (t )
∀t ∈ I
i
where ci ' s are common vector parts and are constant.
Theorem : Prove the following
(i)
A curve α is constant iff its velocity is zero.
(ii)
A non-constant curve is a straight line iff its acceleration is zero. i.e. α " = 0
(iii)
A vector field Y on a curve is parallel iff its derivative is zero. Y ' = 0
Proof : Let α : I → ¡3 be a curve such that
α (t ) = (α1, α2 , α3 )
(i)
Let α be a constant curve ⇒ every α i are constants say α i = pi
⇒ α (t ) = ( p1, p2 , p3 ) ,
∀i = 1,2,3
pi = constant
⇒ α '(t ) = (0,0,0)
⇒ velocity of the curve α is zero.
Conversely,
let
α'=0
⇒ α '( t ) = (α1' , α2' , α 3' )
⇒ α '(t ) = (0,0,0)
⇒ α i = pi constant for each i
⇒ α (t ) = ( p1, p2 , p3 ) , a constant curve
(ii)
Let the curve α be a straight line passing through the point p = α (0) in the direction of q.
⇒ α (t ) = ( p +tq ) = ( p1 + tq1, p2 + tq2 , p3 + tq3 ) , q ≠ 0
where pi qi , are points in ¡ 3 .
The acceleration of the curve α is given by
Differential Geometry
70
α "( t ) = ( α1" , α2" , α3" )
α i = pi + tqi
where
∀i = 1,2,3
⇒ α1' = q1, α2' = q2 , α3' = q3
⇒ α1" = 0, α2" = 0,α3" = 0
⇒ α "( t ) = (0,0,0)
Thus if a non constant curve α is a straight line, then the acceleration of the curve is zero.
Conversely,
Let the curve α be such that its acceleration is zero.
⇒ α "( t ) = (0,0,0) ∀t
⇒ α1" = 0, α2" = 0,α3" = 0
Integrating we get α1' = qi
∀i = 1,2,3
Integrating once again we get, α i = pi + tqi , ∀i , where pi are some constants.
Thus we have
α (t ) = ( p1 + tq1, p2 + tq2 , p3 + tq3 )
which represents a straight line.
(iii)
Let Y be a parallel vector field on a curve α .
⇒ Y ( t ) = ( c1, c2 , c3 )α (t ) = ∑ ci U i
i
where ci are constant ∀i
⇒ Y ( t) = (0,0,0)α (t ) = 0
Thus a vector field Y on a curve is parallel if its derivative is zero.
Conversely,
Let
Y '( t ) = ( y1' ( t), y2' ( t ), y 3' (t ) )
Now
Y '( t ) = 0 ⇒ yi' (t ) = 0
Differential Geometry
∀i = 1,2,3
71
⇒ yi (t ) = ci – constants
⇒ Y ( t ) = ( c1, c2 , c3 )α (t )
⇒ Y (t) is a parallel vector field.
Worked examples .........................................................................................................................•
Example 1 : Show that the curve α (t ) = (t cos t ,t sin t , t ) lies on a cone in ¡ 3 . Find the velocity,,
speed and acceleration of α at the vertex of the cone.
Solution : We know the equation of the cone is given by
x 2 + y2 = z 2
... (1)
The curve α : I → ¡3 is defined by
α (t ) = (t cos t ,t sin t , t )
... (2)
We see that the Euclidean coordinate functions of the curve satisfies the equation of the cone. Hence
the curve lies on the cone. Now the vertex of the cone is (0,0,0) = α (0)
Thus the velocity of α is given by
α '(t ) = ( cos t − t sin t,sin t + t cos t ,1)
Thus the velocity of α at the vertex of the cone is given by
α '(t) |t =0 = (1,0,1)
Hence the speed of the curve α = α '(t )
Thus the speed of at the vertex = α '(0) = 2
Now the acceleration of the curve is α "(t )
Then
α "( t ) = ( −2sin t − t cos t ,2cos t − t sin t , 0 )
α "( t) | y =0 = (0,2,0)
Example 2 : Consider the curve α (t ) = ( 2t , t 2 ,log t ) on I : t > 0. Show that this curve passes through
the points p = (2, 1, 0) and q = (4, 4, log 2) and find its arc length between these points.
Solution : The curve α : I → ¡3 is defined by
α (t ) = ( 2t , t 2 ,log t )
t>0
⇒ α (1) = ( 2,1,0) and α (2) = ( 4,4,log2)
Differential Geometry
72
.
... (1)
This shows that the curve α passes through the points (2, 1, 0) and (4, 4, log 2). The arc length of the
curve α from t = a to t = b is given by
b
s = ∫ α '(t ) dt
... (2)
a
where
1

α '(t ) =  2 , 2t ,  and a = 1 and b = 2.

t
1

⇒ α '(t ) =  2t + 

t
Hence equation (2) becomes
2
1

s = ∫  2t +  dt

t
1
= 3 + log 2
Example 3 : If α is a regular curve show that show that α has constant speed iff the acceleration α "
is orthogonal to α ' .
Solution : Let α : I → ¡3 be a regular curve
⇒ α '( t ) ≠ 0
∀t ∈ I
Let the curve α is of constant speed
⇒ α ' = constant
⇒ α '⋅ α ' = constant
Differentiating we get
α "⋅α ' = 0
Showing that acceleration is orthogonal to velocity.
EXERCISE ......................................................................................................•

t3 
For the curve α =  2t , t 2 , 

3
(i)
Find the velocity, speed and acceleration of α . Also find the values at t = 1.
(ii)
Find the arc length function of α from t = -1 to t = 1.
•.......................................................................................................................•
❏❏❏
Differential Geometry
73
CHAPTER - II
THE FRENET FORMULAS
Unit I
Intorduction :
In this chapter we study the geometry of curves in ¡ 3 by assigning at each point a certain
frame. i.e. a set of three orthogonal unit vectors. The rate of change of these vectors along the curve is
then expressed as in terms of the vectors themselves by the Frenet formula.
Frenet apparatus which is the basic tool in the study of curves. it consist of three vector field
along the given curve - the tangent T , the normal N , and the binormal B and the two scalar valued
functions viz. the curvature κ and the torsion τ . Frenet apparatus does completely determine the
geometry of curves. The principal normal N explains the turning of the curve at every point and
binormal B explains the twisting of the curve at every point of the curve. The curvature and the torsion
influence the shape of the curve.
In the first section we deal only with the unit speed curve and in the next section we extend the
result to arbitrary regular curves.
The Frenet Fromulae for Unit Speed Curve :
Definition : Let β : I → ¡ 3 be a unit speed curve, so that β '( s ) = 1 for each s ∈ I .
We define
(i)
Unit tangent vector field on β :
It is denoted by T and is defined by T = β ' .
... (1)
i.e. velocity vector to the curve β is called the tangent vector field. since β has unit speed
⇒ T =1
⇒ tangent vector field is the unit tangent vector field.
Now from equation (1) we have
T '= β"
Differential Geometry
... (2)
74
This gives the rate of change of unit tangent vector with respect to arc length. Since T ' measures the
way, the curve is changing in ¡ 3 . Hence we call T ' as the curvature vector field of β . The length of
the curvature T ' gives the numerical mesurement of the turning of β . This real valued function is
called the curvature of β and denoted by κ .
Thus
κ ( s ) = T '( s )
∀s ∈ I
... (3)
Further,
T ⋅T = 1
⇒ T ⋅T = 0
⇒ T ' is always orthogonal to T . Hence it must be in the direction of Normal vector to β ' . Thus we
define
(ii)
Principal Normal Vector Field :
The unit vector field in the direction of T ' on β is called the principal normal vector field.
Thus
N=
T'
... (4)
T'
⇒T ' = κ ⋅ N
where
κ = T'
The principal unit normal N of a curve tells the direction in which the curve is turning at each point.
κ is called curvature function. Also κ ≥ 0 .
Note : κ ≥ 0 , the larger the value of κ , sharper is the turning of β .
we assume that κ > 0 . For, if κ = 0 ⇒ T ' = 0 and hence T ' = 0
From equation (2)
⇒ β "=0
⇒ β'= q
a constant
⇒ β = p + sq
This is a straight line. Hence we assume κ > 0 .
Differential Geometry
75
(iii)
Binomial vector field of β :
It is denoted by B and defined as
B =T ×N
... (5)
The magnitude of B is given by
B = T×N
= T ⋅ N ⋅ sin θ
Since T and N are orthogonal and T = 1 = N
⇒ B =1
Remarks :
(i)
In fact the unit orthogonal vectors {T , N , B} is a right handed orthogonal basis.
⇒ T = N × B , B = T × N , N = B ×T
(ii)
The osculating plane to a unit speed curve β at a point β ( s ) is the plane perpendicular to B
hence spaned by T and N .
(iii)
The normal plane of β ( s ) is the plane perpendicular to T .
(iv)
The rectifying plane of β is the plane perpendicular to N .
Rectifying
Plane
Binorm al
Normal
Plane
N
Osculating
Plane
Principal Normal
Result : Let β be a unit speed curve in ¡ 3 , with κ > 0 , then show that the vector field T , N , B
at every point of β forms a Frenet frame field on β .
Proof : Let β : I → ¡ 3 be a unit speed curve.
Differential Geometry
76
⇒ β '( s ) = 1
... (1)
The tangent vector to the curve β is defined by
T = β '( s)
⇒ T =1
... (2)
⇒ T ⋅T = 1
⇒ T '⋅ T = 0
... (3)
T ' is orthogonal to T .
⇒ T ' must be in the direction of normal vector..
⇒T ' =κ N
⇒ N =
... (4)
1
T'
κ
⇒ N =1
as T ' = κ
... (5)
Using (4) in (3) we get for κ ≠ 0
N ⋅T = 0
... (6)
⇒ N is orthogonal to T .
Further B = T × N
⇒ T and N are both orthogonal to B .
⇒ B = T ×N
= T ⋅ N ⋅ sin θ
= sin
π
2
⇒ B =1
... (7)
Thus we have proved that
T = N = B =1
and
T ⋅ N = N⋅ B= B⋅ T = 0
This shows that T , N , B form a frame field on the unit speed curve β .
Differential Geometry
77
Torsion (Twisting) of Curve :
Let β : I → ¡ 3 be a curve in ¡ 3 .
By definition
T =β'
⇒ T ' =β " =κ N
... (1)
As T , N , B is a frame, we have therefore
T ⋅B=0
Differentiating w. r. t. arc length of of the curve, we get
T '⋅ B +T ⋅ B ' = 0
⇒ κ N ⋅ B+ T ⋅ B ' = 0
As
N⋅B =0
⇒T ⋅ B' = 0
... (2)
Also we have B ⋅ B = 1
Differentiating we get
B '⋅ B = 0
.
... (3)
Equations (2) and (3) shows that B ' is orthogonal to both T and B . Hence if must be in the direction
of N .
⇒ B ' = −τ N
... (4)
where τ is called torsion of the curve and negative sign is convensional.
Frenet Formulae :
These formulae tells us that the rate of change of the frame field T , N , B w. r. t. the arc length. These
rates are expressed again in terms of T , N , B .
Theorem : If β : I → ¡ 3 is unit speed curve with κ > 0 and torsion τ then prove that
T'=κN
B ' = −τ N
N ' = −κ T + τ B
Differential Geometry
78
Proof : The first two formulae are proved in above result. We will prove only the last result.
Since we have
N = B ×T
⇒ N ' = B×
' T + B ×T '
= −τ N × T + B × κ N
= τ (T × N ) − κ ( N × B )
N ' =τ B − κT
.... (1)
Aliter :
We will also prove the result by using the formula
N ' = ( N '⋅T ) T + ( N '⋅ N ) N + ( N '⋅B ) B
... (2)
We have
N ⋅ T = 0 ⇒ N '⋅ T + N ⋅T ' = 0
⇒ N '⋅ T = − N ( κ N )
⇒ N '⋅ T = −κ
... (3)
Also
N ⋅ N = 1 ⇒ N '⋅ N = 0
... (4)
Also
N ⋅ B = 0 ⇒ N '⋅ B +N ⋅B ' = 0
N '⋅ B = − N ( −τ ⋅ N )
⇒ N '⋅ B = τ
... (5)
Using equation (3), (4) and (5) in the equation (2) we get
N ' = −κ T + τ B
Remark : T , N , B , κ , τ are called the Frenet apparatus for a given curve. It is independent of the
choice of the parameter.
Example 1 : If a unit speed curve has curvature identically zero, then it is a straight line.
Solution : Let β : I → ¡ 3 be a unit speed curve and has curvature κ = 0 .
Claim : We prove β is a straight line.
Since κ = 0
⇒ κ ⋅N = 0
⇒T ' = 0
Differential Geometry
79
⇒ T = constant
⇒ β ' = c , constant
⇒ β = cs + d
⇒ β is a straight line.
Conversely,
If β is a straight line, then we know that its acceleration β " = 0 .
⇒ β "=T ' = κN = 0
⇒κ = 0
⇒ Curvature of β is zero.
Example 2 : Compute the Frenet apparatus of the unit speed curve
3
4

β ( s) =  cos s,1 − sin s, − cos s 
5
5

Show that this curve is a circle. Find its center and radius.
Solution : Let β : I → ¡ 3 be a unit speed curve defined by
3
4

β ( s) =  cos s,1 − sin s, − cos s 
5
5

... (1)
The velocity vector to the curve β is the tangent vector T to β .
Thus,
3
 4

T = β '( s) =  − sin s , − cos s , − sin s 
 5
5
 β (s )
... (2)
We see that β is a unit speed curve ⇒ β ' = 1 = T .
Differentiating (2) w. r. t. s we get
3
 4

T ' = κ N = β "( s ) =  − cos s,sin s , cos s 
 5
5
 β ( s)
κ ( s ) = T '( s ) =
16
9
cos 2 s + sin 2 s + cos2 s
25
25
⇒ κ =1
Differential Geometry
... (3)
... (4)
80
Hence from (3) we have
3
 4

N =  − cos s,sin s , cos s 
 5
5
 β ( s)
... (5)
Now to find the expression for binormal vector field, we have
U1
U
U3
2
4
B = T × N = − sin s − cos s
5
4
− cos s sin s
5
3
sin s
5
3
cos s
5
... (6)
4
 3
B =  − ,0, − 
 5
5
We see that B( s ) is independent of s.
⇒ B '( s ) = ( 0,0,0 )
⇒τ = 0
... (7)
Since torsion of the curve β is zero ⇒ β is a plane curve.
Further the curvature κ is constant at each point.
⇒ the plane curve must be a circle. to find the radius and the center of the circle, we have from
equation (1)
x=
4
3
cos s , y = 1 − sin s , z = − cos s
5
5
⇒ x 2 + ( y −1) 2 + z 2 = 1
... (8)
This is the sphere with center at (0, 1, 0) and of radius 1.
Also we see that 3x + 4z = 0
... (9)
This is the plane passing through the center of the sphere. Hence the given curve lies on the intersection
of a sphere and a plane and hence it is a circle.
25 x 2 + 16 y2 − 32 y = 0
Differential Geometry
81
EXERCISE .....................................................................................................•
1.
2.
Obtain the unit reparametrization of the curve helix and hence comput Frenet apparatus for he
unit speed helix.
1

3/2 1
3/2 1
s  defined on I, -1 < s < 1 has unit
Show that the curve β ( s) =  3 (1 + s) , 3 (1 − s) ,
2 

speed. Compute its Frenet apparatus.
3.
8
12
5

Show that α ( s ) =  cos s , − sin s, − cos s  is unit speed curve. Compute its Frenet
 13
13
13

apparatus.
ANSWERS .....................................................................................................•
1.
Helix

s
s b 
β ( s) =  a cos   , a sin   , s  ,
c 
 c c 

where c = a 2 + b2
 a  s a
 s  b
T =  − sin   , cos   , 
c c
c  c
 c
κ=
a
c2
b  s b
 s  a
B =  sin  , − cos   , 
c c
 c c 
c
2.
τ=
b
c2
κ=
1
2
1
=
2
2
4 1− s
8 (1 − s )
τ=
1
8 (1 − s
2
)
 (1 + s )1 / 2 − (1 − s )1 / 2 1 
,
,

, T =
2
2
2

 1 (
1/2 1
(1 + s )1 / 2 , 0 
N =
1 + s) ,
2
 2

 1
1/2 1
1/2 1 
B =  − (1 + s ) , (1 − s ) ,

2
2
 2
Differential Geometry
82
Worked examples ........................................................................................................................•
Example 3 : For a unit speed curve c : ( x1 ( s), x2 ( s) ) in ¡ 2 , show that the unit tangent T = ( x1' , x'2 )
and unit normal N = ( − x2' , x1' ) . Further if T ' = κ N , then N ' = −κ T .
Solution : Let α : I → ¡ 2 be a curve c in ¡ 2 defined by
α ( s ) = ( x1 ( s), x2 ( s) )
... (1)
The curve α has unit speed
⇒ α '( s ) = 1
... (2)
The unit tangent T to the curve α is its velocity vector..
Therefore
T = α '( s ) = ( x1' , x'2 )
... (3)
Differentiating equation (3) w. r. t. s, we get
T ' = κ N = ( x1", x"2 )
⇒ = T' =
⇒N=
... (4)
(x 1" ) + (x "2 )
2
1
( ) +( )
" 2
x1
" 2
x2
2
... (5)
( x1", x2" )
... (6)
Since T = 1 ⇒T ⋅T = 1
= ( x1' , x2' ) ⋅ ( x1' , x'2 ) = 1
= ( x1' ) + ( x'2 ) = 1
2
2
... (7)
Differentiating equation (7), we get
x1' x1" + x'2 x2" = 0
... (8)
x'2 "
= − ' x2
x1
... (9)
⇒
x1"
Substituting this in equation (6) we get
1
N=
( − x2' , x1' )
2
2
( x1' ) + ( x2' )
Using (7) we get N = ( − x2' , x1' )
... (10)
Hence N ' = ( − x"2 , x1" )
•...................................................................................................................................................•
Differential Geometry
83
Frenet Approximation of Curve :
Introduction : To get more information in the close of an arbitrary point, one should know how the
curvature and torsion influence the shape of the curve at a given point, hence we study Frenet
approximation of curve.
Lemma : Obtain the Frenet approximation β of near s = 0 for the unit speed curve.
Proof : Let β : I → ¡ 3 be a unit speed curve such that β ( s ) = ( β1, β2 , β3 ) . To find the approximation
of β at β (0) in term of Frenet apparatus, let T 0 , N 0 , B0 , κ 0 and τ 0 be the values of Frenet
apparatus at the origin.
By, Taylor's series expansion we have
β ( s) = β (0) + sβ '(0) +
s2
s3
β "(0) + β '"(0) + ....
2!
3!
... (1)
We know the tangent vector to the curve β is given by
T = β '( s)
⇒ T 0 = β '(0)
Also we have
β "( s ) = T ' = κ N
⇒ β "(0) = κ 0 N 0
Further
β '"(0) = κ ' N + κ N '
= κ ' N + ( −κ T + τ B ) κ
β '"( s ) = κ ' N − κ 2 T + κτ B
⇒ β '"(0) =κ 0' N 0 −κ 02 T 0 + κ 0τ 0 B 0
Assume that κ is very small near origin, hence we neglect κ 0' and κ 02 terms and write
β '"(0) = κ 0τ 0 B0
Substituting these values in (1), we get
β ( s) = β (0) + sT 0 +
Differential Geometry
s2
s3
κ 0 N 0 + κ0τ 0 B 0
2!
3!
84
... (2)
If β̂ is the Frenet approximation of β near origine s = 0 then we write eqution (2) as
s2
s3
βˆ ( s) = β (0) + sT 0 + κ 0 N 0 + κ0τ 0 B 0
2!
3!
... (3)
Which is the required Frenet approximation of the curve β near origin.
Note :
(i)
If s is so small that its power higher than one are neglected, then we have from (3)
βˆ ( s ) = β (0) + sT 0
... (4)
This is the straight line passing through β (0) in the direction of T 0 .
⇒ the linear approximation of β is a tangent line.
(ii)
If power of s upto second order are retained then, we have from (3)
s2
βˆ ( s) = β (0) + sT 0 + κ 0 N 0
2!
... (5)
which is a parabola.
Thus quadratic approximation of the curve β near β (0) is parabolla.
Note that this parabola lies in the plane through β (0) orthogonal to B0 . i.e. parabola lies in
the osculating plane of β ( s ) at β (0) .
Theorem : Let β be a unit speed curve in ¡ 3 with κ > 0 . Then β is a plane curve iff its torsion is
zero.
Proof : Let β : I → ¡ 3 be a unit speed curve in ¡ 3 with κ > 0 .
Let β be a plane curve. i.e. β lies on plane.
⇒ β ( s ) must satisfy the equation of the plane. Therefore there exist points p and q such that
( β (s) − p ) ⋅ q = 0
∀s
... (1)
where p is a fixed point in the plane and q is a tangent vector at p and orthogonal to the plane.
Differentiating equation (1) w. r. t. s we get
β '( s ) ⋅ q = 0
⇒T ⋅q = 0
Differential Geometry
... (2)
85
Differentiating equation (2) again we get
T ⋅q = 0
⇒ κ ⋅ N⋅ q = 0
⇒ N ⋅q = 0
for κ ≠ 0
... (3)
Equations (2) and (3) show that q is orthogonal to both T and N .
⇒ q is the direction B .
⇒ B = αq
⇒ B =α⋅ q
⇒1= α ⋅ q
⇒α =
1
q
1
q
... (4)
⇒ −τ N = 0
... (5)
⇒α = ±
Thus
B =±
q
q
⇒ B' = 0
⇒τ = 0
Thus if β is a plane curve, then its torsion is zero.
Conversly,
Assum that τ = 0
⇒ −τ N = 0
⇒ B' = 0
⇒ B = constant
Now to prove β is a plane curve through β (0) orthogonal to B , consider a function f defined by
f ( s) = ( β ( s ) − β (0) ) ⋅ B
Differential Geometry
∀s
86
... (6)
Differentiating (6) w. r. t. s we get
f '( s) = β '( s ) ⋅ B + ( β ( s ) − β (0) ) ⋅ B '
= β '( s ) ⋅ B
as B = constant
=T⋅B
f '( s ) = 0
⇒ f ( s ) = constant ∀s
Also f (0) = 0
⇒ constant = 0
⇒ f ( s) = 0 , ∀s
Consequently equation (6) becomes
( β (s) − β (0)) ⋅ B = 0
∀s
⇒ β ( s ) lies in a plane i.e. β ( s ) is a plane curve.
Theorem : If β is a unit speed curve with constant curvature κ > 0 and torsion τ = 0 , then prove
that β is a part of a circle with radius
1
.
κ
Solution : Let β : I → ¡ 3 be a unit speed curve such that the curvature κ is constant and torsion τ
is zero.
⇒ β is a plane curve.
Claim : We prove that β is a part of a circle with radius
Consider the curve γ = β +
1
N
κ
... (1)
Differentiating we get
γ ' = β '+
γ ' =T +
Differential Geometry
1
.
κ
N'
κ
1(
τ B − κT )
κ
87
γ '=
1
τB
κ
⇒γ '= 0
⇒ γ = c , constant
Consequently, equation (1) becomes
c =β+
1
N
κ
1
κ
(c − β ) = N
or
Therefore,
c − β ( s) =
1
= constant
κ
c − β ( s) = d ( c , β ( s ) ) = constant
⇒ β ( s ) is always at a constant distance of
radius
1
from the point c . Hence β ( s ) is a part of a circle with
κ
1
and center at c.
κ
Worked examples ........................................................................................................................•
Example 4 : Find the center and radius of the circle
3
4

β ( s) =  cos s,1 − sin x, − cos s 
5
5

Solution : We have proved in example 2 that the curvature κ and torsion τ of the curve β ( s ) are
respectively 1 and 0.
⇒ β ( s ) is the part of the circle; whose radius is given by
⇒ radius of the circle r = 1 , while the center is given by
c =β+
1
N
κ
c =β+N
Differential Geometry
88
1
=1.
κ
where
3
 4

N =  − cos s ,sin s , cos s 
 5
5

⇒ c = (0, 1, 0)
•.....................................................................................................................................................•
Spherical Curves :
Definition : A curve that entirely lies on the sphere is called as a spherical curve.
If β is a spherical curve with center at 0 and radius a, then its equation is β = a
Theorem : If ∑ is a sphere of radius 'a' with center at origin of ¡ 3 . Then every spherical unit speed
curve β has a curvature κ ≥
1
.
a
Proof : Let ∑ be a sphere of radius 'a' with center at origin. The equation of the spherical curve must
satisfy the equation
β (s) − 0 = a
2
⇒ β (s) = a 2
⇒ β ⋅ β = a2
Differentiating we get
... (1)
2 ββ ' = 0
⇒ T ⋅β = 0
... (2)
Differentiating again, we get T '⋅ β + T ⋅ β ' = 0
Using Frenet formula, we have
κβ ⋅ N + T ⋅ T = 0
⇒ κβ ⋅ N = −1
Since κ > 0
⇒κ = −
⇒κ =
1
β ⋅N
1
β ⋅N
... (3)
But by Schwarz inequality, we have
β ⋅N ≤ β ⋅ N =a
Differential Geometry
... by (1)
89
⇒
1
β⋅N
1
a
≥
Hence from equation (3) we have
κ≥
1
a
Example 5 : If A is a vector field such that A = τ T + κ B on a unit speed curve β , then show that
the Frenet formulae become
T ' = A×T ,
N ' = A× N ,
B ' = A× B .
Solution : Given that A = τ T + κ B
... (1)
Consider
A × T = (τ T + κ B ) × T
A×T = κ B ×T
= κN
=T'
⇒ T ' = A×T.
Now consider
A × N = (τ T + κ B ) × N
=τT × N +κB× N
= τ ⋅ B + κ ( −T )
= τ B − κT
= N'
Thus
N ' = A× N
Now consider
A × B = (τ T + κ B ) × B
=τT ×B
Differential Geometry
90
= −τ N
= B'
Thus
B ' = A× B
Example 6 : Let α be a unit speed curve with κ > 0 and τ ≠ 0 . If α lies on a sphere of radius r and
center at c , show that α − c = −} N −}'σ B , where } =
r 2 = } 2 + ( } 'σ )
1
1
and σ = . Thus
κ
τ
2
Solution : Let α : I → ¡3 be a unit speed curve.
⇒ α '(t ) = 1
Since the curve α lies on the sphere of radius r and center at c , it must be satisfy
c −α
2
= r2
⇒ (α − c ) (α − c ) = r 2
Differentiating we get
(α − c ) α ' = 0
⇒ (α − c ) T = 0
... (1)
Differentiating again we get
(α − c ) T '+ α '⋅ T = 0
⇒ (α − c ) κ N + T ⋅T = 0
⇒ (α − c ) κ N = −1
⇒ (α − c ) N = −
1
κ
⇒ (α − c ) N = −}
... (2)
Differentiating again we get
⇒ (α − c ) N '+ α '⋅ N = −}'
⇒ (α − c ) (τ B − κT ) + T ⋅ N = −}'
⇒ (α − c ) (τ B − κT ) = −}'
Differential Geometry
... (3)
91
Equation (1) shows that lies in N − B plane i.e. normal plane. We have
α −c = λN + µ B
... (4)
for some scalar λ and µ are to be determined.
Multiply equation (4) by N , we get
(α − c ) ⋅ N = λ
By eqution (2)
λ = −}
... (5)
Similarly, multiply equation (4) by (τ B − κ T ) , we get
(α − c ) (τ B − κ T ) = ( λ N + µ B ) ( τ B − κ T )
⇒ −}' = µτ
for σ =
⇒ − µ = −}'τ
1
τ
... (6)
Hence equation (4) becomes
α − c = −} N −}'σ B
Consequently,
r 2 = (α − c ) (α − c ) becomes
r 2 = ( } N + }'σ B )( } N − }'σ B )
r 2 = } 2 + ( } 'σ )
2
Example 7 : If κ > 0 , τ ≠ 0 . If } 2 + ( }'σ )2 has constant value r 2 and }' ≠ 0 . Show that α lies on
a sphere of radius r.
Solution : Given that r 2 =}2 + ( }'σ ) 2
... (1)
We prove that α lies on a sphere. Hence we prove that
α − c = (α − c ) (α − c ) = r 2
2
Differentiating equation (1) w.r.t. s we get
}}'+ ( }'σ ) + ( }'σ ) ' = 0
⇒}' ( }+σ ( }'σ ) ') = 0
Differential Geometry
92
⇒} +σ ( }'σ ) ' = 0
⇒
for }' ≠ 0
}
+ ( }' σ ) ' = 0
σ
... (2)
Let γ be a curve given by
γ = α + } N +}' σ B
... (3)
Differentiating (3) we get
γ ' = α '+}' N + } N '+ ( }'σ ) '⋅ B + ( }'σ ) ⋅ B '
⇒ γ ' = T +}' N +}(τ B − κ T ) + ( }'σ ) '⋅ B + ( }'σ ) ⋅ ( −τ N )
Since } =
1
1
,σ =
κ
τ
⇒ γ ' =}τ B + ( }'σ ) '⋅ B
}

⇒ γ ' =  + ( } 'σ ) '  B
σ

Using equation (2) we have
γ '=0
⇒ γ = c - a constant
Substituting in (2) we have
c = α +} N + }' σ B
or
(α − c ) = − } N −}'σ B
Hence
(α − c ) (α − c ) = ( }N + }'σ B )( } N + }'σ B )
= }2 + ( }'σ )
2
= r2
... by (1)
This shows that α lies on a sphere of radius r and center at c .
•.....................................................................................................................................................•
Differential Geometry
93
Frenet Formulae for Arbitrary Speed Curve :
The Frenet formulae where are derived in the earlier section are valid only for the unit speed
curves. They tell us the rate of change of the frame field T , N , B with respect to the arc length.
We shall now determine the Frenet apparatus for a curve that is not a unique speed
parametrization.We shall denote the derivation of a curve with respect to arc length of the curve by
dash and with respect to any other parameter by over head dot.
Lemma : If α : I → ¡3 is regular curve in ¡ 3 then show that
T=
α&
α& × α&&
α& × α&&
α& ⋅ (α&& ×α&&&)
κ
=
τ
=
B=
,
,
,
and
3
2
α&
α& × α&& N = B × T
α&
α& ×α&&
Proof : Let α : I → ¡3 be a regular curve with parameter other than the arc length, and which does
not necessarily have unit speed. Let β be the unit speed
parametrization of α by s, where s is the arc length function
of the curve α .
⇒ α (t ) = β ( s (t ) )
∀t
... (1)
such that β ' = 1
Now we have from equation (1) that
dα d β ds
=
⋅
dt
ds dt
⇒ α& = β '⋅ s&
or
&
⇒ α& = sT
And
& = s& T
⇒ α& = sT
... (2)
= s&
(since s& is a scalar function)
where T is the unit tangent vector to the unit speed curve β , and s& is the speed of the arbitrary curve
α (t ) denoted by v,
i.e.
T = β ' and s& = v
Thus we have α& = vT and α& = v
Differential Geometry
... (3)
94
Thus
⇒ s& (t ) = α& ( t ) = v
... (4)
This gives the speed of the non unit speed curve α .
Thus from equation (2) and (4) we have
α&
α&
T=
... (5)
Differentiating equation (2) w.r.t. t we get
&
α&& = &&
sT + sT
But
dT dT ds
T& =
=
⋅
dt
ds dt
T& = T ' s&
⇒ T& = κ Ns&
Hence we write
α&& = &&
sT + κ s& 2 N
... (6)
Now consider
& × ( &&
α& × α&& = sT
sT + κ s&2 N )
= κ s&3T × N
&& = κ s& 3 B
α& × α
... (7)
⇒ α& × α&& = κ s&3
Consequently
κ=
⇒κ =
α& × α&&
s&3
α& × α&&
α&
... (8)
3
Also from equation (7) we have
B=
α& ×α&&
κ s&3
⇒B=
α& × α&&
α& × α&&
... (9)
Since {T , N , B} is a right handed orthogonal basis
⇒ N = B× T
Differential Geometry
... (10)
95
Now differentiating eqaution (6) w. r. t. t we get
g
&&& = &&&
&& '+ (κ s& 2 ) ⋅ N + κ s& 2 N '
α
s T + sT
g
g
&&& = &&&
&&& '+ (κ s& 2 ) ⋅ N + κ s& 3 N ' ........... Q N = s& N '
α
s T + ssT
From equation (10) we find
N ' = B×
' T + B× T '
= −τ N × T + B × κ N
N ' =τ B − κT
Hence we write
g
&&& = &&&
&&& + (κ s& 2 ) ⋅N + κs&3 ( τB − κT
α
s T +κ ssN
)
⋅
&&& = &&&
⇒α
s − ( κ 2s&3 )  T + κ ss
&&& + (κ s& 2 )  N + κτ s& 3 B
... (11)
Hence
&& ×α
&&&) = α&& ⋅ (α
&&& × α& ) = α
&&& ⋅(α& × α&& ) = τ ( κ s& 3 )
α& ( α
2
&&&) = τ (κ s& 3 )
⇒ α& ⋅(α&& × α
... (12)
2
&&&) = τ α& × α&&
⇒ α& ⋅(α&& × α
... by (7)
2
⇒τ =
α& ⋅ (α&& ×α&&&)
α& ×α&&
... (13)
2
Lemma : For a non unit speed regular curve in ¡ 3 , prove that
 T& 
κv 0   T 
   0
 
 N&  =  −κ v 0 τ v   N 

  
  
 B&   0
−
τ
v
0
 B 
 
Proof : Let α ( t ) be a regular curve in ¡ 3 which does not necesarily have unit speed. Let β be the
unit speed parametrization of α by s, where s is the arc length function of the curve α .
such that
⇒α = β (s)
... (1)
T = β '( s )
... (2)
However, we have T& = sT
& '
Differential Geometry
... (3)
96
where s& is the speed of the curve α , and T ' = κ N ,
Thus we write
T = κ vN
... (4)
Further for N = B × T
we have
N ' =τ B − κT
and
& '
N& = sN
... (5)
⇒ N& = v (τ B − κ T )
or
⇒ N& = vτ B − vκ T
... (6)
Similarly we find from B = T × N
B& = T& × N + T × N&
Using equation (4) and (6) we have
B& = κ vN × N + T × (τ vB − κ vT )
B& = −τ vN
... (7)
Thus we have
T& = κ vN
N& = −κ vT + τ vB
... (8)
B& = −τ vN
Note : We notice that if v = 1 then the equation (8) reduces to the expression for unit speed curve.
Worked examples ............................................................................................................................•
Example 1 : Let α (t ) be a regular curve. Prove that
2
2  dv 
κ 2v4 = α&& −   .
 dt 
Solution : For regular non unit speed curve α (t ) we have
α&& = &&
sT + κ s& 2 N
Differential Geometry
97
where s& = v is the speed of α . Therefore, we have
dv
T + κv2 N
dt
α&& =
dv
⇒ α&& =
T + κ v2 N
dt
2
2
 dv
  dv

=  T + κ v2 N  ⋅  T +κ v2 N 
 dt
  dt

2
2
 dv 
α&& =   + (κ v 2 )
 dt 
2
2
 dv 
⇒ κ v = α&& −  
 dt 
2 4
2
•.....................................................................................................................................................•
Remark : There are two ways to find the Frenet apparatus of a regular curve, either to find the unit
speed reparametrization of the curve and use Frenet formulae or use Frenet formulae for regular curve.
Example 2 : For a regular curve α (t ) with constant speed c > 0, find the Frenet apparatus.
Solution : Let α : I → ¡3 be a regular curve with constant speed c > 0. Let the curve α (t ) be
reparametrized by the arc length s to the curve β ( s (t ) ) .
⇒ α (t ) = β ( s )
... (1)
where β ' = 1 , and
α& = β '⋅ s&
α& = T ⋅ s&
... (2)
Since the curve has constant speed c ⇒ s& = v = c
where α& = c . Hence we write
T=
α&
c
... (3)
Also, differentiating (2) w. r. t. t, we get
α&& = c ⋅ T&
= c ( cT ')
Differential Geometry
98
α&& = c 2κ N
... (4)
⇒ α&& = c 2 κ
⇒κ =
α&&
c2
... (5)
Consequently equation (4) becomes
α&& = α&& N
⇒N=
α&&
α&&
... (6)
Now as B = T × N , using equations (3) and (6) we write
B=
α& ×α&&
c α&&
... (7)
Differentiating equation (4) w. r. t. t we get
&&& = c 2κ N& + c2κ& N
α
where N g = cN '
g
N = c( −κ T +τ B )
Thus
&&& = −κ 2c 3T +τκ c 3 B + c2κ& N
α
... (8)
Now consider the vector triple product
&&& ⋅ (α& ×α&& ) = ( −κ 2 c 3 T +τκ c 3 B + c2κ& N ) ⋅ c α&& B
α
&&& ⋅ (α& ×α&& ) =τκ c 4 α&&
α
= τ c 2 α&&
2
This gives
τ=
&&& ⋅ (α& ×α&& )
α
c2 α&&
... (9)
2
Equation (3), (5), (6), (7) and (9) gives the required Frenet aparatus.
Example 3 : Find the Frenet apparatus for the curve
α (t ) = (1 + t 2 , t , t 3 ) at t = 0.
Differential Geometry
99
Solution : The curve α : I → ¡3 is defined by
α (t ) = (1 + t 2 , t , t 3 )
... (1)
⇒ α& (t ) = ( 2t, 1, 3t 2 )
α&&(t ) = ( 2,0,6t )
&&&(t ) = ( 0,0,6)
α
Hence
U1
U2
α& × α&& = 2t
2
1
0
U3
3t 2 = ( 6t , − 6t 2, −2 )
6t
... (2)
Thus we find
&&&) = α
&&& ⋅(α& ×α&& )
α& ⋅ (α&& ×α
= ( 0,0,6 ) ⋅ ( 6t , − 6t 2 , −2 )
α& ⋅ (α&& ×α&&&) = −12
... (3)
Thus the curvature κ of the curve is given by
α& × α&&
κ=
α&
3
6tU 1 − 6t 2U 2 − 2U 3
⇒κ =
2tU 1 + U 2 + 3t 3U 3
3
1
κ=
Therefore
( 36t 2 + 36t 4 + 4 ) 2
3
( 4t2 + 9t 4 + 1) 2
κ ( o) = ± 2
... (4)
Similarly, torsion of α is given by
τ=
α& ⋅ (α&& ×α&&&)
α& ×α&&
⇒τ =
( 36t
2
−12
4
+ 36t 2 + 4 )
τ (0) = − 3
Differential Geometry
... (5)
100
Similarly tangent vector T to α is given by
α&
T=
α&
⇒T =
( 2t,1,3t 2 )
1
( 9t 4 + 4 t 2 + 1) 2
⇒ T (0) = (0,1,0)
... (6)
α& × α&&
Similarly, binormal B = & && gives
α ×α
B=
( 6t, −6t 2 , −2 )
1
( 36t4 + 36t 2 + 4) 2
⇒ B(0) = (0,0, − 1)
Next N = B × T .
This gives
N=
( −18t 4 + 2, −18t3 − 4t,12t3 + 6t )
1
1
( 36t 4 + 36t 2 + 4 ) 2 ( 9t 4 + 4 t 2 + 1) 2
⇒ N (0) = (1,0,0)
... (7)
EXERCISE ....................................................................................................•
1.
Show that α (t ) = (t ,1 + t −1, t −1 − t ) , t > 0 is a plane curve.
2.
Compute the Frenet apparatus for the curves
(i)
α (t ) = (t cos t , t sin t , t ) at t = 0
(ii)

t3 
α (t ) =  2t , t 2 ,  at t = 0

3
(iii)
α (t ) = ( t , t 2 , t 3 ) , t > 0
(iv)
α (t ) = ( t − cos t ,sin t , t ) at t = 0
(v)
α (t ) = ( et cos t , et sin t , et ) at t = 0
Differential Geometry
101
ANSWERS ....................................................................................................•
1.
Show that τ = 0 , κ ≠ 0
(i)
T=
(ii)
T = (1,0,0) , B = ( 0,0,1) , N = ( 0,1,0) , κ =
(iii)
T=
N=
1
1
(1,0,1) , B =
( −2,0,2) , N = (0,1,0) , κ = 1 , τ = 3
2
2 2
4
1
2 (1 + t
1
(1 + t 2 )
2
)
(1 − t 2 , 2 ,t 1 + t2 ) ,
( −2t,1 − t2 , 0 ) ,
κ=
B=
1
1
, τ=
2
2
1
2 (1 + t
1
3 (1 + t 2 )
2
=τ
2
)
( t 2 − 1, −2t,1 + t 2 ) ,
,
•........................................................................................................................•
Example 4 : If α (t ) = ( x (t) , y (t ),0 ) is a regular curve in ¡ 3 , show that its curvature is given by
κ=
xy
&&& − xy
&&&
( x&
2
+
2
2 3
y&
)
Solution : Consider a regular curve α : ¡ → ¡3 defined by
α (t ) = ( x (t) , y (t ),0 )
... (1)
We can check that it is not unit parametrization. Hence the curvature of the curve α is given by
κ=
α& × α&&
α&
... (2)
3
where from equation (1) we find
α& = ( x& , y& , 0 )
⇒ α& = ( x& +
2
1
2 2
y&
)
... (3)
Now we find
α&& = ( &&
x, &&
y, 0 )
Differential Geometry
... (4)
102
Therefore
U1 U2 U 3
α& × α&& = x&
y&
0
&&
&&
x
y
0
&&& − xy
&&& )
⇒ α& × α&& = ( 0,0, xy
... (5)
&&& − xy
&&& )
α& × α&& = ( xy
... (6)
Thus
Using (3) and (6) in equation (2) we get
κ=
xy
&&& − xy
&&&
( x&
2
+
2
2 3
y&
)
... (7)
Example 5 : Determine the curvature for the ellipse
α (t ) = (a cos t , b sin t , 0 )
and deduce from it the curvature of a circle of radius r.
Solution : Let α : I → ¡3 be a curve defined by
α (t ) = (a cos t , b sin t , 0 )
... (1)
which is the plane curve called ellipse.
The velocity vector to the curve is given by
α '(t ) = ( −a sin t , b cos t , 0 )
... (2)
⇒ α '(t ) = a2 sin 2 t + b 2 cos2 t ≠ 1
... (3)
⇒ The curve (1) is not a unit speed curve.
For such curve, the curvature is given by
κ=
α& × α&&
α&
... (4)
3
We find from (2)
α& (t ) = ( −a sin t , b cos t , 0 )
⇒ α&&(t ) = ( −a cos t , −b sin t , 0 )
Differential Geometry
... (5)
103
U1
⇒ α& ×α&& = −a sin t
−a cost
U2
bcos t
−b sint
U3
0
0
⇒ α& ×α&& = ( 0,0, ab )
... (6)
⇒ α& ×α&& = ab
... (7)
Hence the curvature of the curve is given by
κ=
ab
3
... (8)
( a 2 sin 2 t + b2 cos 2 t ) 2
Now the curve (1) will be circle of radius r if a = b = r.
Substituting this in (8) we get the curvature of the circle of radius r is
κ=
r2
⇒κ =
3
r2 2
( )
1
r
... (9)
Note : Since ellipse is a plane curve ⇒ its torsion is zero.
The torsion of the curve is given by τ =
&&&
α& ⋅ α&& ×α
α& × α&&
... (10)
2
We first find
&&& = ( a sin t , −b cos t , 0 )
α
... (11)
&&& ) = α&& ⋅ (α
&&& ×α& ) = α
&&& ⋅ (α& ×α&& )
α& ⋅ (α&& ×α
= ( a sin t , −b cos t , 0 )( 0,0, ab )
α& ⋅ (α&& ×α&&&) = 0
... (12)
τ =0
... (13)
From this we infere that torsion of circle is also zero.
Example 6 : Compute the Frenet apparatus for the curve helix given by
α (t ) = ( a cos t , a sin t , bt ) ,
b≠0
Solution : Let α : I → ¡3 be a regular curve defined by
α (t ) = ( a cos t , a sin t , bt )
Differential Geometry
... (1)
104
which does not necessarily have unit speed.
For such curve the Frenet apparatus is given by
T=
α&
α& × α&&
α& × α&&
α& ⋅ (α&& ×α&&&)
κ
=
τ
=
B=
,
,
,
and
3
2
α&
α& × α&& N = B × T
α&
α& ×α&&
Now from equation (1) we find
α& (t ) = ( −a sin t , a cos t , b )
⇒ α& = a 2 + b2
⇒T =
α&
gives ⇒ T =
α&
1
a +b
2
2
( −a sin t, a cos t , b )
⇒ T =1
Now,
α&& = ( −a cos t , −a sin t , 0 )
&&& = ( a sin t , − a cos t , 0 )
α
U1
U2
⇒ α& × α&& = − a sin t acos t
−a cost −a sin t
U3
b
0
⇒ α& ×α&& = ( ab sin t , − ab cos t , a 2 )
⇒ α& ×α&& = a b 2 + a 2
Hence binormal vector field gives
α& × α&& ( ab sin t , −ab cos t , a 2 )
=
2
2
α& × α&&
a b +a
B=
⇒ B =1
Curvature of the curve α is given by
κ=
a a2 + b2
3
( a2 + b 2 ) 2
⇒κ =
Differential Geometry
a
a + b2
2
105
Similarly torsion of the curve α is given by
τ=
&&&
α& ⋅ α&& × α
α& × α&&
⇒τ =
⇒τ =
⇒τ =
2
=
&&& ⋅ α& × α&&
α
α& × α&&
2
( a sin t , −a cos t, 0 )⋅ ( ab sin t, − ab cos t , a 2 )
a 2 ( b2 + a 2 )
a 2b ⋅ ( sin 2 t + cos 2 t)
a2 ( a 2 + b 2 )
b
a2 + b2
N = B×T =
Next
U1
U1
U1
b sin t
−b cos t
a
a2 + b 2
−a sin t
a 2 + b2
a cos t
a2 + b2
b
a2 + b 2
a 2 + b2
a2 + b2
⇒ N = ( − cos t , − sin t , 0 )
⇒ N =1
Note : If α : I → ¡3 is a regular curve and β is the unit reparametrization of α then we notice that
the Frenet apparatus is the same for both the curves, only a change of parameter is involved.
Example 7 : Compute the Frenet apparatus of the curve
α (t ) = ( cosh t ,sinh t , t )
Express the curvature and torsion of α as function κ ( s ) and τ ( s ) of arc length s measured from
t = 0.
Solution : Let α : I → ¡3 be a regular curve such that
α (t ) = ( cosh t ,sinh t , t )
... (1)
The velocity vector of α is given by
α '(t ) = ( sinh t ,cosh t ,1)
Differential Geometry
... (2)
106
We see that
1
α '(t )
= ( sinh 2 t + cosh 2 t + 1) 2
⇒ α '(t ) = 2cosh t
as cosh 2 t − sinh 2 t = 1
... (3)
The arc length function s(t) of the curve from t = 0 is given by
t
t
0
0
s( t ) = ∫ α '(u ) du = ∫ 2cosh udu
⇒ s (t ) = 2sinh t
... (4)
Let β be the reparametrization of α given by

s2 s
 s  
β ( s) = α ( t ( s ) ) =  1 + ,
,sinh −1 

2 2
 2 

... (5)
s
1
1


T = β '( s) = 
,
,

2
2
s2 
 2 1+ s
2 1+


2
2 

... (6)
We see that
⇒ β '( s ) = 1
⇒ β ( s ) is the unit speed curve.
For this curve, to find κ and τ we first find from (6)



1
−s
T ' = β "( s ) = κ N = 
,0,
3
3
 
2 2
2 2

s
 2 1 + s 
2 2 1 + 
2

2
 
By definition,




2
1
s


κ = T' =
+
3
3


2
2
 4 1 + s  8 1 + s  


 
 
2

2 
 
Differential Geometry
107







... (7)
⇒ κ (s) =
1
 s2 
21+ 

2 
... (8)
Hence



1
−s
N =
,0,
1
1

2 2
2 2

s
  1 + s 
2 1+ 
2 

2








... (9)
⇒ N =1
Next,
B = T ×N =
U1
U1
U1
s
1
1
2 1+
s2
2
1
2
2
0
s
1+
2
2 1+
s2
2
−s
s2
2 1+
2
1
−1
 −s
⇒ B =
,
,
2
2
s2
 2 1+ s
2
1
+

2
2






... (10)
⇒ B =1
Thus,
−1
s

B ' = −τ N = 
,0,
3
3
2
  s2 2
 s 2
 2 1 + 
2 2 1 + 
2

2
 
−1
s

⇒ −τ N ⋅ N = 
,0,
3
3
2
  s2 2
 s 2
 2 1 + 
2 2 1 + 
2

2
 
Differential Geometry





−1
s
 
,0,
⋅
3
3
2
   s2  2
 s 2
  2 1+ 
2 2 1 + 
2

2
  
108
... (11)





1
⇒ τ (s) =
 s2 
1+ 

2 
Equations (8) and (12)
⇒ κ (s) = τ (s)
... (12)
Now for non unit speed curve (1), we know the Frenet apparatus is given by
T=
α&
α& × α&&
α& × α&&
α& ⋅ (α&& ×α&&&)
κ
=
τ
=
B=
,
,
,
and
3
2
α&
α& × α&& N = B × T
α&
α& ×α&&
... (13)
Thus from equation (2) we find
α&&(t ) = ( cosh t ,sinh t , 0 )
U1
⇒ α& ×α&& = sinh t
cosh t
... (14)
U1 U1
cosh t 1
sinh t 0
⇒ α& × α&& = ( − sinh t ,cosh t ,− 1) ........ as sinh 2 t − cosh 2 t = −1
... (15)
1
⇒ α& ×α&& = ( sinh 2 t + cosh 2 t + 1) 2
⇒ α& ×α&& = 2cosh t
... (16)
Also from equation (14)
&&&(t ) = ( sinh t ,cosh t , 0)
α
... (17)
&&& ⋅(α& ×α&& ) = ( sinh t ,cosh t , 0 ) ⋅ ( − sinh t ,cosh t ,− 1)
⇒α
= − sinh 2 t + cosh2 t
&&& ⋅(α& ×α&& ) = 1
⇒α
... (18)
Thus,
T (t) =
1
( sinh t ,cosh t,1)
2cosh t
... (19)
⇒ T =1
B=
1
( − sinh t,cosh t ,− 1)
2cosh t
⇒ B =1
Differential Geometry
109
... (20)
U1
− sinh t
N = B×T =
2cosh t
sinh t
2cosh t
U1
1
2
1
2
U1
−1
2cosh t
1
2cosh t
sinh t 
 1
N =
,0, −

 cosh t
cosh t 
⇒N=
1
(1,0, − sinh t )
cosh t
... (21)
⇒ N =1
κ=
and τ =
2cosh t
(
2cosh t )
3
⇒κ =
1
2cosh 2 t
1
2cosh 2 t
... (22)
... (23)
We now express κ ( t ) and τ (t ) of α as function of κ ( s ) and τ ( s ) . It follows obviously from
equation (8), (12), (22) and (23).
κ (t) = κ ( s)
τ (t) = τ ( s)
Example 8 : Let σ be the spherical image of a unit speed curve β . Prove that the curvature and
torsion of σ are
d τ 
 
τ 
ds  κ 
κσ = 1 +   , τσ =
  τ 2 
κ 
3
κ 1 +   
 κ  
2
where κ and τ are the curvature and torsion of β .
Solution : Let β : I → ¡ 3 be a unit speed curve whose curvature and torsion are denoted by κ and
τ respectively..
By definition, the velocity vector of β is given by
T = β ' ( s ) ⇒ T = β ' (s ) = 1
Differential Geometry
110
... (1)
We know the curve σ = T as the spherical image of the unit speed curve β .
⇒ σ =1
(by (1))
... (2)
Let κσ and τσ be the curvature and torsion of the spherical image of curve β .
Then
σ '=T'
... (3)
σ ' = T ' = κ N as T ' = κ N
... (4)
where T σ is the tangent vector field of σ .
The speed of the spherical image is given by the length of its velocity vector σ ' .
(equal to the curvature of β )
σ ' =κ
... (5)
⇒ σ is not a unit speed curve.
To compute the curvature κσ of σ we find
σ " = (κ N ) '
σ"=
=
σ"=
dκ
N +κ N '
ds
dκ
N + κ ( −κ T +τ B )
ds
as N ' = −κ T + τ B
dκ
N '− κ 2 T + κτ B
ds
... (6)
From equation (5) the curve σ has not unit speed. In this case we know the curvature and torsion for
such curve σ are given by
κσ =
τ=
σ '× σ "
σ"
... (7)
3
σ '"⋅ (σ '× σ " )
σ '× σ "
... (8)
2
Now we first find
 dκ

σ '× σ " = κ N × 
N − κ 2 T + κτ B 
 ds

= −κ 3 N × T + κ 2τ N × B
= κ 3 ( T × N ) + κ 2τ ( N × B )
Differential Geometry
111
= κ 3 B + κ 2τ T
σ '× σ " = κ 3 B + κ 2τ T
... (9)
1
⇒ σ '×σ " = (κ 6 + κ 4τ 2 ) 2
1
⇒ σ '× σ " = κ
2
(κ 2 + τ 2 )2
... (10)
The curvature κσ of σ is obtained from (7) as
1
κ 2 (κ 2 + τ 2 ) 2
1( 2
κσ =
⇒
κ
=
κ + τ 2 )2
σ
3
κ
κ
1
1
2 2
 τ 
κσ = 1 +    > 1
 κ  
or
... (11)
Now to find the torsion of σ , we first find
σ '" =
d 2κ
dκ
dκ
dκ
dτ
⋅N +
⋅ N '− 2κ
T − κ 2 T '+
τ B +κ
B + κτ B '
2
ds
ds
ds
ds
ds
σ '" =
d 2κ
dκ (
dκ
dκ
dτ
⋅N +
⋅ −κ T + τ B ) − 2κ
T − κ 2 (κ N ) +
τ B +κ
B + κτ ( −τ N )
2
ds
ds
ds
ds
ds
... as N ' = −κ T + τ B ; B ' = −τ N ; T ' = κ N
⇒ σ '" =
d 2κ
dκ
dκ
dτ
⋅ N + 3κ
⋅ T − 2τ
B −κ 3 N + κ
B − κτ 2 N
2
ds
ds
ds
ds
 d 2κ

dκ
dτ 
 dκ
⇒ σ '" =  2 − κ 3 − κτ 2  ⋅ N − 3κ
⋅ T +  2τ
+κ
B
ds

ds
ds 
 ds

  d 2κ

dκ
dτ  
 dκ
σ '"⋅ ( σ '× σ ") =   2 − κ 3 − κτ 2  ⋅ N − 3κ
⋅ T +  2τ
+κ
 B ⋅
ds
 ds
ds  

  ds
⋅ (κ 3 B + κ 2τ T )
= −3κ 3τ
σ '"⋅ ( σ '× σ " ) = −κ 3τ
Differential Geometry
dκ
dκ
dτ
2τκ 3
+κ4
ds
ds
ds
dκ
dτ
+κ4
ds
ds
112
... (12)
dκ 
 dτ
σ '"⋅ ( σ '× σ " ) = −κ 3  κ
−τ

 ds
ds 
σ '"⋅ (σ '× σ " ) = −κ 3
d τ 
 
ds  κ 
... (13)
Hence from equation (8) we find
d τ 
 
τσ = 4 ds2 κ 2
κ (κ + τ )
κ3
d τ 
 
⇒ τσ = ds 2 κ 2
κ (κ + τ )
or
d τ 
 
ds
κ 
τσ =
  τ 2 
3
κ 1 +   
 κ  
•....................................................................................................................................................•
Cylindrical Helix :
u
θ θ
θ
Definition : A regular curve α in ¡ 3 is a cylindrical helix provided
that the unit tangent T of α has constant angle θ with some fixed
unit vector u .
i.e. T ⋅ u = cos θ
∀ t
Theorem : Prove that a regular curve α with κ > 0 is a cylindrical helix iff the ratio
τ
= constant
κ
Solution : Let α : I → ¡3 has a unit speed. Let also α be a cylindrical helix. Then we have
T ⋅ u = cos θ
... (1)
where θ is a constant and u is a fixed unit vector and T the unit tangent vector to α .
Differentiating equation (1) w.r.t. the parameter, we get
(T ⋅ u ) ' = 0
⇒ T '⋅ u = 0
(as u is a fixed unit vector)
⇒κ N ⋅u = 0
Differential Geometry
113
⇒ N ⋅u = 0
as κ > 0
This shows that u lies in the plane determined by T and B .
⇒ u can be expressed as a linear combination of T and B .
Thus we have
B
u = c1T + c2 B ,
u
where c1 and c2 are to be determined.
π −θ
2
θ
T
⇒ u ⋅ B = c1 and u ⋅ T = c2
... (2)
⇒ c1 = cos ( 90 − θ ) = sin θ and c2 = cos θ
... (3)
Thus we have
u = cos θT + sinθ B
... (4)
Differentiating equation (4) we get
0 = T 'cos θ + B 'sin θ
0 = κ N cos θ − τ N sinθ
........... as T ' = κ N , B ' = −τ N
⇒ (κ cos θ − τ sin θ ) N = 0
⇒ κ cos θ − τ sin θ = 0
⇒
τ
= cot θ = constant
κ
... (5)
Conversly,
Let us suppose that
τ
= constant.
κ
We choose θ such that
cot θ =
τ
κ
⇒ τ sin θ −κ sin θ = 0
... (6)
Let U be a vector in the T and B plane. Then we can express U as a linear combination of T and
τ
B by taking fixed θ for which κ = cot θ ,
U = cos θT + sinθ B
Differential Geometry
.
114
... (7)
Differentiating (7) we get, U ' = cos θ T ' + sin θ B ' = κ cosθ N −τ sinθ N
⇒ U ' = (κ cos θ −τ sin θ ) N
⇒U ' = 0
due to (5)
⇒ U = constant
⇒ U is a parallel vector field.
Also
U ⋅ U = ( cos θ T + sin θ B ) ⋅ ( cos θ T + sin θ B )
= cos 2 θ + sin 2 θ = 1
Hence U is a unit vector and
T ⋅ U = T ⋅ ( cos θ T + sin θ B )
⇒ T ⋅U = cos θ
⇒ α is a cylindrical helix.
Summary : We have seen that for a curve α if κ is a curvature and τ the torsion, then
(i)
κ = 0 ⇔ α is a straight line.
(ii)
κ > 0,τ = 0 ⇔ plane curve.
(iii)
κ = constant > 0,τ = 0 ⇔ α is a circle.
(iv)
κ = constant > 0,τ = constant ≠ 0 ⇔ α is a circular helix.
(v)
τ
= constant ⇔ α is a cylindrical helix.
κ
•.....................................................................................................................................................•
Differential Geometry
115
Unit II
Covariant Derivative :
Definition : Let W be a vector field on ¡ 3 and let v p be a tangent vector to ¡ 3 at point p. The
covariant derivative of W with respect to v p is the tangent vector ∇v W at p defined by
∇v W =
d
W ( p + tv ) |t =0
dt
∇v W measures the initial rate of change of of W ( p ) as p moves in the v direction.
Note : ∇ v p W = ∇v W
W
W(p)
W(p+tv)
v
p
p+tv
vW
Worked examples ........................................................................................................................•
Example 1 : If W = x 2 U 1 + yzU 3 and v = ( −1,0,2) , p = ( 2,1,0) then find ∇v W at p.
Solution : The covariant derrivative of a vector field W w.r.t. the tangent vector v is defined by
∇v W =
d
W ( p + tv ) |t = 0
dt
...(1)
where p + tv = ( 2 − t ,1,2t )
Thus
W ( p + tv ) = W ( 2 − t ,1,2t )
W ( p + tv ) = ( 2 − t ) U 1 + 2tU 3
2
Then
W ' ( p + tv ) = −2 ( 2 − t ) U1 + 2U 3
Thus
∇v W =
d
W ( p + tv ) |t = 0
dt
= ( −4,0,2) p
which is a tangent vector.
Differential Geometry
116
(by definition of W )
EXERCISE ....................................................................................................•
Consider a tangent vector v = (1, −1,2) at the point p = (1,3, −1) . Compute ∇v W , for
(i)
W = x2 U 1 + yU 2
(ii)
W = xU 1 + x 2U 2 − z 2U 3
ANSWERS .....................................................................................................•
(i) ∇ v W = ( 2, − 1,0) p
(ii) ∇ v W = (1,2,4) p
•.......................................................................................................................•
Theorem : If W = ∑ wi U i is a vector filed on ¡ 3 and v is tangent vector at p, then prove that
i
∇ v W = ∑ v p [ wi ]U i ( p )
i
where v p [ wi ] is the directional derivative of coordinate functions of W with respect to
v at p.
Proof : Given that W = ∑ wi U i
i
⇒ W ( p + tv ) = ∑ wi ( p +tv ) U i ( p + tv )
i
Differentiating this w.r.t. t we get,
W ' ( p + tv ) = ∑
i
Thus
d
wi ( p + tv ) ⋅ U i ( p + tv )
dt
∇ v W = W ' ( p + tv ) |t =0 = ∑
i
d
wi ( p + tv ) t = 0 ⋅U i ( p )
dt
But by definition of directional derivative, we have
vp[ f ] =
d
f ( p + tv ) |t =0
dt
Using this we write
∇ v W = W ' ( p + tv ) (0) = ∑ v p [ wi ] ⋅ Ui ( p)
i
This proves the theorem.
Differential Geometry
117
Remark :
To find the covariant derivative of a vector field we find the directional derivative of its coordinate
functions in the direction of tangent vector at p.
Now we use the above theorem to find the covariant derivative of a given vector field with
respect to the tangent vector.
Worked examples .......................................................................................................................•
Example 2 : If W = xU 1 + x 2U 2 − z 2U 3 . Find the covariant derivative of W w. r. t. v at point
p = (1,3, −1) , where v = (1, −1,2)
Solution : We have from the above theorem that
∇ v W = ∑ v p [wi ] ⋅U i (p )
i
As
v p [ wi ] = ∑ vk
k
∂wi
( p)
∂xk
Thus
∇ v W = ∑∑ vk
i
k
∂wi
( p ) ⋅ U i ( p)
∂xk
∂w
∂w2
 ∂w

= ∑  vk 1 U 1 + vk
U 2 + vk 3 U 3 
∂xk
∂xk
∂xk
k 

∂w
∂w 
∂w
∂w 
 ∂w
 ∂w
∇ v W =  v1 1 + v2 1 + v3 1  U 1 +  v1 2 + v2 2 + v3 2  U 2 +
∂x2
∂x3 
∂x2
∂x3 
 ∂x1
 ∂x1
∂w
∂w 
 ∂w
+  v1 3 + v2 3 + v3 3  U 3
∂x2
∂x3 
 ∂x1
Since
W = ( xU 1 + x 2U 2 − z 2U 3 ) and v = (1, −1,2) , we have
∇v W = U 1 + 2 xU 2 − 4 zU 3
Thus
∇ v W = (U 1 + 2 xU 2 − 4 zU 3 ) p
∇ v W = (1,2,4) p
•....................................................................................................................................................•
Differential Geometry
118
Theorem : Let v be w the tangent vectors to ¡ 3 at p and let Y and Z be vector fields on ¡ 3 ,
a, b be the scalars and f be a diffenertiable real function, then prove that,
(i)
∇ (av +bw) Y = a∇ v Y + b∇w Y
(ii)
∇ v ( aY + bZ ) = a∇ v Y + b∇v Y
(iii)
∇ v ( f Y ) = v p [ f ] Y ( p) + f ( p )∇v Y
(iv)
v p [ f ] = ∇ v Y ⋅ Z ( p ) + Y ( p ) ⋅∇ v Z
∀b
Proof : Let Y = ∑ yi U i and Z = ∑ zi U i be the vector fields v and w be the tangent vectors
i
i
to ¡ 3 at point p. Then by definition of covariant derivative of Y w. r. t. v is given by
∇ v Y = ∑ v p [ y i ] ⋅U i ( p )
... (1)
i
using equation (1) we write
(i)
∇ (av +bw) Y = ∑ ( av + bw ) p [ yi ] ⋅U i ( p )
i
= a ∑v p [y i ] ⋅U i ( p) +b ∑w
i
p
[y i ]⋅U i (p )
i
⇒ ∇ (av +bw )Y = a∇ v Y + b∇w Y
(ii)
... (2)
Now consider
∇ v ( aY + bZ ) = ∑ v p [ ayi + bzi ] ⋅U i ( p)
i
= ∑ ( av p [ yi ] + bv p [ zi ] ) ⋅U i ( p )
i
= a ∑v p [y i ] ⋅U i ( p ) +b ∑v
i
p
[z i ] ⋅U i (p )
i
∇ v ( aY + bZ ) = a∇ v Y + b∇v Z
(iii)
... (3)
Consider
∇ v ( f Y ) = ∑ v p [ fyi ] ⋅U i ( p)
i
= ∑ ( v p [ f ] yi ( p ) + f ( p) v p [ yi ] ) ⋅U i ( p )
i
Differential Geometry
119
( v p [ fg ] = f
vp [g ] + v p [ f ] g )
= ∑ v p [ f ] yi ( p) ⋅ Ui ( p) + ∑ f ( p )v p [ yi ] ⋅U i ( p )
i
i
⇒ ∇ v ( f Y ) = v p [ f ] ∑ yi ( p) ⋅ Ui ( p ) + f ( p ) ∑ v p [ y i ] ⋅U i ( p )
i
i
⇒ ∇ v ( f Y ) = v p [ f ]Y ( p ) + f ( p )∇v Y
(iv)
... (4)
Consider
Y ⋅ Z = ( y1U 1 + y 2U 2 + y3 U 3 )+ ( z1U 1+ z 2U 2 + z 3U 3 )
= y1 z1 + y2 z 2 + y3z3
Y ⋅ Z = ∑ yi zi
... (5)
i
Consider
v p [Y ⋅ Z ] = v p ∑ yi zi 
 i

= ∑ v p [ yi zi ]
i
v p [Y ⋅ Z ] = ∑ {v p [ yi ] zi ( p) + yi ( p )v p [ zi ]}
i
... (6)
Next consider




∇ v Y ⋅ Z ( p) + Y ( p) ⋅∇ v Z =  ∑ v p [ yi ] ⋅U i ( p)  Z ( p ) + Y ( p )  ∑ v p [ z i ] ⋅U i ( p ) 
 i

 i



 


=  ∑ v p [ y i ] ⋅U i ( p )   ∑ zk U k ( p)  +  ∑ y k U k ( p )   ∑ v p [ z i ] ⋅U i ( p ) 
 i
 k
  k
 i

= ∑ v p [y i ] z k U i U k ( p ) + ∑ yk v p [ zi ]U k U i ( p )
i ,k
i ,k
= ∑ v p [y i ] z kδ ik + ∑ yk v p [ zi ] δ ik
i ,k
i, k
⇒ ∇ v Y ⋅ Z ( p) + Y ( p) ⋅∇ v Z = ∑ v p [ yi ] zi ( p) + ∑ yi ( p )v p [ zi ]
i
i
From equation (6) and (7) we have
v p [Y Z ] = ∇ v Y ⋅ Z ( p) + Y ( p) ⋅ ∇ v Z
Differential Geometry
120
... (7)
Covariant Derivative of Vector Field w.r.t. a Vector Field :
Definition : Let V and W be vector fields in ¡ 3 . The covariant derivative of vector field W w. r.. t.
a vector field V is a vector field ∇v W whose value at each point p is given by
(∇V W ) ( p ) = ∑ V ( p)[ wi ] Ui ( p)
i
(
)
Note : ∇V W ( p ) = ∇V ( p)W
Remark :
(i)
The vector field ∇v W at p measures the rate of change of W as p moves in the V direction.
(ii)
Covariant derivative of a scalar function f is same as its ordinary derivative.
(iii)
(∇V W ) ( p ) = ∇V ( p)W
= ∑V ( p) [wi ]U i ( p)
i
(
)
= ∑ V [ wi ]U i ( p)
i
Worked examples .........................................................................................................................•
Example 3 : Let V = − yU 1 + xU 3 and W = cos xU 1 + sin xU 2 be the vector fields. Express the
following covarient derivatives in terms of U 1 , U 2 ,U 3 .
(i) ∇V W
(
(iv) ∇V ∇V W
)
(ii) ∇V V
(iii) ∇W V
(v) ∇V ( z 2W )
(vi) ∇V ( xV − zW )
Solution : We have by definition
(i)
where
∇V W = ∑ V [ wi ]U i
... (1)
i
V [ wi ] = ( − yU 1 + xU 3 ) [ wi ]
V [ wi ] = − yU 1 [ wi ] + xU 3 [ wi ]
Differential Geometry
121
V [ wi ] = − y
∂wi
∂w
+x i
∂x
∂z
U k [ wi ] =
∂wi
, k = 1,2,3
∂xk
... (2)
Substituting this in equation (1) we get
∂w
 ∂w
∇V W = ∑  − y i + x i
∂x
∂z
i 

U i

Since W = cos xU 1 + sin xU 2
⇒ ∇V W = y sin xU 1 − y cos xU 2
⇒ ∇V W = ( y sin x , y cos x, 0 ) , which is a vector field
(ii)
... (3)
Now we will obtain the value of ∇V V by using the definition
V [ wi ] = ∑ Vk
k
∂wi
∂xk
∇ v V = ∑ V [ vi ]U i
... (4)
i
∂vi
where V [ vi ] = ∑ vk ∂x
k
k
⇒ V [ vi ] = v1
∂vi
∂v
∂v
+ v2 i + v3 i
∂x
∂y
∂z
But by hypothesis V = − yU 1 + xU 3
∂v 
 ∂v
V [ vi ] =  − y i + x i 

∂x
∂z 
Substituting this in equation (4) we get
∂v
 ∂v
∇V V = ∑  − y i + x i
∂x
∂z
i 

U i

∂v 
∂v 
∂v 
 ∂v
 ∂v
 ∂v
⇒ ∇V V =  − y 1 + x 1  U 1 +  − y 2 + x 2  U 2 +  − y 3 + x 3  U 3

∂x
∂z 

∂x
∂z 

∂x
∂z 
⇒ ∇V V = ( 0,0, − y )
Differential Geometry
... (5)
122
(iii)
Similarly we obtain
∇wV = ( − sin x,0,cos x )
(iv)
... (6)
Consider now
(
)
∇V ∇V W = ∇V Z = ∑ V [ zi ]U i
i
Where Z = ∇V W = ( y sin x, − y cos x, 0 )
(
)
⇒ ∇V ∇V W = ∑ ( − yU 1 + xU 3 ) [ zi ]U i
i
∂z 
∂z 
∂z 
 ∂z
 ∂z
 ∂z
=  − y 1 + x 1  U1 +  − y 2 + x 2  U 2 +  − y 3 + x 3  U 3

∂x
∂z 

∂x
∂z 

∂x
∂z 
= − y 2 cos xU 1 − y 2 sin xU 2
(
)
⇒ ∇V ∇V W = ( − y 2 cos x, − y 2 sin x ,0 )
(v)
... (7)
Consider ∇V ( z 2W ) , where z 2W = z 2 cos xU 1 + z 2 sin xU 2
Hence
∇V ( z 2W ) = ∑ V  z 2 wi  U i
i
= ( − yU 1 + xU 3 )  z 2 w1  U 1 + ( − yU 1 + xU 3 )  z 2 w 2  U 2 + ( − yU 1 + xU 3 )  z 2 w3  U 3
∂
∂
∂
∂




=  − y ( z 2 cos x ) + x ( z 2 cos x )  U 1 +  − y ( z2 sin x ) + x ( z 2 sin x )  U 2
 ∂x
∂z

 ∂x
∂z

=  yz 2 sin x + 2 xz cos x  U 1 + − z 2 y cos x + 2 xz sin x  U 2
∇V ( z 2W ) = ( yz 2 sin x + 2 xz cos x, − z 2 y cos x + 2 xz sin x ,0 )
(vi)
... (8)
Let Z = ( xV − zW ) ⇒ Z = − ( xy + z cos x )U 1 − z sin xU 2 + x 2U 3
In the same way we prove that
∇V ( xV − zW ) = ( y 2 − yz sin x − x cos x, yz cos x − x sin x , −2 xy )
Differential Geometry
123
... (9)
EXERCISE ........................................................................................................•
1.
If V = ( y − x)U 1 + xyU 3 , W = x 2 U 1 + yzU 3 be the vector fields in ¡ 3 . Find the covariant
derivative of W w. r. t. V .
ANSWER : ∇V W = ( 2 xy − 2 x 2 ,0, xy 2 )
•........................................................................................................................•
Note : ∇V W ≠ ∇W V
Theorem : Let V , W , Y and Z be vector fields on ¡ 3 , then prove that
(i)
∇V ( aY + bZ ) = a∇V Y + b∇V Z
(ii)
∇ ( f V + gW ) Y = f ∇V Y + g∇W Y
(iii)
∇V ( f Y ) = f ∇V Y + V [ f ] ⋅ Y
(iv)
V [ Y ⋅ Z ] = ∇V Y ⋅ Z + Y ⋅ ∇V Z
Solution : By definition of covariant derivative, we know that
∇V W = ∑ V [ wi ]U i
... (1)
i
(i)
Consider
∇V ( aY + bZ ) = ∑ V [ ayi + bz i ]U i
i
= ∑ aV [ yi ] + bV [ zi ]U i
i
(by linearity of directional derivation)
= a ∑V [ yi ] U i + b ∑V [ zi ]U i
i
i
∇V ( aY + bZ ) = a∇V Y + b∇V Z
(ii)
(by (1))
Consider now
∇ ( f V + gW ) Y = ∑ ( f V + gW ) [ yi ] U i
i
=∑
{( f V )[ yi ] + ( gW )[ zi ]} U i
i
Differential Geometry
(by definition)
124
... (2)
= ∑ f V [ y i ]U i + ∑ gW [ zi ]U i
i
i
= f ∑ V [ y i ]U i + g ∑W [ zi ]U i
i
i
∇ ( f V + gW ) Y = f ∇V Y + g∇W Z
(iii)
(by definition)
... (3)
Next consider
∇V ( f Y ) = ∑ V [ fyi ]U i
... (4)
i
Since we know that
V [ f ⋅ g] ( p ) = V ( p) [ f ⋅ g ]
= f ( p) ⋅V ( p) [ g ] +V ( p ) [ f ] ⋅ g ( p )
V [ f ⋅ g] ( p) = ( f ⋅ V [ g ] + V [ f ] ⋅ g ) ( p)
⇒ V [ f ⋅ g ] = f ⋅V [ g] + V [ f ]⋅ g
Using this in equation (4) we get
(
)
∇V ( f Y ) = ∑ f ⋅ V [ yi ]+ V [ f ]⋅ yi U i
i
= f ∑ V [ y i ]U i + V [ f ] ∑ yi U i
i
i
Y = ∑ yi U i
∇V ( f Y ) = f ∇V Y + V [ f ] ⋅ Y ,
(iv)
i
Now we claim that
(
)
(
V [ Y ⋅ Z ] = ∇V Y ⋅ Z + Y ⋅ ∇V Z
)
Let Y = ∑ yi U i and Z = ∑ zi U i
i
Differential Geometry
i
125
... (6)
Y ⋅ Z = ∑ yi zi
Thus
i
Consider now
V [ Y ⋅ Z ] = V  ∑ yi zi 
 i

= ∑ V [ yi zi ]
(by linearity of directional derivative)
i
(
= ∑ yi V [ zi ] + V [ yi ] zi
)
(by using (5))
i
... (7)
Consider the R.H.S.
(∇V Y ) ⋅ Z + Y ⋅ (∇V Z ) = ( ∑V [ yi ]U i ) ⋅ ( ∑ zk U k )+ (∑ yi U i )⋅ ( ∑V [ zk ]U k )
i
k
i
k
= ∑ V [ yi ] z k U i U k + ∑ yi V [ zk ] U i U k
i ,k
i,k
= ∑ V [ yi ] z kδ ik + ∑ yi V [ zk ] δik
i ,k
i ,k
(∇V Y ) ⋅ Z + Y ⋅ (∇V Z ) = ∑V [ yi ] zi + ∑ yi V [zi ]
i
i
U i U k = δ ik
... (8)
Equation (7) and (8) give
V [ Y ⋅ Z ] = ∇V Y ⋅ Z + Y ⋅ ∇V Z
... (9)
EXERCISE ......................................................................................................•
1.
Use the properties of the covariant derivative in the above theorem to find the value of
(i) ∇V z 2W
(ii) ∇V ( xV − zW )
where V = − yU 1 + xU 3 , W = cos xU 1 + sin xU 2
2.
If X = ∑ xi U i , prove that ∇v X = V
i
3.
Prove that ∇ f v Y = f ∇v Y
4.
If W is a vector field with constant length W , prove that for any vector field V , ∇V W is
orthogonal to W at each point of ¡ 3 .
•........................................................................................................................•
Differential Geometry
126
Unit III
Isometries of ¡ 3 (Rigit Motion) :
Intorduction : An isometry or rigit motion of Euclidean space is a special type of mapping which
preserves the Euclidean distance between points. We study the following types of isometries in this
unit.
(i) Translation
(ii) Rotation
(iii) Orthogonal tramnformation
Isometry :
Definition : An isometry of ¡ 3 is a mapping F : ¡3 → ¡3 such that
d ( F ( p ) ) , ( F ( q ) ) = d ( p, q )
where
... (1)
d ( p, q) = p − q .
This shows that the distance is preserved under F .
Translation : Let a be a fixed point of ¡ 3 . Then the mapping T that adds a to every point
p of ¡ 3 .
1.
i.e.
T ( p) = p + a
∀p ∈ ¡ 3
is called translation by a.
Example : Prove that translation is an isometry.
Solution : Now to prove that T is an isometry, we simply prove that T preserves the Euclidean
distance between points.
Therefore consider
d ( T ( p ) ,T ( q ) ) = d ( p + a, q + a )
=
( p + a ) − ( q + a)
= p −q
d ( T ( p ) , T ( q ) ) = d ( p, q )
⇒ T is an isometry..
Differential Geometry
127
2.
Rotation around coordinate axis :
The rotation of the xy plane through an angle θ carries the point p = ( p1 , p2 ) to the point q = ( q1 , q2 ) .
The rotation through an angle θ is affected by the transformation equations
q1 = p1 cos θ − p2 sinθ
q2 = p1 sin θ + p2 cos θ
q3 = p3
... (2)
Thus the rotation of xy plane through an angle θ around z
axis is a mapping c on ¡ 3 .
i.e.
c : ¡3 → ¡3 , such that
c ( p) = q
∀p , q ∈ ¡ 3
... (3)
where c is the matrix of rotation about z axis given by
 cos θ

c ( p ) =  sin θ
 0

− sin θ
cos θ
0
0  p1 
 
0  p2  = ( q1 q2 q3 )
1   p3 
Thus c ( p ) = q gives
c ( p1, p2 , p3 ) = ( p1 cos θ − p2 sin θ , p1 sin θ + p 2 cos θ , p 3 )
= ( q1, q2 , q3 )
∀p ∈ ¡ 3
... (4)
Example : Show that rotation c around z axis is an isometry..
Proof : Let p and p' be two points in a plane, such that p maps to q and p' maps to q' under the
mapping c .
⇒ c ( p ) = q and c ( p ') = p '
... (5)
To prove c is an isometry, we simply prove that
d ( c ( p) , c ( p ') ) = d ( p , p ' )
Using (5) we have
d ( c ( p) , c ( p ') ) = d (q , q ' )
=
Differential Geometry
( q1 − q '1 ) 2 + ( q2 − q '2 ) 2 + ( q3 − q '3 )1
128
Using transformation of rotation viz. equation (4), we get
d ( c ( p) , c ( p ') ) =
1
( p1 cos θ − p 2sin θ − p ' 1cosθ + p ' 2sin θ )2 +
2


=
2

+ ( p1 sin θ + p2 cos θ − p'1 sin θ − p '2 cos θ ) + ( p3 − p '3 ) 
1
2 2
{
⇒ d ( c ( p) , c ( p ') ) = p12 + p 22 + p ' 12 + p '22 − 2 p1 p '1− 2 p2 p '2 + ( p3 − p '3 )
⇒ d ( c ( p) , c ( p ') ) =
{( p − p ' )
1
2
1
+ ( p2 − p '2 ) + ( p3 − p '3 )
2
}
1
2 2
}
⇒ d ( c ( p) , c ( p ') ) = d ( p , p ' )
⇒ The rotation of xy plane about Z-axis is an isometry..
Theorem : Let F and G are isometries of ¡ 3 , then show that the composite mapping GF is also
an isometry ¡ 3 .
Proof : Let F : ¡3 → ¡3 and G : ¡3 → ¡3 be isometry of ¡ 3 .
⇒ d ( F ( p ) , F ( q ) ) = d ( p, q )
d ( G ( p) ,G ( q ) ) = d ( p, q )
and
... (1)
∀p , q ∈ ¡ 3
... (2)
To prove that composite mapping
G ⋅ F : ¡3 → ¡3
is an isometry, consider the distance from G ( F ( p ) ) to G ( F ( q ) ) given by
d G ( F ( p ) ) , G ( F ( q ) )  = d ( F ( p ) ,F (q ) ) ......... as is an isometry
⇒ d G ( F ( p ) ) , G ( F ( q ) )  = d ( p, q )
......... by (1)
This proves that GF is also an isometry in ¡ 3 .
Theorem : Prove the following
(i)
If S and T are translations, then ST = T S is also a translation.
(ii)
If T is translation by a, then T has an inverse T –1 which is translation by – a.
Differential Geometry
129
(iii)
Given any two points p and q of ¡ 3 , there exists a unique translation T such that
T ( p) = q
Proof :
(i) Let S and T are translations by a and b respectively..
⇒ S ( p) = p + a
and
... (1)
T ( q) = q + b
∀p , q ∈ ¡ 3
... (2)
Consider
ST ( p ) = S ( T ( p) )
Thus
= T ( p)+ a
by (1)
= p+ b + a
by (2)
ST ( p ) = p+ b + a
... (3)
Similarly,
T S ( p ) = T ( S ( p) )
Thus
= S ( p) + b
by (2)
= p+ a + b
by (1)
T S ( p ) = p+ a + b
... (4)
From equation (3) and (4) we have
ST = T S
is a translation by a + b .
(ii) Given that T is a translation by a
⇒ T ( p) = p + a
Let
T ( p) = p+ a = q
... (5)
We see that is one-one, because
T ( p) = T (q )
⇒ p + a = q +b
⇒ p =q
Differential Geometry
130
It is also on-to as ¡ 3 is finite dimensional space
⇒ T –1 exists.
Thus from equation (5) we have
T
T
−1
( q) =
−1
(q ) = q − a
p
This proves that is T –1 a translation by – a.
(iii) Translation T by q − p is definited by
T ( p) = p + (q − p) = q
∀p ∈ ¡ 3
... (6)
This shows that translation by ( q − p ) certainly carries the point p to q. Now if T is a translation by a
then
T ( p) = p + a
But
T ( p) = q
... (7)
(by equation (6))
⇒ p+ a= q
⇒a =q−p
... (8)
This proves uniquness.
Corollary : If T is a translation such that for some point p ∈ ¡ 3 , T ( p ) = p then T = I the identity
translation.
Solution : If p and q are two points of ¡ 3 then there exist a unique translation such that
T ( p) = q
∀p ∈ ¡ 3
T ( p) = p
for some p ∈ ¡ 3
⇒ p = q =T ( p )
∀p ∈ ¡ 3
But given that
⇒ T =I
Differential Geometry
131
Orthogonal Transformation :
Definition : A linear transformation c : ¡3 → ¡3 is said to be orthogonal if
c ( p ) ⋅c ( q ) = p ⋅ q
∀p , q ∈ ¡ 3
This shows that the orthogonal transformation c preserves the dot product.
Worked examples .......................................................................................................................•
Example 1 : Show that the orthogonal transformation is linear.
Solution : Let c : ¡3 → ¡3 be a orthogonal transformation.
To prove c is linear, we simply prove that
c ( ap + bp ') = ac ( p ) + bc ( p ')
Therefore consider
 cos θ

c ( ap + bp ') =  sin θ
 0

− sin θ
cos θ
0
0   ap1 + bp '1 


0   ap2 + bp '2 
1   ap3 + bp '3 
 ( ap1 + bp '1 ) cos θ − ( ap2 + bp '2 ) sin θ ,

=


ap
+
bp
'
sin
θ
+
ap
+
bp
'
cos
θ
,
ap
+
bp
'
(
)
(
)
(
)
1
1
2
2
3
3 

= a ( p1 cos θ − p2 sin θ , p1 sin θ + p 2 cos θ , p 3 ) +
+b ( p '1 cos θ − p '2 sin θ , p '1 sin θ + p '2 cos θ , p '3 )
= ac ( p ) + bc ( p ' )
⇒ c ( ap + bp ') = ac ( p ) + bc ( p ')
This proves that c is linear..
Example 2 : Show that rotation is an orthogonal transformation.
Solution : We know rotation c around z axis through an angle θ is characterized by the equation
c ( p ) = q = ( p1 cosθ − p2 sin θ , p1 sin θ + p 2 cos θ , p 3 )
∀p ∈ ¡ 3
c ( p ') = q ' = ( p '1 cos θ − p '2 sin θ , p '1 sin θ + p '2 cos θ , p '3 ) ∀p ' ∈ ¡ 3
To prove c is orthogonal, we simply prove that c preserves dot product.
Differential Geometry
132
Therefore, we consider
c ( p ) ⋅ c ( p ') = q ⋅ q '
c ( p ) ⋅ c ( p ' ) = ( p1 cos θ − p2 sin θ , p1 sin θ + p 2 cos θ , p 3 )
⋅ ( p '1 cos θ − p '2 sin θ , p '1 sin θ + p '2 cos θ , p '3 )
= p1 p '1 + p2 p '2 + p3 p '3
⇒
= ( p1, p2 , p3 )( p'1 + p '2 + p '3 )
= p1 ⋅ p '
⇒ c ( p ) ⋅c ( p ') = p1 ⋅ p '
⇒ c preserves dot product.
Hence the rotation is an orthogonal transformation.
•.....................................................................................................................................................•
Lemma : If c : ¡3 → ¡3 is an orthogonal transformation, then prove that c is an isometry of ¡ 3 .
Proof : Given that the rotation c : ¡3 → ¡3 is an orthogonal transformation. This implies that it
preserves the dot product.
⇒ c ( p ) ⋅c ( p ') = p1 ⋅ p '
∀p, p ' ∈ ¡ 3
... (1)
∀p ∈ ¡ 3
... (2)
In perticular, if p1 = p ' , then we have
c ( p) ⋅c ( p) = p ⋅ p
i.e.
c ( p) = p
2
2
⇒ c ( p) = p
This proves that the orthogonal transformation preserves the norm.
Now to prove that c is an isometry,,
Consider,
d ( c ( p ) ,c ( q ) ) = c ( p ) − c ( q )
= c ( p − q)
......... as c is linear
= p −q
......... by (2)
⇒ d ( c ( p ) ,c ( q ) ) = p − q
... (3)
This proves that the orthogonal transformation is an isometry in ¡ 3 .
Differential Geometry
133
Theorem : If F is an isometry of ¡ 3 such that F ( 0 ) = 0 , then show that F is an orthogonal
transformation.
Proof : Given F is an isometry of ¡ 3 .
⇒ d ( F ( p ) , F ( q ) ) = d ( p, q )
... (1)
Now to prove F preserves dot product, we first prove that F preserves the norm.
(i) By definition we have
p = 0 − p = d ( 0, p ) .
Therefore,
... (2)
F ( p ) = d ( 0, F ( p ) )
= d ( F ( 0 ), F ( p ))
Q F (0 ) = 0
= d ( 0, p )
= p
(by equation (2))
⇒ F ( p) = p
... (3)
⇒ F preserves the norm.
(ii)
Now we claim that F preserves the dot product.
Since F is an isometry,,
⇒ d ( F ( p ) , F ( q ) ) = d ( p, q )
⇒ F ( p ) − F (q )
2
= p −q
2
⇒ ( F ( p ) − F ( q ) ) ⋅( F ( p ) − F ( q ) ) = ( p − q ) ⋅ ( p − q )
⇒ F ( p ) F ( p ) + F ( q ) F ( q ) − 2 F ( p ) F ( q ) = p⋅ p + q ⋅ q − 2 p ⋅ q
⇒ F ( p ) + F ( q ) − 2 F ( p ) ⋅F ( q ) = p + q − 2 p ⋅ q
2
2
2
2
Using equation (3) we have
F ( p) ⋅ F ( q ) = p ⋅ q
... (4)
This proves that dot product is preserved.
Differential Geometry
134
(iii)
Next we prove that F is a linear transformation.
i.e.
F ( ap + bq ) = a F ( p ) + bF ( q )
Take
p = ( p1, p2 , p3 ) ∈¡3
We write
p = ( p1, p2 , p3 ) = ∑ pi U i
3
... (5)
i =1
Where U i is the natural frame field on ¡ 3 . Such that
U i ⋅U j = δ ij
... (6)
Since F preserves dot product, this implies that
F (U i ) ⋅ F ( U j ) = U i ⋅U j = δij
... (7)
⇒ F (U 1 ) , F (U 2 ) , F (U 3 ) are orthogonal unit frame.
Any vector F ( p ) ∈ ¡3 can be expressed in terms of orthogonal unit frames F (U i ) as
(
)
F ( p ) = ∑ F ( p) ⋅ F (U i ) ⋅ F (U i )
i
⇒ F ( p ) = ∑ ( p ⋅U i ) ⋅ F (U i )
by equation (4)
F ( ap + bq ) = ∑ ( ap + bq ) ⋅U i  ⋅ F (U i )
by (8)
i
... (8)
Now consider,
i
= ∑  ap ⋅ U i + bq ⋅U i  ⋅ F (U i )
i
F ( ap + bq ) = a ∑ ( p ⋅ U i ) ⋅ F (U i ) + b ∑ ( q ⋅ U i ) ⋅ F (U i )
i
i
F ( ap + bq ) = a F ( p ) + bF ( q )
... (9)
⇒ F is linear..
Thus we have shown that F is linear and
F ( p ) F ( q) = p ⋅ q
Hence F is Orthogonal transformation.
Differential Geometry
135
Theorem : Every isometry of ¡ 3 can be uniquely described as orthogonal transformations followed
by translation.
Solution : Let F be an isometry of ¡ 3 . If T is a translation by F ( 0 ) , then T –1 is also translation
by − F ( 0 ) . As translation is an isometry
⇒ T –1 is an isometry..
.... (1)
Since F is an isometry. Therefore the composite map T –1 F is also an isometry..
... (2)
Now we claim that T –1 F is an orthogonal transformation.
Therefore consider
T
−1
−1
F ( 0) = T F ( 0 )
= F (0) −F (0)
T
−1
Since T –1 is translation by − F ( 0 )
F ( 0) = 0
... (3)
Since T −1 F is an isometry and ( T −1 F ) ( 0 ) = 0 shows that T −1 F is an orthogonal transformation.
Let this orthogonal transformation be denoted by c .
i.e.
T
−1
F =c
⇒ T (T
−1
⇒ (T T
−1
F ) = Tc
) F = Tc
⇒ 1F = Tc
⇒ F = Tc
... (4)
⇒ An isometry of ¡ 3 is expressed as an orthogonal transformation followed by translation.
c is called the orthogonal part of F and T the translation part of F and Tc ≠ cT .
Uniquness : Suppose F can also be expressed as T ' c ' .
i.e.
F = T 'c '
... (5)
where T ' is translation and c ' is orthogonal transformation.
We prove that T = T ' and c = c ' .
Now from equation (4) and (5) we have
Tc = T ' c '
⇒ (T T ) c = T
−1
Differential Geometry
... (6)
−1
T 'c '
136
−1
⇒ c =T T ' c '
−1
⇒ c =T T ' c '
... (7)
Since c and c ' are linear transformation
⇒ c ( 0 ) = 0 and c ' ( 0 ) = 0
Thus from equation (7) we have
−1
c ( 0) = T T ' c '( 0)
⇒ 0 = (T T ') 0
−1
⇒ (T T ') 0 = 0
−1
... (8)
Since T −1 T ' is a translation and for some one point 0 ∈ ¡3
( T −1T ') ( 0 ) = 0
−1
⇒T T ' = I
⇒T =T '
Consequently from equation (6) we have
c =c '
Thus every isometry of ¡ 3 can be uniquely described as an orthogonal transformations followed by a
translation.
Worked examples .......................................................................................................................•
Example 3 : Decide whether F is an isometry of ¡ 3 . If so find its translation and orthogonal part.
(i) F ( p ) = − p
(ii) F ( p ) = ( p3 − 1, p2 − 2, p1 − 3)
(iii) F ( p ) = ( p1, p2 ,1)
Solution :
(i) Let F : ¡3 → ¡3 be a function defined by
F ( p) = −p
∀p ∈ ¡ 3
To prove F is an isometry of ¡ 3 , we simply prove that it preserves the distance.
Differential Geometry
137
... (1)
Consider
d ( F ( p ) ,F ( q ) ) = F ( p ) − F ( q )
= −p + q
... by (1)
= −( p − q)
= p −q
⇒ d ( F ( p ) , F ( q ) ) = d ( p, q )
∀p , q ∈ ¡ 3
... (2)
⇒ F ( p ) = − p is an isometry..
We know every isometry F of ¡ 3 is uniquely expressed as an orthogonal transformation followed
by translation.
i.e.
F = Tc
... (3)
where T is translation by a and c the orthogonal translation respectively given by
T ( p) = p + a
and
 c11

c ( p ) =  c21
c
 31
... (4)
c12
c22
c32
c13   p1 
 
c23   p2 
c33   p3 
... (5)
Thus from equation (3) we have
F ( p ) = Tc ( p )
F ( p) = c ( p ) + a
i.e.
 − p1   c11 c12

 
 − p2  =  c21 c22
 −p  c
 3   31 c32
... by (4)
c13  p1   a1 
   
c23  p2  +  a2 
  
c33 
 p3   a3 
 − p1   c11 p1 + c12 p2 + c13 p3 + a1 

 

⇒  − p2  =  c21 p1 + c22 p2 + c23 p3 + a2 
−p  c p + c p + c p + a 
 3   31 1 32 2 33 3 3 
 −1 0 0 


⇒ c =  0 −1 0 
and a = ( 0,0,0)
 0 0 − 1


Differential Geometry
138
... by (1) and (5)
 −1 0 0 


⇒ Translation is by zero vector and rotation part is c =  0 −1 0 
0
0 − 1

(ii)
Here F ( p ) = ( p3 − 1, p 2 − 2, p1 − 1)
... (6)
To prove F is an isometry, consider
d ( F ( p) , F ( p ) ) = F ( p ) − F ( q )
Using equation (6) we have
d ( F ( p) , F ( q ) ) = ( p3 − 1, p2 − 2, p1 − 1) − ( q3 − 1, q2 − 2, q1 −1)
=
=
( p3 − q3 , p2 − q2 ,
p1 − q1 )
( p3 − q3 )2 + ( p2 −q2 ) 2 + ( p1 − q1 )2
= p −q
d ( F ( p ) ,F ( q ) ) = d ( p , p )
⇒ F defined by equation (6) is an isometry of ¡ 3 .
Now from equation (3) we have
F ( p ) = Tc ( p )
= T ( c ( p ))
F ( p) = c ( p ) + a
Using (6) and (5) we write
 p3 − 1   c11 p1 + c12 p2 + c13 p3 + a1 

 

 p2 − 2  =  c21 p1 + c22 p2 + c23 p3 + a2 
 p −3  c p +c p + c p + a 
 1
  31 1 32 2 33 3 3 
 0 0 1


⇒ a = ( −1, −2, −3) and c =  0 1 0 
 1 0 0


Similarly one can obtain for F ( p ) = ( p1, p2 ,1)
Differential Geometry
139
Example 4 : If F = T ac where T a is a translation by a = (1,3, −1) and



c =



1
2
0
1
2
0 −
1
0
1 
2 
0 
1 

2 
If p = ( 2, −2,8) , find the coordinate of the point q for which
(i) F ( p ) = p
(ii) F
−1
(p ) =q
Solution : Given that F = T ac



where c = 



1
2
0
1
2
0 −
1
0
... (1)
1 
2 
0  is a matrix of transformation.
1 

2 
(i) T a is a translation by a given by a = (1,3, −1)
⇒ F ( p ) = (T ac ) ( p )
= T a ( c ( p) )
⇒ F ( p)= c ( p ) + a
... by definition of translation
⇒ q = c ( p) + a

q
 1 

⇒  q2  = 
q  
 3 


1
2
0
1
2
0 −
1
0
1 
2   2   1 
0   −2  +  3 
1   8   −1

2 
 q1   2 − 4 2 + 1 


⇒  q2  =  −2 + 3 
q  

 3   2 +4 2 + 1 
Differential Geometry
140
 q1  1 −3 2 


⇒  q2  =  1 
q  

 3   5 2 − 1
⇒ q = ( q1 , q2 , q3 ) = (1 − 3 2,1,5 2 − 1)
(b) Now F is defined by
F
−1
(p ) =q
Thus we have from equation (1)
( F ) −1 = ( T ac ) −1
= c −1Ta −1
⇒F
−1
−1
( p ) = c −1T a ( p )
= c −1 ( p − a)
= c −1 ( ( 2, −2,8) −(1,3, −1) )
= c − 1 (1, −5,9)



−1
⇒ F (p) = 



1
2
0
1
2
0 −
1
0
1 
2   1 
0   −5 
1   9 

2 
 q1   5 2 


⇒  q2  =  −5 
q  

 3  4 2
⇒ q = ( q1 , q2 , q3 ) = ( 5 2, −5,4 2 )
Differential Geometry
141
... (2)
EXERCISE ....................................................................................................•
1.
 2
− 3

2
If c = 
 3

 1
 3
2
3
1
3
2
3
1
− 
3

2
− and p = ( 3,1, −6 ) , q = (1,0,3) . Show that c is orthogonal, then
3

2 

3 
compute c ( p ) and c ( q ) and check c ( p ) ⋅c ( q ) = p ⋅ q .
2.
3.
In each case where F is an isometry in ¡ 3 . Find its translation and orthogonal transformation
part.
(i)
1
 1
F ( p) = 
( p1 − p3 + 1) , ( p2 + 1) , ( p1 + p3 + 1) 
2
 2

(ii)
F ( p ) = ( 2 p2 ,3 p1, − p 3 )
(iii)
F ( p ) = (1, p3 , p1 )
Show that an isometry F = T ac has an inverse mapping F −1 , which is also isometry. Find
translation and orthogonal part of F −1 .
•.........................................................................................................................•
❏❏❏
Differential Geometry
142
CHAPTER - III
CALCULUS ON A SURFACE
Unit I
Introduction :
To study calculus we need linear spaces. But like cylinder, sphere which are not linear and
hence calculus cannot be developed. To make calculus available on the sphere, cylinder etc. we need
coordinate patches. Coordinate patches are used to define surfaces. In this unit we study some properties
of patches that will be useful in studying surfaces. Some methods of constructing surfaces are discussed.
A surface in ¡ 3 is a subset of ¡ 3 . But not all subsets of ¡ 3 are surfaces. It is required that the surface
be smooth and two dimensional. We shall express the requirements in mathematical terms.
Preliminaries :
Regular Mappings :
Definition : A mapping F : ¡n → ¡ m is said to be regular if the Jacobian matrix of F has rank 'n',
where 'n' is the dimension of the domain of the function F.
If F : ¡n → ¡ m
⇒ F ( p ) = ( f1, f 2 ,...., f m ) ∈ ¡m
∀p = ( p1 , p2 ,...., pn ) ∈¡ n
Then the Jacobian matrix of F at p is given by
 ∂f1
 ∂x ( p )
 1
 ∂f1
 ∂x ( p )
 2
 L

 ∂f1 ( p )
 ∂xn
∂f 2
∂f m

( p) L
( p) 
∂x1
∂x1

∂f 2
∂f m

( p) L
( p) 
∂x2
∂x2

L
L
L 

∂f 2
∂f m
( p) L
( p) 

∂xn
∂xn
Rank of a matrix :
A matrix A is said to have rank r if and only if it contains a non zero determinant of order r and
every determinant of order r +1 is zero.
Differential Geometry
143
Coordinate Patch :
Definition : Let D be an open subset of ¡ 2 . Consider a mapping
F : D → ¡3
If X satisfies the conditions
(i)
X is one-one
(ii)
X is regular,
and
then X is called a coordinate patch.
z
v
X(D)
D
¡
E
X
2
2
O
O
u
x
3
E3
¡
y
Note :
(1)
The definition of coordinate patch resembles that of a regular curve, where the open interval I
is replaced by an open disc D and the regularity condition is replaced by Xu × Xv ≠ 0 .
Later we will prove that Xu × Xv ≠ 0 is the regularity condition for the surface. This makes
sense in geometry, because curve is one dimensional, since α '(t ) is a tangent to the curve α insisting
that α '( t ) ≠ 0 ⇒ {α '( t )} is linearly independent set.
A surface is two dimensional, therefore the regularity condition for surfaces should be that we
have two linearly independent vectors at each points.
(2)
Regularity condition gives smoothness and the condition one-one avoids cutting of the surface.
The image X(D) is a smooth 2-dimensional subset of ¡ 3 .
Proper Coordinate Patch :
Definition : A coordinate patch X : D → ¡3 is called as a proper patch if the inverse function
X − 1 : X ( D) → D is continuous.
Differential Geometry
144
The function X −1 is continuous means the coordinate functions of X −1 are continuous.
Remark : Geometrically, if D is considered to be a thin sheet of rubber we can get X(D) by bending
and stretching D or squeezed without being torn apart or glued together.
Worked examples ........................................................................................................................•
Example 1 : Let X : D → ¡3 be defined by
X (u , v) = ( u 2 , uv, v 2 )
where D = {( u, v )∈ ¡2 | u > 0, v > 0}
Is X a coordinate patch ? Is it proper ?
Solution : The mapping X : D → ¡3 is defined by
X (u , v) = ( u 2 , uv, v 2 )
(u , v) ∈ D
To prove X is a coordinate patch, we prove the following.
(i)
X is one-one
Therefore consider
X (u , v) = X ( u1 , v1 )
⇒ ( u 2 , uv, v 2 ) = ( u12 , u1v1, v12 )
⇒ u 2 = u12 , uv = u1v1, v 2 = v12
⇒ u = ±u1 , v = ±v1
⇒ u = u1 ,
v = v1
as u > 0, v > 0
X is one-one.
(ii)
Since
X is regular.
X = ( x1 , x2 , x 3 ) = ( u 2 , uv, v 2 )
⇒ x1 = u 2 , x2 = uv , x3 = v2
 ∂x1
 ∂u

Therefore the Jacobian matrix =  ∂x1

 ∂v
Differential Geometry
∂x2
∂u
∂x2
∂v
∂x3 
∂u  =  2u v 0 
∂x3   0 u 2v 

∂v 
145
. . . (1)
2u v
2
We see that 0 u = 2u ≠ 0
⇒ The rank of Jacobian matrix is 2.
⇒ X is regular. This proves X is a coordinate patch.
Now to prove x is a proper, we prove the inverse function X −1 is continuous.
Define a function
Y : X ( D) → D
by
Y :( x , y, z ) = ( x , z )
where
. . . (2)
y = xz
We see that
X oY (x , y , z ) = X
(
x, z )
X oY (x , y , z ) = ( x, y , z )
. . . (3)
Also
Y o X ( u, v ) = Y ( u2 , uv, v2 )
Y o X ( u, v ) = ( u, v )
.......... by (2)
. . . (4)
From equation (3) and (4) we have
Y o X is an identity map.
⇒ Y = X −1
Since the coordinate functions
x,
z of X −1 are continuous, it follows that X −1 is continuous.
This proves that X is a proper patch.
•.......................................................................................................................................................•
Surface :
Definition : A surface in ¡ 3 is a subset M of ¡ 3 such that for each point P of M there exist a proper
patch in M whose image contains a neighborhood of P in M.
Differential Geometry
146
X
P
X(D)
M
D
O
O
i.e. A surface is a set of points in space which resembles a portion of a plane in the neighborhood of its
points.
Note that neighborhood of P in M is a set of all points of M whose Euclidean distance from P is less
than ∈> 0 .
If NP is a neighborhood of P then
N P = {q ∈ M | p −q <∈}
Worked examples ........................................................................................................................•
Example 2 : Compute a proper patch in a unit sphere ∑ covering a neighborhood of the north pole
(0, 0, 1).
Solution : ∑ is a sphere which consists of all points at a unit distance from the origin.
⇒ ∑ = {( x , y , z ) ∈ ¡ 3 | x 2 + y2 + z 2 = 1}
We compute a proper patch in ∑ covering the neighborhood of the north pole (0, 0, 1).
z
N(0,0)
X(u,v)=(u,v, 1 - u2- v2 )
X(D)
U
D
22
¡
E
(q , q , q )
1 2 3
O(0, 0) ( q , q )
1 2
1 - u2- v2
y
u
v
x
Differential Geometry
147
(u,v)
For each point ( q1, q2 , q3 ) of the northern hemisphere of ∑ , consider its projection ( q2 , q2 ,0 ) in the
xy-plane. Therefore we get a function
f :∑ → D
defined by
f ( q1 , q2 , q3 ) = ( q1, q2 , 0 )
Clearly this function is one-one and onto, where is northern hemisphere and D is a unit disc in xy-plane
of radius 1 and centered at origin. If we identify this plane ¡ 2 by means of natural association
( q1, q2 , 0 ) = ( q1, q2 ) , then D becomes the disc in
¡ 2 consisting of all points (u, v) such that
u2 + v2 < 1.
i.e. the disc D = {( u, v ) ∈ ¡2 | u 2 + v 2 < 1} .
Define the inverse function
(
X : D → ¡3 by X (u , v) = u , v , 1− u 2 − v2
)
We claim that X is a proper coordinate patch.
(i)
X is one-one :
Consider
X (u , v) = X ( u1 , v1 )
) (
(
⇒ u , v , 1 − u 2 − v 2 = u1, v1, 1 − u12 − v12
)
⇒ u = u1 and v = v1
⇒ X is one-one.
(ii)
X is regular :
We show that the rank of the Jacobian of the transformation is 2. The Jacobian of
X (u , v) = ( x1, x2 , x3 ) is given by
 ∂x1

J =  ∂u
 ∂x1

 ∂v
∂x2
∂u
∂x2
∂v
∂x3 
∂u 
∂x3 

∂v 
where x1 = u , x2 = v , x3 = 1− u 2 − v 2
Differential Geometry
148
∂x3 

1 0 ∂u 
J =

 0 1 ∂x3 



∂v 
This shows that the rank of J is 2.
⇒ The mapping X is regular..
(iii) X −1 is continuous :
Now the mapping X − 1 : X ( D) → D , defined by
X − 1 ( x , y , z ) = ( x , y ) where z = 1 − x 2 − y 2
is clearly continuous. Since co-ordinate functions are continuous.
⇒ X is proper patch. Also X (0, 0) = (0, 0, 1). Hence X covers the neighborhood of (0, 0, 1).
Note : The proper patch X(D) in the above example contains the neighbourhood of each point of the
northern hemisphere. Similarly we can find proper patches covering the points (0, 0, –1), (0, 1, 0),
(0, –1, 0), (1, 0, 0) and (–1, 0, 0). Hence neighbourhood of every point is covered by a proper patch
which shows that spheres in ¡ 3 are surfaces.
•.....................................................................................................................................................•
Counter Examples :
1.
Cone M : z 2 = x 2 + y 2 :
At the vertex of the cone, it is impossible to find a proper patch that
will cover a neighborhood of p of M. Hence is not a surface.
2.
Closed disc :
At all points on the circumference of a circle, it is impossible to find a
proper patch that will cover a neighborhood of p of M. Hence closed
disc
p
Differential Geometry
M : x 2 + y 2 ≤ 1 , z = 0 is not a surface.
149
3.
Folded Planes M : xy = 0 , x ≥ 0 , y ≥ 0 :
p
At all points of the intersection of two planes, it is impossible to find a
proper patch that will cover a neighborhood of p of M. Hence folded
planes is not a surface.
Remark : Not all subsets of ¡ 3 are surfaces.
Monge Patch :
Definition : If f is a real valued differentiable function on an open set D in ¡ 2 , then the function
X : D → ¡3 such that
X (u , v) = ( u , v, f (u , v ) )
is a proper patch called as Monge Patch.
A surface M is called a simple surface if there is a Monge Patch X : D → ¡3 such that X(D) = M.
Example : The surface ∑ N , northern hemisphere in above example is a simple surface for,,
X ( D) = ∑ N
Note : Any surface in ¡ 3 can be constructed by gluing together simple surfaces.
Example : Every differentiable function real valued function f on ¡ 3 determines a surface in ¡ 3 .
M = {( x, y, z ) / z = f (x ,y )}
Then M is the image of a Monge Patch
X : ¡2 → M
X (u , v) = ( u , v, f (u , v ) )
Then X is a proper patch with X ( ¡2 ) = M .
∴ M : z = f (x , y ) is a simple surface.
Differential Geometry
150
Note : (1) If g is any real valued function on ¡ 3 and c is number then the set M = { p ∈ ¡3 / g( p) = c}
is subset of ¡ 3 . The following theorem gives a condition when such a subset is surface of ¡ 3 .
(2)
Spheres are not simple surfaces. But saddle surface defined by X (u , v) = ( u, v, uv) ,
(u , v) ∈ ¡ 2 is a simple surface.
Theorem : Let g be a differentiable real valued function on ¡ 3 and c be a number. The subset
M = {( x , y , z ) / g( x, y, z) = c} of ¡ 3 is a surface if the differential dg ≠ 0 at any point of M.
Proof : Given that
g (x, y, z) = c and dg ≠ 0
⇒ dg =
∂g
∂g
∂g
dx +
dy + dz ≠ 0 on M.
∂x
∂y
∂z
⇒ At least one of the partial derivatives is not zero at any point P of M.
Let
 ∂g 
  ≠ 0.P = ( p1, p2 , p3 )
 ∂z  P
.
Then by implicit function theorem, the equation g (x, y, z) = c can be solved for z. Therefore there exist
a differentiable real valued function h defined on the neighborhood D of ( p1 , p2 ) such that,
(i)
For each point ( u , v ) ∈ D the point (u, v, h (u, v)) lies in M.
i.e. g (u, v, h (u, v)) = c.
(ii)
The set {( u , v , h( u, v) ) /( u, v) ∈ D} is a neighborhood of the point P in M.
Z
X (D)
M
P
Y
O
X
Differential Geometry
D
(P,1 P,2 0) = (P,
P)
1
2
151
Thus a function X : D → ¡3 defined by X (u, v) = (u, v, h (u, v)) is a Monge patch.
Since P was an arbitrary point of M, for every point P∈ M, there exist a proper patch in M whose
image contains a neighborhood of P in M.
Hence M is a surface.
Worked examples .......................................................................................................................•
Example 6 : Show that the sphere in ¡ 3 are surfaces.
Solution : Let ∑ be a sphere in ¡ 3 with center c = ( c1 , c2 , c3 ) and radius r. Then
∑=
{( x , x , x ) / ( x − c )
2
1
2
3
1
1
+ ( x2 − c2 ) + ( x3 − c3 ) = r 2
2
3
⇒∑: g =r
2
2
}
where g = ∑ ( xi − ci )
2
i =1
3
Then dg = ∑ 2 ( xi − c i ) dx i
i =1
3
If dg = 0 then
or
∑ 2 ( xi − ci ) = 0
i =1
which implies
xi − ci = 0
i = 1, 2, 3
xi = ci
i = 1, 2, 3
But then ( x1, x2 , x3 ) = ( c1, c2, c3 ) which is center of the sphere. Hence dg ≠ 0 for any point on ∑ .
Therefore by theorem ∑ is a surface.
Example 7 : Show that cylinders in ¡ 3 are surfaces.
Solution : As a line perpendicular to a plane P moves along a curve C it sweeps out a cylinder.
In particular, let P be an XY-plane and the line L is always parallel to the Z-axis and let C be any curve
in the plane.
The curve in XY-plane is given by
C : f (x, y) = c
Let f% be a function on ¡ 3 such that f% ( p1, p2 , p3 ) = f ( p1, p 2 ) .
Differential Geometry
152
Then the cylinder generated by the curve C is given by
M : f% (x , y , z) = c ,
c is some constant
Now,
∂f%
∂f%
∂f%
df% = dx + dy + dz
∂x
∂y
∂z
But
f% (x , y , z ) = f (x , y )
⇒
∂f% ∂f ∂%f ∂ f ∂ f% ∂ f
= ,
= ,
=
=0
∂x ∂x ∂y ∂y ∂z ∂z
Hence
∂f
∂f
df% =
dx + dy = df
∂x
∂y
Since C : f = c is a curve,
∂f
∂f
and
are not zero simultaneously on C.
∂y
∂x
Hence df ≠ 0 on C.
Therefore df% ( p1 , p2 , p3 ) = df ( p1, p 2 ) ≠ 0 for any point P ∈ M
Thus M is surface.
•....................................................................................................................................................•
Note :
(i)
When C = x 2 + y2 = r 2 is a circle in XY-plane then M = x 2 + y 2 = r 2 is a right circular
cylinder in ¡ 3 .
(ii)
Cylinders are the surfaces obtained by translating a line along the curve.
Now we will find surface by rotating a curve about a given axis.
Surface of Rotation :
Definition : A surface traced by a plane curve when it revolves about a fixed line without touching it is
called a surface of revolution M in ¡ 3 .
Note :
A curve which rotates about a line is called profile curve or generator guiding curve and the fixed line
is called the axis of revolution.
Differential Geometry
153
Remarks :
(i)
In the case of sphere, the profile curve is a semi circle, however it meets the axis of revolution
into two points (south and north pole). Hence it is not a surface of revolution. (such surface is
called augmented surface of revolution)
(ii)
In the case of torus, profile curve is a circle which does not meet the axis. Hence torus is a
surface of revolution.
(iii)
Let M be a surface of revolution. The circles generated by each point of the curve are called
parallels of M.
z
Singular point
Parallels
(Latitudes)
y
Guiding Curve
x
Greate circles or
meridians
(Longitudes)
Singular point
(4)
Meridians : Different positions of the profile curve as it is rotated are called as the meridians of
the surface M.
(5)
The points of the surface which lie on the axis are called singular points. At singular points
parallel degenerates into a point.
(6)
Parallels and meridians on cone :
z
Parallels
Meridians
y
x
Note : Cone is not a surface of revolution.
Differential Geometry
154
Worked examples .........................................................................................................................•
Example 8 : Show that a surface obtained by rotating a curve is a surface. (surface of revolution)
Solution : Let C be a curve in a plane P and let A be a line in the plane which does not meet C. If this
profile curve C is revolved around the axis A, it sweeps out a surface called as surface of revolution M
in ¡ 3 .
Y
C
Q
A
Pv
O
X
Z
Let A be the X-axis and let C be the curve in XY plane. Let Q = ( q1 , q2 , 0) be a point on C. Then Q
describes a circle. This circle gives is of radius q 2 and is in a plane parallel to YZ-plane .
Hence its equation is given by
y 2 + z 2 = q22
If P is any point on the circle then it has coordinates
( q1, q2 cos v, q2 sin v ) ,
0 ≤ v ≤ 2π
On the other hand if P = ( p1, p2 , p3 ) is a point of the surface then corresponding point on the curve
is given by
(
P = p1, p22 + p32 , 0
)
Thus if C : f (x, y) = c is a profile curve then the surface of revolution generated by C is given by
(
M : g (x, y, z) = c, where g ( x , y , z ) = f x ,
y2 + z 2
Now
(
g ( x , y, z) = f x ,
⇒ dg =
Differential Geometry
y2 + z 2
)
∂g
∂g
∂g
dx +
dy +
dz
∂x
∂y
∂z
155
)
=
∂f
∂f
∂f
dx + dy + dz
∂x
∂y
∂z
≠0
Since f is not a constant,
∂f ∂f ∂f
, ,
are not zero simultaneously on M.
∂x ∂ y ∂ z
Therefore M : g = c is a surface.
EXERCISE .....................................................................................................•
1.
If y 2 = 4 x is a curve in XY place, obtain a surface of revolution. (Paraboloid)
2.
If
x2 y 2
+
= 1 is revolved about X-axis, find the surface of revolution (Ellipsoid).
a2 b2
•........................................................................................................................•
Example 9 : Show that a rectangular strip bent into figure - 8 is not a surface.
Solution : Consider a rectangular strip D and is bend to form a figure of 8. Then we show that the
configuration M of ¡ 3 is not a surface.
v
z-axis
A
D
−π
O
π
u
We show that the formation of this figure - 8 is not a surface.
Let D be a rectangle defined by −π < u < π , 0 ≤ v ≤ 1 in ¡ 2 .
Define X : D → ¡3 by X (u, v) = (sin u, sin 2u, v)
We show that X is a co-ordinate patch.
X is one-one. For,
X (u , v) = X ( u1 , v1 )
Differential Geometry
156
⇒ ( sin u,sin2u , v ) = ( sin u1 ,sin2u1, v1 )
⇒ v = v1 , sin u = sin u1 , sin2u = sin2u1
⇒ u = u1 , v = v1
⇒ ( u, v) = ( u1 , v1 )
Next, the Jacobian matrix of X,
 ∂x1

J =  ∂u
 ∂x1
 ∂v
∂x2
∂u
∂x2
∂v
∂x3 
∂u  =  cos u

∂x3   0
∂v 
2cos2u 0 
0
1 
where x1 = sin u , x2 = sin2u , x3 = v
Therefore rank of J = 2 = dim of D. This X is a regular. And hence X is a co-ordinate patch.
But X is not a proper patch, since X −1 : M → D is not continuous through the axis A (Z-axis).
X −1( x , y, z ) = ( sin −1 x, z )
For,
1

=  sin −1 y, z 
2

1
π
π
lim sin −1 y = − ,0, and is not unique.
y→0 2
2
2
⇒ X −1 is not a continuous on −π < x < π .
Therefore the surface of figure - 8 is not a surface.
Example 10 : Which of the following mapping X : ¡ 2 → ¡3 is a patch ?
(i)
X ( u, v ) = ( u , u 2 , v + v 3 )
(ii)
X ( u, v ) = ( cos2uπ ,sin2uπ , v )
Solution :
(i)
We know that X is a patch if X is one-one and regular.
Consider
X ( u, v ) = X ( u1 , v1 )
Differential Geometry
157
⇒ ( u , u 2 , v + v 3 ) = ( u1 , u12 , v1 + v13 )
⇒ u = u1 and
v + v3 = v 1 + v13
v + v3 = v 1 + v13
Now
⇒ ( v − v1 ) + v3 − v13 = 0
⇒ ( v − v1 ) ⋅ (v 2 − vv1 + v12 + 1) = 0
If v ≠ v1 then ( v 2 − vv1 + v12 + 1) = 0
But this is a quadratic in v (or v1 ) and therefore
v=
−v1 ± −3v12 − 4
2
which is not real number. Hence v = v1
X ( u, v ) = X ( u1 , v1 ) ⇒ ( u , v ) = ( u1 , v1 )
Thus
which shows that X is one-one.
Next
 ∂x1

J =  ∂u
 ∂x1
 ∂v
∂x2
∂u
∂x2
∂v
∂x3 
∂u 

∂x3 
∂v 
where x1 = u , x2 = u 2 , x3 = v + v3
Therefore
1 2u 0

J =
2
0 0 1 + 3 y 
Therefore rank of J = 2 = dim of ¡ 2 .
Hence X is regular.
Thus X is a patch.
(ii)
Consider
X ( u, v ) = X ( u1 , v1 )
⇒ ( cos2uπ ,sin2uπ , v ) = ( cos2u1π ,sin2u1π , v1 )
Differential Geometry
158
⇒ v = v1 but u need not be equal to u1 .
Hence X is not one-one.
Therefore X is not a patch.
Example 11 : Prove that M : ( x 2 + y 2 ) + 3z 2 = 1 is a surface.
2
Solution : Let g = ( x 2 + y 2 ) + 3z 2
2
Therefore M : g = c and
⇒ dg =
∂g
∂g
∂g
dx +
dy +
dz
∂x
∂y
∂z
= 2 ( x 2 + y 2 ) (2 x) dx + 2 ( x2 + y 2 ) (2 y ) dy + 4 zdz
∴ dg = 0 ⇒ 4 x ( x 2 + y 2 ) dx + 4 y ( x2 + y 2 ) dy + 4 zdz = 0
⇒ 4 x ( x2 + y 2 ) = 0 , 4 y ( x 2 + y 2 ) = 0 , 4 z = 0
⇒ x = 0, y = 0, z = 0
But ( 0,0,0 ) ∉ M , hence dg ≠ 0 on M. Thus M : g = c is a surface.
Example 12 : For what values of c is M : z (z – 2) + xy = c is a surface ?
Solution : Let g = z ( z − 2) + xy = z 2 + xy − 2 z
Then
dg =
∂g
∂g
∂g
dx +
dy +
dz
∂x
∂y
∂z
= ydx + xdy + (2 z − 2) dz
Therefore, dg = 0 if x = 0, y = 0, z = 0
But at (0, 0, 1), c = 0 (From the equation of surface)
Thus for c = 0, (0, 0, 1) lies in the surface and dg = 0 at this point. Hence M is not a surface when
c = 0. i.e. if c ≠ 0 then M : z (z – 2) + xy = c is a surface.
Differential Geometry
159
Example 13 : Let X : ¡ 2 → ¡3 be the mapping defined by X (u, v) = (u + v, u – v, uv). Show that
X is a proper patch and that the image of X is the surface z =
Solution : X (u, v) = (u + v, u – v, uv)
If
X ( u, v ) = X ( u1 , v1 )
⇒ ( u + v, u − v, uv ) = ( u1 + v1 , u1 − v1 , u1v1 )
⇒ u + v = u1 + v1
u − v = u1 − v1
uv = u1v1
⇒ u = u1 , v = v1
This shows that X is one-one.
Next
 ∂x1

J =  ∂u
 ∂x1
 ∂v
∂x2
∂u
∂x2
∂v
∂x3 
∂u 

∂x3 
∂v 
1 1 v 
∴J = 

1 −1 u 
Hence rank of J = 2 = dim of ¡ 2 .
Therefore X is regular.
Now
X (u, v) = (x, y, z)
⇒ (u + v, u v, uv) = (x, y, z)
⇒ x = u + v,
⇒u=
y = u – v,
z = uv
x+ y
x− y
, v=
2
2
Therefore
 x + y x−
X
,
 2
2
Differential Geometry
y
 = ( x , y, z )

160
x2 − y2
.
4
 x+ y x− y 
⇒ X − 1( x , y , z ) = 
,

 2
2 
Hence X −1 : ¡3 → ¡2 is continuous.
Thus X is one-one, regular and X −1 is continuous mapping. Therefore X is a proper patch.
If ( x , y , z ) ∈ M is arbitrary point then
X (u, v) = (x, y, z)
⇒ (u + v, u – v, uv) = (x, y, z)
⇒ x 2 − y 2 = (u + v)2 − ( u − v )2 = 4uv = 4 z
or
z=
x2 − y2
4
This shows that every point of image of
i.e.
Image of X = M : z =
X ( ¡ 2 ) lies in the surface
x2 − y2
M :z=
4
x2 − y2
4
This surface M is hyperbolic paraboloid in ¡ 3 .
EXERCISE ..............................................................................................................................•
1.
Let X : ¡ 2 → ¡3 defined by
X ( u, v ) = ( u 2 , v 2 , uv )
show that X is not a coordinate patch.
However, X : D → ¡3 defined by
X ( u, v ) = ( u 2 , v 2 , uv )
is a patch where D = {( u, v )∈ ¡2 | u > 0, v > 0} .
2.
Prove that M : z =
Differential Geometry
x2 − y2
is a surface.
4
161
3.
Prove or disprove that the mapping
X : ¡ 2 → ¡3
defined by
X ( u, v ) = ( u , uv, v )
is a patch. Is it a coordinate patch ? Justify.
4.
x2 y2
Let M : z = 2 − 2 and let X : ¡ 2 → ¡3 defined by
a
b
X ( u, v ) = ( a( u + v ), b(u − v),4uv )
Show that X is a proper patch covering all of M.
•.........................................................................................................................................................•
Differential Geometry
162
Unit II : Patch Computation
Introduction :
Coordinate patches are used to define a surface. Now we consider some properties of patches
which are useful in studying properties of surfaces.
Definition : Let X : D → ¡3 be a coordinate patch. For each point ( u0 , v0 ) in D the curve
α : u → ( u, v0 ) is called the u-parameter curve, denoted by v = v0 of X. Similarly the curve
β : v → ( u 0 , v ) is called the v-parameter curve u = u0 .
v
u = u0
X(D)
v0
(u0 , v0)
v = v0
D
O
v = v0
u-parameter
curve
u
u0
E
u = u0 v-parameter
curve
3
2
E
Thus the region X (D) is covered by these families of curves u = u0 and v = v0 which are images of
horizontal and vertical lines in D.
Definition : Let X : D → ¡3 be a coordinate patch. For each point ( u0 , v0 ) in D
(i)
The velocity vector at u0 of u-parameter curve v = v0 is denoted by Xu ( u0 , v0 )
(ii)
The velocity vector at v0 of the v-parameter curve u = u0 is denoted by Xv ( u0 , v0 )
The vectors Xu ( u0 , v0 ) and Xv ( u0 , v0 ) are called partial velocities of X at ( u0 , v0 )
u=u
0
Xv
Xu
X (u0 , v0)
Differential Geometry
163
v=v
0
Note : Xu and Xv are tangent vectors to ¡ 3 at a point X ( u0 , v0 ) .
If
X ( u, v ) = ( x1( u, v), x2 (u , v ), x3 (u , v) )
then
 ∂x ∂x ∂x 
 ∂x ∂x ∂x 
X u =  1 , 2 , 3  and X v =  1 , 2 , 3 
 ∂u ∂ u ∂ u 
 ∂v ∂ v ∂ v 
Result : A mapping X : D → ¡3 is regular iff Xu × Xv ≠ 0 , ∀(u , v) ∈ D .
Proof : Let a mapping X : D → ¡3 defined by
X ( u, v ) = ( x1, x2 , x3 ) ∈ ¡3
be regular.
⇒ the rank of the Jacobian matrix is 2.
 ∂x1
 ∂u

i.e. the rank of the matrix  ∂x1
 ∂v
∂x2
∂u
∂x2
∂v
∂x3 
∂u 

∂x3  is 2.
∂v 
Claim : we prove that X u × X v ≠ 0
Let us assume that X u × X v = 0
⇒
U1
U2
∂x1
∂u
∂x1
∂v
∂x2
∂u
∂x2
∂v
∂x2
⇒ U 1 ∂u
∂x2
∂v
Differential Geometry
U3
∂x3
=0
∂u
∂x3
∂v
∂x3
∂u − U
2
∂x3
∂v
∂x1
∂u
∂x1
∂v
∂x3
∂u + U
3
∂x3
∂v
∂x1
∂u
∂x1
∂v
164
∂x2
∂u = 0
∂x2
∂v
. . . (1)
⇒ Each coefficient determinant is zero separately. Consequently, this implies that the rank of the
 ∂x1
 ∂u

matrix  ∂x1
 ∂v
∂x2
∂u
∂x2
∂v
∂x3 
∂u 

∂x3  is not 2.
∂v 
This contradicts that the mapping X is regular.
⇒ Our assumption is wrong.
⇒ Xu × Xv ≠ 0
Conversely, let us assume that X u × X v ≠ 0 .
Now we claim that the mapping
X : D → ¡3
is regular.
Since X u × X v ≠ 0
∂xi
∂u
⇒ at least one of the determinants ∂xi
∂v
∂x j
∂u
∂x j in equation (1) is not zero
∂v
for i, j = 1, 2, 3; i ≠ j .
⇒ rank of the Jacobian matrix is 2.
⇒ The patch X (u, v) is regular..
Worked examples .........................................................................................................................•
Example 2 : For a patch X : D → ¡3 , if E = X u ⋅ X u , F = X u ⋅ X v and G = X v ⋅ X v then prove
that X is regular iff EG − F 2 ≠ 0 .
Solution :Let X : D → ¡3 be a patch.
Since
E = Xu ⋅ Xu = Xu
2
G = X v ⋅ Xv = X v
2
F = X u ⋅ X v = X u X v cos θ
where θ is the angle between Xu and Xv .
Differential Geometry
165
Then we get
2
EG − F 2 = X u
2
X v sin2 θ
. . . (1)
Now
Xu × Xv
2
= ( Xu × Xv )⋅ ( X u × X v )
= ( X u ⋅ X v ⋅ sin θ ⋅ uˆ )( X u ⋅ X v ⋅ sin θ ⋅ uˆ )
where û is the unit vector perpendicular to both Xu and Xv .
⇒ X u × Xv
2
= Xu
2
2
X v sin2 θ
. . . (2)
From equation (1) and (2) we have
EG − F 2 = X u × X v
Now
2
. . . (3)
X u × X v ≠ 0 ⇔ EG − F 2 ≠ 0
⇒ X is regular iff EG − F 2 ≠ 0 .
Example 3 : Let X : ¡ 2 → ¡3 defined by
X ( u, v ) = ( u 2 , v 2 , uv )
Show that X is not a coordinate patch. However, X : D → ¡3 , defined by
X ( u, v ) = ( u 2 , v 2 , uv )
is a patch, where D = {( u, v )∈ ¡2 | u > 0, v > 0}
Solution :
X ( u, v ) = ( u 2 , v 2 , uv )
⇒ X u = ( 2u,0, v )
and
X v = ( 0,2v, u )
Consider
U1
X u × X v = 2u
0
Differential Geometry
U2 U 3
0
v
2v u
166
X u × X v = −2v 2U 1 − 2u 2U 2 + 4uvU 3
∴ X u × X v = 0 ⇒ u = 0, v = 0
Thus at (0,0) ∈ ¡ 2 , Xu × Xv = 0
Hence X is not regular and hence it is not a coordinate patch.
Note that the function X ( u, v ) = ( u 2 , v 2 , uv ) is not one-one.
For
X ( u, v ) = X ( u1 , v1 )
⇒ ( u 2 , v 2 , uv ) = ( u12 , v12 , u1v1 )
⇒ u 2 = u12 , v 2 = v12
⇒ u 2 − u12 = 0 , v 2 − v12 = 0
⇒ ( u − u1 )(u + u1 ) = 0 ,
( v − v1 )(v + v1 ) = 0
⇒ u = ±u1 , v = ±v1
Hence X is not one-one.
If X is defined on D then (0,0) ∉ D . Then Xu × Xv ≠ 0 .
X is regular and also X is one-one.
Hence X is a coordinate patch.
Parametrization of the Region X(D) in M :
Definition : A regular mapping X : D → ¡3 whose image lies in a surface M is called a parametrization
of X(D) in M. A parametrization which is one-one is a patch.
The Geographical patch in the sphere :
Example : Describe geographical patch X : D → ¡3 in the sphere ∑ . Show that X is a parametrization
of X(D) in ∑ . where ∑ is a sphere of radius r with centre at the origin.
Solution : Let ∑ be a sphere of radius r > 0 and centered at origin of ¡ 3 .
Let
{
D = ( u, v) | −π < u < π , −
Differential Geometry
π
π
<v<
2
2
167
}
Define X : D → ¡3 by
X ( u, v ) = (r cosu cos v , r sin u cos v , r sin v )
Then the image X(D) is the entire sphere ∑ excepts one semicircle from north pole to south pole
through - ve X-axis.
Z
π/2
X(u , v )
0
v=v
0
0
−π
0
π
v
u r
− π/2
u=u
Y
0
X
The u-parameter curve v = v0 is a circle (parallel of latitude v0 ) and the v-parameter curve u = u0 is
a semicircle (meridian of longitude u0 ).
To prove that X is a parametrization of X(D) in ∑ , we prove that X is regular..
We have,
X ( u, v ) = (r cosu cos v , r sin u cos v , r sin v )
Therefore
X u = ( −r cosv sin u , r cos v cos u, 0 )
and
X v = ( −r sin v cos u, −r sin v sin u, r cos u )
Consider
U1
U2
U3
X u × X v = − r cos v sin u r cos v cos u
0
−r sin v cos u −r sin v sin u r cos u
X u × X v = r 2 {( cos 2 v cos u ) U 1 − ( − cos 2 v sin u ) U 2 +
+ ( sin 2 u cos v sin v + cos 2 u sin v cos v ) U 3}
Thus
X u × X v = r 2 {cos u cos 2 vU 1 + sin u cos 2 vU 2 + sin v cos vU 3}
≠0
Differential Geometry
168
π
π
Since sin u and cos u are never zero simultaneously and − < v <
implies that cos v ≠ 0 .
2
2
Hence X is a regular mapping. This proves that X is parametrization of spherical surface.
To show that X is a patch, we prove that X is one-one.
Now
X ( u, v ) = X ( u1 , v1 )
⇒ r ( cos v cos u,cos v sin u ,sin v ) = r ( cos v1 cos u1,cosv1 sin u1,sin v1 )
⇒ cos v cos u = cos v1 cos u1
cos v sin u = cos v1 sin u1
sin v = sin v1
π
π
π
π
⇒ v = v1 for − < v <
and − < v1 <
2
2
2
2
and hence u = u1
⇒ ( u , v ) = ( u1 , v1 )
Hence X is one-one.
Therefore X is a patch in ∑ . This patch X is called Geographical patch in ∑ .
Note : Because of this Geographical patch, the map of the earth can be obtained as a plane rectangle.
Parametrization of Cylinder :
Example : Let M be a cylinder over a curve C : f (x, y) = a in xy-plane. If a = ( a1 , a2 ,0 ) is a
parametrization of C then show that the mapping X : D → ¡3 defined by
X ( u, v ) = ( a1 ( u), a2 (u ),v )
where D = I × R , I is an interval on which is defined, is a parametrization of M.
Solution : Consider the mapping X : D → ¡3 defined by
X ( u, v ) = ( a1 ( u), a2 (u ),v )
Differential Geometry
. . . (1)
169
v-parameter curve
(rulings) or elements
I ×R
(u,v)
X
o
o
X(u,v)
u
I
C
u-parameter curve
(translates of C or
cross sectional curves)
v
α
where D is the strip given by
D = {(u , v) | u ∈ I and v is arbitrary}
Now we prove the function X : I × ¡ → ¡3 is regular..
Therefore at each point ( u , v ) ∈ D we find
 da da

X u =  1 , 2 ,0 
 du du 
Xv = (0 , 0, 1)
U1
da
∴ X u × Xv = 1
du
0
= U1
U2
da2
du
0
U3
0
1
da2
da
−U 2 1
du
du
≠0
Since x is a regular curve.
The u - parameter curves are translates of C and are called cross sectional curves of the cylinder. The
v - parameter curves are straight lines called the rulings of the cylinder.
If the curve C is not closed then X is one to one and hence X is a patch. But if C is closed, X wraps D
number of times around C.
Differential Geometry
170
Parametrization of surface of revolution :
Example : Find the parametrization of surface of revolution obtained by revolving a curve C which is
in the upper half of the xy-plane about x axis.
Solution : Suppose that a surface is obtained by revolving a curve C in the upper half of xy-plane
about x axis.
Let α : I → ¡3 be a curve whose parametrization is α (u ) = ( g (u ), h( u),0 ) .
u
Y
(u,v)
α:I→E
, 0)
h (u)
,
)
u
( g(
3
v
u
O
I
u
X
O
D
X(u,v)
Z
When the curve is rotated the point (g(u), h(u), 0) on the profile curve C has been rotated through an
angle v and the point on the curve reaches to X (u, v) with same X coordinate but new Y and Z
coordinates and is given by
0
 x  1
  
X (u , v) =  y  =  0 cos v
 z   0 sin v
  
i.e.
 g 
 
− sin v   h 
cos v   0 
0
X (u , v) = ( x , y , z ) = ( g (u ), h(u )cos v, h( u)sin v )
Y
X(u,v)
h(u) h(u)
v v h(u) cos v = y
h(u) sinv = z Z
Differential Geometry
171
Clearly X defines a mapping from ( u , v ) ∈ D onto the surface of revolution M.
i.e. X : D → M
X (u , v) = ( x , y , z ) = ( g (u ), h(u )cos v, h( u)sin v )
Now
X u = ( g '(u), h '( u)cos v , h '( v)sin v )
X v = ( 0, −h(u )sin v , h( u )cos v )
 U1
U2
U3 


X u × X v =  g '(u ) h '(u )cos v h '(u )sin v 
 0
−h (u )sin v h (u )cos v 
= U 1  h( u) h'(u ) ( cos 2 v + sin 2 v ) − U 2 [ g '(u )h (u )cos v ]
+ U 3 [ − g '(u )h (u )sin v ]
= h( u) h '(u )U 1 − g '(u )h (u )cos vU 2 − g '(u )h (u )sin vU 3
≠0
Since g (u) and h (u) are not constant simultaneously and sinv and cosv are not zero simultaneously.
This implies X is regular. Hence x is a parametrization of surface of revolution.
Torus of Revolution :
Example : Find a parametrization of the torus of revolution obtained on revolving the circle of radius
r with centered at (R, 0, 0) in the xz-plane about z axis if R > r.
Solution : It is given that the profile curve is the circle in the xz plane with radius r > 0 and centered at
(R, 0, 0), where R > 0.
z
z
(x, z)
r
o
u
(R,0, 0)
x
y
(h, g)
(x - R)2 + z2 = r 2
x
Differential Geometry
172
u
The natural parametrization of the profile curve C in the xz-plane is given by
α (u ) = ( R + r cos u,0, rsin u)
. . . (1)
Let (h(u), 0, g(u)) = (R + cos u, r sin u) be any point on the profile curve C. Let the point (h, 0, g) on
the profile curve be rotated through an angle v about z axis and let it be reached to a point
x (u, v) = (x, y, z) in space, we get the surface
X :D → M
defined by
 x   cos v − sin v 0   h(u ) 
  


X (u , v) =  y  =  sin v cos v 0   0 
z  0
0
1   g (u ) 
  
. . . (2)
⇒ X (u , v) = ( x , y , z ) = ( h (u )cos v , h( u )sin v, g( u) )
⇒ X (u , v) = ( x , y , z ) = ( ( R + r cos u ) cos v, ( R + r cos u ) sin v, r sin u ) . . . (3)
Thus the image of (3) is a surface of revolution and is called the torus.
Worked examples ........................................................................................................................•
Example 1 : Show that M :
(
)
x2 + y 2 − 4 + z 2 = 4 is a torus of revolution. Find a profile circle and
the axis of revolution. hence find the parametrization of the torus.
Solution : The surface M ⊂ ¡ 3 is given by
{
M = ( x , y , z ) ∈ ¡3 |
(
)
2
}
x2 + y 2 − 4 + z 2 = 4
Since the profile curve is a circle in a plane, this circle is either XZ-plane or YZ-plane. If the profile
curve is XZ-plane then y = 0. Hence equation of the profile circle is given by
( x − 4) 2 + z 2 = 4
which is a circle with center at (4, 0, 0) and radius 2. Its parametric representation in given by
x – 4 = 2 cos u
z = 2 sin u
⇒
x = 4 + 2 cos u
z = 2 sin u
y=0
This circle is rotated about z-axis. To parametrize this surface
Let
h (u) = 4 + 2 cos u
Differential Geometry
173
g (u) = 2 sin u
Then any point on the profile circle is given by (h(u), 0, g(u)).
If this profile circle is rotated about Z axis through an angle v then the point on the circle becomes a
point and X (u, v) ∈ M is given by
 x   cos v − sin v 0   h(u ) 
  


X (u , v) =  y  =  sin v cos v 0   0 
z  0
0
1   g (u ) 
  
 x   h(u )cos v 
  

 y  =  h( u)sin v 
 z   g ( u) 
  

⇒ ( x , y , z ) = ( (4 + 2cos u )cos v,(4 + 2cos u )sin u,2sin u )
This is the parametrization of the torus.
⇒ x = (4 + 2 cos u) cos v,
y = (4 + 2 cos u) sin v,
z = 2 sin u
2
⇒ x 2 + y 2 = ( 4 + 2cos u )
x 2 + y 2 = 4 + 2cos u
x 2 + y 2 − 4 = 2cos u
Therefore
Also z = 2 sin u.
Therefore, we get
(
)
2
x2 + y2 − 4 + z 2 = 4
This is the torus obtained by rotating the circle ( x − 4 )2 + z 2 = 4 .
•.....................................................................................................................................................•
Note : If we take the profile circle in YZ-plane then, its equation is obtained by putting x = 0, the
equation becomes
( x − 4 )2 + z 2 = 4
and the axis of revolution is Z axis.
Differential Geometry
174
EXERCISE ....................................................................................................•
Example 1 : Find the parametrization for the surface obtained by revolving the profile curve
C : ( z − 2) + y 2 = 1
2
around y-axis.
Example 2 : Find the parametrization of the entire surface obtained by revolving the curves.
1.
C : y = cosh x around the x-axis.
2.
2
C : ( x − 2 ) + y 2 = 1 around the y-axis.
3.
C : z = x 2 around the z-axis.
•.........................................................................................................................•
Worked examples ........................................................................................................................•
Example 2 : Show that M :
x2 y2 z 2
+
+
= 1 is a surface and
a 2 b2 c 2
X(u, v) = (a cos u cos v, b cos u sin v, c sin u)
defined on D is a parametrization, where D = −
Solution :
x2 y2 z 2
M : 2 + 2 + 2 =1
a
b
c
Let
x2 y2 z2
g= 2+ 2+ 2
a
b
c
π
π
< u < , v∈ ¡
2
2
Then M : g = 1and
dg =
Now dg = 0
i.e.
2x
2y
2z
dx + 2 dy + 2 dz
2
a
b
c
⇒
2x
2y
2z
=0, 2 =0, 2 =0
2
a
b
c
⇒
x = 0,
y = 0,
z=0
dg = 0 at O (0, 0, 0).
Differential Geometry
175
But O (0, 0, 0) ∉ M since the point O does not satisfy the equation g = 1.
Thus dg ≠ 0 on M.
Therefore M : g = 1 is a surface.
It is given that X : D → ¡3 defined by
X (u, v) = (a cos u cos v, b cos u sin v, c sin u)
Let
x = a cos u cos v
y = b cos u sin v
z = c sin u
Consider
x2 y2 z2
+ 2 + 2 = cos 2 u cos 2 v+ cos 2 u sin 2 v + sin 2 u
2
a
b
c
= cos 2 u ( cos 2 v + sin 2 v ) + sin 2 u
=1
Thus X (u, v) ∈ M for all (u, v) ∈ D
Next to prove that X is regular we compute X u and X v .
X u = ( −a sin u cos v, −b sin u sin v, c cos u )
X v = ( −a cosu sin v, −b cos u cos v, 0 )
Consider
U1
U2
U3
X u × X v = −a sin u cos v −b sin u sin v c cos u
−a cos u sin v − bcos u cos v
0
= U 1 [bc cos2 u cos v ] − U 2 [ ac cos2 u sin v] +
+U 3 [ −ab cos 2 v sin u cos u − ab sin u cos u sin 2 v]
= bc cos 2 u cos vU 1 − ac cos 2 u sin vU 2 − ab sin u cos uU 3
Since −
π
π
< u < , cos u ≠ 0
2
2
and sin v and cos v are not zero for any v simultaneously
⇒ Xu × X v ≠ 0
Therefore X is regular and X(D) lies in M.
Hence X is a parametrization of M.
Differential Geometry
176
EXERCISE ...................................................................................................•
1.
Show that M : z =
x2 y2
+
is a surface and
a2 b2
X (u , v) = ( au cos v, bu sin v, u 2 )
is a parametrization of M.
2.
Show that M : z =
x2 y2
−
is a surface and
a2 b2
X(u, v) = (a (u + v), b (u – v), 4uv)
is a proper patch covering all of M.
•......................................................................................................................•
Differentiable functions and Tangent Vectors :
Definition : Let M be a surface and f be a real valued function defined on M.
Let X : D → M be a coordinate patch in M. Then the composite function f o X or f ( x ) is called
a coordinate expression for f.
f
M
D
X
R
X(D)
f o x = f(x)
f o X : D → ¡ is a real valued function. The function f defined on a surface M is called differentiable
if all its coordinate expressions are differentiable.
Lemma : If α is a curve in a surface M whose route lies in the image X(D) of a single patch X, then
there exist unique differentiable function a1 (t ) and a2 (t ) on I such that
α (t ) = X ( a1 ( t) , a2 (t ) )
i.e.
α = X ( a1 , a2 )
Differential Geometry
177
Proof :
Let
α : I → M and X : D → M .
Then
X − 1 : X ( D) → D is a function.
Therefore
( X −1 o α ) : I → D
.
v
X(D)
X
(a(t),
a(t))
1
2
α(t)
(a1, a2)
α
u
O
M
t
I
Since α is differentiable function, X −1 o α is also differentiable. Let a1 and a 2 be Euclidean coordinate
functions of X −1 o α .
Therefore
( X −1 o α ) ( t) = ( a1 ( t) , a2 (t ) )
Since X −1 o a is differentiable, a1 and a 2 are also differentiable functions on I.
And
( X −1 o α ) = ( a1 , a2 )
⇒ α = X ( a1 , a2 )
Further, if α = X (b1 , b 2 ) where b1 and b2 are differentiable functions on I then
( a1, a2 ) = ( X −1 o α ) = X −1 o X (b1,b2 )
⇒ ( a1 , a2 ) = ( b1 , b2 )
⇒ a1 = b1 , a2 = b2
Hence a1, a2 are unique functions such that a = X ( a1 , a2 ) .
Remark : The real valued functions ( a1 , a2 ) are called the coordinate functions of the curve w. r. t.
patch X.
Differential Geometry
178
Tangent and Normal Vector Field on surface :
Definition : Let D be a point of the surface M in ¡ 3 . A tangent vector v to ¡ 3 at point P is a tangent
to M if v is velocity vector of some curve in M.
v = α’ |P
v
The set of all tangent vectors to M at the point P is
called the tangent plane of M at P, denoted by T p ( M ) .
P
M
T(M)
P
T p ( M ) is a two dimensional subspace of ¡ 3 .
P
M
Lemma : Let P be a point of a surface M in ¡ 3 and let X be a patch in M such that X ( u0 , v0 ) = P .
A vector v of ¡ 3 at a point P is tangent to M if and only if v can be written as a linear combination
of X u (u 0 , v 0 ) and X v (u 0 , v 0 ) .
Proof : First suppose that v is tangent to surface M at a point P ∈ M . Therefore there is a curve α
in M such that α (0) = P and α '(0) = v .
By lemma for a given coordinate patch X, α = X ( a1 ( t), a2 ( t) ) where a1, a2 are differentiable
functions on I.
i.e.
α (t ) = X ( a1 ( t) , a2 (t ) )
⇒ α '(t ) =
d
X ( a1( t), a2 ( t) )
dt
=
d
X (u , v )
dt
=
∂X du ∂X dv
+
∂u dt ∂v dt
= Xu
Differential Geometry
u = a1(t ) , v = a2 (t )
da1
da
+ Xv 1
dt
dt
179
At t = 0,
α (0) = P = X ( u0 , v0 ) = ( a1 (0), a2 (0) )
Hence
a1 (0) = u0 and a2 (0) = v0
Therefore
 da 
 da 
α '(t) |t =0 = X u |t =0  1  + X v |t =0  2 
 dt t = 0
 dt  t =0
But
α '(t ) = v , a1 (0) = u0 , a2 (0) = v0
Thus
 da 
 da 
v = X u ( u0 , v0 )  1  + X v ( u0 , v0 )  2 
 dt t =0
 dt t =0
Which shows that v is a linear combination of Xu and Xv at ( u0 , v0 ) .
Conversely,
let a tangent vector v in ¡ 3 be written as
v = c1 Xu ( u0 , v0 ) + c2 Xv ( u0 , v0 )
Define a curve α : I → M in M by
α (t ) = X ( u0 + tc1 , v0 + tc2 )
α is a curve in M such that
α (0) = X ( u0 , v0 ) = P
Differentiate α w. r. t. t
α '(t ) =
=
d
 X ( u0 + tc1 , v0 + tc2 ) 
dt 
∂X du ∂X dv
+
∂u dt ∂v dt
u = u0 + tc1 , v = v0 + tc2
= X u ⋅ c1 + X v ⋅c 2
α '(t ) = c1 ⋅ Xu ( u0 + tc1, v0 + tc 2 ) + c2 ⋅ X v ( u0 + tc1, v0 + tc2 )
At t = 0
α '(0) = c1 ⋅ Xu (u 0 , v0 ) + c2 ⋅ X v (u 0 ,v0 )
⇒ α '(0) = v
Thus v is a velocity vector of a curve α at a point P. Hence v is tangent to M at point P..
Differential Geometry
180
Definition :
A Euclidean vector field Z on a surface M in ¡ 3 is a function
Z(P)
that assigns to each point P of M a tangent vector Z ( P) at P..
P
M
Definition :
A Euclidean vector field V for which each vector V ( P) is
tangent to M at P is called a tangent vector field on M.
v(P)
P
M
TP(M)
Definition : A Euclidean vector Z at a point P of M is normal to M if it is a orthogonal to the tangent
plane TP (M ) .
A Euclidean vector field Z on M is a normal vector field on M if every vector Z ( P) is normal
to M.
Following result gives a technique to construct a normal vector field on a surface.
Lemma : If M : g = c is a surface in ¡ 3 then the gradient vector field
∂g
Ui
i =1 ∂xi
3
∇g = ∑
defined on M is a nonvanishing normal vector field on the entire surface M.
Proof : g is a differentiable function defined on ¡ 3 and since M is a surface, dg ≠ 0 on M.
⇒
Now
∂g
are not all zero on M for i = 1, 2, 3.
∂xi
 ∂g ∂g ∂g 
∇g =  , , 
 ∂x ∂y ∂z 
∂g
Ui
i =1 ∂xi
3
=∑
Differential Geometry
181
where U i is natural frame field.
Since
∂g
≠0
∂xi
∀ i = 1, 2, 3
∇g ≠ 0
on M.
Now if α = ( a1, a2 , a3 ) is a curve in M then g (α ) = c .
Differentiate w. r. t. t, we get
∂g da1 ∂g da2 ∂g da3
⋅
+
⋅
+
⋅
=0
∂a1 dt
∂a 2 dt
∂a3 dt
∂g
da (t )
(a ) ⋅ i = 0
dt
i =1 ∂ai
3
⇒∑
Now if v is a tangent vector to M at a point P then v is velocity of some curve α in M.
i.e.
v = α '(0) for some curve α in M.
 da (0) da (0) da (0) 
⇒ ( v1, v2 , v3 ) =  1 , 2 , 3 
 dt
dt
dt 
⇒ vi =
da1 (0)
dt
i = 1, 2, 3
∂g ( a )
⋅ vi = 0
i =1 ∂ai
3
⇒∑
 ∂g ∂g ∂g 
⇒
,
,
 ⋅ ( v1, v2 , v3 ) = 0
 ∂a1 ∂a2 ∂a3 
⇒ ∇g ⋅ v = 0
This shows that ∇g is a perpendicular vector to v where α '(0) = v and hence ∇g is normal to the
surface M at P.
Therefore ∇g is a normal vector field on M.
Worked examples ........................................................................................................................•
Example 1 : Find a normal and tangent vector fields on the sphere ∑ .
Solution : If ( x1, x2 , x3 ) ∈∑ then
Differential Geometry
182
x12 + x22 + x32 = r2
3
i.e.
∑ xi2 = r 2
Thus
∑: g =r ,
i =1
3
2
g = ∑ xi2
i=1
Therefore
 ∂g ∂g ∂g 
∇g = 
,
,

 ∂x1 ∂x2 ∂x3 
= ( 2 x1, 2 x2 , 2 x3 )
= 2 ( x1 , x2 , x3 )
1
⇒ ∇ g = ( x1, x2 , x3 )
2
3
Let X = ( x1, x2, x 3 ) = ∑ xi U i
i =1
Then X is normal vector field on ∑ .
If V = ( 0, − x3 , x2 ) is a vector field on ∑ then
Then X ( P) ⋅V ( P) = ( x1, x2 , x3 ) ⋅ ( 0, − x3 , x2 )
X(P)
P
X
O
= − x2 x3 + x2 x3
=0
Thus V is orthogonal to normal vector field.
Hence V is a tangent vector field.
Let W be another vector field on ∑ defined by W = ( x3 ,0, − x1 )
Then
X ⋅W = 0
⇒ W is a tangent vector field on ∑ .
Differential Geometry
183
Example 2 : Let X be a geographical path in the sphere ∑ . Find the coordinate expression f (x)
where f : ∑ → ¡ is a function defined by
(1)
f ( P) = P12 + P22
(2)
f ( P) = ( P1 − P2 ) + P32
2
Solution : X is a geographical patch on the sphere ∑ .
X :D→ ∑
Thus
defined by
X (u , v) = (r cos u cos v , r sin u cos v , r sin v )
D : −π < u < π , −
where
(1)
π
π
<v<
2
2
Consider
f ( X )( u, v) = f ( X (u ,v ))
= f ( r cos u cos v , r sin u cos v , r sin v )
= r 2 cos2 u cos2 v + r 2 sin 2 u cos2 v
f ( X )( u, v) = r 2 cos2 v
(2)
f ( X )( u, v) = f ( X (u ,v ))
= f ( r cos u cos v , r sin u cos v , r sin v )
2
= f ( r cos u cos v − r sin u cos v) + r 2 sin 2 u
= r 2 cos2 v ( cos u − sin u ) + r 2 sin2 v
2
f ( X )( u, v) = r 2 cos 2 v (1 − sin u ) + r 2 sin 2 v
2
Example 3 : Find a non vanishing normal vector field on M : z = xy. And also find two linearly
independent tangent vector fields at each point of M.
Solution : Given
M : z = xy : z – xy = 0
Let
g = z – xy
Therefore
 ∂g ∂g ∂g 
∇g =  , , 
 ∂x ∂x ∂x 
= (– y, – x, 1)
Differential Geometry
184
Let
Z = ( − y , − x,1)
Then Z is a normal vector field on M.
If v is a tangent vector field on M thenv ( p ) ⋅Z ( p ) = 0
Take
z
v (p) = (x, 0, z), v ' (p) = (x, – y, 0), v '' (p) = (0, y,, z)
Then
v ( p ) ⋅Z ( p ) = 0
v '( p ) ⋅Z ( p ) = 0
v "( p ) ⋅ Z ( p ) = 0
⇒ v , v ' and v '' are linearly independent tangent vector fields on M.
Note that the surface M is called Saddle surface.
Example 4 : Prove that a vector v is a tangent vector to a surface M : z = f (x , y ) at a point P of M
if and only if
v3 =
where
∂f
∂f
( p1, p2 ) v1 + ( p1, p2 ) v2
∂x
∂x
v = ( v1, v2 , v3 )
Solution : Define a function X by
X (x, y) = (x, y, f (x, y))
Then X is a patch in M.
If v is any tangent vector to the surface M then by theorem
⇒ v = c1 X x + c2 X y
Now
X = (x, y, f (x, y))
Therefore
∂f 

X x =  1,0, 

∂x 
and
∂f 

X y =  0,1, 
∂y 

Hence
∂f 
∂f 


v = c1  1,0,  + c2  0,1, 
∂x 
∂y 


Differential Geometry
185
∂f
∂f 

=  c1, c2 , c1 + c2
∂x
∂y 

⇒ v1 = c1 ,
v2 = c2
and
v3 = c1
∂f
∂f
+ c2
∂x
∂y
v3 = v1
∂f
∂f
+ v2
∂x
∂y
v3 = v1
∂f
∂f
( p1, p2 ) + v2 ( p1 , p2 )
∂x
∂y
Thus,
p = ( p1, p2 , p3 )
where
Example 5 : Let c be a right circular cone parametrized by X (u, v ) = (v cos u, v sin u, v). If α is the
curve α (t ) = X ( 2t , et ) ,
(i)
Express α ' in terms of Xu and Xv .
(ii)
Show that at each point of α , the velocity α ' bisects the angle between Xu and Xv.
Solution :
α (t ) = X ( 2t , et )
α ( t ) = X (u , v )
where u = 2t , v = et
Differentiating above equation w. r. t. t, we get
α '(t ) =
∂X du ∂X dv
⋅ +
⋅
∂u dt ∂v dt
α '(t ) = X u ⋅ 2 + X v ⋅ et
. . . (1)
If t = 0 then ( u0 , v0 ) = (0,1)
α '(0) = 2 X u ( u0 , v0 ) + X v (u 0 , v 0 )
. . . (2)
X (u, v) = (v cos u, v sin u, v)
. . . (3)
Now
Differential Geometry
186
Differentiating equation (3) w. r. t. u and v, we get
X u = ( −v sin u, v cos u, 0 )
X v = ( cos u,sin u,1)
Therefore
Xu = v 2 sin 2 u + v2 cos 2 u
=v
= et
and
Xv = cos 2 u + sin 2 u + 1
= 2
Observe that, X u ⋅ X v = 0 .
Hence X u ⊥ X v .
If φ1 is the angle between α ' and Xu then it is given by
α '⋅ X u
cos φ1 =
=
α ' ⋅ Xu
=
=
(
)
2 X u + et X y ⋅ X u
α ' ⋅ et
2 ⋅ X u ⋅ X u + et X y ⋅ X u
α ' ⋅et
2 Xu
2
α ' ⋅ et
2e 2t
=
α ' ⋅ et
cos φ1 =
2e t
α'
. . . (4)
Similarly, if φ2 is the angle between α ' and Xy then
2 X u + et X v ) ⋅ X v
(
α '⋅ X v
cos φ 2 =
=
α' ⋅ Xv
α' ⋅ 2
=
cos φ 2 =
Differential Geometry
et X v ⋅ X v
α' ⋅ 2
2et
α'
187
=
et X v
α'⋅ 2
et ( 2 )
2
2
=
α' ⋅ 2
. . . (5)
Hence from equation (4) and (5) we have
cos φ1 = cos φ2
⇒ φ1 = φ2
This shows that α ' bisects the angle between Xu and Xv.
Note :
α '(t ) = 2 Xu( u, v) + et Xv( u, v)
u = 2t , v = et
α '(t ) = et ( − 2sin 2t + cos 2t , 2cos 2t + sin 2t ,1)
Hence
α '( t ) = 2et
Then
cos φ1 =
∴φ1 =
This implies
2et
2e t
1
=
=
t
α'
2e
2
π
4
π
.
4
φ2 =
Example 6 : Find the equation of the form ax + by + cz = d the plane TP (M ) for the surface
M : x 2 + y 2 + ( z − 1) 2 = 1 at the point P = (0, 0, 0).
Solution : Equation of the surface is
M : x 2 + y 2 + ( z − 1) 2 = 1
Let
g = x 2 + y 2 + ( z −1) 2 = 1
Then
 ∂g ∂g ∂g 
∇g =  , , 
 ∂x ∂y ∂z 
= ( 2 x, 2 y,2( z − 1) )
Differential Geometry
188
∇g is the normal vector field on M and at the point P = (0, 0, 0) the normal vector to the surface M
is given by
n = ( 2 x, 2 y ,2( z − 1) ) p
= (0, 0, –2)
This vector n is also normal to the plane TP (M ) and hence equation of the plane is given by
a ( x − x0 ) + b ( y − y0 ) + c ( z − z0 ) = 0
( a, b, c ) ≡ ( 0,0, −2 ) and P ≡ ( x0 , y0 , z ) ≡ ( 0,0,0)
Therefore equation of the plane is given by
0 + 0 − 2( z ) = 0
i.e. z = 0
EXERCISE .....................................................................................................•
1.
Find the equation of the form ax + by + cz = d for the plane TP (M ) for the following surfaces
(i)
x2 y2 z 2
M: +
+
= 1 at P = (1, –2, 3)
4 16 18
(ii)
 π
The surface M parametrized by X (u , v) = (u cos v ,u sin v, 2v ) at the point P = X  2,  .
 4
•.......................................................................................................................•
❏❏❏
Differential Geometry
189
CHAPTER - IV
SHAPE OPERATOR
Unit I
Introduction :
In the case of a curve there is only one direction to move and its shape is measured by taking
the rate of change of tangent vector T to the curve. In other words, the shape of the curve is measured
by its curvature and torsion.
However, for a surface there is a plane of directions to move and hence tangent vectors to M
at P alone will not determine the change of the surface in all possible direction. Hence for a surface a
unit normal vector field U is determined and the rate of change of U in the v direction tells how the
tangent planes of M are varying in the v -direction. The rate of change of the unit normal vector field
U is described by a special type of linear operator defined on the tangent planes of the surfaces. Such
operators are called linear shape operators. With the help of shape operators we are able to measure
how rapidly surface M is pulling itself away from the tangent vector plane. This amounts the change in
the shape of the surface in the neighbourhood.
Shape Operator of Surfaces :
Definition : If Z is a vector field on a surface M of ¡ 3 then covariant derivative of Z w. r. t. a
tangent vector v is defined as
∇v Z = Z ' ( p + tv ) |t =0
∇v Z is the rate of change of Z in v direction.
If Z is expressed in terms of natural frame fields then,
3
Z = ∑ zi U i
i=1
3
and
∇ v Z = ∑ v [ zi ]U i
Differential Geometry
i=1
190
where
v [ zi ] =
dzi
( p + tv ) |t =0
dt
∇g
Note : If M ⊂ ¡ 3 is a surface defined by M : g = c then ∇g is a normal vector field on M and ∇g
is a unit normal vector field on M.
Definition : If p is any point of ¡ 3 then for each tangent vector v to M at p the function defined by
S p ( v ) = −∇v U
where U is a unit normal vector field in the neighbourhood of p in M, is called the shape operator of
M at p.
U(P)
v
SP(v)
P
∇v ( U)(P)
M
Note : ∇v U represents the rate of change of U in v direction. It tells how the tangent plane of M is
changing in v - direction and gives information about the way by which M is curving.
Lemma : For each point P of M ⊂ ¡ 3 , the shape operator is a linear operator..
Proof : Since U is a unit normal vector field, by definition of shape operator, we have
S p (U ) = − ∇ v U , ∀ tangent vector v to M at P..
............ (1)
This shows that the shape operator is a function of the tangent vector v to M at P. Clearly the domain
of Sp is
S p = {v p / v p is a tangent vector to M at P}
Differential Geometry
191
⇒ domain of S p is T p ( M ) .
Now to find the range of S p , we have in equation (1) U is the unit normal vector field on M.
⇒ U ⋅U = 1
Differentiating covariantly we get,
⇒ ∇ v (U ⋅U ) = 0
⇒ ∇v U ⋅ U + U ⋅ ∇v U = 0
⇒ 2∇v U ⋅U = 0
⇒ ∇v U ⋅ U = 0
⇒ ∇v U ⊥ U
⇒ − S p (v ) ⊥ U
⇒ S p (v ) ⊥ U
⇒ S p ( v ) ∈ Tp (M )
where v ∈T p (M )
⇒ The range of S p is T p ( M ) .
Hence S p :T p ( M ) → Tp ( M ) is an operator..
Next
S p ( av + bw ) = −∇( av +bw )U
= −  a∇ v U + b∇w U 
= aS p ( v ) + bS p ( w )
Hence S p is a linear operator..
Worked examples .......................................................................................................................•
Example 1 : Find the shape operator on a sphere in ∑ .
Solution : Let ∑ be a sphere of radius r..
∴ ∑ = { p ∈ ¡3 / p = r }
= {( x, y , z ) / x 2 + y 2 + z 2 = r 2}
Differential Geometry
192
Let U be an outward unit normal vector field on ∑ .
Then
∇g ,
∇g
U=
Now
∇g = ( 2 x, 2 y , 2 z ) and
∇g = 4 x 2 + 4 y2 + 4z2
= 2 x2 + y2 + z 2
= 2 r2
=2r
Thus unit normal vector field on ∑ is given by
2 ( x, y , z ) 1
∇g
=
= ( x, y , z )
∇g
2r
r
U=
U=
Therefore
1 3
xi U i , U i are natural frame fields.
∑
r i =1
∇v U =
1 3
∑ v [ xi ]U i
r i=1
where v = ( v1, v2 , v3 ) is a tangent vector to M at P..
3
And
v [ xi ] = ∑ vj
Therefore
∇v U =
j =1
=
∇v U =
Hence S p ( v ) = −
∂xi
∂x
= vi i = vi
∂x j
∂xi
1 3
∑ viU i
r i =1
1
( v1, v2 , v3 )
r
v
r
v
for all tangent vectors v at P to ∑ .
r
Differential Geometry
193
Thus the shape operator S is just a scalar multiplication by − 1 . This uniformity in S reflects the
r
roundness of sphere.
i.e. the surface bend the same way in all directions at all points.
Example 2 : Show that the shape operator of a plane surface is zero.
Solution : If P is a plane of ¡ 3 then unit normal vector field gives parallel vectors at all points.
U
i.e.
U = constant
∴∇v U = 0
P
S p (v ) = 0
∀ v and ∀p ∈ P
Thus the shape operator is zero for a plane surface. This shows there is no bending in the plane in any
direction.
Example 3 : Show that the shape operator describes the cylindrical surface as half flat and half round.
Solution : Let C be a circular cylinder of ¡ 3 based on a circle x 2 + y 2 = r 2 .
At any point P of right circular cylinder let e1 and e2 be the two unit tangent vectors with e1 tangent to
the ruling of the cylinder through P and e2 be the tangent to the cross sectional circle.
Let U be an outward unit normal vector field on C. When U moves from P in e1 direction, U
remains parallel to itself just as a plane.
Hence
S p ( e1 ) = 0
When U moves from P in e2 direction then it topples
forward exactly as on sphere of radius r. Hence
1
S p ( e2 ) = − e2
r
i.e. bending of the cylinder along e2 direction is uniform.
Thus the shape operator describes the cylinder as half flat half round.
Differential Geometry
194
Example 4 : Let α be a curve in M ⊂ ¡ 3 . If U is a unit normal of M restricted to the curve α .
Show that S (α ') = −U ' .
Solution : α is a curve in M. Therefore the velocity vector α ' is tangent to α .
By definition
S (α ') =−∇α 'U
where U is a normal vector field on α .
∇α 'U = ( U ( α (u ) + tα '(u ) ) ) ' |t =0
Now
= U ' (α (u ) )
But
S (α ') =−∇α 'U
⇒
S (α ') = −U '
P = α (u ) is a point on the curve.
on α
Example 5 : Consider the surface M : z = f (x , y ) where f (0,0) = fx (0,0) = fy (0,0) = 0 show
that
(a)
The vectors u1 = U 1(o) and u2 = U 2 (o) are tangent to M at the origin o and
U=
− f x U 1 − f y U 2 +U 3
1 + f x2 + f y2
is a unit normal vector field on M.
(b)
S ( u1 ) = f xx (0,0)u1 + f xy (0,0)u2
S ( u2 ) = f yx (0,0) u1 + f yy (0,0) u2
Solution : For a surface M : z = f (x , y ) the normal vector field N is given by
∂g
 ∂g ∂g ∂g 
Ui =  , , 
 ∂x ∂ y ∂ z 
i =1 ∂xi
3
N=∑
Here
g = z − f (x , y )
Therefore
∂g
∂g
= − f y , ∂g = 1
= − fx ,
∂y
∂x
∂z
⇒ N = ( − f x , − f y ,1)
= − f x U 1 − f y U 2 +U 3
Differential Geometry
195
Hence unit normal vector field on M is given by
U=
(b)
− f x U 1 − f y U 2 +U 3
N
=
N
f x2 + f y2 + 1
We know that ∇U i ( p )W =
∂W
( p)
∂xi
where U i are natural frame field vectors at P..
Now
u1 = U 1(o) ,
u2 = U 2 (o)
Therefore
S ( u1 ) = − ∇U U = −
1
∂ ( )
U
∂x
S ( u2 ) = − ∇U U = −
2
∂ ( )
U
∂y
where U is a unit normal vector field on the surface.
We have
U=
− f x U 1 − f y U 2 +U 3
1 + f x2 + f y2
Therefore
S ( u1 ) = −
=
∂  − f x U 1 − f y U 2 +U 3 


∂x 

1 + f x2 + f y2


f xxU 1 + f xy U 2
1+
f x2
+
f y2
−
2 ( f x f xx + f y f xy )
f x2 + f y2 + 1
× ( f x , f y ,1)
At the origin o (0, 0, 0)
S ( u1 ) =
f xx (0,0)U 1 + f xy (0,0)U 2
1 + f x2 (0,0) + f y2 (0,0)
−
2  f x (0,0) f xx (0,0) + f y (0,0) f xy (0,0) 
− 
× ( f x (0,0), f y (0,0),1)
f x2 (0,0) + f y2 (0,0) + 1
Differential Geometry
196
Given that
f x (0,0) = f y (0,0) = f (0,0) = 0
Hence
S ( u1 ) = −
f xx (0,0)U 1 (0,0) + f xy (0,0)U 2 (0)
1+ 0 + 0
−0
S ( u1 ) = f xx (0,0)U 1 + f xy (0,0)U 2
S ( u2 ) = f yx (0,0)U 1 + f yy (0,0)U 2
Similarly
.
Example 6 : For a saddle surface M : z = xy Show that S ( au1 + bu 2 ) = au2 + bu1 where
u1 = U 1(o) and u2 = U 2 (o) .
Solution : M : z = xy
Let
f (x , y ) = xy
⇒ f x = y, f y = x , f xx = 0, f xy = 1, f yy = 0
If u1 = U 1(o) and u2 = U 2 (o) are two tangent vectors to the surface at the origin, then
S ( u1 ) = f xx (0,0)U 1 + f xy (0,0)U 2
= 0 ⋅ U 1 + 1⋅ U 2
S ( u1 ) = U 2 = u2
Similarly
S ( u2 ) = f xy (0,0)U 1 + f yy (0,0)U 2
= 1 ⋅U 1 + 0 ⋅ U 2
⇒ S ( u2 ) = U 1 = u1
Since S is linear operator,
S ( au1 + bu 2 ) = au2 + bu1
Example 7 : For a surface M : z = ( x + y ) 2 , show that S ( au1 + bu2 ) = 2( a + b ) ( u1 + u2 )
Solution :
M : z = (x + y )2
Differential Geometry
197
f (x , y ) = ( x + y) 2
Let
∴ f x = 2( x + y)
f y = 2( x + y)
f xx = 2
f yy = 2
f xy = 2
f yx = 2
Let u1 = U 1(o) and u2 = U 2 (o) be two tangent vectors to the surface M at the origin.
At the origin o (0, 0, 0)
f x (0,0) = 0 = f y (0,0)
.
f xx (0,0) = 2
f yy (0,0) = 2
But we have
S ( u1 ) = f xx (0,0)U 1 + f xy (0,0)U 2
= 2 ⋅U 1 + 2 ⋅U 2
S ( u2 ) = f xyxx (0,0)U 1 + f yy (0,0)U 2
And
= 2 ⋅U 1 + 2 ⋅U 2
Therefore
S ( au1 + bu 2 ) = aS ( u1 ) + bS ( u2 )
⇒ S ( au1 + bu 2 ) = a ( 2 ⋅ U 1 + 2 ⋅ U 2 ) + b ( 2 ⋅U 1 + 2 ⋅U 2 )
⇒ S ( au1 + bu 2 ) = 2 ( a + b ) ( u1 + u2 )
EXERCISE ........................................................................................................•
1.
Obtain the value of S ( au1 + bu 2 ) in terms of u1 and u2 for the surface
(a) M : z = 2x 2 + y 2
(2) M : z = xy 2
•.......................................................................................................................•
Gauss Map
Definition : Let M be a surface in ¡ 3 oriented by a unit normal vector field
U = g1U 1 + g 2U 2 + g 3U 3 . The Gauss Map G : M → ∑ is defined by
Differential Geometry
198
G( p) = ( g1 ( p), g2 ( p), g3 ( p) ) , P ∈ M .
where ∑ is a unit sphere centered at the origin.
Note : 1. The unit normal at P ∈ M , U ( p) is moved by parallel motion to the origin.
2. The map G describes the turning of U as it traverses M.
Example : Find the Gauss Map for the saddle surface M : z = xy .
Solution : Let g = z − xy = 0 . Then ∇g = ( − y, − x,1) .
Hence ∇g = 1 + x 2 + y2 ⋅∇ g is a normal vector field on M. Therefore unit normal vector field on
M is given by,
U=
=
Let
g1 =
∇g
( − y , − x,1)
=
∇g
1 + x2 + y2
−y
1+ x2 + y2
−y
1 + x2 + y2
U1 −
x
1+ x 2 + y 2
, g2 =
U2 +
−x
1+ x2 + y2
1
1+ x 2 + y 2
, g3 =
U3
1
1+ x2 + y2
Let G : M → ∑ be the Gauss Map. Then,
G( p) = ( g1 ( p), g2 ( p), g3 ( p) )
−y
−x
1

G ( p) = 
,
,
 1 + x 2 + y 2 1 + x 2 + y 2 1 + x2 + y 2



p
Image of the point P ∈ M is a point on the unit sphere ∑ .
EXERCISE ........................................................................................................•
Obtain the Gauss map for the following surfaces.
(i) Cylinder M : x 2 + y 2 = r 2
(ii) Cone M : z = x 2 + y 2 ,
(iii) Sphere M : ( x −1) 2 + y 2 + ( z + 2 ) 2 = 1 ,
(iv) Plane M : x + y + z = 0
(v) Torus M : X ( u , v ) = ( ( R + r cos u ) cos v, ( R + r cos u) sin v , r sin u )
•.......................................................................................................................•
Differential Geometry
199
Unit II : Normal curvature
The shapes of a surface in ¡ 3 influences the shape of the curves in M.
Lemma : If α is a curve in M ⊂ ¡ 3 then α "⋅ U = S (α ') ⋅ α ' where U is a unit normal to the
surface M.
Proof : α is a curve in M. Therefore its velocity vector α ' is tangent vector to M.
Hence α '⋅U = 0 .
Differentiating we get,
α '⋅U '+ α "⋅ U = 0
But
S (α ') = −U '
.
This gives
−α '⋅ S (α ') +α "⋅ U = 0
⇒ α "⋅ U = S (α ') ⋅ α '
Thus at each point on the curve α , α "⋅ U is the component of the acceleration α " in normal
direction. And the above result shows that that this normal component depends on α ' and shape
operator S in the direction of α ' .
Definition : Let u be the unit tangent vector to the surface M of ¡ 3 at a point p. Then the normal
curvature of M in the direction of u is defined as
κ (u ) = S (u ) ⋅ u
S(- u )
-u
Pp
u
S(u)
Note that κ ( u ) = S ( u ) ⋅ u = S (− u ) ⋅ ( −u ) = κ ( −u )
Note : For a unit tangent vector u to a surface M at a point p, let α be a unit speed curve in M with
α '(0) = u
Differential Geometry
200
By Frenet apparatus on α ,
κ is curvature function of α .
α ' =T, α " =T ' =κ N ,
Then normal curvature
κ (u ) = S (u ) ⋅ u
= S (α ' ) ⋅ α '
= α "⋅U
= α "(0) ⋅U ( p)
= κ ( 0 ) N (0)U ( p)
= κ ( 0 ) ⋅ N (0) ⋅ U ( p ) ⋅ cos θ
κ ( u ) = κ ( 0 ) ⋅ cos θ
where θ is the angle between N (θ ) and U ( P ) .
Thus the normal curvature of M in u direction is κ ( 0 ) ⋅ cos θ , where κ ( 0 ) is the curvature of the
curve α at α (0) = p and θ is the angle between the principal normal N ( 0) and the surface normal
U(p).
Definition : If P is the plane determined by u and U ( p) , then the plane p cuts the surface near the
point p into a curve σ called as normal section of M in u - direction.
If we give a unit speed parametrization to this curve with σ '(0) = u then
N (0) = ±U ( p)
U(P)
U
( p)
P
u
pP
p
σ(0) = P
σ
M
N(0)
Differential Geometry
201
σ '(0) = T (0) ⇒ T '(0) = σ "(0)
Also
= κ (0) N (0)
Since T (0) ⊥ N (0) , N (0) lies in a plane P. Therefore for a normal section curve, the normal curvature
becomes,
κ ( u ) = κ (0) N (0)U ( P)
= κσ (1 ⋅ 1 ⋅ cos θ )
= κσ cos θ
where θ = 0 or π
κ ( u ) = ±κσ
Thus normal curvature of the surface in u direction is the curvature of the normal section curve in
u direction.
Meaning of the Sign of the Normal Curvature :
(i)
If κ ( u ) > 0 then for normal section curve N (0) = U ( p ) . So that the curve is bending towards
U ( p) at p. Thus in u direction the surface M is bending towards the vector U ( P ) at P. The surface
is concave in u direction.
U(P)
N(0) = U (P)
u
σ
P
(ii)
P
u
σ
N(0)
If κ ( u ) < 0 then N (0) = −U ( P) . Therefore the normal section curve σ is bending away
from U ( P) at point P. Thus in direction the surface M is bending away from the U ( P) . The surface
is convex in u direction.
(iii)
If κ ( u ) = 0 then κσ (0) = 0 which shows that the normal section curve σ is not turning at
σ (0) = p . Therefore the surface is flat in u direction.
Note that in different directions at a given point the surface may have different types of bending.
Differential Geometry
202
Worked examples .........................................................................................................................•
Example 1 : Saddle Surface
z
M : z = xy
Y
P
x=y
−Y
X
x =− y
Consider a saddle surface M : z = xy.
Let p (0, 0, 0) be a point on the surface. i.e. p = 0
The normal curvature along the X-axis and Y-axis is zero.
The normal curvature at point p (0, 0, 0) along the line y = x is positive and the normal curvature along
the line y = – x is negative.
Thus at p (0, 0, 0) on M : z = xy
κ ( u ) = 0 if u is along the coordinate axes.
κ ( u ) > 0 if u is along the line y = x.
κ ( u ) < 0 if u is along the line y = – x.
•....................................................................................................................................................•
Principal Curvature :
Definition : Let p be a point of a surface M ⊂ ¡ 3 . The maximum and minimum values of the normal
curvature κ ( u ) at p are called Principal curvatures of M at point p.
The corresponding maximum and minimum values of the curvatures are denoted by κ1 and κ 2 . The
directions in which these extreme values occur are called the principal directions; denoted by e1 and
e2 .
Differential Geometry
203
Worked examples .........................................................................................................................•
Example 2 : Find the principal curvatures for a right circular cylinder.
Solution : Consider a circular cylinder of radius r given by M : x 2 + y 2 = r 2 ; z is arbitrary..
For a right circular cylinder if U is an outward normal vector field on the surface, then every normal
section curve bend away from the normal. Hence κ ( u ) is negative. And in the direction of ruling the
κ ( u ) is zero.
k1 = 0
e1
e2 k2 = − 1
r
U
Ruling
Hence the maximum value of the normal curvature is zero along the ruling and the minimum value of the
normal curvature occurs only in the direction of tangent to a cross section.
Thus
κ1 = 0 and κ 2 = −
1
r
•.....................................................................................................................................................•
Umbilic Points :
Definition : A point p of M is called Umbilic point if the normal curvature κ ( u ) is constant on all unit
tangent vectors u at p.
Worked examples .........................................................................................................................•
Example 3 : For a surface M : x 2 + y2 + z 2 = r 2 , show that every point is umbilic.
1
Solution : M : x 2 + y2 + z 2 = r 2 is a spherical surface. For every point P ∈ M , κ ( u ) = − where
r
u is any tangent vector to the surface M at p. Thus κ ( u ) = constant for every direction at p. Since p
is arbitrary the normal curvature is constant at every point p of M. Hence every point of M is umbilic.
•.....................................................................................................................................................•
Differential Geometry
204
Theorem :
(i)
If p is an umbilic point of a surface M ⊆ ¡3 then the shape operator S at p is just a scalar
multiplication by κ = κ1 = κ2
(ii)
If p is non umbilic point, κ1 ≠ κ2 then there are exactly two principal directions which are
orthogonal. Further if e1 and e2 are principal vectors in these directions, then
S ( e1 ) = κ1e1 and S ( e2 ) = κ 2 e2
Proof :
(1)
Let κ1 be the maximum value of κ along e1 at point P..
κ1 = κ ( e1 ) = S ( e1 ) ⋅ e1 where S is the shape operator.
Let e2 be the unit tangent vector perpendicular to e1 at P. We will show that it is also a principal vector..
e2
If u is any other tangent vector at p then we have
u
u ( θ ) = e1 cos θ + e2 sin θ = Ce1 + Se2 , where
θ
pP
C = cos θ and S = sin θ .
e1
Thus normal curvature κ at p becomes a function of θ .
κ (θ ) = κ ( u ( θ ) )
= S ( u (θ ) ) ⋅ u (θ )
= S ( Ce1 + Se2 ) ⋅ ( Ce1 + Se2 )
=  CS ( e1 ) + SS ( e2 )  ⋅ ( Ce1 + Se2 )
= C 2 S ( e1 ) ⋅ e1 + CS ( e1 ) ⋅ e2 + CS ( e2 ) ⋅ e1 + S 2 S ( e2 ) ⋅ e2
Let Sij = S ( ei ) ⋅ e j . Then S ( e1 ) ⋅ e 2 = S12 and S ( e2 ) ⋅ e1 = S 21 . By symmetry of shape operator,,
S12 = S 21 . Therefore, κ (θ ) = C 2 S11 + 2CSS12 + S 2 S22
Differentiating κ (θ ) w.r.t. θ , we get,
dκ ( θ )
= −2CSS11 + 2 S12 [ C2 − S 2 ] + 2SCS22
dθ
Differential Geometry
205
We know that κ (θ ) is maximum when u ( θ ) = e1 i.e. θ = 0
Hence
dκ (θ )
=0
dθ
when θ = 0
Therefore we get,
dκ (θ )
= 0 and S = 0, C = 1
dθ θ =0
⇒ 0 = 2S12 . Hence S12 = 0. And then normal curvature κ ( u (θ ) ) becomes,
κ ( u (θ ) ) = C 2 S11 + S 2 S22
Now e1 and e2 are orthogonal vectors and hence forms a basis of TP(M). Also S ( e1 ) ∈ TP ( M ) and
S ( e2 ) ∈ TP ( M ) . Therefore by orthonormal expansion we get
S ( e1 ) = ( S ( e1 ) ⋅ e1 ) e1 + ( S ( e1 ) ⋅ e2 ) e2
= S1 1e1 + S12e2
S ( e1 ) = S1 1e1
(Q S12 = 0 )
Similarly we can show that S ( e2 ) = S 22 e2 .
Now if p is an umbilic point then normal curvature at p in any direction is constant.
i.e.
κ ( e1 ) = κ ( e2 ) = κ , constant
⇒ S ( e1 ) ⋅ e1 = S ( e2 ) ⋅ e2 = κ
⇒ S11 = S 22 = κ
Hence we get S ( e1 ) = S11 ⋅ e1 = κ e1 and S ( e2 ) = S 22 ⋅ e2 = κ e2
In any other direction at point p, if u is a unit tangent vector at p, u = α e1 + β e2 where α , β are
scalars. Then,
S ( u ) = S (α e1 + β e2 )
= α S ( e1 ) + β S ( e2 )
= ακ e1 + βκ e2
= κ ( α e1 + β e2 )
= κ ⋅u
Differential Geometry
206
This shows that the shape operator S at an umbilic point p is a scalar multiple of the constant
κ ( = κ1 , = κ 2 ) .
(2)
Let P be non umbilic point.
The normal curvature κ in u direction at P is given by
κ ( u ( θ ) ) = κ ( θ ) = C 2S11 + S 2 S22
Differentiating w.r.t. θ we get,
dκ (θ )
= −2CSS11 + 2CSS 22
dθ
= 2CS [ S 22 − S11 ]
For maxima or minima
Therefore
dκ (θ )
=0
dθ
2CS [ S22 − S11 ] = 0
⇒ 2CS = 0
⇒ C = 0 or S = 0
⇒θ =
π
or θ = 0
2
If sin θ = 0 i.e. θ = 0 then u ( θ ) = e1 which is a principal direction and the normal curvature κ ( e1 )
is maximum in this direction and κ ( e1 ) = κ1 (max. value).
Hence κ (θ ) is minimum when θ =
π
π 
i.e. when u   = e2 . And the minimum value of the normal
2
2
curvature is given by κ ( e2 ) = S ( e2 ) ⋅ e2 = S22 = κ 2 , say..
But we have S ( e1 ) = S1 1e1 and S ( e2 ) = S 22 e2 .
i.e. S ( e1 ) = κ1e1 and S ( e2 ) = κ 2 e2 where κ1 and κ 2 are maximum and minimum values of normal
curvature at P.
Note : If κ1 and κ 2 are principal curvatures at point p then normal curvature at p in any directionθ
given by
κ (θ ) = cos 2 θ ⋅ κ1 + sin 2 θ ⋅ κ 2
This formula is called Eulers formula. for normal curvature at p.
Differential Geometry
207
Corollary : If κ1 and κ 2 are principal curvatures and e1 and e2 are principal vectors of a surface
M ⊆ ¡3 , then for any unit tangent vector u at P
u = e1 cos θ + e2 sin θ
And the normal curvature of M in u direction is given by
κ ( u ) = κ1 ⋅ cos 2 θ + κ 2 ⋅ sin2 θ
Proof : Since e1 and e2 are linearly independent vectors of TP (M ) , the set { e1 , e2 } forms a basis
TP (M ) . Hence any tangent vector u at p is given by
u = e1 cos θ + e2 sin θ
where θ is the angle between u and e1 .
Also
κ ( u ( θ ) ) = κ (θ ) = cos2 θ ⋅ S11 + sin 2 θ ⋅ S22
But
S11 = S ( e1 ) ⋅ e1 = κ1 , maximum value of κ
and
S22 = S ( e2 ) ⋅ e 2 = κ 2 , minimum value of κ
Hence
κ ( u ) = κ1 ⋅ cos 2 θ + κ 2 ⋅ sin2 θ
Quadratic Approximation of a Surface :
Let M be a surface in ¡ 3 . We assume that
(1)
The point p is at the origin
(2)
The XY - plane is tangent to the surface at p.
(3)
The X-axis and the Y-axis are the principal directions.
Near the point p, M can be expressed as M : z = f (x, y).
By Taylor's theorem about origin,
f (x , y ) = f (0,0) + xf x (0,0) + yf y (0,0) +
+
Differential Geometry
1 2
 x f xx (0,0) + 2 xyf xy (0,0) + y 2 f yy (0,0)  + ...

2! 
208
z
(x, y, f(x, y))
y
u2
u1
(x, y)
x
Let
f x (0,0) = f x0 ,
f y (0,0) = f y0 ,
f xx (0,0) = f xx0 ,
f yy (0,0) = f yy0 ,
f xy (0,0) = f xy0
By neglecting higher order derivative terms from the equation we get,
1
f (x , y ) : f (0,0) + xf x0 + yf y0 +  x 2 f xx0 + 2 xyf xy0 + y 2 f yy0 
2
The surface M is tangent to XY- plane at the origin (p).
f y0 = 0
Therefore
f x0 = 0 ,
And also
f (0,0) = 0
Hence
f (x , y ) :
.
1 2 0
 x f xx + 2 xyf xy0 + y 2 f yy0 

2
Now for tangent vectors u1 ≡ (1,0,0) and u2 ≡ ( 0,1,0) at p = 0, we have
S ( u1 ) = − ∇ u U = f xx0 u1 + f xy0 u2
1
S ( u2 ) = − ∇ u U = f yx0 u1 + f yy0 u2
2
But u1 and u2 are principal vectors, therefore
S ( u1 ) = κ1u1 and S ( u2 ) = κ 2 u2
Differential Geometry
209
Hence we get
κ1u1 = f xx0 u1 + f xy0 u2
κ 2u2 = f yx0 u1 + f yy0 u2
f yy0 = κ 2 ,
⇒ f xx0 = κ1 ,
f xy0 = f yx0 = 0
Thus
1 2
 x κ1 + y 2κ 2 
2
f (x , y ) :
1
⇒ f (x , y ) = κ1 x2 + κ 2 y 2 
2
This is called quadratic approximation of f.
The shape of the surface M near p is same as that of the surface,
¶ : z = 1 (κ x 2 + κ y 2 )
M
1
2
2
¶ is called quadratic approximation of M near p.
M
Worked examples .......................................................................................................................•
Example 4 : Find the quadratic approximation near origin for the surface
M : z = ex
2
+ y2
−1
Solution :
M : z = ex
2
+ y2
−1
Let
f (x , y ) = e x
2
+ y2
−1
Differentiating w. r. t. x, y we get
f x = 2 xe x
+ y2
2
f xx = 2 xe x
2
2
= 2e x
+ y2
⇒ f x0 = 0 ,
Differential Geometry
+ y2
,
2
f y = 2 ye x
(2 x) + 2e x
2
+ y2
+ y2
[ 2 x 2 + 1]
f xx0 = 2
210
2
+y2
Similarly
f yy = 2e x
And
f y0 = 0 ,
Hence
κ1 = f xx0 = 2
( 2 y 2 + 1)
f yy0 = 2
κ 2 = f yy0 = 2
Therefore the quadratic approximation is
¶ : z = 1 (κ x 2 + κ y 2 )
M
1
2
2
=
1
(2 x2 + 2 y2 )
2
¶ : z = x2 + y2
M
EXERCISE ....................................................................................................•
Find the quadratic approximation near the origin for the surfaces
(i)
M : z = logcos x − logcos y
(ii)
M : z = ( x + 3y )
(iii)
M : z = ( x + 3y )
3
2
ANSWER .......................................................................................................•
(i)
2
2
¶:z= y −x
M
2
(ii)
¶:z=0
M
(iii)
¶ : z = x2 + 9 y2
M
•.............................................................................................................................•
Differential Geometry
211
Unit III : Gaussian Curvature and Mean Curvature
Introduction :
A linear transformation can be expressed as a matrix whose rows represents linearly independent set
of vectors called basis vectors. For a given linear transformation the value of the determinant of the
matrix and the sum of the diagonal elements of the matrix is constant irrespective of basis. Hence for a
given transformation the determinant and the trace are unique. Now we discuss the geometrical meaning
of the characteristic values and characteristic vectors of the shape operator.
Definition : The shape operator S p : TP ( M ) → TP ( M ) is a linear transformation. For each point
p ∈ M the Gaussian curvature K(P) of M at p is defined as the determinant of the shape operator M
at p.
i.e.
K (p) = det S
The mean curvature of M at p is defined as
H=
1
trace of S
2
Gaussian and Mean Curvatures in terms of Principal Curvatures :
Lemma : If κ1 and κ 2 are principal curvatures at a point p ∈ M then the Gaussian curvature and
mean curvature are respectively given by
K ( p ) = κ1 ⋅ κ 2 and H ( p ) =
κ1 + κ 2
2
Proof : If e1 and e2 are principal vectors at a point P ∈ M then { e1 , e2 } forms a basis of TP ( M ) .
Also
S ( e1 ) = κ1 ⋅ e1 and S ( e2 ) = κ 2 ⋅ e2
We write these equations in matrix form as
 S ( e1 )   κ1 0  e1 

=
 
 S ( e2 )   0 κ 2  e2 
Therefore matrix for S is
0
κ
S= 1

 0 κ2 
⇒ det S = κ1 ⋅ κ 2
Differential Geometry
212
By definition K ( p ) = κ1 ⋅ κ 2 .
Similarly trace of S = κ1 + κ 2 .
Thus the mean curvature of M becomes
H ( p) =
κ1 + κ 2
2
Note : Sign of the Gaussian curvature
Let p ∈ M be a point of a surface M ⊆ ¡3
(i)
If K ( p ) > 0 then κ1( p) and κ 2 ( p) have same sign. Therefore the normal curvature
κ ( u ) > 0 or κ ( u ) < 0 for all unit vectors u at p. Thus M has similar bending from the tangent plane
TP ( M ) in all directions.
UU(P)
( p)
U
U(P)
( p)
e22
e2
P
ee11
e11
κ1(P) < 0, κ2(P) < 0,
κ1(P) > 0, κ2(P) > 0,
The quadratic approximation of M near P is the paraboloid,
2 z = κ1 ( p ) x 2 + κ2 ( p) y 2
(ii)
If K ( p ) < 0 then the principal curvatures κ1( p) and κ 2 ( p) have opposite signs. The quadratic
approximation of M near p is hyperboloid.
U
( p)
U(P)
κ1( p )< 0
κ2 ( p) > 0
P
ee22
ee11
Differential Geometry
213
(iii) If K ( p ) < 0 then there are two cases
(i)
Both κ1( p ) = 0 and κ 2 ( p ) = 0 .
In this case quadratic approximation near p is z = 0. which is XY-plane.
(ii)
Only one principal curvature is zero, say
κ1( p ) ≠ 0 and
κ2 ( p) = 0
1
In this case quadratic approximation of M near p is z = κ1 x2 .
2
So M is trough-shaped near p.
UU(P)
( p)
e22
P
e11
Example : For a torus of revolution T, at points on the outer half of T, the surface bends away from
its tangent plane and hence K > 0. Near each point of inner half plane the surface is saddle shaped and
hence K < 0. And near each point on the two circles (top and bottom) the torus is trough-shaped,
hence K = 0.
Top Circle
The following lemma is useful in computation of Gaussian and mean curvature.
Lemma : If v and w are linearly independent tangent vectors at a point p of M then.
(i)
S ( v ) × S ( w ) = K ( p) ( v × w)
(ii)
S ( v ) × w + v × S ( w) = 2 H ( p) ( v× w)
Differential Geometry
214
Proof : Since v and w are linearly independent tangent vectors, they forms a basis for TP ( M ) . And
therefore we can write for a vector S ( v ) ∈ TP ( M ) .
S ( v ) = av + bw
S ( w) = cv + dw
 S ( v )   a b  v 
⇒
=
 
 S ( w )   c d  w 
a
⇒ matrix of S =  c

And det s = ad – bc,
b
d  w. r. t. the basis v , w .
trace S = a + d
Hence K ( p ) = ad − bc and H ( p) =
1
(a + d )
2
Now
S ( v ) × S ( w) = ( av + bw)× ( cv + dw )
= adv × w + bcw × v
= ( ad − bc ) v × w
⇒ S ( v ) × S ( w) = K (p )v × w
Now consider
S ( v ) × w + v × S ( w ) = ( av + bw) ×w + v × ( cv + dw)
= av × w + dv + w
= (a + d)v ×w
S ( v ) × w + v × S ( w) = 2 H ( p) ( v× w)
Note : If V and W are tangent vector fields that are linearly independent at each point of an oriented
region, then we have vector field equations
S (V ) × S (W ) = K (V × W )
S (V ) ×W + V × S (W ) = 2 HV × W
Differential Geometry
215
Taking dot product with (V × W ) we have,
 S (V ) × S (W ) ⋅ ( V × W ) = K (V ×W ) ⋅ (V × W )
And
 S (V ) × W  ⋅ ( V × W ) + W × S (W ) ⋅ (V × W ) = 2 H ( V × W )⋅ (V × W )
We have Lagrange's identity
(V ×W ) ⋅ ( a ×b ) = V ⋅ a V ⋅ b
W ⋅ a W ⋅b
Using this identity we get
K=
S (V ) ⋅ V
S (V ) ⋅W
S (W ) ⋅ V
S (W ) ⋅W
V ⋅V
V ⋅W
and
W ⋅ V W ⋅W
K=
S (V ) ⋅ V
S (V ) ⋅W
W ⋅V
W ⋅W
2
V ⋅V
+
V ⋅V
V ⋅W
S (W ) ⋅ V S (W ) ⋅W
V ⋅W
W ⋅ V W ⋅W
Once K is and H are known we can calculate the principal curvatures κ1 and κ 2 .
Corollary : On an oriented region in M, the principal curvatures κ1 and κ 2 are given by
κ1 , κ 2 = H ± H 2 − K
Proof : We know that κ1 ⋅ κ 2 = K and κ1 + κ 2 = 2H .
This shows that κ1, κ 2 are the roots of the quadratic equation
t 2 − (κ1 + κ2 ) t + κ1κ 2 = 0
t 2 − 2 Ht + K = 0
Differential Geometry
216
The roots are,
t=
2 H ± 4H 2 − 4K
2
⇒ t = H ± H 2 −K
⇒ κ1 , κ 2 = H ± H 2 − K
Flat Space and Minimal Space :
Definition (1) : A surface M in ¡ 3 is called flat surface if its Gaussian curvature is zero.
Definition (2) : M is called minimal surface if its mean curvature is zero.
Note :
(i)
For a plane surface κ1 = κ2 = 0 . Hence κ = 0 .
(ii)
For cylinder, κ1 = 0 . Hence K = 0. Thus cylinders are also flat surfaces.
(iii)
For minimal surfaces H = 0 ⇒
κ1 + κ2
=0
2
⇒ κ1 = −κ 2
Therefore K = κ1 ⋅κ 2 = −κ12 < 0 .
Thus for minimal surface the Gaussian curvature is always negative.
(iv)
1
For a sphere of radius r, κ1 = κ2 = − 1 . Hence K = 2 .
r
r
Thus the sphere has constant positive Gaussian curvature. Smaller the sphere, the larger its Gaussian
curvature.
Worked examples ............................................................................................................................•
Example 1 : The characteristic polynomial of an arbitrary operator S is defined as
P(κ ) = det ( A − κ I ) ,
where A is any matrix of S. Show that the characteristic polynomial of shape operator S is given by
P(κ ) = κ 2 − 2H κ + K
Differential Geometry
217
Solution : The principal vectors e1 and e2 are linearly independent and
S ( e1 ) = κ1 ⋅ e1 and S ( e2 ) = κ 2 ⋅ e2
0
κ
⇒ Matrix of S =  1

 0 κ2 
det S = κ1κ 2
and
trace S = κ1 + κ2
⇒ K = κ1κ2
and
2H = κ1 + κ 2
Now characteristic polynomial of S,
P(κ ) = det ( S − κ I )
 1 0 
 κ1 0 
= det  
−κ 


 0 1  
  0 κ2 
κ − κ
= det  1
 0
0 
κ 2 − κ 
= ( κ1 − κ )( κ2 − κ)
= κ1κ 2 − (κ1 + κ 2 ) κ + κ 2
= κ 2 − ( κ1 +κ 2 )κ + κ1κ 2
P(κ ) = κ 2 − 2H κ + K
•....................................................................................................................................................•
Computational Techniques :
Patchwise computation :
Let S be the shape operator of a surface M of ¡ 3 . Let K be the Gaussian curvature and H be the
mean curvature and κ1, κ 2 be the principal curvatures along the principal directions e1 and e2 .
(I)
If X : D → M is a patch in M then we define
E = Xu ⋅ Xu
F = X u ⋅ Xv = X v ⋅ Xu
and
G = Xv⋅ Xv
where Xu and Xv are velocity vectors of u-parameter curve and v-parameter curves of X.
E, F, G are called warping functions of the patch X. They measures the way D is transformed to X(D)
in M.
Differential Geometry
218
(II)
If v and w are any tangents at a point of X(D) then
v = v1 X u + v2 X v ,
w = w1 X u + w2 X v
v ⋅ w = ( v1X u + v2 X v ) ⋅( w1 Xu + w2 Xv )
Then
= v1 w1 X u ⋅ X u + ( v1 w2 + w1 v2 ) Xu ⋅ X u + v2 w2 X v ⋅ X v
v ⋅ w = v1w1 E + F ( v1 w2 + w1v2 ) + Gv2 w2
Thus E, F, G completely determines the dot product of tangent vectors.
(III) Since Xu and Xv are tangent vectors to the surface M at X (u, v) then X u × X v is normal to the
surface M at X (u, v).
Consider,
Xu × Xv
2
= ( Xu × Xv )⋅ ( X u × X v )
= Xu
2
Xv
2
− ( Xu ⋅ Xv )
2
= ( X u ⋅ X u )⋅ ( X v ⋅ X v ) − ( X u ⋅ X v )
= EG − F 2
If X is regular then X u × X v ≠ 0
⇒ EG − F 2 ≠ 0
Hence a unit normal function on M is given by
U=
(IV)
Xu × X v
Xu × X v
If X ( u, v ) = ( x1 ( u, v ) , x2 ( u, v ) , x3 ( u, v ) )
then
 ∂ 2 x ∂2 x ∂ 2 x 
Xuu =  21 , 22 , 23 
 ∂u ∂ u ∂ u 
 ∂ 2 x1 ∂ 2 x2 ∂ 2 x3 
Xuv = Xvu = 
,
,

 ∂u∂v ∂u∂v ∂u∂v 
Differential Geometry
219
2
 ∂ 2 x ∂2 x ∂ 2 x 
Xvv =  21 , 22 , 23 
 ∂v
∂v
∂v 
Xuu and Xvv gives acceleration of u-parameter and v-parameter curves and Xuv = Xvu, gives both
covariant derivative of Xu in Xv direction and Xv in Xu direction.
(V)
If S is a shape operator then we define
l = S ( Xu ) ⋅ Xu
m = S ( X u ) ⋅ Xv = S ( X v ) ⋅ X u
n = S ( Xv ) ⋅ X v
Since X u , X v forms a basis of TP ( M ) the shape operator S can be determined uniquely in terms of
l, m, n.
The following lemma gives simple expression for Gaussion and mean curvature for a patch.
Lemma : If X is a patch in M ⊂ ¡ 3 then
ln − m2
K(X ) =
EG − F 2
H (X ) =
and
Gl − En − 2 Fm
2 ( EG − F 2 )
Proof : For tangent vectors v and w at a point p, we have
S ( v ) × S ( w) = Kv × w
. . . (1)
S ( v ) × w + v × S ( w) = 2 Hv × w
. . . (2)
Taking dot product with v × w and using the identity
( a × b ) ⋅( v ×w ) =
a ⋅v a ⋅ w
b ⋅v b ⋅w
we obtain
[ S ( v ) × S ( w) ] (v × w) = K ( v × w) ⋅ ( v × w)
S (v ) ⋅ v
S ( w) ⋅ v
Differential Geometry
S (v ) ⋅ w
v ⋅v
=K
w⋅ v
S (w ) ⋅ w
v ⋅w
w⋅w
220
K=
S ( v ) ⋅v
S ( w) ⋅ v
S (v ) ⋅ w
S ( w) ⋅ w
v ⋅v
v ⋅w
w ⋅v
w⋅ w
Replace v and w by X u and X v respectively
Therefore
S ( Xu ) ⋅ Xu
S ( X v ) ⋅ Xu
K(X ) =
Xu ⋅ Xu
Xv ⋅ Xu
S ( Xu ) ⋅ Xv
S ( Xv ) ⋅ Xv
Xu ⋅ X v
Xv ⋅ Xv
l m
m n
=
E F
F G
K(X ) =
ln − m2
EG − F 2
Similarly taking dot product of with equation (2) we get
[ S ( v ) × w + v ] ⋅( v × w) + [v × S ( w )] ⋅ ( v × w ) = 2 H ( v × w )⋅ ( v × w)
S ( v ) ⋅v
S ( w) ⋅ v
S (v ) ⋅ w
v ⋅v
v ⋅w
v ⋅v
+
= 2H
w⋅ v
S ( w) ⋅ w S ( w ) ⋅ v S ( w) ⋅ w
v ⋅w
w⋅ w
Replace v and w by X u and X v respectively
S ( Xu ) ⋅ Xu
S ( X v ) ⋅ Xu
⇒
Xu ⋅ Xu
S ( Xu ) ⋅ Xv
+
S ( X v ) ⋅ X v S ( X v ) ⋅ Xu
l m E
+
F G m
F
E
= 2H
n
F
F
G
⇒ Gl − Fm + En − Fm = 2 H ( EG − F 2 )
⇒H=
Gl + En − 2 Fm
2 ( EG − F 2 )
Differential Geometry
221
Xu ⋅ Xv
S ( Xv ) ⋅ X v
= 2H
Xu ⋅ Xu
Xv ⋅ Xu
Xu ⋅ Xv
Xv ⋅ X v
Lemma : If X is patch in M ⊂ ¡ 3 and U is a unit normal vector field to M, then
l = U ⋅ X uu
m = U ⋅ X uv
n =U ⋅ X vv
Proof : Since U is a unit normal vector field to M,
U ⋅ X u = 0 and U ⋅ X v = 0
Differentiating w. r. t. v and u respectively, we get
U v ⋅ Xu + U ⋅ X uv = 0
U v ⋅ Xv + U ⋅ X uv = 0
But
U v = ∇ Xv U = −S ( X v )
U u = ∇ XuU = −S ( X u )
Also
X uv = X vu
Hence we get,
U ⋅ X uv = −U v ⋅ X u
= S ( Xv ) ⋅ X u
U ⋅ X vu = −U u ⋅ X v
= S ( Xu ) ⋅ Xv
But
X uv = X vu .
Hence we get,
U ⋅ X uv = S ( X u ) ⋅ X v = S ( X v ) ⋅ X u
⇒ U ⋅ X uv = m
Next
U ⋅ Xu = 0
Differential Geometry
222
Differentiating this equation w. r. t. u we get,
U u ⋅ Xu + U ⋅ X uu = 0
⇒ U ⋅ X uu = −U u ⋅ X u
⇒ U ⋅ X uu = S ( X u ) ⋅ X u
⇒ U ⋅ X uu = l
Similarly
U ⋅ Xv = 0
Differentiating this equation w. r. t. v we get,
U v ⋅ Xv + U ⋅ X vv = 0
⇒ U ⋅ X vv = −U v ⋅ X v
⇒ U ⋅ X vv = S ( X v ) ⋅ X v
⇒ U ⋅ X vv = n
Thus we get,
l = U ⋅ X uu
m = U ⋅ X uv
n =U ⋅ X vv
Worked examples ........................................................................................................................•
Example 2 : A patch X in M is orthogonal provided F = 0. Show that in this case
S ( Xu ) =
l
M
Xu +
Xv
E
G
S ( Xv ) =
m
n
Xu + Xv
E
G
Solution : Let X : D → M be a patch.
If S is the shape operator, then the real valued functions defined on D are given by
l = S ( Xu ) ⋅ Xu
Differential Geometry
223
m = S ( X u ) ⋅ Xv
n = S ( Xv ) ⋅ X v
. . . (1)
where X u and X v give a basis for TP ( M ) at each point of X(D).
Since S ( X u ) , S ( X v ) ∈ TP ( M ) , therefore it can be expressed as a linear combination of the basis
vector X u and X v . Thus we have,
S ( Xu ) = α Xu + β Xv
. . . (2)
Taking scalar product of (2) with X u we get
S ( Xu ) ⋅ Xu = α X u ⋅ X u + β Xv ⋅ Xu
Since the patch X in M is orthogonal ⇒ X u ⋅ X v = F = 0 and
E = Xu ⋅ Xu
Now using (1) we write equation (3) as
l = αE
⇒α =
l
E
Similarly, on multiplying equation (2) by X v , we obtain
β=
n
G
Substituting these values of α and β in the equation (2) we get
S ( Xu ) =
l
m
X u + Xv
E
G
Similarly we obtain
S ( Xv ) =
Differential Geometry
m
n
Xu + Xv
E
G
224
. . . (3)
Example 3 : Show that the curve
α (t ) = X (α1 ( t) ,α 2 ( t) )
a≤t≤b
has length
b
L (α ) = ∫
1
 Eα '2 + 2 Fα 'α ' + Gα '2  2 dt
 1
1 2
2 
a
Solution : Let α be a curve defined by
where
α (t ) = X ( u, v )
. . . (1)
u = α1 (t ) , v = α 2 (t )
. . . (2)
The velocity vector to the curve α is given by
α '(t ) = X u
du
dv
+ Xv
dt
dt
⇒ α '(t ) = X uα1' + X vα 2'
. . . (3)
⇒ α '(t ) = α ' ⋅α ' = ( X uα1' + X vα 2' ) ⋅ ( X uα1' + X vα 2' )
2
2
2
⇒ α '(t ) = X u ⋅ X uα1' + α1'α 2' ⋅ ( X u ⋅ X v + X v ⋅ X u ) + α 2' X v ⋅ X v
2
2
2
2
⇒ α '(t ) = α1' E + 2 Fα1'α 2' + Gα 2'
. . . (4)
The length of the curve α is defined by
b
L (α ) = ∫ α '(t ) dt
2
a
b
⇒ L (α ) = ∫
1
 Eα '2 + 2 Fα 'α ' + Gα '2  2 dt
 1
1 2
2 
. . . (5)
a
Example 4 : Compute Gaussian curvature and mean curvature H for the surface given by
X ( u, v ) = (u cos v ,u sin v, bv ) ,
b≠0
Solution : The patch X is given by,
X ( u, v ) = (u cos v ,u sin v, bv )
Differential Geometry
225
Differentiating w. r. t. u and v we get
X u = ( cos v,sin v, 0 )
X v = ( −u sin v ,u cos v, b )
Therefore
E = X u ⋅ X u = cos 2 v + sin 2 v = 1
F = X u ⋅ Xv = 0
G = X v ⋅ X v = u 2 sin 2 v + u 2 cos 2 v + b 2 = u2 + b2
Now to find unit normal vector field on a surface, we find
U1
U2
U3
X u × X u = cos v
sin v
0
−u sin v u cos v b
= b sin vU 1 − bcos vU 2 + U 3
= ( b sin v , −b cos v, u )
And
X u × X u = b 2 sin 2 v + b2 cos 2 v + u 2
= b2 + u2
Thus unit normal vector field on surface is given by
Xu × Xu
Xu × Xu
U=
⇒U =
Now
( b sin v, −b cos v, u )
b +u
2
2
X u = ( cos v,sin v, 0 )
Differentiating X u w. r. t. u we get
X uu = ( 0,0,0)
and differentiating X u w. r. t. v we get
X uv = ( − sin v ,cos v, 0 )
Differential Geometry
226
Similarly,
X v = ( −u sin v ,u cos v, b )
Differentiating w. r. t. v we get
X vv = ( −u cos v, −u sin v, 0 )
Next
l = U ⋅ X uu
(since X uu = 0 )
⇒l =0
m = U ⋅ X uv
⇒m =
( b sin v, −b cos v , u )
b +u
2
2
⋅ ( − sin v,cos v, 0 )
−b
⇒m =
b2 + u 2
n =U ⋅ X vv
⇒n=
( b sin v, −b cos v, u )
b +u
2
2
⋅ ( −u cos v, −u sin v, 0 )
⇒n =0
Thus we have
E = 1, F =0 , G = b2 + u 2 , l = 0, m =
Therefore Gaussian curvature,
K=
ln − m2
EG − F 2
 −b 
−

b2 + u 2 

=
b2 + u2
K=
Differential Geometry
2
− b2
( b 2 + u 2 )2
227
−b
b +u
2
2
,n = 0
And mean curvature,
H=
Gl + En − 2 Fm
2 ( EG − F 2 )
=0
(since l = m = n = 0)
Since H = 0, the helicoid surface is minimal surface.
Example 5 : The saddle surface M : z = xy is described by Monge patch
X ( u, v ) = ( u , v , uv )
.
Find and H.
Solution :
X ( u, v ) = ( u , v , uv )
⇒ X u = (1,0, v )
X v = ( 0,1, u )
X uu = ( 0,0,0)
X vv = ( 0,0,0)
X uu = ( 0,0,1)
Therefore,
E = X u ⋅ X u = 1 + v2 , F = X u ⋅ X v = uv , G = X u ⋅ X v = 1 + u 2
Next
U1
Xu × X v = 1
0
U2 U 3
0
v
1
u
= −vU 1 − uU 2 + U 3
X u × X v = ( −v, −u,1)
Therefore
X u × X v = ( −v, −u,1)
X u × X v = v2 + u2 + 1
Differential Geometry
228
Hence unit normal vector field on the surface is given by
U=
=
Xu × Xu
Xu × Xu
( −v, −u ,1)
1+u +v
2
2
Now
( Since X uu = 0 )
l = U ⋅ X uu = 0
m =U ⋅ X uv =
⇒m=
( −v, −u,1) ⋅ ( 0,0,1)
1+ u + v
2
2
1
1+ u2 + v2
( Since X vv = 0 )
n =U ⋅ X vv = 0
Therefore
K=
ln − m
2
EG − F 2
⇒K=
=
−
1
1+ u2 + v2
(1 + u 2 )(1 + v 2 ) − u 2v 2
1
(1 + u 2 + v 2 )2
Similarly
H=
Gl + En − 2 Fm
2 ( EG − F 2 )
−
=
H=
Differential Geometry
2uv
1 + u2 + v2
2  (1 + u 2 )(1 + v 2 ) − u 2v 2 
−uv
3
(1 + u 2 + v 2 ) 2
229
Example 6 : Show that the Gaussian curvature of a surface of revolution M, obtained by revolving the
profile curve.
α (u ) = ( g (u ), h( u),0 )
in the upper half of the xy plane about x-axis is given by.
K=
g ' ( h ' g "− g ' h ")
h ( g '2 + h '2 )
2
Find the value of K if the profile curve is a unit speed curve.
Solution :
The profile curve in the xy plane is given by
α (u ) = ( g (u ), h( u),0 )
... (1)
where (g, h) any arbitrary point on the profile curve. Let this point (g, h) on the profile curve be rotated
through an angle v about x-axis and let it be reached to the point (x, y, z) in space. The surface of
revolution is given by
X :D → M
define by
0
1

X (u , v) =  0 cos v
 0 sin v

  g ( u) 


− sin v   h (u) 
cos v   0 
0
 x   g (u ) 
  

X (u , v) =  y  =  h cos v 
 z   h sin v 
  

⇒ X (u , v) = ( x, y , z ) = ( g (u ), h( u)cos v, h( u)sin v )
... (2)
The straight forward calculation give
X u = ( g '(u), h '( u)cos v , h '( u)sin v )
... (3)
X v = ( 0, −h(u )sin v , h( u )cos v )
... (4)
Thus from the definitions of warping functions of the patch, we obtain,
E = ( g '2 + h '2 ) , F = 0, G = h2
Differential Geometry
230
... (5)
Now we find from equations (3) and (4)
X u × X v = ( hh ', − g ' h cos v, − g ' h sin v )
... (6)
X u × X v = h h '2 + g '2
... (7)
This gives
Hence the unit normal vector field to the surface M is given by
U=
Xu × Xv
( h ', − g 'cos v, − g 'sin v )
=
Xu × Xv
h '2 + g '2
... (8)
Now the acceleration vectors to the u-parameter curve and v-parameter curve are obtain by
differentiating equations (3) and (4) with respect to u and v we get
X uu = ( g "(u ), h "( u)cos v , h "sin v ) ,
X vv = ( 0, − h(u )cos v, − h(u )sin v ) ,
X uv = ( 0, −h '( u)sin v , h '(u )cos v ) .
The definitions
l = U ⋅ X uu , m = U ⋅ X uv , n =U ⋅ X vv ,
give
l=
1
h '2 + g '2
( h ' g "− g ' h ")
m =0
n=
hg '
... (9)
h '2 + g '2
Substituting these values in the expression for Gaussian curvature of the surface M
K=
ln − m2
EG − F 2
We obtain
K=
Differential Geometry
g ' ( g ' g "− g ' h ")
h ( g '2 + h '2 )
... (10)
2
231
If however, the profile curve has unit speed, then we have α '(u ) = 1 . This gives
g '2 + h '2 = 1 .
Hence Gaussian curvature of the unit speed curve becomes
K=−
h"
h
... (11)
Example 7 : Show that the maximum and minimum values of Gaussian curvature of torus are
respectively
1
1
and −
.
r(R + r )
r(R + r )
Solution : We know the torus of revolution M is parametrized by
X (u , v) = ( h(u )cos v, h( u)sin v, g( u) )
where,
... (1)
h (u ) = R + r cos u ,
g ( u) = r sin u
... (2)
Thus the tangent vectors to the u-parameter curve and v-parameter curve of X are obtain as
X u = ( −r sin u cos v, −r sin u sin v , r cos u )
X v = ( − ( R + r cos u ) sin v, ( R + r cos u ) cos v, 0)
... (3)
Similarly, the acceleration vectors are given by
X uu = ( − r cos y cos v,− r cos u sin v, − r sin u ) ,
X vv = ( − ( R + r cos u ) cos v, − ( R + r cos u ) sin v,0 ) ,
Also we find
X uv = ( r cos u sin v, − r cos u cos v, 0)
... (4)
From these equations we find
Xu ⋅ Xu = E = r2
X u ⋅ X v = F = 0 and
X v ⋅ X v = G = 0r 2 ( R + r cos u )
2
From equation (3) we find
X u × X v = −r ( R + r cos u ) ( cos u cos v,cos u sin v,sin u )
Differential Geometry
232
... (5)
The norm of this vector gives
X u × X v = r ( R + r cos u )
... (6)
The unit vector normal to the surface M is
U=
Xu × Xv
= − ( cos u cos v,cos u sin v,sin u )
Xu × Xv
... (7)
Now the quantities l, m, n are obtain as
l = U ⋅ X uu = r ,
m = U ⋅ X uv = 0 ,
n =U ⋅ X vv = cos u ( R + r cos u )
... (8)
Hence the Gaussian curvature of the surface M given by
ln − m2
K=
EG − F 2
becomes,
K=
cos u
r ( R + r cos u )
... (9)
Note on the outer equator u = 0, hence cos u = 1, and on the inner equator u = π , hence
cos u = –1. Since K is positive on the outer half of the torus and negative on the inner half. Therefore
K has maximum value when cos u = 1 and minimum value at cos u = – 1.
Hence
Kmax =
and
1
r(R + r )
Kmin = −
−1
r( R − r)
Note : K is zero on the top and bottom circle i.e. when u = ±
Differential Geometry
233
π
.
2
Example 8 : Show that M : z = f ( r ) , where r = x 2 + y 2 > 0 and f is a differentiable function, is
a surface of revolution and that its Gaussian curvature is
K=
f '(r ) f "( r )
r (1 + f '2 (r ) )
2
Solution : The surface of revolution is given by
M : z = f (r)
where r = x 2 + y 2 > 0 is the usual polar coordinate function.
{
This gives
M = (x , y, z) | z = f
i.e.
M = x, y, f
(
(
(
x2 + y2
x2 + y2
)}
))
... (1)
The surface M has parametrization
X (u , v) = (u cos v ,u sin v , f ( u) )
... (2)
Here we have used u for r and v for θ .
Differentiating equation (2) partially with respect to u and v respectively we get
X u = ( cos v,sin v, f '(u ) )
X v = ( −u sin v ,u cos v, 0 )
... (3)
Consequently, we find
E = X u ⋅ X u = 1 + f '2 (u ) ,
F = X u ⋅ Xv = 0 ,
G = X v ⋅ Xv = u2 .
This gives
EG − F 2 = u 2 (1 + f '2 )
... (4)
Now we compute
X u × X v = u ( − f '(u )cos v, − f '(u )sin v ,1)
The norm of this vector becomes
X u × X v = u 1 + f '2 (u )
Differential Geometry
234
The unit vector normal to the surface M becomes
U=
Xu × X v
Xu × X v
⇒U =
1
( − f '(u )cos v , − f '( u)sin v,1)
1 + f ' (u )
2
Differentiating equation (3) partially with respect to u and v we obtain
X uu = ( 0,0, f "(u ) ) ,
X vv = ( −u cos v, −u sin v, 0 ) ,
X uv = ( − sin v ,cos v, 0 ) .
The definition of numbers l, m,n gives
f "(u )
l = U ⋅ X uu =
1 + f '2 (u )
m = U ⋅ X uv = 0 ,
n =U ⋅ X vv =
1
1 + f '2 (u )
( uf '( u) ) .
Using these values in the expression for Gaussian curvature
ln − m2
K=
,
EG − F 2
we get
K=
f '( u ) f "(u )
u (1 + f '2 (u) )
2
Now replacing u by r we get
K=
Differential Geometry
f '(r ) f "( r )
r (1 + f '2 (r ) )
2
.
235
... (5)
EXERCISE ...................................................................................................•
1.
If X is a geographical patch on a sphere of radius r, defined by
X (u , v) = (r cos v cos u , r cosv sin u , r sin v ) ,
prove that K =
2.
1
.
r2
Show that the image of a patch
X (u , v) = ( u , v,logcos v − logcos u )
is a minimal surface with Gaussian curvature
K=
− sec 2 u sec 2 v
w4
where w2 = 1 + tan 2 u + tan 2 v .
3.
For a Monge patch
X (u, v) = (u, v, f (u, v))
show that
E = 1 + fu2 , F = fu fv , G = 1 + fv2 , l =
fuu
f
f
, m = uv , n = vv
w
w
w
1
where w = (1 + f 2 + f 2 ) 2
u
v
.
Find and H. Show also that X is flat iff f uu f vv − f vv2 = 0 .
X is minimal iff (1 + f u2 ) f vv + (1 + f v2 ) f uu − 2 fu fu fuv = 0
4.
Show that the maximum and minimum values of the Gaussian curvatures of the torus
M:
(
are respectively
Differential Geometry
)
2
x2 + y 2 − 4 + z 2 = 4
1
1
and − .
12
4
236
5.
Prove that for the surface


u3
v3
2
X (u , v) =  u − + uv , v − + u 2v, u 2 − v 2 

3
3

the principal curvatures are
κ1 =
2
(1 + u 2 + v )
2 2
, κ2 = −
2
(1 + u 2 + v 2 ) 2
and thus it is a minimal surface. Also show that its asymptotic curves are u + v = constant,
u – v = constant.
6.
Show that the curvature of the monkey saddle M : z = x 3 − 3 xy 2 is given by
K=
−36r 2
(1 + 9r 4 )2
,
where r = x 2 + y 2 .
7.
Compute the Gaussian curvature at an arbitrary point of the torus given by paramertrization
X (u , v) = ( ( a + b cos u ) cos v, ( a + b cos u ) sin v, b sin u ) ,
0 < u < 2π , 0 < v < 2π .
cos u
Ans. : K = (
b a + b cos u )
•.......................................................................................................................•
Differential Geometry
237
Unit II :
Formula for the Shape Operator when the Surface is given in the form
M : g(x, y, z) = C
Introduction :
We have seen that to compute the Gaussian and mean curvature of a surface M : g = c using patches
one must begin by explicity finding enough of patches to cover all of M. Computation of Gaussian
curvature K may thus be tedious even when the surface g (x, y, z) = c is simpler. Hence in the following
we find the formula for Gaussian curvature K and mean curvature H when M : g (x, y, z) = c. For this
purpose, in this unit, we first establish the formula for the shape operator when the surface is of the
form
g (x, y, z) = c.
Lemma : Find the expression for the shape operator on a surface M given by g (x, y, z) = c.
Proof : Let M :g (x, y, z) = c be a surface. The normal vector field on M is given by
∇g = ∑
i
∂g
Ui
∂xi
. . . (1)
Let us denote Z = ∇g .
Hence the normal vector field U on M becomes
U=
Z
. . . (2)
Z
If V is a tangent vector field to M, then by definition of the shape operator, we have
S (V ) =−∇V U
where
. . . (3)
 Z 
∇V U = ∇V 

 Z 
∇V U =
1
 1 
∇V Z +V 
⋅ Z
Z
 Z 
. . . (4)
(∇ ( f Y ) = V [ f ]Y + f ∇ Y )
V
Differential Geometry
238
V
U
1 ∇ z
||z|| V
⇒ ∇V U =
1
Z
∇ V Z − NV
 1 
where − N V = V 
⋅Z
 Z 
∇VU
. . . (5)
is a normal vector field.
Using (5) and (3) we obtain
TP(M)
S (V ) = −
1
V ||z|| z
1
Z
∇ V Z + NV
. . . (6)
This is the required expression for the shape operator.
Proposition : Let Z be a non vanishing normal vector field on M. If V and W are tangent vector
fields such that V ×W = Z . Then the expression for K and H are in the form
K=
( Z ⋅∇V Z × ∇W Z )
Z
H=
and
4
(
−Z ⋅ ∇V Z × W + V × ∇W Z
2 Z
)
3
Proof : Let V and W be tangent vector fields on M. Then the Gaussion curvature K and the mean
curvature H of a surface in terms of the shape operator are defined by
S (V ) × S (W ) = K ⋅(V ×W )
. . . (1)
S (V ) ×W + V × S ( W ) = 2 H ⋅ (V ×W )
. . . (2)
Multiply equation (1) and (2) by (V × W ) and using
Z =V × W
we obtain
K=
S (V ) × S (W ) ⋅ Z
Z
H=
and
2
Z ⋅  S (V ) ×W +V × S (W )
2 Z
Differential Geometry
2
239
. . . (3)
. . . (4)
However, we know the expression for the shape operator on the surface g(x, y, z) = c is given by
S (V ) = −
1
Z
∇ V Z + NV
. . . (5)
where − N V = V  1  ⋅ Z a normal vector field.
 Z 
. . . (6)
Using equation (5) in (3) we obtain
 1
 − Z ∇ V Z + NV
K = 
 ×  − 1 ∇ Z + N  ⋅ Z
W 
  Z W
 

Z
2
1
1
1
Z ⋅  2 ∇V Z ×∇W Z −
∇V Z × N W − NV ×
∇ W Z + N V × N W 
Z
Z
 Z

K=
2
Z
Since Z is non vanishing normal vector field on M and N V is also normal.
⇒ Z || N V and Z || N W
⇒ N V || N W
⇒ N V ×NW = 0
(
)
as Z N W
(
)
as Z NV
Also
Z ⋅ ∇V Z × NW = ∇V Z ⋅ ( N W × Z ) = 0
Similarly
Z ⋅ N V ×∇W Z = ∇W Z ⋅ ( Z × N V ) = 0
Hence we have
K=
Z ⋅∇V Z × ∇W Z
Z
. . . (7)
4
where Z = ∇g .
Now using equation (5) we write equation (4) as

1
Z ⋅  −
∇ Z + NV
Z V


H=
 ×W + V ×  − 1 ∇ Z + N  
W 

 Z W



2 Z
Differential Geometry
2
240
1
 1

Z ⋅ −
∇ V Z × W + N V ×W − V ×
∇W Z + V × N W 
Z
 Z

H=
2
2 Z
Since Z ⋅ ( N V × W ) = 0 and Z ⋅ ( V × N V ) = 0 as Z NV and Z N W
(
−Z ⋅ ∇V Z × W + V × ∇W Z
H=
2 Z
)
3
. . . (8)
Equation (7) and (8) give the required expression for the Gaussian curvature and mean curvature.
Worked examples .......................................................................................................................•
Example 1 : Find the Gaussian curvature for the ellipsoid given by
M:g=
x2 y2 z 2
+
+
=1
a 2 b2 c 2
Solution : The surface is ellipsoid defined by
M : g =1
where
g=
x2 y2 z2
+
+
a 2 b2 c 2
. . . (1)
Now the normal vector field to the ellipsoid M is its gradient and is given by
∇g = ∑
i
∂g
Ui
∂xi
= 2∑
i
⇒
xi
Ui
ai
1
x
∇g ≡Z = ∑ i U i
2
i ai
 x y z 
⇒ Z = 2 , 2 , 2 
a b c 
1
2
x
y
z
⇒ Z = 4 + 4 + 4 
a
b
c 
2
Differential Geometry
2
2
. . . (2)
241
If V = ∑ vi U i and W = ∑ wi U i are tangent vector fields to the surface, then we have
i
i
x 
∇V Z = ∑ V  2i  Ui
i
 ai 
1
V [ xi ] U i
ai2
⇒
=∑
⇒
= ∑∑
1
∂xi
v
Ui
2 j
∂x j
ai
⇒
= ∑∑
1
v jδ ij U i
ai2
⇒
i
i
i
∇V Z = ∑
i
j
j
1
v v v 
v U =  12 , 22 , 32 
2 i i
ai
a b c 
. . . (3)
1
w w w 
w U =  21 , 22 , 23 
2 i i
ai
a b c 
. . . (4)
Similarly we find
∇W Z = ∑
i
Now the expression for Gaussian curvature is given by
K=
Z ⋅∇V Z × ∇W Z
Z
where
. . . (5)
4
x
a2
v
Z ⋅ ∇V Z ×∇W Z = 12
a
w1
a2
y
a2
v2
a2
w1
a2
z
a2
v3
a2
w1
a2
x
1
= 2 2 2 v1
a b c
w1
⇒ Z ⋅ ∇V Z ×∇W Z =
Differential Geometry
1
a b c
2 2 2
y
z
v2
w2
v3
w3
( Z ⋅ V ×W )
242
. . . (6)
But we choose the vector fields V and W such that Z =V ×W ,
⇒ Z ⋅ ∇V Z ×∇W Z =
⇒ Z ⋅ ∇V Z ×∇W Z =
1
a b c
2 2 2
1
2 2 2
a b c
( X ⋅Z )
for X = (x, y, z)
( x, y , z ) 
x
a
2
,
y
b
2
⇒ Z ⋅ ∇V Z ×∇W Z =
 x 2 y2 z 2 
+ 
 +
a2 b2 c 2  a 2 b 2 c 2 
⇒ Z ⋅ ∇V Z ×∇W Z =
1
a 2b2 c 2
,
z 

c2 
1
Substituting this in equation (5) we get
h2
K= 2 2 2
a b c
where h =
. . . (7)
1
Z
•.....................................................................................................................................................•
Note : If a = b = c = r, then ellipsoid is a sphere of radius r and Gaussian curvature K becomes
K=
1
.
r2
EXERCISE ....................................................................................................•
1.
Show that the Gaussian curvature of hyperboid with one sheet
x2 y2 z2
+ − =1
a 2 b2 c 2
is
K =−
1
2 2 2
x 2 y2 z 2 
a b c  4+ 4− 4
a
b
c 
.
2
Show also that it is minimal surface.
Differential Geometry
243
2.
Show that the hyperboid of two sheets
x2 y2 z 2
− − =1
a2 b2 c2
has Gaussian curvature K =
3.
h4
1
, where h =
.
2 2 2
a b c
Z
Show that the Gaussian curvature of the Monkey saddle M : z = x 3 − 3 xy 2 is given by
K=
−36r 2
(1 + 9r 4 )
where r = x 2 + y 2
•.......................................................................................................................•
Differential Geometry
244
Unit III :
Special curves in a Surface :
We shall discuss some geometrically significant types of curves in a surface M ⊂ ¡ 3 . The main purpose
is to illustrate some of the ideas already introduced.
Principal Curves (Line of Curvature):
Definition : A regular curve α in a surface M of ¡ 3 is called a principal curve (or a line curvature) if
the velocity vector α ' of α always points in a principal direction.
Thus principal curve always travel in directions in which the bending of the surface M takes its extreme
values.
α’
α’
e1
P
β’
β
α
M
e2
Thus there are exactly two principal curves through each non umbilic point of M and these curves cut
orthogonally across each other.
The following lemma gives simple criterion for a curve to be principal and also gives principal curvature
along this curve.
Lemma : Let α be a regular curve in a surface M of ¡ 3 , and let U be a unit normal vector filed
restricted to α . Then
(i)
The curve α is principal if and only if U ' and α ' are collinear at each point.
(ii)
If α is principal curve, then the principal curvature of M, κ i in the direction of α ' is given by
κi =
Differential Geometry
α "⋅ U
α '⋅α '
245
Proof :
(i)
If α is a regular curve then we have
S (α ') = −U '
Therefore U ' and α ' are collinear ifff S (α ') and α ' are collinear. If S (α ') and α ' are collinear
than S (α ') = κα ' for some scalar k which implies that α ' is a principal direction and hence α is a
principal curve. Thus α is a principal curve if and only if α ' and U ' are collinear..
(ii)
Since α is principal curve the unit vector
α'
is a principal vector along the curve the curve.
α'
If κ i denote the principal curvature then
α'
 α' 
S
 = κi ⋅
α'
 α' 
Taking dot product of the equation with
α'
we get,
α'
α' α '
 α'  α '
S
= κi ⋅
⋅
⋅
α' α '
 α'  α '
⇒
⇒
S (α ') ⋅ α '
α'
S (α ') ⋅ α '
α'
⇒ κi =
But
2
2
= κi ⋅
= κi ⋅
α '⋅ α '
α'
2
α'
2
α'
2
S ( α ') ⋅ α '
α'
2
S (α ') ⋅ α ' = α "⋅U
where U is a unit normal vector field on α .
Hence
κi =
Differential Geometry
α "⋅ U
α'
2
246
Lemma : Let α be a curve cut from a surface M ⊂ ¡ 3 by a plane P. If the angle between M and P
is constant along then is a principal curve of M.
Proof : Let U be a unit normal vector field along the curve α .
Let V be the unit normal vector field to the plane P..
v
U
M
v
U
v
α
As P is a plane surface the normal vector field V is
constant. Hence V ' = 0. Since angle between M and
P is constant, the angle between U and V is also
constant. Hence at any point on the curve α ,
U
U ⋅ V = U ⋅ V cos θ = cos θ = constant
P
U ⋅ V = constant
Differentiating we get
U '⋅V +U V
⋅ ' =0
(as V ' = 0 )
⇒ U '⋅ V = 0
⇒U ' ⊥V
Also U ' ⊥ U . Similarly the curve α lies in M and P implies α ' ⊥ U and α ' ⊥ V . Hence α ' and U '
are collinear. Hence by lemma α is a principal curve.
Note : Meridians and Parallels of a surface of revolution M are its principal curves.
Asymptotic Directions :
Definition : The directions which are tangent to M for which normal curvature is zero are called
asymptotic directions.
Thus the tangent vector v is asymptotic provided κ ( v ) = S ( v ) ⋅ v = 0 . Hence in an asymptotic
direction the surface M is not bending away from its tangent plane.
Following lemma gives nature of asymptotic directions in terms of Gaussian curvature.
Differential Geometry
247
Lemma : Let P be a point of a surface M of ¡ 3 .
(i)
If K(P) > 0, there are no asymptotic directions at P.
(ii)
If K(P) < 0, there are exactly two asymptotic directions at P which are bisected by the principal
directions at an angle θ given by,,
tan 2 θ = −
(iii)
κ1 ( P )
κ2 ( P)
If K(P) = 0, there every direction is asymptotic if P is planer point, otherwise there is only one
asymptotic direction which is principal.
Proof : We know that the normal curvature is given by Euler's formula
κ ( u ) = κ1 ( P ) cos 2 θ + κ 2 ( P ) sin2 θ
Also Gaussian curvature K = κ1 ⋅ κ 2
(i)
If K(P) > 0, then κ1 , κ 2 have same sign. Hence κ ( u ) is never zero. Therefore there is no
asymptotic directions.
(ii)
If K(P) < 0, then κ1 , κ 2 have opposite signs and if u is asymptotic direction then
κ (u ) = 0
⇒ κ1 cos 2 θ + κ 2 sin 2 θ = 0
⇒ κ1 cos 2 θ = −κ 2 sin2 θ
⇒ tan 2 θ = −
κ1
κ2
Since κ1 and κ 2 have opposite signs, −
κ1
> 0 and hence tan 2 θ > 0 .
κ2
This shows that there are two real values of θ which gives two directions in which κ ( u ) = 0 . Thus
there are two asymptotic directions if K(P) <0.
(iii)
If K(P) = 0, then (i) κ1 = 0 or κ 2 = 0 or (ii) both κ1 and κ 2 are zero.
If P is a planer point then κ1 = κ2 = 0 , therefore κ ( u ) = 0 for every u at P. Thus every direction is
asymptotic.
Next if either κ1 = 0 or κ 2 = 0 .
Differential Geometry
248
Suppose κ 2 = 0 and κ1 ≠ 0 .
Therefore
κ ( u ) = κ 1 cos2 θ + κ 2 ( P ) sin 2 θ
⇒ κ ( u ) = κ 1 cos 2 θ
Hence κ ( u ) = 0 implies cos θ = 0 ⇒ θ =
π
.
2
This shows that u = e2 . Thus asymptotic direction is principal.
Similarly if κ1 = 0 and κ 2 ≠ 0 , then we get u = e1 which is principal direction.
Asymptotic Curves :
Definition : A regular curve α in M is called asymptotic curve if its velocity vector α ' always points
in asymptotic direction.
Thus a curve α is asymptotic if and only if κ (α ') = S ( α ') ⋅ α ' = 0
But S (α ') = −U ' where U is normal vector field on α .
Therefore the curve α is asymptotic iff U '⋅ α ' = 0 .
Note : A curve α in M is asymptotic iff U '⋅ α ' = 0 .
Now U is normal to M and α ' is tangent to M. Therefore α '⋅U = 0 .
Differentiating we get,
α '⋅U '+ α "⋅ U = 0
⇒ α "⋅U = 0
(Since α '⋅U ' = 0 )
Thus α is asymptotic iff α "⋅ U = 0 .
α " is acceleration of α , which is always tangent to the surface for asymptotic curves.
Geodesics :
Definition : A curve α in M ⊂ ¡ 3 is a geodesic of M if its acceleration α " is always normal to M.
Differential Geometry
249
Note :
(i)
Since acceleration α " is normal to the surface, α " ⊥ α ' .
⇒ α "⋅α ' = 0
⇒ 2α "⋅α ' = 0
⇒ α "⋅ α '+α '⋅α " = 0
⇒ d (α '⋅ α ') = 0
⇒ α '⋅ α ' = constant
⇒ α'
2
= constant
⇒ α ' = constant
⇒ α has constant speed.
Thus if α is geodesic then its speed is constant.
(2)
A straight line α ( t ) = p + tq is always a geodesic of M since its acceleration α " = 0 which is
trivially normal to M. Also speed α ' 2 = α ' ⋅α ' = q ⋅ q = constant.
(3)
Also
α"
If α is a unit speed curve in M then T = α ' , T ' = α " =κ N ⇒ N =
.
κ
S (α ') =−∇α 'U
(α ' = T )
S (T ) =−∇α ' N
(Since N is normal to M, for α is geodesic.)
= −N '
= −( −κ T + τ B )
S (T ) = κ T − τ B
where α is geodesic on M.
(4)
Geodesic on a surface are the shortest routs of travel. There is a geodesic through every point
of M in every direction.
Differential Geometry
250
Worked examples ........................................................................................................................•
Examples : Geodesics on some surfaces of ¡ 3 .
(1)
Planes : Geodesics on plane P are straight lines.
If α is a geodesic in a plane P then α '⋅U = 0 where U is normal vector field on P. Hence
U = constant.
Differentiating we get,
α '⋅U '+ α "⋅ U = 0
⇒
α "⋅ U = 0
(Since U ' = 0 )
But α " normal to P. Hence α "⋅ U = 0 ⇒ α " = 0
⇒α '= q
⇒ α = p + tq where p and q are constant vectors or fixed points of ¡ 3 .
⇒ α is a straight line through p in the direction of q .
Thus α is a geodesic in P then α is a straight line.
⇒ Geodesic of P are all straight lines in P..
(2)
Spheres : Geodesics on a sphere are arcs of the great circles.
Let α be a unit speed curve on a surface of a sphere ∑ .
If α is a geodesic then we have
S (T ) = κ T − τ B
where T = α ' .
But for sphere
1
S (T ) = ± T
r
α
γ
This gives
β
1
± T = κT − τ B
r
∑
α , β , γ are geodesics of ∑
⇒κ =±
⇒κ =
Differential Geometry
251
1
r
1
and τ = 0
r
(since κ ≥ 0 )
1
hence α lies on a great circle. These geodesics are the arcs of the
r
great circles. The geodesics of ∑ are the constant speed parametrizations of its great circles.
On the sphere curvature is κ ≥
(3)
Cylinders : Geodesics on cylinder is helix.
Any curve in M : x 2 + y 2 = r 2 is written as
α (t ) = (r cos v(t ), r sin v (t ), h (t ) )
And any vector whose Z - coordinate is zero is normal to M. Therefore if α is a geodesic then h " = 0
and hence h(t) = ct + d
α '(t ) = ( − r sin v ⋅ v ', r cos v ⋅ v', h ')
∴ α '(t ) = r 2v '2 + h '2
∴
Hence
(h ' = c)
= r 2 v '2 + c 2
α '(t ) = constant
⇒ r 2v '2 + c 2 = constant
⇒ r 2v '2 + c 2 = constant
α
⇒ v '2 = constant
γ
⇒ v ' = constant = a
⇒ v = at + b
Hence the curve α becomes
β
α (t ) = ( r cos( at + b ), r sin( at + b ), ct + d )
Case (I) :
a ≠ 0 , c ≠ 0 then α is a helix on the surface.
M : x2 + y2 =r2
(I) α a ≠ 0 , c ≠ 0
Case (II) :
(II) γ a = 0
a = 0 then α is a ruling of M.
(III) β c = 0
Case (III) :
c = 0 then α is a cross sectional circle.
•........................................................................................................................................................•
Differential Geometry
252
Note : The essential properties of the three types of curves are as follows.
(i)
Principal curves : κ (α ') = κ1 or κ 2 , S (α ') is collinear with α ' .
(ii)
Asymptotic curves : κ (α ') = 0 , S (α ') is orthogonal to α ' and α " is tangent to the
surface.
(iii)
Geodesic : α " is normal to the surface.
Worked examples ........................................................................................................................•
Example 1 : Prove that a curve α in a surface M is a straight line of ¡ 3 if and only if α is both
geodesic and asymptotic.
Solution : Let α be a straight line.
⇒ α = p + tq
⇒α '= q , α" =0
⇒ U ⋅α " = 0 ,
U is normal vector field on M.
⇒U ⊥α "
⇒ α is asymptotic curve.
Since α is a straight line, it is also geodesic.
Conversely,
If α is both asymptotic and geodesic then for any normal U to the surface,
α"⊥U
and
α "|| U
⇒ α "⋅U = 0 , and
α "× U = 0
⇒α" = 0
(since U is not zero)
⇒ α = p + tq
which is a straight line.
Differential Geometry
253
Example 2 : If X is a patch in M, prove that α curve α (t ) = X ( a1 ( t) , a2 (t ) ) is
(i)
asymptotic iff la1'2 + 2m a1' a2' + na2' 2 = 0 , and
(ii)
a2' 2
principal iff E
l
− a1' a2'
F
m
a1' 2
G =0.
n
Solution : We represent the curve
α (t ) = X ( a1 ( t) , a2 (t ) )
... (1)
α ( t ) = X (u , v ) ,
... (2)
u = a1(t ) , v = a2 (t )
... (3)
by the equation
where
Differentiating equation (2) with respect to t we get
α '(t ) = X u
∂u
∂v
+ Xv
∂t
∂t
... (4)
α '(t ) = X u a1' (t ) + X v a2' ( t)
... (5)
Differentiating equation (5) with respect to t again we get
α "( t ) = X uu ( a1' ) + 2 X uv a1' a2' + X vv ( a2' ) + X u a1" + X v a"2
2
2
... (6)
Let the curve α be asymptotic, then we have by definition α " is tangent to the surface M. If
U is the vector normal to the surface, then we have
(i)
α "⋅ U = 0
⇒ (U ⋅ X uu ) ( a1' ) + 2 (U ⋅ X uv ) a1' a'2 + U ⋅ X vv ( a'2 ) + (U ⋅ X u ) a1" + (U ⋅ X v ) a"2 = 0
2
2
i.e.
l ( a1' ) + 2ma1' a2' + n ( a2' ) = 0
where
U ⋅ X u = 0 =U ⋅ X v , and l = U ⋅ X uu , m = U ⋅ X uv , n =U ⋅ X vv .
(ii)
2
2
We know if the α is principal curve then α ' points along the principal direction and conversely..
i.e., α is principal curve iff U ' and α ' are collinear..
i.e., α is principal curve off S (α ') and α ' are collinear..
Differential Geometry
254
Equivalently, it means that
α is principal curver iff S (α ') × α ' = 0 .
...(8)
Using equation (5) we have
α is principal curve iff S (a 1' X u + a'2 X v ) × ( a1' X u + a'2 X v ) = 0 .
SInce the shape operator S is linear, hence we have
'
'
'
'
α is principal curve iff  a1S ( X u ) + a2S ( X v )  × ( a1 X u + a2 X v ) = 0
On simplifying we get,
α is principal curve iff
a1'2 S ( X u ) × X u + a1' a2' (S ( X u ) × X v + S ( X v ) × X u ) + a2' 2S ( X v ) × X v = 0
... (9)
Since S ( X u ) ∈ TP ( M ) is tangent vector to the surface M. Therefore, it can be expressed as a linear
combination of the basis vector { X u , X v } of TP ( M ) . This implies
S ( Xu ) = α Xu + β Xv
... (10)
where α , β are to be determined. Now taking the cross product of the equation (10) with X u we get
S ( Xu ) × Xu = −β Xu × Xv
... (11)
This gives
−β =
(S ( X u ) × X u ) ⋅ ( X u × X v )
⇒β=
or
β=
Xu × Xv
2
− ( S ( X u )⋅ X u ) ( X u ⋅ X v ) − ( S ( X u ) ⋅ X v ) ( X u ⋅ X u ) 
EG − F 2
Em − Fl
EG − F 2
... (12)
Substituting this in the equation (11) we get,
 Fl − Em 
S ( Xu ) × Xu = 
 Xu × Xv
 EG − F 2 
... (13)
Similarly, now taking the vector product of the equation (10) with X v we get,
S ( Xu ) × X v = α Xu × X v
Differential Geometry
... (14)
255
Similar straight forward calculations give
α=
Gl − Fm
EG − F 2
... (15)
Hence from equation (14) we find
 Gl − Fm
S ( Xu ) × X v = 
 EG − F 2

 Xu × Xv

... (16)
Similarly, we express the vector S ( X v ) ∈TP ( M ) in terms of the basis vectors of TP ( M ) we get
S ( Xv ) = γ Xu + δ Xv
... (17)
Where γ and δ are similarly determined as
γ=
Gm − Fn
En − Fm
δ=
2 and
EG − F
EG − F 2
... (18)
Consequently, from equation (17) and (18) we find
and
 Fm − En 
S ( X v )× X u = 
 Xu × Xv
 EG − F 2 
... (19)
 Gm − Fn 
S ( X v )× X v = 
 X u × Xv
 EG − F 2 
... (20)
Now using equations (13), (16), (19) and (20) in the equation (9) we get after simplifying
α is principal curve iff ( a1' ) ( Fl − Em) + a1' a'2 ( Gl − En ) + ( a'2 ) ( Gm − Fn ) = 0
2
2
This is equivalent to
a2' 2
α is the principal curve iff E
l
− a1' a2'
F
m
a1' 2
G =0.
n
Example 3 : Prove that a tangent vector v = v1 X u + v2 X v is a principal vector iff
v22
E
l
Differential Geometry
−v1 v2
F
m
v1 2
G = 0.
n
256
Solution : The tangent vector defined by
v = v1 X u + v2 X v
is similar to the tangent vector α '(t ) to the curve α (t ) = X ( a1 ( t) , a2 (t ) ) defined in the example (2).
Hence replacing a1' by v1 and a '2 by v2 in the above example we get the required result.
Reference Books :
1.
Elementary Differential Geometry by B. O'Neil. - Academic Press, Revised Second edition,.
2006.
2.
Differential Geometry - An integrated Approach by Nirmala Prakash
- Tata McGraw Hill Publi. com. 2000.
❏❏❏
Differential Geometry
257