Page 311 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes Thus, the second fundamental form is:
kdu 2

The principal curvatures are 0 and k. We have for the mean and Gauss curvatures:
1
H   k, K  0.
2
A surface on which K = 0 is called developable.
3
24.4.3.2. Tangent Surfaces. Let :[a,b]   be a curve with nonzero curvature k  0. Its tangent
surface is the ruled surface:

X(u,v)   (u) v (u).
 

 
Since     0, the curve  is the line of striction of its tangent surface. We have X = e + vk e and
2
u
1
X = e , hence the first fundamental form is:
v
1
2

g
     1 v k 2 1 . 
ij
 1 1 
The unit normal is N = –e , and clearly N = 0. Thus,
3
v
3

2
1
24.4.3.3. Hyperboloid. Let : (0,2 )    be the unit circle in the x x -plane: (t)  cos(t), sin(t),0 .
3

Define a ruled surface X : (0,2 )     by:
X(u,v)   (u)   v (u) e   3  cos(u) v sin(u), sin(u) vcos(u),v .   
Note that (x ) + (x ) – (x ) = 1 so that X(U) is a hyperboloid of one sheet. A straightforward
3 3
1 2
2 2
calculation gives:
1
N    cos(u) vsin(u),sin(u) vcos(u), v ,   
1 2v 2

and
2
N v 2  1 4v  4v 4 .
2

It follows from Proposition 7 that X has Gauss curvature K < 0.
24.5 Summary
A parametric surface patch is a smooth mapping:

X :U   3 ,
where U   is open, and the Jacobian dX is non-singular.
2
Write X = (x , x , x ), and each x = x (u , u ), then the Jacobian has the matrix representation:
1
2
i
i
2
3
1
1
 x 1 1 x 
2
 
dX   x 2 1 x 2 2 
 x 3 x 3 
 1 2 
304 LOVELY PROFESSIONAL UNIVERSITY