Page 349 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes
Then it is not difficult to see that  D, g ij  and  U, h ij  are isometric with the isometry given by:

 2v 1 u  v 2 
2

 :(u,v)  (x,y)   2 2 , 2 2 

 (1 u)  v 1 u  v 
In fact, a good bookkeeping technique to check this type of identity is to compute the differentials:
v(1 u) (1 u)  v 2
2


dx   4 2 du 2 2 dv

(1 u) 2  v 2  (1 u) 2  v 2 
2


dy   2 (1 u)  v 2 2 du 4 v(1 u) 2 dv,

(1 u) 2  v 2  (1 u) 2  v 2 
substitute into
2
dx  dy 2 ,
y 2
and then simplify using du dv = dv du to obtain (2). It is not difficult to see that this is equivalent
to checking (1).
Definition 2. Let (U, g) be a Riemannian surface. The Christoffel symbols of the second kind of
g are defined by:
1 mn
m
  2 g  g ni,j  g nj,i  g ij,n  . ...(3)
ij
The Gauss curvature of g is defined by:

1
n
m
K  g ij   m    m    m  . ...(4)
2 ij,m ij nm im nj
If (U, g) is induced by the parametric surface X : U   then these definitions agree with those
,
3
studied earlier.
28.2 Lie Derivative

.

Here, we study the Lie derivative. We denote the standard basis on  2 by , Let f be a
1
2
i
smooth function on U, and let Y  y   T U be a vector at u  U. The directional derivative of
u
i
f along Y is:
i
i
f
 Y f  y   y f . i ...(5)
i
1
i
Since y   Y u where u ,u 2  are the coordinates on U, we see that Y = Z follows from    Z
Y
i
as operators. The next proposition shows that the directional derivative of a function is
reparametrization invariant.





Proposition 1. Let : U  U be a diffeomorphism, and let Y be a vector at u U. Then for any

smooth function f on U, we have:

 f     f  .
    d Y  Y 
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