Page 352 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Unit 28: Local Intrinsic Geometry of Surfaces

z
,
i
i
Now, setting Y  y  and Z   we compute: Notes
i
i
i
k
j k
j
k
j
j k

 (Y,Z)   i g y z   m ji g km y z   m g y z  g y zk g y z ,i
ki
jk
ik
mi
jk
,i
ig
k
 g jk  y j ,i   j mi y m  z  g y j  z k ,i   k mi z m   g Y , Z  g Y, Z .  ;i 
;i
ik
This completes the proof of (13) and of the proposition.
i
Definition 4. Let Y  y  be a vector field on the Riemannian surface (U, g). Its divergence is the
i
function:
i
i
div Y   i y   i y   i ij y . j
Note that:
1 1
i
im
  2 g im  g mi,j  g mj,i  g ij,m   2 g g im,j   j log det g.
ij
Thus, we see that:
1
div Y   i  det g y i  ...(15)
det g
Observe that this implies

2
i
1

 U div Y dA   i  det g y du du .
U
Thus, Green’s Theorem in the plane implies the following proposition.
Proposition 6. Let Y be a compactly supported vector field on the Riemannian surface (U, g).
Then, we have:
 U div Y dA  0.
Definition 5. If f : U   is a smooth function on the Riemannian surface (U, g), its gradient f
is the unique vector field which satisfies  g  f,Y    Y f.
The Laplacian of f if the divergence of the gradient of f:
f = div f.
,
ij
It is easy to see that f  g f  hence
j
j
1
 f   i  g ij det g f . ...(16)
 j
det g
Thus, in view of Proposition 6, if f is compactly supported, we have:
 U  f dA  0.

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