Page 329 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes
g
Definition 5. Let X : U   be a parametric surface and let   be its first fundamental form.
3
ij
The surface area element of X is:
  du du .
dA  det g ij 1 2
If V  U is open then the surface area of X over V is:

  du du
AX(V)  V  dA  V  det g ij 1 2 ...(4)
By Proposition 5, the surface area of X over V is reparametrization invariant, and we can thus
speak of the surface area of X(V).

Definition 6. Let X : U   be a parametric surface, and let V  U be open. The total curvature
3
of X over V is:

K (V)   V K dA.
X
It is easy to show, as in the proof of Proposition 5 that the total curvature of X over V is invariant
under reparametrization. We now introduce the signed surface area, a variant of Definition 5
which allows for smooth maps Y into a surface X, with Jacobian dY not necessarily everywhere
non-singular, and which also accounts for multiplicity.

3
Definition 7. Let X : U   be a parametric surface, and let X : U  X(U) be a smooth map.
Define (u) to be 1, –1, or 0, according to whether the pair Y (u), Y (u) has the same orientation
2
1
as the pair X (u), X (u), the opposite orientation, or is linearly dependent, and let h = Y  Y. If
j
i
2
1
ij
V  U is open then the signed surface area of Y over V is:
ˆ
  du du
A (V)  V   det h ij 1 2
Y
For a regular parametric surface, this definition reduces to Definition 5. Next, we prove that the
total curvature of a surface X over an open set U is the area of the image of U under the Gauss
map counted with multiplicity.
Theorem 3. Let X : U   be a parametric surface, and let V  U be open. Let N : U   be the
2
3
Gauss map of X, then:
ˆ
K (V)  A (V).
X
N
Proof. We first derive a formula which is of independent interest:
j
N   k X j ...(5)
i
i
To verify this formula, it suffices to check that the inner product of both sides with the three
j
linearly independent vectors X ,X ,N are equal. Since N  N = 1, we have N N i  0   k X N  0,

i
j
2
1
j

and k X X i j j  l   k g   k   N X . In particular, if h = N  N , then we find:
j
i
ij
i
ij
k
i
jl
m
n
m
h  k X m   k X n j n   k k g mn  k k g mn .
i
l
ij
jn
im
j
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