Page 328 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY
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Unit 26: Lines of Curvature
Now, in any simply connected subset of U, we can always find analytic functions G = x +iy Notes
j j j
G F .
1
3
satisfying j j We let X x , x , x . Then X is conformal and harmonic. Indeed, x being
2
j
the real parts of complex analytic functions, are harmonic, and hence X is harmonic. Furthermore,
x
x , and by the Cauchy-Riemann equations y n . j Thus, we see
we have j j j j
u v u
that
3 2 2 2
v
X X X X j 0,
u
v
u
j
j 1
and
3
X X 0,
v
j
j
u
j 1
det
3
and hence, X is conformal. Since X is real analytic, the zeroes of X X j are isolated. Removing
i
the set Z of those zeroes from U, we get that X : U \Z is a harmonic and conformal
3
parametric surface, hence, X is a minimal surface .
4
If we carry out this procedure starting with the complex analytic functions f() = 1 and
h() = 1/, then X is another parametrization of the catenoid.
26.3 Surface Area
In this section, we will give interpretations of the Gauss curvature and the mean curvature. Both
of these involve the concept of surface area. Before introducing the definition, we first prove a
proposition which will show that the definition is reparametrization invariant.
Proposition 5. Let X : U be a parametric surface with first fundamental form , and
g
3
ij
3
g
V U. Let X : U be a reparametrization of X, let V 1 (V), and let be the coordinate
ij
representation of the first fundamental form of X. Then, we have:
du du .
du du
det g 1 2 V det g 1 2 ...(3)
V ij ij
Proof. Now, we have
det
det g ij det g ij j i
i
j
i
where u / u . Thus, for any open subset V U, and V 1 (V), we have:
j
det
du du
det g 1 2 det g du du i 1 2 V det g 1 2
du du
V ij V ij j ij
Thus, the integral on the right-hand side of (3) is reparametrization invariant. This justifies the
following definition.
3 Of course, Y = (y , y , y ) is also conformal.
2
1
3
4 X is also said to be a branched minimal surface on U. The zeroes of det(g ) are called branched points.
ij
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