Chapter 1 DIFFERENTIAL GEOMETRY OF REAL MANIFOLDS

1.4. GEOMETRY OF SURFACES 325
Exercise 1.4.12 Show that harmonicity of f can be expressed by the equation
where z = x + iy, w = u + iv, ∂
∂¯z
∂2f ∂ρ ∂f ∂f
+ 2
∂z∂¯z ∂w ∂z ∂¯z
= 1
2
( ∂
∂x
+ i ∂
∂y
= 0,
) and ∂
∂z
= 1
2
( ∂
∂x
∂ − i ) relative to isothermal
∂y
coordinates given by (1.4.22). Show that for a harmonic mapping the quadratic differential
is holomorphic.
2ρ(f(z)) ∂f ∂
Ψ = e
∂z
¯ f
∂z dz2
Since we have not developed the basic facts regarding complex manifolds even in dimension
one, we cannot yet exploit the implications of holomorphy of Ψ. For example,
anticipating the elementary fact that that there are no holomorphic quadratic differentials
on CP (1) S 2 (which will be discussed in volume 2) we see that for every harmonic map
f : S 2 → S 2 we have Ψ = 0. Consequently f is either holomorphic or antiholomorphic.
This maybe regarded as an analogue of the fact that harmonic functions in the plane are
representable as a sum of a holomorphic and an antiholomorphic function.
Example 1.4.3 The existence of isothermal coordinates allows us to extend the isoperimetric
inequality to negatively curved surfaces (at least locally). We assume the metric is in the
form ds 2 = e −2ρ (dx 2 + dy 2 ) and note that the curvature of the surface is given by e 2ρ ∆ρ < 0
by assumption. Let Γ be a simple closed curve on M enclosing a relatively compact open set
D. Since our considerations are local at this point we may assume D ⊂ C with piece-wise
smooth boundary Γ. The area of D and length of γ are given by
S = e −2ρ(x,y)
dx ∧ dy, L =
D
e −ρ dx 2 + dy 2 . (1.4.25)
Let ϕ1 be the harmonic function on D with boundary values given by rho. Let ϕ2 be the
conjugate harmonic function so that ϕ(z) = ϕ1(z) + iϕ2(z) is holomorphic. Consider the
mapping F : D → C defined by
F (z) =
z
z◦
e −ϕ(z) dz,
where z◦ ∈ D is any fixed point. Then F ′ is nowhere vanishing and we obtain a diffeomorphism
of D onto a region D ′ bounded by a curve Γ ′ . The area of D ′ and length of Γ ′ are
given by
S ′
=
D
e −2ϕ1(x,y)
dx ∧ dy, L =
e −ϕ1
dx2 + dy2 . (1.4.26)