'''Koebe's Theorem.'
A compact planar Riemann surface X
is conformally equivalent to the Riemann sphere. A non-compact planar Riemann surface X'' is conformally equivalent either to the complex plane or to the complex plane with finitely many closed intervals parallel to the real axis removed. •
The harmonic function U. If
X is a Riemann surface and
P is a point on
X with local coordinate
z, there is a unique real-valued harmonic function
U on
X \ {
P} such that
U(
z) – Re
z−1 is harmonic near
z = 0 (the point
P) and
dU is square integrable on the complement of a neighbourhood of
P. Moreover, if
h is any real-valued smooth function on
X vanishing in a neighbourhood of
P of
U with ||
dh||2 = ∫
X dh∧∗
dh X dU ∧ *
dh = 0. ::This is an immediate consequence of
Dirichlet's principle in a planar surface; it can also be proved using
Weyl's method of orthogonal projection in the space of square integrable 1-forms. •
The conjugate harmonic function V. There is a harmonic function
V on
X \ {
P} such that ∗
dU =
dV. In the local coordinate
z,
V(
z) − Im
z−1 is harmonic near
z = 0. The function
V is uniquely determined up to the addition of a real constant. The function
U and its harmonic conjugate
V satisfy the
Cauchy-Riemann equations Ux =
Vy and
Uy = −
Vx. ::It suffices to prove that ∫
C ∗
dU = 0 for any piecewise smooth Jordan curve in
X \ {
P}. Since
X is planar, the complement of
C in
X has two open components
S1 and
S2 with
P lying in
S2. There is an open neighborhood
N of
C made up of a union of finite number of disks and a smooth function 0 ≤
h ≤ 1 such that
h equals 1 on
S1 and equals 0 on
S1 away from
P and
N. Thus (
dU,
dh) = 0. By Stokes' theorem, this condition can be rewritten as ∫
C ∗
dU = 0. So ∗
dU is exact and therefore has the form
dV. •
The meromorphic function f. The meromorphic differential
df =
dU +
idV is holomorphic everywhere except for a double pole at
P with singular term
d(
z−1) at the local coordinate
z. • '''Koebe's separation argument.
Let φ and ψ be smooth bounded real-valued functions on R''' with bounded first derivatives such that φ'(
t) > 0 for all
t ≠ 0 and φ vanishes to infinite order at
t = 0 while ψ(
t) > 0 for
t in (
a,
b) while ψ(
t) ≡ 0 for
t outside (
a,
b) (here
a = −∞ and
b = +∞ are allowed). Let
X be a Riemann surface and
W an open connected subset with a holomorphic function
g =
u +
iv differing from
f by a constant such that
g(
W) lies in the strip
a (dh,dh)=\int_X dh\wedge \star dh = \int_W (\varphi^\prime(u)^2 \psi(v)^2 + \varphi(u)^2 \psi^\prime(v)^2) \, du \wedge \star du \le 2M^2 \, \|dU\|^2 :where
M = sup (|φ|, |φ'|, |ψ|, |ψ'|), and :: (dU,dh)=\int_X dU \wedge \star dh = \int_W \varphi^\prime(u)\psi(v)\, du\wedge \star du >0, :contradicting the orthogonality condition on
U. •
Connectivity and level curves. (1) A level curve for
V divide
X into two open connected regions. (2) The open set between two level curves of
V is connected. (3) The level curves for
U and
V through any regular point of
f divide
X into four open connected regions, each containing the regular point and the pole of
f in their closures. ::(1) Since
V is only defined up to a constant, it suffices to prove this for the level curve
V = 0, i.e. that
V = 0 divides the surface into two connected open regions. If not, there is a connected component
W of the complement of
V = 0 not containing
P in its closure. Take
g =
f and
a = 0 and
b = ∞ if
V > 0 on
W and
a = −∞ and
b = 0 if
V 0 and ±
v > 0. If one of these open sets is not connected, then it has an open connected component
W not containing
P in its closure. If
v > 0 on
W, take
a = 0 and
b = ÷∞; if
v –1 as a local coordinate. •
Univalence of f at regular points. The function
f takes different values at distinct regular points (where
df ≠ 0). ::Suppose that
f takes the same value at two regular points
z and
w and has a pole at ζ. Translating
f by a constant if necessary, it can be assumed that
f(
z) = 0 =
f(
w). The points
z,
w and ζ lies in the closure of each of the four regions into which the level curves
U = 0 and
V = 0 divide the surface. the points
z and
w can be joined by a Jordan curve in the region
U > 0,
V > 0 apart from their endpoints. Similarly they can be joined by a Jordan curve the region
U 0,
V 0 is disjoint from γ and intersects a neighbourhood of ζ, both must be contained in
Y. On the other hand using
f to define coordinates near
z (or
w) the curve lies in two opposite quadrants and the other two open quadrants lie in different components of the complement of the curve, a contradiction. •
Regularity of f. The meromorphic function
f is regular at every point except the pole. ::If
f is not regular at a point, in local coordinates
f has the expansion
f(
z) =
a +
b zm (1 +
c1
z +
c2
z2 + ⋅⋅⋅) with
b ≠ 0 and
m > 1. By the
argument principle—or by taking the
mth root of 1 +
c1
z +
c2
z2 + ⋅⋅⋅ —away from 0 this map is
m-to-one, a contradiction. •
The complement of the image of f. Either the image of
f is the whole Riemann sphere
C ∪ ∞, in which case the Riemann surface is compact and
f gives a conformal equivalence with the Riemann sphere; or the complement of the image is a union of closed intervals and isolated points, in which case the Riemann surface is conformally equivalent to a horizontal slit region. ::Considered as a holomorphic mapping from the Riemann surface
X to the Riemann sphere,
f is regular everywhere including at infinity. So its image Ω is open in the Riemann sphere. Since it is one-one, the inverse mapping of
f is holomorphic from the image onto the Riemann surface. In particular the two are homeomorphic. If the image is the whole sphere then the first statement follows. In this case the Riemann surface is compact. Conversely if the Riemann surface is compact, its image is compact so closed. But then the image is open and closed and hence the whole Riemann sphere by connectivity. If
f is not onto, the complement of the image is a closed non-empty subset of the Riemann sphere. So it is a compact subset of the Riemann sphere. It does not contain ∞. So the complement of the image is a compact subset of the complex plane. Now on the Riemann surface the open subsets
a 1 =
u1 +
iv1 and
w2 =
u2 +
iv2 are points in
C \ Ω with
v1 2. Take a point in the strip
v1 2, say
w. By compactness of
C \ Ω, this set is contained in the interior of a circle of radius
R centre
w. The points
w ±
R lie in the intersection of Ω and the strip, which is open and connected. So they can be joined by a piecewise linear curve in the intersection. This curve and one of the semicircles between
z +
R and
z −
R give a Jordan curve enclosing
w1 with
w2 in its exterior. But then
w1 and
w2 lie on different connected components of
C \ Ω. Finally the connected components of
C \ Ω must be closed, so compact; and the connected compact subsets of a line parallel to the
x axis are just isolated points or closed intervals. Since does not contain the infinity at ∞, the construction can equally be applied to taking \mathbb{C} with horizontal slits removed to give a uniformizer . The uniformizer now takes to \mathbb{C} with parallel slits removed at an angle of to the -axis. In particular = leads to a uniformizer for \mathbb{C} with vertical slits removed. By uniqueness = . == Classification of simply connected Riemann surfaces ==