In mathematics, a planar Riemann surface (or schlichtartig Riemann surface) is a Riemann surface sharing the topological properties of a connected open subset of the Riemann sphere. They are characterized by the topological property that the complement of every closed Jordan curve in the Riemann surface has two connected components. An equivalent characterization is the differential geometric property that every closed differential 1-form of compact support is exact. Every simply connected Riemann surface is planar. The class of planar Riemann surfaces was studied by Koebe who proved in 1910, as a generalization of the uniformization theorem, that every such surface is conformally equivalent to either the Riemann sphere or the complex plane with slits parallel to the real axis removed.
Elementary properties
A closed 1-form ω is exact if and only if ∫γ ω = 0 for every closed Jordan curve γ. This follows from the Poincaré lemma for 1-forms and the fact that ∫δ df = f(δ(b)) – f(δ(a)) for a path δ parametrized by [a, b] and f a smooth function defined on an open neighbourhood of δ([a, b]). This formula for ∫δ df extends by continuity to continuous paths, and hence vanishes for a closed path. Conversely if ∫γ ω = 0 for every closed Jordan curve γ, then a function f(z) can be defined on X by fixing a point w and taking any piecewise smooth path δ from w to z and set f(z) = ∫δ ω. The assumption implies that f is independent of the path. To check that df = ω, it suffices to check this locally. Fix z0 and take a path δ1 from w to z0. Near z0 the Poincaré lemma implies ω = dg for some smooth function g defined in a neighbourhood of z0. If δ2 is a path from z0 to z, then f(z) = ∫δ1 ω + ∫δ2 ω = ∫δ1 ω + g(z) − g(z0), so f differs from g by a constant near z0. Hence df = dg = ω near z0.
A closed Jordan curve γ on a Riemann surface separates the surface into two disjoint connected regions if and only if ∫γ ω = 0 for every closed 1-form ω of compact support.
If the closed Jordan curve γ separates the surface, it is homotopic to a smooth Jordan curve δ (with non-vanishing derivative) that separates the surface into two halves. The integral of dω over each half equals ± ∫δ ω by Stokes' theorem. Since dω = 0, it follows that ∫δ ω = 0. Hence ∫γ ω = 0.
Conversely suppose γ is a Jordan curve that does not separate the Riemann surface. Replacing γ by a homotopic curve, it may be assumed that γ is a smooth Jordan curve δ with non-vanishing derivative. Since γ does not separate the surface, there is a smooth Jordan curve δ (with non-vanishing derivative) which cuts γ transversely at only one point. An open neighbourhood of γ ∪ δ is diffeomorphic to an open neighbourhood of corresponding Jordan curves in a torus. A model for this can be taken as the square [−π,π]×[−π,π] in R2 with opposite sides identified; the transverse Jordan curves γ and δ correspond to the x and y axes. Let ω = a(x) dx with a ≥ 0 supported near 0 with ∫ a = 1. Thus ω is a closed 1-form supported in an open neighbourhood of δ with ∫γ ω = 1 ≠ 0.
A Riemann surface is planar if and only if every closed 1-form of compact support is exact.
Let ω be a closed 1-form of compact support on a planar Riemann surface. If γ is a closed Jordan curve on the surface, then it separates the surface. Hence ∫γ ω = 0. Since this is true for all closed Jordan curves, ω must be exact.
Conversely suppose that every closed 1-form of compact support is exact. Let γ be closed Jordan curve. Let ω be closed 1-form of compact support. Because ω must be exact, ∫γ ω = 0. It follows that γ on separates the surface into two disjoint connected regions. So the surface is planar.
Every connected open subset of a planar Riemann surface is planar.
This is immediate from the characterization in terms of 1-forms.
Every simply connected Riemann surface is planar.
If ω is a closed 1-form of compact support, the integral ∫γ ω is independent of the homotopy class of γ. In a simply connected Riemann surface, every closed curve is homotopic to a constant curve for which the integral is zero. Hence a simply connected Riemann surface is planar.
If ω is a closed 1-form on a simply connected Riemann surface, ∫γ ω = 0 for every closed Jordan curve γ.
This is the so-called "monodromy property." Covering the path with disks and using the Poincaré lemma for ω, by the fundamental theorem of calculus successive parts of the integral can be computed as f(γ(ti)) − f(γ(ti − 1)). Since the curve is closed, γ(tN) = γ(t0), so that the sums cancel.
Uniformization theorem 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 < ∞, then (dU,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.See: 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 < Im z < b. Define a real-valued function by h = φ(u)ψ(v) on W and 0 off W. Then h, so defined, cannot be a smooth function; for if so
where M = sup (|φ|, |φ'|, |ψ|, |ψ'|), and
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 on W. The boundary of W lies on the level curve V = 0. Take g = f in this case. Since ψ(v) vanishes to infinite order when v = 0, h is a smooth function, so Koebe's argument gives a contradiction.
(2) It suffices to show that the open set defined by a < V < b is connected. If not, this open set has a connected component W not containing P in its closure. Take g = f in this case. The boundary of W lies on the level curves V = a and V = b. Since ψ(v) vanishes to infinite order when v = a or b, h is a smooth function, so Koebe's argument gives a contradiction.
(3) Translating f by a constant if necessary, it suffices to show that if U = 0 = V at a regular point of f, then the two level curves U = 0 and V = 0 divide the surface into 4 connected regions. The level curves U = 0, V = 0 divide the Riemann surface into four disjoint open sets ±u > 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 < 0 on W, set a = −∞ and b = 0. Take g = f in this case. The boundary of W lies on the union of the level curves U = 0 and V = 0. Since φ and ψ vanish to infinite order at 0, h is smooth function, so Koebe's argument gives a contradiction. Finally, using f as a local coordinate, the level curves divide an open neighbourhood of the regular point into four disjoint connected open sets; in particular each of the four regions is non-empty and contains the regular point in its closure; similar reasoning applies at the pole of f using f(z)–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 apart from their endpoints, where the curve is transverse to the boundary. Together these curves give a closed Jordan curve γ passing through z and w. Since the Riemann surface X is planar, this Jordan curve must divide the surface into two open connected regions. The pole ζ must lie in one of these regions, Y say. Since each of the connected open regions U > 0, V < 0 and 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 + c1z + c2z2 + ⋅⋅⋅) with b ≠ 0 and m > 1. By the argument principle—or by taking the mth root of 1 + c1z + c2z2 + ⋅⋅⋅ —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.