construction of Riemann surface using paths

Note: All arcs and curves are assumed to be smooth in this entry.

Let f be a complex function defined in a disk D about a point z0. In this entry, we shall show how to construct a Riemann surface such that f may be analytically continued to a function on this surface by considering paths in the complex plane.

Let 𝒫 denote the class of paths on the complex plane having z0 as an endpointMathworldPlanetmath along which f may be analytically continued. We may define an equivalence relationMathworldPlanetmath on this set — C1C2 if C1 and C2 have the same endpoint and there exists a one-parameter family of paths along which f can be analytically continued which includes C1 and C2.

Define 𝒮 as the quotient of 𝒫 modulo . It is possible to extend f to a function on 𝒮. If C𝒫, let f(C) be the value of the analytic continuation of f at the endpoint of C (not z0, of course, but the other endpoint). By the monodromy theoremMathworldPlanetmath, if C1C2, then f(C1)=f(C2). Hence, f is well defined on the quotient 𝒮.

Also, note that there is a natural projectionMathworldPlanetmath map π:𝒮. If C is an equivalence classMathworldPlanetmath of paths in 𝒮, define π(C) to be the common endpoint of those paths (not z0, of course, but the other endpoint).

Next, we shall define a class of subsets of 𝒮. If f can be analytically continued from along a path C from z0 to z1 then there must exist an open disk D centered about z1 in which the continuation of f is analyticPlanetmathPlanetmath. Given any zD, let C(z) be the concatenation of the path C from z0 to z1 and the straight line segment from z1 to z (which lies inside D). Let N(C,D)𝒮 be the set of all such paths.

We will define a topology of 𝒮 by taking all these sets N(C,D) as a basis. For this to be legitimate, it must be the case that, if C3 lies in the intersectionMathworldPlanetmath of two such sets, N(C1,D1) and N(C2,D2) there exists a basis element N(C3,D3) contained in the intersection of N(C1,D1) and N(C2,D2). Since the endpoint of C3 lies in the intersection of D1 and D2, there must exist a disk D3 centered about this point which lies in the intersection of D1 and D2. It is easy to see that N(C3,D3)N(C1,D1)N(C2,D2).

Note that this topology has the Hausdorff property. Suppose that C1 and C2 are distinct elements of 𝒮. On the one hand, if π(C1)π(C2), then one can find disjoint open disks D1 and D2 centered about C1 and C2. Then N(C1,D1)N(C2,D2)= because π(N(C1,D1))π(N(C2,D2))=D1D2=. On the other hand, if π(C1)=π(C2), then let D3 be the smaller of the disks D1 and D2. Then N(C1,D3)N(C2,D3)=.

To completePlanetmathPlanetmathPlanetmathPlanetmathPlanetmath the proof that 𝒮 is a Riemann surface, we must exhibit coordinate neighborhoodsMathworldPlanetmath and homomorphismsPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath. As coordinate neighborhoods, we shall take the neighborhoods N(C,D) introduced above and as homomorphisms we shall take the restrictionsPlanetmathPlanetmathPlanetmathPlanetmath of π to these neighborhoods. By the way that these neighborhoods have been defined, every element of 𝒮 lies in at least one such neighborhood. When the domains of two of these homomorphisms overlap, the compositionMathworldPlanetmath of one homomorphism with the inverseMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath of the restriction of the other homomorphism to the overlap region is simply the identity map in the overlap region, which is analytic. Hence, 𝒮 is a Riemann surface.

Title construction of Riemann surface using paths
Canonical name ConstructionOfRiemannSurfaceUsingPaths
Date of creation 2013-03-22 14:44:23
Last modified on 2013-03-22 14:44:23
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 21
Author rspuzio (6075)
Entry type Proof
Classification msc 30F99