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∈ℂ. 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 — C1∼C2 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 C∈𝒫, 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 C1∼C2, 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→ℂ. 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 z∈D′, 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))=D1∩D2=∅. 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