PlanetPhysics/Construction of Riemann Surface Using Paths

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

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

Let denote the class of paths on the complex plane having as an endpoint along which may be analytically continued. We may define an equivalence relation on this set --- if and have the same endpoint and there exists a one-parameter family of paths along which can be analytically continued which includes and .

Define as the quotient of modulo . It is possible to extend to a function on . If , let be the value of the analytic continuation of at the endpoint of (not , of course, but the other endpoint). By the monodromy theorem, if , then . Hence, is well defined on the quotient .

Also, note that there is a natural projection map . If is an equivalence class of paths in , define to be the common endpoint of those paths (not , of course, but the other endpoint).

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

We will define a topology of by taking all these sets as a basis. For this to be legitimate, it must be the case that, if lies in the intersection of two such sets, and there exists a basis element contained in the intersection of and . Since the endpoint of lies in the intersection of and , there must exist a disk centered about this point which lies in the intersection of and . It is easy to see that .

Note that this topology has the Hausdorff property. Suppose that and are distinct elements of . On the one hand, if , then one can find disjoint open disks and centered about and . Then because . On the other hand, if , then let be the smaller of the disks and . Then .

To complete the proof that is a Riemann surface, we must exhibit coordinate neighborhoods and homomorphisms. As coordinate neighborhoods, we shall take the neighborhoods introduced above and as homomorphisms we shall take the restrictions of to these neighborhoods. By the way that these neighborhoods have been defined, every element of lies in at least one such neighborhood. When the domains of two of these homomorphisms overlap, the composition of one homomorphism with the inverse of the restriction of the other homomorphism to the overlap region is simply the identity map in the overlap region, which is analytic. Hence, is a Riemann surface.