PlanetPhysics/Construction of Riemann Surface Using Paths

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

Let $f$ be a complex [[../Bijective/|function]] defined in a disk $D$ about a point $z_0\in ℂ$. In this entry, we shall show how to construct a [[../RiemannSurface/|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 $z_0$ as an endpoint along which $f$ may be analytically continued. We may define an [[../TrivialGroupoid/|equivalence relation]] $\sim$ on this set --- $C_1\sim C_2$ if $C_1$ and $C_2$ have the same endpoint and there exists a one-parameter family of paths along which $f$ can be analytically continued which includes $C_1$ and $C_2$.

Define $𝒮$ as the quotient of $𝒫$ modulo $\sim$. It is possible to extend $f$ to a function on $𝒮$. If $C\in 𝒫$, let $f(C)$ be the value of the analytic continuation of $f$ at the endpoint of $C$ (not $z_0$, of course, but the other endpoint). By the monodromy [[../Formula/|theorem]], if $C_1\sim C_2$, then $f(C_1)=f(C_2)$. Hence, $f$ is well defined on the quotient $𝒮$.

Also, note that there is a natural projection map $\pi :𝒮\to ℂ$. If $C$ is an equivalence class of paths in $𝒮$, define $\pi (C)$ to be the common endpoint of those paths (not $z_0$, 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 $z_0$ to $z_1$ then there must exist an open disk $D^{\prime }$ centered about $z_1$ in which the continuation of $f$ is analytic. Given any $z\in D^{\prime }$, let $C(z)$ be the concatenation of the path $C$ from $z_0$ to $z_1$ and the straight line segment from $z_1$ to $z$ (which lies inside $D^{\prime }$). Let $N(C,D^{\prime })\subset 𝒮$ be the set of all such paths.

We will define a topology of $𝒮$ by taking all these sets $N(C,D^{\prime })$ as a basis. For this to be legitimate, it must be the case that, if $C_3$ lies in the intersection of two such sets, $N(C_1,D_1)$ and $N(C_2,D_2)$ there exists a basis element $N(C_3,D_3)$ contained in the intersection of $N(C_1,D_1)$ and $N(C_2,D_2)$. Since the endpoint of $C_3$ lies in the intersection of $D_1$ and $D_2$, there must exist a disk $D_3$ centered about this point which lies in the intersection of $D_1$ and $D_2$. It is easy to see that $N(C_3,D_3)\subset N(C_1,D_1)\cap N(C_2,D_2)$.

Note that this topology has the Hausdorff property. Suppose that $C_1$ and $C_2$ are distinct elements of $𝒮$. On the one hand, if $\pi (C_1)\ne \pi (C_2)$, then one can find disjoint open disks $D_1$ and $D_2$ centered about $C_1$ and $C_2$. Then $N(C_1,D_1)\cap N(C_2,D_2)=\mathrm{\varnothing }$ because $\pi (N(C_1,D_1))\cap \pi (N(C_2,D_2))=D_1\cap D_2=\mathrm{\varnothing }$. On the other hand, if $\pi (C_1)=\pi (C_2)$, then let $D_3$ be the smaller of the disks $D_1$ and $D_2$. Then $N(C_1,D_3)\cap N(C_2,D_3)=\mathrm{\varnothing }$.

To complete the proof that $𝒮$ is a Riemann surface, we must exhibit coordinate neighborhoods and [[../TrivialGroupoid/|homomorphisms]]. As coordinate neighborhoods, we shall take the neighborhoods $N(C,D)$ introduced above and as homomorphisms we shall take the restrictions of $\pi$ 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 [[../Bijective/|domains]] of two of these homomorphisms overlap, the [[../Cod/|composition]] of one homomorphism with the inverse of the restriction of the other homomorphism to the overlap region is simply the [[../Cod/|identity]] map in the overlap region, which is analytic. Hence, $𝒮$ is a Riemann surface.