# Constructing Riemann surfaces using the covering spaces

In the paper "On the dynamics of polynomial-like mappings" of Adrien Douady and John Hamal Hubbard, there is a way of constructing Riemann surfaces. I recite it as follow:

A polynomail-like map of degree d is a triple $(U,U',f)$ where $U$ and $U'$ are open subsets of $\mathbb{C}$ isomorphic to discs, with $U'$ relatively compact in $U$, and $f: U'\rightarrow U$ a $\mathbb{C}$-analytic mapping, proper of degree $d$. Let $L \subset U'$be a compact connect subset containing $f^{-1}\left(\overline{U'}\right)$ and the critical points of $f$, and such that $X_0=U-L$ is connected. Let $X_n$ be a covering space of $X_0$ of degree $d^n$, $\rho_n:X_{n+1}\rightarrow X_n$ and $\pi_n:X_n\rightarrow X_0$ be the projections and let $X$ be the disjoint union of the $X_n$. For each $n$ choose a lifting $$\widetilde{f}_n\colon \pi_n^{-1}(U'-L)\rightarrow X_{n+1},$$ of $f$. Then $T$ is the quotient of $X$ by the equivalence relation identifying $x$ to $\widetilde{f}_n(x)$ for all $x\in \pi_n^{-1}(U'-L)$ and all $n=0,1,2,\ldots$. The open set $T'$ is the union of the images of the $X_n, n=1,2,\ldots$, and $F:T'\rightarrow T$ is induced by the $\rho_n$.

Why $T$ is a Riemann surface and isomorphic to an annulus of finit modulus? Is there anything special about the $\pi_n,\rho_n$? What kind of background do I need?

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## 1 Answer

I will give an informal answer. Your covering space is just a collection of holed spaces right? The equivalence relation just projects the holed space down onto a space isomorphic to $U'-L$. It does this by pasting the image and the preimage of $f$ together. (In an informal sense with each iteration the covering space gets bigger. To see this consider the map $z \mapsto z^2$ on the punctured disk (disk without the origin) with radius 1 from the origin. With 1 iteration you cover the disk twice. Applying it another time on the double covering of the disk you cover it four times etc. Quotienting by all these iterations you get the punctured disk.) Hence you get something isomorphic to a space with an annulus of finite modulus. The latter space is just a classical (Parabolic) Riemann surface (Google it!).

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