Covering in Riemann Surfaces

The main goal of this chapter is to realize the relationship among the following concepts:

1. covering maps, proper maps, local homeomorphisms, para-compact spaces…
2. nonconstant holomorphic map, branched covering maps, (un)ramified…

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Topological side

Let’s start with some definitions in topology.

Def A topological space $X$ is called locally compact, if it is Hausdorff and that every point has a compact neighborhood.

Def Let $f:X\rightarrow Y$ be a continuous map between topological spaces. If for every compact subset $K\subset Y$, we have $f^{-1}(K)$ is also compact, then $f$ is called proper.

In Chinese, “proper” refers to “逆紧”, which means that the inverse of a compact set is compact.

Def (Covering) Let $X$ and $Y$ be two topological spaces. A continuous map $f:X\rightarrow Y$ is a covering if $\forall y\in Y$, $\exists V\subset Y$ an open neighborhood of $y$, and $U_x$ an open neighborhood of $x\in f^{-1}(y)$, such that

1. $f^{-1}(V)=\bigsqcup\limits_{x\in f^{-1}(y)}U_x$,
2. $f|_{U_x}:U_x\rightarrow V$ is a homeomorphism.

The above two statements are equivalent to $f^{-1}(V)\cong f^{-1}(y)\times V$.

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Riemann surfaces: Ramification/Branch

When Riemann surfaces show up, complex analysis has an active and elegant performance. Recall that local normal form theorem tells us that any nonconstant, holomorphic map $f:X\rightarrow Y$ between Riemann surfaces has a local normal form $z^k$ near $x\in X$, given that the local coordinates are well-chosen. First we recall the definition of ramification.

Def. Let $f:X\rightarrow Y$ be a nonconstant holomorphic map. Suppose $x\in X$ and the local normal form of $f$ near x is $f=z^k$. Define

$$v_x(f):=k$$

to be the ramification index of $x\in X$. If $v_x(f)\geq 2$, then $x$ is said to be a ramification point. If $f$ contains no ramification point, then $f$​ is said to be unramified. In addition, the set of ramification points is denoted by

$$R:=\{a\in X:v_a(f)\geq 2\}.$$

The idea of ramification lies in the domain.

Def. Let $f:X\rightarrow Y$ be a nonconstant holomorphic map. Define

$$B:=f(R)$$

to be the set of branched points of $f$.

The idea of branched lies in the codomain.

Fact $\forall x\in X$ and $U\subset X$ be an open neighborhood of $x$ in $X$. Set $y=f(x)$. Then $\exists U'\subset U$ another open neighborhood of $x$ such that

$$|f^{-1}(y)\cap U'|=v_a(f)$$

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Riemann surfaces have both sides

First, let’s play with some naive cases.

e.g. Let $f:X\rightarrow Y$ be a holomorphic map between Riemann surfaces.

1. If $f$ is constant, then $f$ is proper $\Leftrightarrow$ $X$ is compact.
2. If $Y$ is compact and $f$ is proper, then $X$ is also compact.

Proof. For (1), notice that singleton is compact. For (2), notice that $X=f^{-1}(Y)$ is the inverse of a compact set. $\square$

Lemma. Let $f:X\rightarrow Y$ be a nonconstant holomorphic map between Riemann surfaces. If $X$ is compact, then $f$ is proper.

Proof. Let $K\subset Y$ be a compact subset of $Y$. Then $Y$ is closed. Since $f$ is continuous, then $f^{-1}(Y)$ is also closed. As a closed subset of a compact space $X$, $f^{-1}(Y)$ is also compact. $\square$

Lemma. Let $f:X\rightarrow Y$ be a nonconstant holomorphic proper map between Riemann surfaces. Then

1. $|f^{-1}(y)|<\infty$, $\forall y\in Y$.
2. Let $y\in Y$, $U\subset X$ be an open neighborhood of $f^{-1}(y)$. Then there exists an open neighborhood $V\subset Y$, such that $f^{-1}(V)\subset U$.

Proof. (1) Since $f$ is holomorphic, then by identity theorem, $f^{-1}(y)$ is discrete, which means that $\forall x\in f^{-1}(y)$, $\exists U_x\subset X$ an open neighborhood of $x$, such that $f\neq 0$ in $U_x-x$ . Additionally, since $f^{-1}(y)$ is compact, and $f^{-1}(y)$ can be covered by disjoint $\{U_x\}_{x\in f^{-1}(y)}$, then $f^{-1}(y)$ is finite.

(2) Step 1: Proper holomorphic map is closed. Let $A\subset X$ be a closed set. We cover $Y$ by closed “disks”, namely holomorphic preimages of closed disks under local coordinate charts. Denote

$$\bar{B}=\varphi^{-1}(\overline{B_r(z)})\subset X.$$

It suffices to show that $f(A)\cap\bar{B}$ is closed.

Then $\forall x\in\partial f(A)$, there exists a sequence $x_i\in f(A)$ converging to $x$. Denote $x_i=f(a_i)$ where $a_i\in A$. There is a subsequence of $a_i$ lies in $\bar{B}$ since $f(A)\cap\bar{B}$ is closed.

Since $\bar{B}$ is compact and $f$ is proper, then $f^{-1}(\bar{B})$ is compact and thus closed. Then $A':=A\cap f^{-1}(\bar{B})$ is a closed subset of $f^{-1}(\bar{B})$ and thus compact. Using that continuous map takes compact set to compact set, we have $f(A\cap f^{-1}(\bar{B}))=f(A)\cap\bar{B}$ is compact and thus closed.

Step 2: Construction of $V$. Let $Z=X-U$ closed. Then $f(Z)$ is closed and $y\notin f(Z)$. Then there exists an open neighborhood $V$ of $y$, such that $V\cap f(Z)=\emptyset$. Therefore, $f^{-1}(V)\cap Z=\emptyset$ and $f^{-1}(V)\subset U$. $\square$​

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Coverings are ubiquitous

Def. Let $f:X\rightarrow Y$ be a nonconstant holomorphic map of Riemann surfaces. Say $f$ is a branched covering if $\exists X'\subset X$ open and dense, s.t.

$$f|_{X'}:X'\rightarrow Y$$

is covering.

Cor. Let $f:X\rightarrow Y$ be a proper, nonconstant, holomorphic map between Riemann surfaces. Set

$$Y'=Y-B(f)$$

$$X'=f^{-1}(Y')$$

Then $f|_{X'}:X'\rightarrow Y'$ is a covering.

Proof. Since $f$ is proper, then $f':=f|_{X'}$ is also proper. Then the set of ramification points $R(f)\subset X$ and the set of branched points $B(f)\subset Y$ are discrete. Thus, $f'$ is unramified $\Rightarrow$ $f'$ is a local homeomorphism. By the lemma above, $\exists U\subset X$ an open neighborhood of $f^{-1}(y)$, s.t.

$$f^{-1}(V)\subset U.$$

Hence $f'$ is a covering. $\square$