Riemann Surfaces: Construction via Group Action

Sep 21—2025

Proposition: $f: X \rightarrow Y$ be a non-constant holomorphic map between compact connected Riemann surfaces. For each $q \in Y$ , define

$d_q(f) := \sum_{p \in f^{-1}(q)} mult(f, p)$

where $mult(f, p)$ is how we defined in the previous post Riemann Surfaces: Multiplicity of a Map at a Point.

Claim: $d_q(f)$ is independent of $q \in Y$ , i.e. the degree is an invariant of the map.

Idea: show that $d_q(f)$ is locally constant. Since Y is connected, it follows that $d_q(f)$ must be a constant function.

Step 1: $d_q(f)$ is locally constant.

Let $q \in Y$ be arbitrary, and let $f^{-1}(q) = {p_k}_{\{ k \in \mathcal{K} \} }$ . For each $p_k$ , define $m_k := mult(f, p_k)$ .

By the theorem in the previous post, for each $p_k$ , there exists a chart $(U_k, \phi_k)$ (with $p_k \in U_k$ , we can ensure $U_k$ ‘s are pairwise disjoint because the set of ramification points are discrete, as shown in the previous post) and correspondingly there exists a chart $(V_k, \psi_k)$ with $q \in V_k$ such that $\phi_k(p_k) = 0$ , $\psi_k(q) = 0$ , and

$F_k := \psi_k \circ f \circ \phi_k^{-1}: \phi_k(U_k) \rightarrow \psi_k(V_k)$ is given by

$F_k(z) = z^{m_k}$ .

Goal: We wish to find a neighborhood $V$ of $q$ such that $f^{-1}(V) \subset \cup_{k \in \mathcal{K}} U_k =: U$ .

Why do we wish to find such $V$ ?

Suppose such $V$ exists, then for any $y \in V$ and $k \in \mathcal{K}$ :

if $y = q$ , then $f^{-1}(y) \cap U_k = \{ q_k \}$ , which has multiplicity $m_k$
if $y \neq q$ , then $f^{-1}(y) \cap U_k$ has $m_k$ points of multiplicity 1 (why? Recall the corresponding/equivalent map $F_k$ is $z \mapsto z^{m_k}$ , for any $\xi \in F_k(\text{small neighborhood of 0})$ , $\xi$ has $m_k$ pre-images, namely $\{ \xi e^{\frac{2\pi i \cdot j}{m_k}} : 0 \leq j < m_k \}$ .)

Thus, for any point $y \in V$ , $d_y(f) = d_q(f)$ , i.e. $f$ is locally constant in $V$ .

Step 2: So now, let’s show the existence of $V$ .

Suppose no such that $V$ exists.

Consider $\mathbb{D}_{1/n} \subset \mathbb{C}$ to be the the open disk of radius $1/n$ centered at 0.

Then, for any $k \in \mathbb{K}$ and $n \in \mathbb{N}$ , $f^{-1}({\psi_k}^{-1}(\mathbb{D}_{1/n})) \not\subset U$ (*). (Recall: $U = \cup_{k \in \mathbb{K}} U_k$ ) Otherwise, ${\psi_k}^{-1}(\mathbb{D}_{1/n})$ is our $V$ .

By (*), there exists $y_n \in {\psi_k}^{-1}(\mathbb{D}_{1/n})$ and $x_n \in X \backslash U$ such that $f(x_n) = y_n$ .

It follows that $\{ y_n \} \rightarrow q$ because $\{ \mathbb{D}_{1/n} \} \rightarrow \{ 0 \}$ and ${\phi_k}^{-1}(0) = \{ q \}$ .

Since $\{ x_n \}$ is a sequence on X, a compact Riemann Surface, it follows that it has a convergent subsequence, say $\{ x_{n_l} \}_{l \in \mathbb{N}}$ . Suppose $x_{n_l} \rightarrow x$ .

Claim: $x \not\in U$ .

Suppose not. Then, let $x \in U_j$ for some $j \in \mathcal{K}$ . It follows that $\phi_j(x) \in \phi_j(U_j)$ . Since $\phi_j(U_j)$ is open, $\phi_j(x)$ is contained in an open disk, say $\mathbb{D}_{\phi_j(x)}$ contained in $\phi_j(U_j)$ .

But since $x_{n_l} \not\in U_j$ , $\phi_j(x_{n_l})$ is always outside of $\phi_j(U_j)$ and so, there is no way for $\phi_j(x_{n_l})$ to approach $\phi_j(x)$ as close as possible! This shows that the claim is true.

So now that we know $x \not\in U$ , this means $f(x) \neq q$ (since $f^{-1}(q) \subset U$ ). It follows that $\{ f(x_{n_l}) \} = \{ y_{n_l} \} \not\rightarrow q$ , i.e. a subsequence of $\{ y_n \}$ does not converge to $q$ . This contradicts the fact $\{ y_n \} \rightarrow q$ .