Riemann Surfaces: Multiplicity of a Map at a Point (19/09/2025)

This post is based upon my understanding of what I am learning in my Riemann Surface class.

Intuition: Given a holomorphic function on an open neighborhood U of 0, if $f(0) = 0$ and $f'(0) \neq 0$ , then f will locally behave like its linear term. if $f(0) = 0$ and $f^{(k)}(0) \neq 0$ (where $k \in \mathbb{N}$ is the smallest), then f will locally behave like its $z^k$ -term.

Lemma: f is holomorphic on an open neighborhood U of 0, $f(0) = 0$ and it is not a constant function. There is a unique integer $k \geq 1$ such that on some smaller neighborhood of 0 $U' \subset U$ , we can find a holomorphic function g with $g(0) =0, g'(0) \neq 0$ and

$f(z) = g(z)^k$ on $U'$ .

Proof:

Key: we use the fact that f has a Taylor expansion at $z = 0$ .

$f(z) = a_k z^k + a_{k+1} z^{k+1} + ...$ , $a_k \neq 0$

We can re-write the above as:

$f(z) = a_k z^k (1 + b_1 z + b_2 z^2 + ...)$ , where $b_j = \frac{a_{k+j}}{a_k}$ .

Define $h(z) := (1 + b_1 z + b_2 z^2 + ...)^{1/k}$ . This is a multi-valued function, but that’s okay. Let’s see why. Since $b_1$ is a finite value, there exists a small neighborhood, say $U'$ of 0, such that for any $z \in U'$ , $|1 + b_1 z + b_2 z^2 +...| > 0$ . Hence, $h$ is well-defined on $U'$ .

Let $g(z) := {a_k}^{1/k} \cdot z \cdot h(z)$ .

Notice that $g(0) = 0, g'(0) = {a_k}^{1/k} \neq 0$ , which is what we want. $\square$

Corollary: Let X, Y be two Riemann surfaces and $f: X \rightarrow Y$ be non-constant and holomorphic. Then, for any point $p \in X$ , there is a unique integer $k = m_{p, f}$ such that there exists charts $(U, \phi)$ and $(V, \psi)$ (with $p \in U$ and $f(p) \in V$ ) such that $\phi(p) = 0$ , $\psi(f(p)) = 0$ , and $\psi \circ f \phi^{-1}: z \rightarrow z^k$ .

Picture proof:

Definition: The integer k given in the corollary above is the multiplicity of f at p, and is denoted by $mult(f, p)$ .

Definition: A point $x \in X$ is called a ramification point for f if $mult(f, x) \geq 2$ .

(note: ramify means to divide into 2 or more branches)

A point $y \in Y$ is a branch point of f if it is the image point of a ramification point for f.

Claim: the set of ramification points of f is discrete on X. (Recall: a subset S of X is discrete if each $p \in S$ is contained in an open neighborhood U such that $S \cap U = \{ p \}$ )

Key: A ramification point is a point where the derivative of the map is 0. Notice that in a neighborhood of a ramification point, the derivative of $\tilde{f}: z \rightarrow z^k$ is non-zero. Then, we use the fact that transition maps are biholomorphic to show that no matter which charts we choose, the points in a neighborhood of the ramification point cannot be a ramification point.

Proof: Given X, Y are Riemann surfaces and $f: X \rightarrow Y$ is a non-constant holomorphic map. Let $p \in X$ be a ramification point. Then, there exists a neighborhood $U$ of p along with the map $\phi: U \rightarrow \mathbb{C}$ and a neighborhood $V$ of $f(p)$ (such that $f(U) \subset V$ ) and a map $\psi: V \rightarrow \mathbb{C}$ such that:

$F := \psi \circ f \circ \phi^{-1}: \phi(U) \rightarrow \psi(V)$ is given by

$F: z \mapsto z^k$ (for some $k \in \mathbb{N}_{\geq 2}$ )

Claim: no points in $U \backslash \{ 0 \}$ is a ramification point.

Suppose for a contradiction there exists $q \in U$ such that it is a ramification point. Then, we can find $U_q \subset U$ a neighborhood of $q$ and $\phi_q : U_q \rightarrow \mathbb{C}$ and $V_{f(q)} \subset V$ a neighborhood of $f(q)$ (with $f(U_q) \subset V_{f(q)}$ ) and $\psi_{f(q)} : V_{f(q)} \rightarrow \mathbb{C}$ such that:

$F_q := \psi_{f(q)} \circ f \circ {\phi_q}^{-1}: \phi_q(U_q) \rightarrow \psi_{f(q)}(V_{f(q)})$ is given by

$F_q: z \rightarrow z^m$ for some $m \in \mathbb{N}_{\geq 2}$

It follows that $F_q'(\phi_q(q)) = 0$ . We know that the transition maps $\phi_q \circ \phi^{-1}$ and $\psi \circ {\psi_{f(q)}}^{-1}$ are biholomorphic.

Since $F = \psi \circ f \circ \phi^{-1}$ and $F_q = \psi_{f(q)} \circ f \circ {\phi_q}^{-1}$ , it follows that:

$F = (\psi \circ {\psi_{f(q)}}^{-1}) \circ F_q \circ (\phi_q \circ \phi^{-1})$

This is equivalent to:

$(\psi_{f(q)} \circ \psi^{-1}) \circ F = F_q \circ (\phi_q \circ \phi^{-1})$

Since $\phi(q) \in \phi(U) \backslash \{ 0 \}$ , it follows that $F'(\phi(q)) \neq 0$ .

But by (*), we have:

$\frac{d (\psi_{f(q)} \circ \psi^{-1}) \circ F}{dz}|_{z = \phi(q)} = \frac{d F_q \circ (\phi_q \circ \phi^{-1})}{dz}|_{z = \phi(q)}$

Expanding the above gives:

$\frac{d(\psi_{f(q)} \circ \psi^{-1})}{dz}|_{z = F(\phi(q))} \cdot \frac{d F}{dz}|_{z = \phi(q)} = \frac{d F_q}{dz}|_{z = F(\phi_q \circ \phi^{-1} (\phi(q)) = F(\phi_q)} \cdot \frac{d \phi_q \circ \phi^{-1}}{dz}|_{z = \phi(q)}$

The LHS = 0.

Since $\psi_{f(q)} \circ \psi^{-1}$ is biholomorphic, its derivative does not vanish anywhere. We also know from above $\frac{d F}{dz}|_{z = \phi(q)} \neq 0$ . Hence, RHS is non-zero, a contradiction.