TL;DR

Why complex analysis on a compact Riemann surface turns into algebraic geometry? It is Riemann–Roch. It converts geometric input (a divisor / line bundle and the genus) into a computable formula for the dimension of meromorphic sections. Riemann–Roch is the hinge connecting residues/meromorphic functions, sheaves/cohomology, Serre duality, and Jacobians.

PDF: https://arxiv.org/pdf/2601.06868

Why Riemann–Roch

Riemann–Roch is not just a dimension formula. It is a machine:

  • It tells you when meromorphic functions exist with prescribed zeros/poles.
  • It explains the size of spaces of differentials and linear systems.
  • It is the formal point where analysis + topology (genus) becomes the algebraic geometry of divisors, line bundles, and cohomology.

Setup and notation

Let \(X\) be a compact Riemann surface of genus \(g\).

  • A divisor is \(D=\sum_p n_p\,p\) (finite sum, \(n_p\in\mathbb{Z}\)).
  • The associated line bundle is \(\mathcal{O}(D)\).
  • Define the dimension of meromorphic sections with pole divisor bounded by \(D\):
\[\ell(D) := \dim H^0(X,\mathcal{O}(D)),\]
  • Let \(K\) be a canonical divisor.

The Riemann–Roch theorem (core statement)

For any divisor \(D\) on \(X\),

\[\ell(D) - \ell(K-D) = \deg(D) + 1 - g.\]

Interpretation:

  • \(\ell(D)\) counts meromorphic functions/sections with poles bounded by \(D\).
  • \(\ell(K-D)\) is the “correction/obstruction term.”
  • \(\deg(D)+1-g\) is the predicted Euler characteristic.

The theorem as a workflow (how I use it)

To decide whether a nonconstant meromorphic function exists with poles bounded by \(D\):

  1. Compute \(\deg(D)\).
  2. Compare \(\deg(D)\) to \(g\).
  3. Use Riemann–Roch and control \(\ell(K-D)\).

A common practical case: if \(\deg(D)\) is large enough that \(\ell(K-D)=0\), then

\[\ell(D)=\deg(D)+1-g.\]

High-impact corollaries

1) Existence of meromorphic functions with prescribed poles

If \(D\) is effective and \(\deg(D)\) is sufficiently large, Riemann–Roch implies \(\ell(D)\ge 2\), so there exists a nonconstant meromorphic function with poles bounded by \(D\).

2) Dimension of holomorphic differentials

Take \(D=0\):

\[\ell(0) - \ell(K) = 1 - g.\]

Since \(\ell(0)=1\) (constants), we get

\[\ell(K)=g,\]

so the space of holomorphic \(1\)-forms has dimension \(g\).

3) Canonical divisor degree

A standard consequence is

\[\deg(K)=2g-2.\]

4) Sanity-check examples (genus 0 and genus 1)

Genus 0: \(X=\mathbb{P}^1\)

  • \(g=0\) and \(\deg(K)=-2\).
  • For \(\deg(D)\ge 0\) one gets
\[\ell(D)=\deg(D)+1,\]

matching the fact that rational functions with pole order \(\le n\) form an \((n+1)\)-dimensional space.

Genus 1: elliptic curve / complex torus

  • \(g=1\) and \(\deg(K)=0\).
  • Riemann–Roch becomes
\[\ell(D)-\ell(-D)=\deg(D).\]

For effective \(D\) with \(\deg(D)\ge 1\), typically \(\ell(-D)=0\), giving \(\ell(D)=\deg(D)\).

How this connects to the rest of the notes

Riemann–Roch is naturally expressed as an Euler characteristic identity:

  • \(\ell(D)=\dim H^0(X,\mathcal{O}(D))\).
  • By Serre duality,
\[H^1(X,\mathcal{O}(D))^\* \cong H^0(X, K\otimes\mathcal{O}(-D)).\]

So

\[\dim H^1(X,\mathcal{O}(D)) = \ell(K-D),\]

and Riemann–Roch is equivalent to

\[\chi(\mathcal{O}(D)) := \dim H^0(X,\mathcal{O}(D)) - \dim H^1(X,\mathcal{O}(D)) = \deg(D)+1-g.\]

Citation (BibTeX)

@misc{cho2026complex_analysis_riemann_surfaces,
  title        = {Complex Analysis and Riemann Surfaces: A Graduate Path to Algebraic Geometry},
  author       = {Cho, Gunhee and collaborators},
  year         = {2026},
  eprint       = {2601.06868},
  archivePrefix= {arXiv},
  primaryClass = {math.CV},
  note         = {Version 1}
}