---
tag: J7K42A9F8A
kind: definition
name: Genus of a curve
msc: [14H99]
defines: 6W8TDV9VVN
uses: [XJM2MT4S3Y, 6VC6HQTFSA, M2FAMD9HZN, H03GKRVBJ1, 0PF6TM8TWT]
sources:
  - url: https://en.wikipedia.org/wiki/Geometric_genus
---

## Definition

Let $k$ be an [[algebraically closed field|0PF6TM8TWT]] and let $C$ be a smooth projective connected [[curve|XJM2MT4S3Y]] over $k$.
Let $\Omega_C$ be its [[sheaf of differentials|6VC6HQTFSA]].
The *genus* of $C$ is
$$g(C) = \dim_k H^0(C, \Omega_C),$$
the dimension of the space of regular differential $1$-forms on $C$.
Let $C$ be an integral curve, possibly singular or not projective.
The *geometric genus* of $C$ is the genus of the smooth projective curve with the same function field.
This curve is the [[normalization|M2FAMD9HZN]] of a projective closure of $C$.

## Remarks

For a smooth projective connected curve, Serre duality gives $g(C) = \dim_k H^1(C, \mathcal{O}_C)$, so the genus equals the [[arithmetic genus|H03GKRVBJ1]].
Example: on $\mathbb{P}^1$ one has $\Omega = \mathcal{O}(-2)$, which has no nonzero global sections, so $g(\mathbb{P}^1) = 0$.
Example: the smooth projective curve $E$ with affine equation $y^2 = x^3 - x$, in characteristic not $2$, has the form $\omega = dx/y$, regular and nowhere zero on $E$, and $g(E) = 1$.
Non-example of the equality with the arithmetic genus: the nodal cubic $y^2 = x^2(x + 1)$ is singular, its normalization is $\mathbb{P}^1$, so its geometric genus is $0$, while its arithmetic genus is $1$.
Over $\C$, the genus equals the topological genus of $C(\C)$ as a closed orientable surface.