If $\deg D > 2g - 2$, then $\ell(K_C - D) = 0$ so:

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

This is a complete, explicit formula. For instance:

• $g = 0$: $\ell(D) = \deg D + 1$ for all $\deg D \geq 0$. So $\ell(n \cdot [P]) = n + 1$, the space of polynomials of degree $\leq n$ on $\mathbb{P}^1$.
• $g = 1$: $\ell(D) = \deg D$ for $\deg D \geq 2$. In particular, $\ell(2P) = 2$, $\ell(3P) = 3$.
• $g = 2$: $\ell(D) = \deg D - 1$ for $\deg D \geq 3$.

A smooth projective curve $C$ has genus $0$ if and only if $C \cong \mathbb{P}^1$ (assuming $C(k) \neq \emptyset$).

Proof: If $g = 0$ and $P \in C(k)$, then $\ell(P) = 1 - 0 + 1 = 2$ by Riemann–Roch. So there exists a non-constant $f \in L(P)$, i.e., a rational function with a simple pole at $P$ and no other poles. This $f$ gives a morphism $C \to \mathbb{P}^1$ of degree $1$, hence an isomorphism.

Conversely, on $\mathbb{P}^1$, $\deg K = -2 = 2g - 2$ gives $g = 0$.

Let $(E, O)$ be an elliptic curve ($g = 1$), with $P = O$.

• $\ell(O) = 1$: only constants (since $\deg(K - O) = -1 < 0$... wait, $\deg K = 0$ so $\deg(K - O) = -1$, giving $\ell(K-O) = 0$. Thus $\ell(O) = 1 - 1 + 1 = 1$).
• $\ell(2O) = 2$: there is a function $x$ with a double pole at $O$. This gives $E \to \mathbb{P}^1$ of degree $2$ (the hyperelliptic map).
• $\ell(3O) = 3$: adding $y$ (triple pole at $O$).
• $\ell(4O) = 4$: adding $x^2$.
• $\ell(5O) = 5$: adding $xy$.
• $\ell(6O) = 6$: adding $x^3$ and $y^2$. But $\dim = 6$ with $7$ functions $\{1, x, y, x^2, xy, x^3, y^2\}$... so there must be a relation:

$y^2 + a_1 xy + a_3 y = x^3 + a_2 x^2 + a_4 x + a_6.$

This is the Weierstrass equation! Riemann–Roch naturally produces the equation of the curve.

Let $C$ have genus $2$. The canonical divisor satisfies $\deg K = 2$ and $\ell(K) = 2$.

The canonical map $\phi_K: C \to \mathbb{P}^1$ is a degree-$2$ morphism (since $\ell(K) = 2$ gives a $1$-dimensional linear system of degree $2$). Therefore:

Every genus-$2$ curve is hyperelliptic.

More explicitly, $|K|$ is a $g^1_2$ (a linear system of dimension $1$ and degree $2$). Taking a point $P$ with $\ell(P) = 1$ (which is true for "most" points since $\ell(P) \geq 1 - 2 + 1 = 0$, and $\ell(P) = 1$ iff $P$ is not a base point of $|K|$):

By Riemann–Roch, $\ell(K - P) = \ell(P) + g - 1 - \deg P = \ell(P)$... Let's be more careful. $\ell(K-P) = \ell(P) + \deg(K-P) - g + 1 - (\ell(P) - ?)$. Better to just use RR directly:

$\ell(2P)$: RR gives $\ell(2P) - \ell(K - 2P) = 2 - 2 + 1 = 1$. If $P$ is a Weierstrass point, $\ell(2P) = 2$ (and $\ell(K - 2P) = 1$). If not, $\ell(2P) = 1$ (and $\ell(K - 2P) = 0$). There are exactly $6$ Weierstrass points (over $\bar{k}$, in char $\neq 2$).

If $|D|$ has no base points (i.e., $\bigcap_{D' \in |D|} \operatorname{Supp}(D') = \emptyset$), then $|D|$ defines a morphism:

$\phi_D: C \to \mathbb{P}^{\ell(D) - 1}.$

• $g = 0$, $D = n \cdot [\infty]$: $\phi_D: \mathbb{P}^1 \to \mathbb{P}^n$ is the Veronese embedding (degree $n$, $\ell(D) = n+1$).

• $g = 1$, $D = 3O$: $\phi_D: E \hookrightarrow \mathbb{P}^2$ is the Weierstrass embedding as a plane cubic.

• $g = 2$, $D = K$: $\phi_K: C \to \mathbb{P}^1$ is the hyperelliptic double cover.

• $g = 3$, non-hyperelliptic, $D = K$: $\deg K = 4$, $\ell(K) = 3$, so $\phi_K: C \hookrightarrow \mathbb{P}^2$ embeds $C$ as a smooth plane quartic. This is the canonical embedding for non-hyperelliptic genus 3.

• $g = 4$, non-hyperelliptic, $D = K$: $\deg K = 6$, $\ell(K) = 4$, so $\phi_K: C \hookrightarrow \mathbb{P}^3$ embeds as a curve of degree $6$ in $\mathbb{P}^3$.

• $D = 0$: $i(0) = \ell(K) = g$, so $\ell(0) = 0 - g + 1 + g = 1$ ✓.
• $D = K_C$: $i(K) = \ell(0) = 1$, so $\ell(K) = (2g-2) - g + 1 + 1 = g$ ✓.
• $\deg D > 2g - 2$: $i(D) = 0$ (non-special), $\ell(D) = \deg D - g + 1$.
• On a hyperelliptic curve of genus $g$, the $g^1_2$ divisor $D$ (degree 2, $\ell(D) = 2$) satisfies $i(D) = 2 - 2 + g - 1 + ?$... By RR: $2 - i(D) = 2 - g + 1$, so $i(D) = g - 1$. Very special!

For a general curve of genus $g$, the following table shows $\ell(nP)$ for a general point $P$:

$g = 0$ ($\mathbb{P}^1$): $\ell(nP) = n + 1$ for all $n \geq 0$.

$g = 1$ (elliptic): $\ell(0) = 1$; $\ell(nP) = n$ for $n \geq 1$.

$g = 2$: $\ell(0) = 1$; $\ell(P) = 1$; $\ell(2P) = 1$ (general $P$); $\ell(nP) = n - 1$ for $n \geq 3$.

$g = 3$: $\ell(0) = 1$; $\ell(nP) = 1$ for $n = 1, 2, 3$ (general $P$); $\ell(nP) = n - 2$ for $n \geq 5$; $\ell(4P) = 2$ (by RR: $4 - 3 + 1 = 2$ since $i(4P) = 0$ for general $P$ on non-hyperelliptic curve).

The Weierstrass gap sequence at $P$ is the set of $n > 0$ with $\ell(nP) = \ell((n-1)P)$, i.e., $n$ is not a pole order of any function with pole only at $P$. By Riemann–Roch, there are exactly $g$ gaps, contained in $\{1, 2, \ldots, 2g\}$.