On $\mathbb{P}^1$ with $D = n \cdot [\infty]$ for $n \geq 0$:

• $L(D) = \{f \in k(t) \mid \operatorname{div}(f) + n[\infty] \geq 0\} = \{$polynomials of degree $\leq n\}$.
• A basis is $\{1, t, t^2, \ldots, t^n\}$, so $\ell(D) = n + 1$.
• $|D| \cong \mathbb{P}^n$: each element is a divisor $[a_1] + [a_2] + \cdots + [a_n]$ (roots of a degree-$n$ polynomial, counted with multiplicity).
• For $n = 1$: $|[\infty]| \cong \mathbb{P}^1$ parametrizes single points of $\mathbb{P}^1$.
• For $n = 2$: $|2[\infty]| \cong \mathbb{P}^2$ parametrizes unordered pairs of points (including coincident pairs).

Let $E$ be an elliptic curve with origin $O$.

• $|O|$: $\ell(O) = 1$, so $|O| = \{O\}$, a single point ($\mathbb{P}^0$).
• $|2O|$: $\ell(2O) = 2$, so $|2O| \cong \mathbb{P}^1$. The basis $\{1, x\}$ gives effective divisors $\operatorname{div}(a + bx) + 2O$ for $[a:b] \in \mathbb{P}^1$. These are the fibers of the double cover $x: E \to \mathbb{P}^1$.
• $|3O|$: $\ell(3O) = 3$, so $|3O| \cong \mathbb{P}^2$. Each element is a group of $3$ points cut out by a line in the Weierstrass embedding $E \hookrightarrow \mathbb{P}^2$.
• $|nO|$ for $n \geq 1$: $\ell(nO) = n$, so $|nO| \cong \mathbb{P}^{n-1}$.

Let $C$ have genus $2$ and let $P \in C$ be a non-Weierstrass point.

• $|P|$: $\ell(P) = 1$ (by Riemann--Roch, since $P$ is not a Weierstrass point), so $|P| = \{P\}$. Trivially, $P$ is a base point.
• $|2P|$: $\ell(2P) = 1$ for a general $P$ (since $\ell(K - 2P) = 1$ when $2P \sim K$ fails). So $|2P| = \{2P\}$ and $P$ is a base point of multiplicity $2$.
• $|K_C|$: $\ell(K_C) = 2$ and $\deg K_C = 2$. For any point $P$, $\ell(K_C - P) = 1$ by Riemann--Roch ($= \ell(P) + 1 - 2 + 1$... more directly, since $K_C$ is the $g^1_2$, every point appears in some fiber). So $|K_C|$ is base-point-free.

Now if $W$ is a Weierstrass point: $\ell(2W) = 2$, so $|2W| \cong \mathbb{P}^1$ and $2W \sim K_C$. In the canonical system, $W$ still appears in every divisor? No -- $|K_C| = |2W| \cong \mathbb{P}^1$, and the divisor corresponding to $x - x(W)$ is $2W$ itself, but other divisors in $|K_C|$ are pairs $Q + Q'$ not involving $W$. So $W$ is not a base point.

• $\mathbb{P}^1$: the system $|n \cdot [\infty]|$ is a $g^n_n$. In particular, $|[\infty]|$ is a $g^1_1$ (the identity map $\mathbb{P}^1 \to \mathbb{P}^1$).
• Elliptic curve: $|nO|$ is a $g^{n-1}_n$. The system $|2O|$ is a $g^1_2$ (double cover of $\mathbb{P}^1$). The system $|3O|$ is a $g^2_3$ (plane cubic embedding).
• Genus-2 curve: $|K_C|$ is a $g^1_2$ (the hyperelliptic pencil).
• Non-hyperelliptic genus-3 curve: $|K_C|$ is a $g^2_4$ (canonical embedding as plane quartic).
• Non-hyperelliptic genus-4 curve: $|K_C|$ is a $g^3_6$ (canonical embedding in $\mathbb{P}^3$, intersection of a quadric and a cubic).
• General genus-$g$ curve ($g \geq 2$): $|K_C|$ is a $g^{g-1}_{2g-2}$.

A key question in curve theory is: what is the smallest $d$ such that $C$ carries a $g^1_d$?

• $g = 0$: $d = 1$ (isomorphism to $\mathbb{P}^1$).
• $g = 1$: $d = 2$ (every elliptic curve is a double cover of $\mathbb{P}^1$).
• $g = 2$: $d = 2$ (every genus-$2$ curve is hyperelliptic).
• $g = 3$: $d = 2$ if hyperelliptic; $d = 3$ if not (project from a point on the canonical quartic).
• $g = 4$: $d = 2$ if hyperelliptic; $d = 3$ for a general curve (the canonical curve lies on a quadric surface in $\mathbb{P}^3$; rulings give a $g^1_3$).

The Brill--Noether number $\rho(g, r, d) = g - (r+1)(g - d + r)$ governs when $g^r_d$'s exist on a general curve: if $\rho \geq 0$, a general curve of genus $g$ carries a $g^r_d$; if $\rho < 0$, it does not.

On $\mathbb{P}^1$, the system $|n \cdot [\infty]|$ with basis $\{1, t, t^2, \ldots, t^n\}$ gives:

$\phi_{n[\infty]} : \mathbb{P}^1 \to \mathbb{P}^n, \quad [s:t] \mapsto [s^n : s^{n-1}t : \cdots : t^n].$

• $n = 1$: the identity $\mathbb{P}^1 \to \mathbb{P}^1$.
• $n = 2$: the Veronese conic $\mathbb{P}^1 \hookrightarrow \mathbb{P}^2$, image is $\{xz = y^2\}$.
• $n = 3$: the twisted cubic $\mathbb{P}^1 \hookrightarrow \mathbb{P}^3$, parametrized by $(s^3, s^2t, st^2, t^3)$.

All are embeddings since the $g^n_n$ separates points and tangent vectors on $\mathbb{P}^1$.

Let $E: y^2 = x^3 + ax + b$ with origin $O$ at infinity.

The $g^1_2$ from $|2O|$: basis $\{1, x\}$, giving $\phi : E \to \mathbb{P}^1$, $P \mapsto [1 : x(P)]$. This is the double cover $(x,y) \mapsto x$, branched at the four roots of $x^3 + ax + b = 0$ and at $\infty$.

The $g^2_3$ from $|3O|$: basis $\{1, x, y\}$, giving $\phi : E \hookrightarrow \mathbb{P}^2$, $P \mapsto [1 : x(P) : y(P)]$. The image is the Weierstrass plane cubic. This is an embedding because $3O$ is very ample on $E$.

The $g^3_4$ from $|4O|$: basis $\{1, x, y, x^2\}$, giving $\phi : E \hookrightarrow \mathbb{P}^3$, image is a curve of degree $4$, the intersection of two quadric surfaces.

For an elliptic curve $E$ ($g = 1$):

• $D = O$ ($\deg = 1$): $\ell(D) = 1$, not base-point-free (point map to $\mathbb{P}^0$).
• $D = 2O$ ($\deg = 2$): $\ell(D) = 2$, base-point-free (gives the $2:1$ cover $E \to \mathbb{P}^1$), but not very ample (does not separate the two preimages of each point).
• $D = 3O$ ($\deg = 3 = 2g + 1$): $\ell(D) = 3$, very ample. Embeds $E$ as a plane cubic.
• $D = 4O$ ($\deg = 4$): $\ell(D) = 4$, very ample. Embeds $E$ in $\mathbb{P}^3$ as the complete intersection of two quadrics.
• $D = nO$ for $n \geq 3$: always very ample, embedding $E$ in $\mathbb{P}^{n-1}$ as a curve of degree $n$.

For a genus-$2$ curve $C$ ($g = 2$, $2g + 1 = 5$):

• $|K_C|$ is a $g^1_2$ ($\deg 2$). Base-point-free but not very ample -- gives a $2:1$ map to $\mathbb{P}^1$.
• A general divisor $D$ of degree $3$: $\ell(D) = 2$ (since $3 - 2 + 1 = 2$ and $D$ is non-special for general $D$). This $g^1_3$ maps $C$ to $\mathbb{P}^1$, so not very ample.
• A general divisor $D$ of degree $4$: $\ell(D) = 3$, giving a map $\phi_D : C \to \mathbb{P}^2$. One can check this is birational onto a plane quartic with one node (since $g = 2$ but a smooth quartic has $g = 3$). Not an embedding, so not very ample.
• $D$ of degree $5$: $\ell(D) = 4$, and $D$ is very ample. Embeds $C \hookrightarrow \mathbb{P}^3$ as a curve of degree $5$.

Every genus-$2$ curve is hyperelliptic. The canonical system $|K_C|$ is a $g^1_2$:

$\phi_K : C \xrightarrow{2:1} \mathbb{P}^1.$

If $C : y^2 = f(x)$ with $\deg f = 5$ or $6$, a basis of $H^0(\omega_C)$ is $\{dx/y,\, x\, dx/y\}$, and the canonical map is just $(x, y) \mapsto [1 : x]$, i.e., the projection to the $x$-coordinate. The six branch points are the roots of $f$ (plus $\infty$ if $\deg f = 5$).

Let $C$ be a non-hyperelliptic curve of genus $3$. Then $|K_C|$ is a $g^2_4$:

$\phi_K : C \hookrightarrow \mathbb{P}^2$

embedding $C$ as a smooth plane quartic. Conversely, every smooth plane quartic is a canonically embedded genus-$3$ curve, since for a plane curve of degree $d = 4$, adjunction gives $K_C \sim (4 - 3)H|_C = H|_C$, i.e., the canonical class is the hyperplane class.

Explicit example: Let $C = V(x^4 + y^4 + z^4) \subset \mathbb{P}^2$ (the Fermat quartic).

• Genus: $g = \frac{(4-1)(4-2)}{2} = 3$.
• The canonical system is cut out by lines: $|K_C| = \{C \cap L \mid L \text{ a line in } \mathbb{P}^2\}$.
• Each canonical divisor consists of $4$ points (the intersection of a line with $C$).
• The differentials $\omega_1 = x\,\frac{dz}{F_y}, \omega_2 = y\,\frac{dz}{F_y}, \omega_3 = z\,\frac{dz}{F_y}$ (up to scaling) give the identity embedding $C \hookrightarrow \mathbb{P}^2$.

Let $C: y^2 = f(x)$ with $\deg f = 7$ or $8$ be a hyperelliptic genus-$3$ curve. A basis of differentials is $\{dx/y,\, x\, dx/y,\, x^2\, dx/y\}$, so:

$\phi_K : C \to \mathbb{P}^2, \quad (x, y) \mapsto [1 : x : x^2].$

The image is the conic $\{xz = y^2\} \cong \mathbb{P}^1$ (the Veronese image of $\mathbb{P}^1$), and $\phi_K$ is $2:1$ onto this conic. The canonical map is not an embedding -- it collapses the pairs of conjugate points $(x, y)$ and $(x, -y)$.

Let $C$ be a non-hyperelliptic curve of genus $4$. The canonical map gives:

$\phi_K : C \hookrightarrow \mathbb{P}^3$

with image of degree $2g - 2 = 6$. By the Enriques--Babbage theorem, the canonical curve $C \subset \mathbb{P}^3$ lies on a unique quadric surface $Q$:

• If $Q$ is smooth ($Q \cong \mathbb{P}^1 \times \mathbb{P}^1$), then $C$ is a $(3,3)$-curve on $Q$, and $C = Q \cap S$ where $S$ is a cubic surface.
• If $Q$ is a quadric cone, then $C$ is a divisor of type $3H$ on $Q$ (where $H$ is the hyperplane class).

In either case, $C$ is a complete intersection of a quadric and a cubic in $\mathbb{P}^3$.

Every curve admits a birational plane model (a $g^2_d$ for some $d$):

• Genus $0$: $\mathbb{P}^1 \hookrightarrow \mathbb{P}^2$ as a line ($g^2_1$) or conic ($g^2_2$).
• Genus $1$: plane cubic ($g^2_3$), the Weierstrass embedding.
• Genus $2$: project from $\mathbb{P}^3$ (degree $5$ embedding) to $\mathbb{P}^2$: get a plane quintic with $\binom{2}{2} = 1$ node... actually, by the degree-genus formula $g = \frac{(d-1)(d-2)}{2} - \delta$, a genus-$2$ curve maps birationally to a plane quartic with $\delta = 1$ node. So the minimal plane model has degree $4$ with one node.
• Genus $3$, non-hyperelliptic: smooth plane quartic ($g^2_4$, no singularities needed).
• Genus $3$, hyperelliptic: plane model of degree $5$ with $2$ nodes (since $3 = 6 - 3 = \frac{4 \cdot 3}{2} - 3$... more carefully, degree $5$ with $\delta = 3$ nodes).

Let $C \subset \mathbb{P}^2$ be a smooth plane quartic (canonical genus $3$, so $g - 1 = 2$ and $\mathbb{P}^{g-1} = \mathbb{P}^2$).

Take $D = P_1 + P_2 + P_3$ where $P_1, P_2, P_3$ are collinear (they lie on a line $L$). Since $L \cap C$ consists of $4$ points, $L$ cuts out a canonical divisor $K \sim P_1 + P_2 + P_3 + P_4$.

• $\dim \overline{P_1 P_2 P_3} = 1$ (a line in $\mathbb{P}^2$), not $\min(2, 2) = 2$.
• Geometric RR: $r(D) = 3 - 1 - 1 = 1$, so $\ell(D) = 2$, i.e., $D$ moves in a $g^1_3$.
• Alternatively: $i(D) = g - 1 - \dim \overline{P_1 P_2 P_3} = 2 - 1 = 1$, so $\ell(D) = 3 - 3 + 1 + 1 = 2$. Confirmed.

This $g^1_3$ is cut out by the pencil of lines through $P_4$: each line through $P_4$ meets $C$ in $P_4$ plus three more points, giving a $1$-parameter family of degree-$3$ divisors.

On a non-hyperelliptic curve of genus $g = 4$ canonically embedded in $\mathbb{P}^3$:

General $D$ of degree $3$: three points $P_1, P_2, P_3$ in general position span a plane ($\dim = 2$). So $r(D) = 3 - 1 - 2 = 0$, meaning $\ell(D) = 1$ and $D$ does not move: it is a $g^0_3$.

Special $D$ of degree $3$: three collinear points $P_1, P_2, P_3$ (lying on a line $L \subset \mathbb{P}^3$). Then $\dim \overline{P_1 P_2 P_3} = 1$, and $r(D) = 3 - 1 - 1 = 1$, so $\ell(D) = 2$. This is a $g^1_3$. Such a line $L$ meets the canonical curve (degree $6$) in at most $6$ points, and if it meets in exactly $3$ points of $D$, the residual intersection gives a $g^1_3$.

A non-hyperelliptic genus-$5$ curve $C \hookrightarrow \mathbb{P}^4$ has $\deg = 2g - 2 = 8$.

Consider $D = P_1 + P_2 + P_3 + P_4$ (four points).

Case 1: the four points are in general position, spanning a $\mathbb{P}^3 \subset \mathbb{P}^4$. Then $r(D) = 4 - 1 - 3 = 0$, so $\ell(D) = 1$: the divisor is rigid.

Case 2: the four points lie on a plane $\Pi \cong \mathbb{P}^2$. Then $\dim \overline{D} = 2$ and $r(D) = 4 - 1 - 2 = 1$, so $\ell(D) = 2$: this is a $g^1_4$.

Case 3 (unlikely for general $C$): all four points are collinear. Then $\dim \overline{D} = 1$ and $r(D) = 4 - 1 - 1 = 2$, so $\ell(D) = 3$: a $g^2_4$ (plane model of degree $4$, giving $C \to \mathbb{P}^2$ with degree $4$ image). By Brill--Noether: $\rho(5, 2, 4) = 5 - 3(5 - 4 + 2) = 5 - 9 = -4 < 0$, so a general genus-$5$ curve has no $g^2_4$.

Let $C$ be a non-hyperelliptic genus-$3$ curve, canonically embedded as a quartic $C \subset \mathbb{P}^2$.

Bitangent lines: a bitangent to $C$ is a line $L$ meeting $C$ at two points $P, Q$ each with multiplicity $\geq 2$, so $L \cdot C = 2P + 2Q$. Since $L \cdot C \sim K_C$, we get $K_C \sim 2P + 2Q$, hence $P + Q$ is a theta-characteristic ($2(P+Q) \sim K_C$). A smooth quartic has exactly $28$ bitangent lines.

Flex points: a flex is a point where $L \cdot C \geq 3P$. If $L \cdot C = 3P + Q$, then $\ell(3P) \geq 2$. A smooth quartic has $24$ flex points.

Linear systems: $|K_C| = g^2_4$ (lines). A pencil of lines through a point $Q \in C$ cuts out a $g^1_3$ (after removing $Q$ from each divisor): the residual system $|K_C - Q|$ is a $g^1_3$.

A non-hyperelliptic genus-$4$ curve $C \hookrightarrow \mathbb{P}^3$ has degree $6$ and lies on a unique quadric $Q$.

Case 1: $Q$ smooth. Then $Q \cong \mathbb{P}^1 \times \mathbb{P}^1$ (over $\bar{k}$), and $C$ has type $(3,3)$ on $Q$. The two rulings of $Q$ cut out two $g^1_3$'s on $C$. Thus every non-hyperelliptic genus-$4$ curve with smooth quadric carries exactly two $g^1_3$'s.

Case 2: $Q$ a quadric cone. The rulings through the vertex cut out a single $g^1_3$. Thus $C$ carries a unique $g^1_3$ and is called trigonal.

Using Brill--Noether: $\rho(4, 1, 3) = 4 - 2(4 - 3 + 1) = 4 - 4 = 0$, consistent with finitely many $g^1_3$'s on a general genus-$4$ curve.

Existence of $g^1_d$ on a general genus-$g$ curve: $\rho(g, 1, d) = g - 2(g - d + 1) = 2d - g - 2$. This is $\geq 0$ when $d \geq (g + 2)/2$, i.e., $d \geq \lceil g/2 \rceil + 1$.

• $g = 2$: $d \geq 2$. Every genus-$2$ curve has a $g^1_2$ (hyperelliptic). $\rho(2,1,2) = 0$: finitely many.
• $g = 3$: $d \geq 3$. $\rho(3,1,3) = 0$: finitely many $g^1_3$'s on a general genus-$3$ curve.
• $g = 4$: $d \geq 3$. $\rho(4,1,3) = 0$: finitely many $g^1_3$'s.
• $g = 5$: $d \geq 4$. $\rho(5,1,4) = 0$: finitely many $g^1_4$'s.
• $g = 6$: $d \geq 4$. $\rho(6,1,4) = 0$.

Existence of $g^2_d$: $\rho(g, 2, d) = g - 3(g - d + 2) = 3d - 2g - 6$. For $g = 5$: $3d - 16 \geq 0$ requires $d \geq 6$, and $\rho(5, 2, 6) = 2$. A general genus-$5$ curve has a $2$-dimensional family of plane models of degree $6$.