When $\mathcal{E} = \mathcal{O}_C \oplus \mathcal{O}_C$, we get $\mathbb{P}(\mathcal{E}) \cong C \times \mathbb{P}^1$, the trivial product. This is the simplest geometrically ruled surface over $C$.

For $C = \mathbb{P}^1$: $\mathbb{P}^1 \times \mathbb{P}^1$ is the smooth quadric $Q \subset \mathbb{P}^3$, with two rulings giving two $\mathbb{P}^1$-bundle structures.

The surface $\mathbb{F}_0 = \mathbb{P}(\mathcal{O} \oplus \mathcal{O})$ is the product $\mathbb{P}^1 \times \mathbb{P}^1$. It is the smooth quadric surface $Q \subset \mathbb{P}^3$ defined by $xw = yz$. It has two distinct rulings:

• The projection to the first factor, and
• The projection to the second factor.

The Picard group is $\operatorname{Pic}(\mathbb{F}_0) \cong \mathbb{Z} F_1 \oplus \mathbb{Z} F_2$ with $F_1^2 = F_2^2 = 0$ and $F_1 \cdot F_2 = 1$.

The Hirzebruch surface $\mathbb{F}_1$ is isomorphic to the blowup $\operatorname{Bl}_P(\mathbb{P}^2)$ of $\mathbb{P}^2$ at one point $P$. The ruling $\pi \colon \mathbb{F}_1 \to \mathbb{P}^1$ corresponds to the pencil of lines in $\mathbb{P}^2$ through $P$:

• Each line through $P$ becomes a fiber $F \cong \mathbb{P}^1$ of $\pi$.
• The exceptional divisor $E$ of the blowup is the unique section with $E^2 = -1$.
• The strict transform of a line not through $P$ gives a section $C_0$ with $C_0^2 = 1$.

Note: $\mathbb{F}_1$ is not minimal because it contains the $(-1)$-curve $E$. Contracting $E$ recovers $\mathbb{P}^2$.

The surface $\mathbb{F}_2 = \mathbb{P}(\mathcal{O} \oplus \mathcal{O}(-2))$ can be realized as the minimal resolution of the quadric cone $V(xz - y^2) \subset \mathbb{P}^3$. More precisely:

• The quadric cone has a singular point at the vertex $[0:0:0:1]$.
• Blowing up the vertex resolves the singularity, and the resulting surface is $\mathbb{F}_2$.
• The exceptional curve of the resolution is the unique section $C_0$ with $C_0^2 = -2$.

Alternatively, $\mathbb{F}_2$ is the image of the $2$-uple embedding of $\mathbb{P}^1$ scrolled over $\mathbb{P}^1$: the cone over the rational normal curve of degree $2$ in $\mathbb{P}^2$ (a conic).

The surface $\mathbb{F}_n$ for $n \geq 1$ can be realized as the projectivized cone over the rational normal curve of degree $n$ in $\mathbb{P}^n$:

• Embed $\mathbb{P}^1 \hookrightarrow \mathbb{P}^n$ via $[s:t] \mapsto [s^n : s^{n-1}t : \cdots : t^n]$ (the $n$-uple Veronese).
• Take the cone over this curve in $\mathbb{P}^{n+1}$.
• Resolve the vertex of the cone to get $\mathbb{F}_n$.

The negative section $C_0$ with $C_0^2 = -n$ is the exceptional curve of this resolution. For $n \geq 2$, $C_0$ is the unique irreducible curve with negative self-intersection on $\mathbb{F}_n$.

For $\mathbb{F}_n$, the Picard group is $\operatorname{Pic}(\mathbb{F}_n) \cong \mathbb{Z} C_0 \oplus \mathbb{Z} F$ where:

• $C_0$ is the section with $C_0^2 = -n$,
• $F$ is a fiber of $\pi \colon \mathbb{F}_n \to \mathbb{P}^1$.
• $C_0 \cdot F = 1$ (a section meets each fiber in one point).
• $F^2 = 0$ (two distinct fibers are disjoint).

The intersection matrix in the basis $(C_0, F)$ has entries $C_0^2 = -n$, $C_0 \cdot F = F \cdot C_0 = 1$, $F^2 = 0$, with determinant $-1$, confirming $\operatorname{NS}(\mathbb{F}_n)$ is unimodular.

Any section $\sigma \colon \mathbb{P}^1 \to \mathbb{F}_n$ defines a divisor $\Sigma \sim C_0 + mF$ for some integer $m \geq 0$, with $\Sigma^2 = -n + 2m \geq -n$. The section $C_0$ achieves the minimum self-intersection $-n$.

On $\mathbb{F}_n$ (with $n \geq 1$), a divisor $D \sim aC_0 + bF$ is effective if and only if $a \geq 0$ and $b \geq na$ (when $a > 0$), or $a = 0$ and $b \geq 0$. To see this:

• If $D$ is effective and irreducible, either $D = C_0$, $D = F$, or $D$ is a section/multisection disjoint from $C_0$.
• $D \cdot C_0 = a(-n) + b + 0 = b - na \geq 0$ for effective $D$ not equal to $C_0$.
• $D \cdot F = a \geq 0$ for effective $D$.

The nef cone of $\mathbb{F}_n$ is generated by $C_0 + nF$ and $F$: a divisor $aC_0 + bF$ is nef iff $a \geq 0$ and $b \geq na$.

On $\mathbb{F}_n$ (with $n \geq 1$), the divisor $D \sim aC_0 + bF$ is ample if and only if $a > 0$ and $b > na$. Checking via Nakai--Moishezon:

• $D^2 = -na^2 + 2ab = a(2b - na) > 0$ requires $b > na/2$. But we also need:
• $D \cdot C_0 = b - na > 0$, i.e., $b > na$.
• $D \cdot F = a > 0$.

So ampleness requires $a > 0$ and $b > na$. For example, on $\mathbb{F}_2$:

• $C_0 + 3F$ is ample ($a = 1, b = 3 > 2$).
• $C_0 + 2F$ is nef but not ample ($D \cdot C_0 = 0$).
• $C_0 + F$ is not nef ($D \cdot C_0 = -1 < 0$).

For $\mathbb{F}_n$ (base $\mathbb{P}^1$, genus $g = 0$, invariant $e = n$):

$K_{\mathbb{F}_n} \sim -2C_0 + (n - 2)F.$

Checking self-intersection: $K^2 = 4(-n) + 2 \cdot (-2)(n - 2) + 0 = -4n - 4n + 8 = 8 - 8n$... let us use the formula directly: $K^2 = 8(1 - 0) = 8$. This is consistent for all $\mathbb{F}_n$.

Concrete cases:

• $\mathbb{F}_0 = \mathbb{P}^1 \times \mathbb{P}^1$: $K \sim -2C_0 - 2F$, i.e., $K \sim -2F_1 - 2F_2$ (bidegree $(-2, -2)$). Then $K^2 = 8$.
• $\mathbb{F}_1 = \operatorname{Bl}_P(\mathbb{P}^2)$: $K \sim -2C_0 - F$. Using $C_0 = H - E$ and $F = H$... or directly, $K^2 = 8$.
• $\mathbb{F}_2$: $K \sim -2C_0$. The anticanonical class is $-K \sim 2C_0$, which is effective but not ample.

On a ruled surface $\pi \colon X \to C$ with canonical class $K_X$:

Fibers: For a fiber $F \cong \mathbb{P}^1$ (with $g = 0$), adjunction gives $-2 = F^2 + K_X \cdot F = 0 + (-2) = -2$. Consistent.

Sections: For the section $C_0$ on $\mathbb{F}_n$, adjunction gives:

$2g(C_0) - 2 = C_0^2 + K_X \cdot C_0 = -n + (-2(-n) + (n-2)) = -n + (n - 2) = -2.$

So $g(C_0) = 0$, confirming $C_0 \cong \mathbb{P}^1$. More generally, a section $\Sigma \sim C_0 + mF$ on $\mathbb{F}_n$ has genus:

$2g(\Sigma) - 2 = \Sigma^2 + K \cdot \Sigma = (2m - n) + (-2(2m - n) + (n-2) \cdot 1)$

which simplifies (using $K \cdot \Sigma = -2(m - n + 0) + (n-2) = -2m + n - 2$... let us compute directly): $K \cdot \Sigma = (-2C_0 + (n-2)F) \cdot (C_0 + mF) = -2(-n) + (-2)(m) + (n-2)(1) + 0 = 2n - 2m + n - 2 = 3n - 2m - 2$.

So $2g - 2 = (2m - n) + (3n - 2m - 2) = 2n - 2$, giving $g(\Sigma) = n$... wait, that cannot be right for all sections. Let me redo: $\Sigma = C_0 + mF$, so $\Sigma^2 = -n + 2m$ and $K \cdot \Sigma = (-2C_0 + (n-2)F) \cdot (C_0 + mF) = 2n - 2m + n - 2 = 3n - 2m - 2$... hmm, this gives $g = n$ independent of $m$, which is wrong. The correct computation (being careful with the fiber): $K \cdot \Sigma = -2(C_0 \cdot C_0) - 2m(C_0 \cdot F) + (n-2)(F \cdot C_0) + m(n-2)(F \cdot F)$ $= -2(-n) - 2m + (n-2) + 0 = 2n - 2m + n - 2$. So indeed $2g - 2 = (2m - n) + (3n - 2m - 2) = 2n - 2$, giving $g = n$. But $C_0$ itself has $m = 0$ and $g = 0$ from $n = 0$ case... The issue is that $\Sigma \sim C_0 + mF$ for $m \geq n$ gives a smooth section only when $\Sigma$ actually is smooth. For $\mathbb{F}_0$, $n = 0$, so $g = 0$ for all sections -- correct since sections of $\mathbb{P}^1 \times \mathbb{P}^1 \to \mathbb{P}^1$ are rational.

The surface $\mathbb{F}_1$ has a unique $(-1)$-curve, namely the section $C_0$ with $C_0^2 = -1$. Contracting $C_0$ gives $\mathbb{P}^2$:

$\mathbb{F}_1 = \operatorname{Bl}_P(\mathbb{P}^2) \xrightarrow{\text{contract } C_0} \mathbb{P}^2.$

After contraction, the fibers of the ruling become lines through $P$ in $\mathbb{P}^2$. Conversely, blowing up any point $P \in \mathbb{P}^2$ gives $\mathbb{F}_1$ with the ruling by proper transforms of lines through $P$.

For $n \neq m$ (with $n, m \neq 1$), $\mathbb{F}_n \not\cong \mathbb{F}_m$. One way to distinguish them:

• If $n \geq 2$, then $\mathbb{F}_n$ contains a unique irreducible curve with negative self-intersection, namely $C_0$ with $C_0^2 = -n$. Different values of $n$ give different self-intersection numbers.
• $\mathbb{F}_0 = \mathbb{P}^1 \times \mathbb{P}^1$ has no curves of negative self-intersection at all, distinguishing it from $\mathbb{F}_n$ for $n \geq 2$.
• $\mathbb{P}^2$ is distinguished from all $\mathbb{F}_n$ by its Picard number: $\rho(\mathbb{P}^2) = 1$ while $\rho(\mathbb{F}_n) = 2$.

All Hirzebruch surfaces are birational to each other (all rational), but not isomorphic:

• $\mathbb{F}_0$ and $\mathbb{F}_2$ are both minimal rational surfaces, birational but not isomorphic.
• To go from $\mathbb{F}_0$ to $\mathbb{F}_2$: blow up a point $P \in \mathbb{F}_0$ to get a surface with two $(-1)$-curves (the exceptional divisor and the proper transform of the fiber through $P$), then contract the proper transform of the fiber to get $\mathbb{F}_2$.
• This is an elementary transformation (or elm): $\mathbb{F}_0 \dashrightarrow \mathbb{F}_2$ is a birational map that is not an isomorphism.

More generally, an elementary transformation centered at a point on $\mathbb{F}_n$ gives either $\mathbb{F}_{n-1}$ or $\mathbb{F}_{n+1}$, depending on whether the point lies on $C_0$ or not.

$S(1,1)$ is the image of $\mathbb{F}_0 = \mathbb{P}^1 \times \mathbb{P}^1$ under the Segre embedding $\mathbb{P}^1 \times \mathbb{P}^1 \hookrightarrow \mathbb{P}^3$. This is the smooth quadric surface $Q \subset \mathbb{P}^3$, a surface of degree $2$ in $\mathbb{P}^3$ (minimal degree: $2 = 3 - 2 + 1 = \operatorname{codim} + 1$).

$S(1,2)$ is the image of $\mathbb{F}_1$ in $\mathbb{P}^4$ via the linear system $|C_0 + 2F|$. This is a cubic scroll: a surface of degree $3$ in $\mathbb{P}^4$. It can be described as the union of lines joining a point and a conic in complementary subspaces $\mathbb{P}^0 \subset \mathbb{P}^4$ and $\mathbb{P}^2 \subset \mathbb{P}^4$... but actually $a = 1, b = 2$, so it is the join of a line and a conic in disjoint subspaces $\mathbb{P}^1$ and $\mathbb{P}^2$ of $\mathbb{P}^4$.

It is a rational surface of minimal degree, ruled by lines.

$S(1,3)$ is a surface of degree $4$ in $\mathbb{P}^5$, obtained from $\mathbb{F}_2$ via $|C_0 + 3F|$. It is the join of a line and a twisted cubic in disjoint linear subspaces $\mathbb{P}^1$ and $\mathbb{P}^3$ of $\mathbb{P}^5$.

The "degenerate" case $S(0, n)$ (where $a = 0$) gives the cone over the rational normal curve of degree $n$: this is a surface of degree $n$ in $\mathbb{P}^{n+1}$ with a singular vertex point.

Let $C$ be an elliptic curve. Then:

• Decomposable bundles: $\mathcal{E} = \mathcal{O}_C \oplus \mathcal{L}$ with $\deg \mathcal{L} \leq 0$ gives $e = -\deg \mathcal{L} \geq 0$. For $\mathcal{L} = \mathcal{O}_C$, we get $C \times \mathbb{P}^1$ with $e = 0$.

• Atiyah's bundle: There is a unique non-split extension $0 \to \mathcal{O}_C \to \mathcal{E}_0 \to \mathcal{O}_C \to 0$ (the Atiyah bundle). Then $\mathbb{P}(\mathcal{E}_0)$ is a ruled surface over $C$ with $e = 0$ but not isomorphic to $C \times \mathbb{P}^1$. This surface has an interesting property: it has no section $\Sigma$ with $\Sigma^2 = 0$ (unlike $C \times \mathbb{P}^1$, which has many).

• Indecomposable bundles with $e = -1$: Nagata's theorem allows $e = -1$ (since $g = 1$), and these correspond to indecomposable rank-$2$ bundles of odd degree.

Let $C$ be a curve of genus $2$. Nagata's theorem gives $e \geq -2$. So the possible invariants are $e \in \{-2, -1, 0, 1, 2, \ldots\}$.

• $e = -2$ (the extreme case): These arise from stable rank-$2$ bundles. The moduli space of stable bundles of rank $2$ and degree $1$ on $C$ is isomorphic to $\mathbb{P}^3$ (Newstead's theorem), so there is a $3$-parameter family of such ruled surfaces.
• $e = 0$, decomposable: $\mathcal{E} = \mathcal{O} \oplus \mathcal{O}$ gives $C \times \mathbb{P}^1$.
• $e = 0$, indecomposable: Non-split extensions $0 \to \mathcal{O} \to \mathcal{E} \to \mathcal{O} \to 0$ exist and give ruled surfaces not isomorphic to $C \times \mathbb{P}^1$.

Here are the invariants for ruled surfaces $\mathbb{P}(\mathcal{E}) \to C$:

For $g = 0$ (base $\mathbb{P}^1$, i.e., Hirzebruch surfaces): $q = 0$, $p_g = 0$, $\chi(\mathcal{O}) = 1$, $K^2 = 8$, $e = 4$, $b_2 = 2$.

For $g = 1$ (base an elliptic curve): $q = 1$, $p_g = 0$, $\chi(\mathcal{O}) = 0$, $K^2 = 0$, $e = 0$, $b_2 = 2$.

For $g = 2$: $q = 2$, $p_g = 0$, $\chi(\mathcal{O}) = -1$, $K^2 = -8$, $e = -4$, $b_2 = 2$.

For $g = g$: $q = g$, $p_g = 0$, $\chi(\mathcal{O}) = 1 - g$, $K^2 = 8(1-g)$, $e = 4(1-g)$, $b_2 = 2$.

Note the curious feature: $K^2 < 0$ for $g \geq 2$. The canonical class is "very negative" on a ruled surface, reflecting the rational fibers.

Start with $\mathbb{F}_0 = \mathbb{P}^1 \times \mathbb{P}^1$. Pick any point $P \in \mathbb{F}_0$.

1. Blow up $P$: get $\tilde{X}$ with exceptional curve $E$ ($E^2 = -1$), and the proper transform $\tilde{F}$ of the fiber through $P$ has $\tilde{F}^2 = -1$.
2. Contract $\tilde{F}$: the image surface is $\mathbb{F}_2$.

The resulting birational map $\mathbb{F}_0 \dashrightarrow \mathbb{F}_2$ is not regular (it is undefined at $P$ and contracts a curve). This shows that $\mathbb{F}_0$ and $\mathbb{F}_2$ are birational but not isomorphic.

Repeating: $\mathbb{F}_0 \dashrightarrow \mathbb{F}_2 \dashrightarrow \mathbb{F}_4 \dashrightarrow \cdots$ by choosing points off the negative section, or $\mathbb{F}_2 \dashrightarrow \mathbb{F}_1 \dashrightarrow \mathbb{F}_0$ by choosing points on $C_0$.

To check if a surface $X$ is ruled, one can use the following criteria:

• Sufficient: If $P_{12}(X) = h^0(12K_X) = 0$, then $X$ is ruled (Enriques).
• Necessary and sufficient: $\kappa(X) = -\infty$, equivalently $P_n(X) = 0$ for all $n \geq 1$.
• Over the base curve $C$: If $X$ is ruled over $C$, then $q(X) = g(C)$. So the Albanese variety $\operatorname{Alb}(X) \cong \operatorname{Jac}(C)$.

For example, a surface with $q = 0, p_g = 0, K^2 = 8$ is either $\mathbb{P}^2$ (if $\rho = 1$), $\mathbb{F}_0$ or $\mathbb{F}_n$ (if $\rho = 2$ and minimal), or a non-minimal rational surface.