# 2024-09-12-12-46-17

## Classification of ruled surfaces

:::{.remark}
Recalling ruled surfaces: let $f:S\to C$ where all fibers are isomorphic to $\PP^1$.
We proved it is a locally free $\PP^1$ bundle, so $\forall p\in C$ there is some $U\ni p$ such that $f\inv(U) \cong U\times \PP^1$.
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:::{.remark}
Classification of locally free rank $r$ sheaves $\mce$ and locally free $\AA^1$ bundles (vector bundles).
Letting $X = \Union U_i$, there is an isomorphism $g_i: \ro{\mce}{U_i} \iso \OO_{U_i}^{\oplus r}$.
Different choices of isomorphisms yield elements $g_i\in \GL_r(\OO(U_i)) = \Aut(\OO_{U_i}^{\oplus r})$.
The coefficients are in the ring of regular functions on $U_i$.
How do these isomorphisms glue?
We specify transition functions $g_{ij}\in \GL_r(\OO(U_i \intersect U_j))$ satisfying the 1-cocycle condition on triple intersections $g_{ij}g_{jk}g_{ki} = 1 \in \GL_r(\OO(U_i \intersect U_j \intersect U_k))$.
We mod out by the relation $g_{ij}' = g_j\inv g_{ij} g_i$, the 1-coboundary condition.
This yields Čech cohomology $\check{H}^1(U, \GL_r(\OO)) \cong H^1(X, \GL_r(\OO))$.
This works generally: global versions on locally free objects live in $\check{H}^1(U, \Aut(\mce))$.
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:::{.remark}
Vector bundles are locally free $\AA^r$ bundles.
Let $f:Y\to X$ have fibers $\AA^1$, so $U_i \subseteq X$ lifts to $f\inv(U_i) = U_i \times \AA^r$.
What are the transition functions?
One looks for automorphisms of $(U_{i} \intersect U_j) \times \AA^r\ni (x,y_1,\cdots, y_r)$, which correspond to $g_{ij}(x)\in \GL_r(\OO(U_i \intersect U_j))$.
What is the corresponding sheaf?
Let $\mce(U)$ be sections over $U$, so morphisms $s: U\to f\inv(U)$.
Locally this is given by $r$ regular functions, so $\mce(U_i) = \OO(U_i)^{\oplus r}$.
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:::{.remark}
For ruled surfaces, we consider $\check{H}^1(C, \PGL_2(\OO))$.
More generally, take $\PGL_r(\OO)$.
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:::{.theorem}
All geometrically ruled surfaces are of the form $\PP_C(\mce)$ where $\mce$ is a locally free sheaf of rank 2.
Moreover, $\PP_C(\mce)\iso \PP_C(\mce')$ over $C$ iff $\mce' \cong \mce \oplus L$ for $L$ some line bundle.
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:::{.remark}
Thus $\mce$ on $C$ is in bijection with an $\AA^2$ bundle $\AA(\mce)\to C$.
What is $\PP_C(\mce)$?
This is the bundle of lines in the vector bundle.
Dually, one could define this as rank 1 quotients of the vector bundle.
Passing back and forth: replace $\mce$ by $\mce\dual$.
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:::{.proof}
There is a sequence $1\to \CC^2\to \GL_2(\CC)\to \PGL_2(\CC)\to 1$ which globalizes to $1\to \OO\units\to \GL_2(\OO)\to \PGL_2(\OO)\to 0$.
Taking the LES yields $\Pic(C)\to H^1(\GL_2(\OO))\to H^1(\PGL_2(\OO))$ where the middle corresponds to rank 2 locally free bundles and the right corresponds to locally free $\PP^1$ bundles.
It suffices to show this is surjective, so consider $H^2(\OO_C\units)$ -- this is trivial by Grothendieck vanishing since $\dim C = 1$.
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:::{.remark}
Goal: understand locally free sheaves on curves, e.g. $\PP^1$ (easy), elliptic curves (harder), and curves of general type (generally hard).
Next: relate birationally and geometrically ruled surfaces.
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:::{.remark}
A (classically) minimal surface is a smooth projective surface with no $-1$ curves.
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:::{.remark}
Ruled surface: $S\birational C\times \PP^1$ for $C$ a curve.
Take a sequence of blowups to obtain $\tilde S\to C\times \PP^1$ a regular morphism.
Consider the composite $\tilde S\to C\times \PP^1\to C$; most fibers are $\PP^1$ but some are more complicated:

\includesvg{inkscape/2024-09-12_13-22.svg}

Claim: either $\tilde S = \PP_C(\mce)$ or there exists a $-1$ curve in a fiber.
In the first case, $\tilde S\to \PP_C(\mce)\to C$ is a relatively minimal model.
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:::{.proof}
Consider a reducible fiber $F = \sum m_i D_i$.
Then every $D_i^2 < 0$.
Use that $F^2 = 0$.
Consider $D_{i_0}\cdot F = 0$ one one hand, and is equal to $D_{i_0}\qty{m_{i_0} D_{i_0} + \sum_{j\neq i} m_j D_j}$.
Then $D_{i_0}D_j \geq 0$ for all $j$, so we must have $D_{i_0}^2 < 0$ for this to equal zero.

Moreover $(K_S+F)\mid_F = K_F = K_{\PP^1}$, and $F\mid_F = 0$, so $K_S F = -2$.
Thus there exists a $D_i \subseteq F$ with $K_S D_i < 0$.
If $D_i$ is a connected curve, one can show $p_a(D_i) = {1\over 2}(KD_i + D_i^2) \geq 0$.
Since $D_i ^2 < 0$ and $KD_i < 0$, this forces $D_i \cong \PP^1$.

> Missed last part of this argument.

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:::{.remark}
Fact: there are no covers $\PP^1\to C$ if $g(C) > 0$.
By Riemann-Hurwitz, $K_{\PP^1} = f^* K_C + \sum (m_i - 1)p_i$, and taking degrees yields $-2$ on the LHS and $2g-2 > 0$ and a positive contribution on the RHS, a contradiction.

Alternatively, consider $\pi: C'\to C$ and consider the induced map $\pi^*: H^0(\Omega_C)\to H^0(\Omega_{C'})$.
Note $y=f(x) \implies dy = f'(x) dx$.
The claim is that this map is injective, so the genus increases.
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## Classification of $\PP_C(\mce)$

:::{.theorem}
(Grothendieck) On $\PP^1$, every locally free sheaf is uniquely isomorphic to \( \bigoplus_i \OO(n_i) \).
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:::{.corollary}
$L \otimes \mce = \OO \oplus \OO(n)$ for some $n\geq 0$.
Thus geometrically ruled surfaces over $\PP^1 = C$ are Hirzebruch surfaces $\FF_n \da \PP_{\PP^1}(\OO \oplus \OO(n))$.
Note that $\FF_0 = \PP^1\times \PP^1$.
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:::{.theorem}
(Atiyah) If $C$ is an elliptic curve, then for any rank $r$ there exists a unique indecomposable rank $r$ locally free sheaf $\mce$, modulo twisting by a degree zero line bundle $L\in \Pic^0(C)=C$.
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:::{.corollary}
If $\deg \mce$ is even, then there exists an $L$ such that $\deg(\mce \otimes L) \deg(\mce) + 2\deg(\mcl) = 0$.
Thus either $\mce = L_1 \oplus L_2$ or $\OO \oplus L_3$.
If $\deg \mce$ is odd, $\deg(\mce \otimes L) = 1$.
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:::{.remark}
Note $\Ext(A, B) = \Ext(\OO, B \otimes A\inv) = H^1(B \otimes A\inv)$.
Considering an elliptic curve and $0\to \OO\to \mce\to \OO\to 0$, one has $H^1(\OO\tensor \OO\inv) = H^1(\OO) = g(C) = 1$, so there is a nontrivial extension.
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:::{.remark}
Next time: $\deg \mce = \deg \det \mce$ as a line bundle.
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