# 2024-09-17-12-46-29

:::{.remark}
We proved that every geometrically ruled surface is of the form $\PP_C(\mce)$ with $\rank \mce = 2$.
Recall that locally free $\OO_C$ modules are in bijection with $\AA^n$ bundles over $C$.
:::

:::{.lemma}
For all rank 2 locally free sheaves $\mce$, there is a SES
\[
0\to L\to \mce \to M \to 0
\]
where $L, M$ are line bundles.
If $h^0(\mce) > 0$, then $L = \OO(D)$ for $D$ an effective divisor.
Note that $\deg \mce \da \deg \det(\mce) = \deg \wedge^2 \mce$.
:::

:::{.proof}
Sections $s\in H^0(C; \mce)$ correspond to morphisms $\OO_C \mapsvia{s} \mce$ where $1\mapsto s$.
Note that the cokernel of a morphism of locally free sheaves may not be locally free.
Taking duals yields $\mce\dual \mapsvia{s\dual} \OO_C$.
Note $\mce\dual\surjects Q = \OO(-D)$ with kernel $K$, which is locally free.
This yields $0\to K\to \mce\dual\to Q\to 0$, so dualize to get the result.

This holds under the hypothesis $h^0(\mce) > 0$, so there is a section.
By Serre, $\mce(n) \da \mce \tensor \OO_C(n)$ is generated by global sections for $n\gg 0$.
So $H^0(\mce(n))\tensor \OO_C \surjects \mce(n)$, and the previous argument applies, then untwist.
:::

:::{.remark}
Recall Riemann-Roch for line bundles on a curve: $\chi(L) = \deg L + 1 - g$.
Note that $\det(\mce) \cong L\tensor M$, so $\deg(\mce) = \deg(L) + \deg(M)$.
Thus 
\[
\chi(\mce) = \chi(L) + \chi(M) = (\deg L + 1-g) + (\deg M + 1-g) = \deg(\mce) = 2(1-g)
.\]
:::

:::{.remark}
When does this extension split?
Twist $L\to \mce\to M$ by $M\inv$ to get $L\tensor M\inv \to \mce\tensor M\inv\to \OO$.
A section of the former is the same as a map $\OO\to \mce\tensor M\inv$.
A section $s\in H^0(\mce \tensor M\inv)$ maps to $1\in H^0(\OO)$ which further maps to some $e\in H^1(L\tensor M\inv)$ in the associated long exact sequence.
The extension splits iff the extension class $e=0$.
As a corollary, if $H^1(L\tensor M\inv) = 0$ then the extension splits.
:::

:::{.remark}
On $\PP^1$, any rank $r$ vector bundle $E$ splits as a sum of $\OO(n_i)$.
:::

:::{.proof}
Compute $\deg(E\tensor A) = \deg E + 2\deg A$ when $\rank A = 1$.
Replace $E$ by some $E(n)$ so that $\deg E = 0$ or $-1$.
By Riemann-Roch, $\chi(E) = \deg E + 2 = 2$ or $1$.
Thus $h^0(E) \geq 1$.
There is then a SES $\OO(n) \to E\to \OO(m)$ where $m = \deg E - n$ with $n\geq 0$.
The extension class $e\in H^1(\OO(2n-\deg E)) = 0$, since $2n-\deg E \geq 0$.
This uses that $H^1(\OO(d)) = 0$ if $d\geq -1$, since this is dual to $h^0(\OO(-2-d))$.
:::

:::{.remark}
Why this decomposition is unique: this minimum of $n_1, n_2$ is unique as a function of $E$.
The $\max(n_1, n_2)$ is the maximal $n$ such that $h^0(E(-n)) > 0$, since twisting by $-n$ yields $\OO \oplus \OO(n_2-n)$ where $n_2 - n \leq 0$.
:::

:::{.remark}
For an elliptic curve $C$, the only indecomposable rank 2 bundles are extensions $E$ in $\OO\to E\to \OO$ where $e\in H^1(\OO) = \CC$, or $\OO\to E\to \OO(p)$ with $e\in H^1(\OO(-p)) = H^0(\OO(p))\dual = \CC$ for $p\in C$ a point.
One can take either of these and tensor by a line bundle.
:::

:::{.remark}
For curves of genus $g\geq 2$, there is a moduli space of indecomposable rank 2 vector bundles of degrees 0 or 1.
The dimension is $3g-3$.
:::

:::{.corollary}
The geometrically ruled surfaces over $\PP^1$ are $\PP(\OO \oplus \OO(n)) = \FF_n$, the Hirzebruch surfaces, for $n\geq 0$.
:::

## Tautological line bundles on $\PP_C(E)$

:::{.remark}
On $S$, there is a natural SES $N\to p^* E\to \OO_{\PP(E)}(1)$ where $p: S\to C$, $p^* E$ is rank 2, and $N, \OO_{\PP(E)}(1)$ are line bundles.
There is a similar sequence on fibers $\OO(-1)\to \OO \oplus \OO \to \OO(1)$.
Regard a point in a fiber as an $\AA^1 \subseteq \AA^2$.
We thus call $\OO_{\PP(E)}(1)$ the tautological bundle.
:::

:::{.remark}
Sections $s: C\to \PP_C(E)$ are in bijection with rank 1 locally free quotients of $E$ on $C$.
:::

:::{.remark}
Intersection theory on a deeper level, toward the enumerative geometry of $\PP_C(E)$.
Let $X$ be a variety, and $F$ a rank $r$ vector bundle on $X$.
Then there exist Chern classes $c_i(F)\in A^i(X)$ for $0\leq i \leq \dim X$.
Recall that $A^i(X)$ are codimension $i$ cycles $\sum n_k Z_k$ modulo linear equivalence, so $A^1(X) = \Pic(X)$.
Recall linear equivalence:

\includesvg{figures/2024-09-17_13-55.svg}

The total Chern class is $c(F) \da \sum_i c_i(F)$.
It is additive in SESs: for $A\to B\to C$, one has $c(B) = c(A)\cdot c(C)$.
A map $f: X\to Y$ induces $f^*: A^i(Y) \to A^i(X)$ where $c_i(f^* F) = f^*(c_i(F))$.
If $L$ is a line bundle, then $c(L) = 1 + L$.
:::

:::{.remark}
By general principles, $c_2(p^* E) = 0$.
Next time $[N] \cdot [\PP_C(E)] = 0$.
:::