# 2024-08-29-12-46-42

:::{.remark}
Recall the construction of the blowup at a point $\pi: \tilde S\to S$.
We proved $E^2= -1$.
:::

:::{.lemma}
Some facts:

- $\Pic \tilde S = \Pic S \oplus \ZZ E$
- $\NS(\tilde S) = \NS(S) \oplus \ZZ E$

Thus every line bundle on $\tilde S$ is of the form $\pi^* L + mE$ for some $m$ and some $L\in \Pic(S)$.
:::

:::{.proof}
Write $\Pic(S)$ as divisors modulo principal divisors, where the latter are $(\varphi) = (\varphi)_0 - (\varphi)_\infty$ for some \( \varphi \in \CC(S) \).
The numerator includes an additional divisor $E$, while the denominator is the same since they are birational.
:::

:::{.lemma}
If $p\in C$, then $\pi^* C = C' + mE$ where $C'$ is the strict transform of $C$, $C' = \overline{\pi\inv(C\setminus p)}$.
:::

:::{.proof}
Cover $\tilde S$ by $S\times \AA^1_u$ and $S\times \AA^1\times v$.
On the first we have coordinates $x,u$ where $y=xu$, on the second we have $y,v$ where $x=yv$.
Then the map is $(x, u) \mapsto (x, xu)$, the affine blowup.
Note that $x=0 \mapsto (0, 0)$ is the exceptional curve.
If $x\neq 0$ this map is invertible, since $u=y/x$.
Write $\pi^* f_m(x,y) = f_m(x,xu) = x^m g(x, u)$, and this is $mE + C'$.
:::

:::{.lemma}
\[
\tilde K_S = \pi^* K_S + E
.\]
:::

:::{.proof}
Write $\OO(K_S) = \Omega^2_S = \gens{dx\wedge dy}_{\OO_S}$.
Check $dx,dy$ in the coordinates $y=xu$ to get
\[
dx \wedge dy = dx\wedge d(xu) = dx\wedge \qty{udx + xdu} = xdx\wedge du
.\]
Conclusion: $\pi^* \Omega_S^2 \subseteq \Omega^2_{\tilde S}$, and \( \pi^* \Omega_S^2 = \Omega^2_{\tilde S}(E) \) since these forms vanish along $E$.
This says $\pi^* K_S = K_{\tilde S} - E$.
:::

:::{.lemma}
Claim: 

- $(C')^2 = C^2 - m^2$.
- $C_1' C_2' = C_1 C_2 - m_1 m_2$.
- $p_a(C') = p_a(C) - {m(m-1) \over 2}$.

:::

:::{.proof}
We have $\pi^* C_1 \pi^* C_2 = C_1 C_2$ since $\pi$ is generically 1-to-1.
By the projection formula, $\pi^* C E = C\pi_* E = 0$.

Recall 
\[
p_a(C) = {1\over 2}(K_S C + C^2) + 1
,\]
so write 
\[
p_a(C') = { (\pi^* K_S + E)(\pi^* C - mE) + (\pi^* C - mE)^2 \over 2} +1
.\]

:::

:::{.remark}
\includesvg{inkscape/2024-08-29_13-06.svg}
:::

:::{.remark}
Blow up the nodal curve to get $p_a' = p_a - 1$.
Blow up the cuspidal curve to get....something.

Note that all cubics have the same $p_a$, since it only depends on the linear equivalence class of $C$.

Consider resolving $C$ to $C'$ by $\nu$ and use the fundamental SES
\[
0\to \OO_C \to \nu^* \OO_{\tilde C} \to F\to 0
.\]
Then $\chi(\OO_{\tilde C}) = \chi(\OO_C) + \chi(F)$ where $\chi(F) = h^0(F) = d > 0$.
Recall $p_a = 1 - \chi(\OO_C)$.
If $C$ is smooth then $p_a(C) = g$.
:::

## Castelnuovo

:::{.theorem}
Suppose $\tilde S$ is a smooth surface containing $E\cong \PP^1$ where $E^2=-1$.
Then $\exists \pi: \tilde S\to S$ where $E\mapsto \pt$ such that $\tilde S\sm E \cong S\sm\pt$ and $\tilde S \cong \Bl_p S$.
:::

:::{.remark}
On linear systems: let $V \subseteq H^0(L)$ where $L = \OO(C)$.
Let $s\in V$, then $(s) = D$ an effective divisor.
Write $\PP V = \ts{ (s) \st s\neq 0}$, this is equivalent to the set of effective divisors $D\sim C$.
Recall $|C| = H^0(\OO(C)) = \ts{ \phi \text{ rational } \st ( \varphi) + C = D \geq 0}$, i.e. those $(\phi)$ such that $(\phi) + C$ has no poles.
This is a complete linear system.
Note that $( \varphi_1) = (\varphi_2) \iff \qty{ \varphi_1\over \varphi_2} = 0$.
Here $s$ and \( \varphi \) are in bijection.
:::

:::{.remark}
Let $f: S \to \PP^n$ such that $f(S)$ is not contained in a hyperplane.
This yields a linear system.
In projective coordinates $x_i$ of $\PP^n$, write a hyperplane $H$ as $\sum c_i x_i \in \PP V$ where $\PP^n = \PP V\dual$.
Then $f^* H$ is a divisor on $S$.
Define a line bundle $L = f^* \OO_{\PP^n}(1)$ on $S$.
Each $H$ defines a section $s$.
This yields an $(n+1)$-dimensional vector space $V \subseteq L$.

\includesvg{inkscape/2024-08-29_13-33.svg}
:::

:::{.remark}
Write the base locus of $\PP V$ as \( \Intersect _{D\in \PP V} D \).
A base component is a divisor in the base locus.
Say $\PP V$ is basepoint free if $\mathrm{Base}(\PP V) = \emptyset$.
We then get a bijection between linear systems without base components for smooth varieties $S$ and rational maps $S\rational \PP^n$.
:::

:::{.lemma}
If $C$ is a smooth curve and $C\rational \PP^n$, then this map can be extended to a regular map.
:::

:::{.remark}
Write the uniformizer of $\OO_{C, p}$ as $t$, let $\phi_0, \cdots, \phi_n$ be rational functions.
Write $\phi_i = u t^{m_i}$, take $d=\min m_i$, and divide by $t^d$.
:::

:::{.example}
Consider the linear system $L$ of lines through a point in $\PP^2$.
Then $L \cong \PP^1$, since each such line is specified by a point in $\PP^1$.
Write $H^0(\OO_{\PP^2}(1)) = \gens{x_0, x_1, x_2}$, so $L = \gens{x_1, x_2}$.
Then take the rational map 
\begin{align*}
\PP^2 &\to \PP^1 \\
[x_0: x_1: x_2] &\mapsto [x_1: x_2]
\end{align*}
which is undefined at $[1:0:0]$.
This single point is the base locus of $L$.
:::

:::{.example}
Take $H^0(\OO_{\PP^2}(2)) \contains L = \gens{x_1^2, x_1x_2}$.
Write $L' = L + E$ where $E = \ts{x_1=0}$.
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:::{.example}
Let $S = \PP^2$ and $L = |2H-p|$, quadrics passing through the origin.
A basis for all quadrics is given by $\gens{x_0^2, x_1^2,x_2^2, x_0x_1, x_0 x_2, x_1 x_2}$.
Imposing $x_0^2=0$ yields $L\cong \PP^4$.
What is the image?
It is the blowup of $\PP^2$ at the origin.
Note that there are no basepoints on the blowup.
The pullbacks of all such quadrics contain $E$, generically with multiplicity one.
The pulled back linear system on $\tilde S$ is $|2H-E|$; this gives an embedding $\tilde S\to \PP^4$.
:::

:::{.remark}
Next time:

- $f$ is injective if $L$ separates points: $\forall p,q, \exists D$ such that $p\in D$ but $q\not\in D$.
- $f$ is a closed embedding if $L$ separates tangent vectors.

This prevents the following situation, where a tangent vector is sent to zero:

\includesvg{inkscape/2024-08-29_14-00.svg}
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