# Monday, April 19

:::{.remark}
Recall that we have the following:
\[
H^2(\hat{S}; \ZZ) = \pi^* H^2(S; \ZZ) \oplus \ZZ[E]
\]
where $E$ is the exceptional curve, which follows from Mayer-Vietoris.
We can write $\hat{S} = S \# \bar{\CP^2}$, and by excision $H^2(S\sm \BB^4) = H^2(S)$.
So we get a LES

\begin{tikzcd}
	{H^3(S, S\sm B)} \\
	\\
	{H^2(S\sm \BB^4)} && {H^2(S)} && \textcolor{rgb,255:red,214;green,92;blue,92}{H^2(S, S\sm B)=0} \\
	\\
	&& {H^1(S)} && \textcolor{rgb,255:red,214;green,92;blue,92}{H^1(S, S\sm B)=0}
	\arrow[from=5-3, to=5-5]
	\arrow[from=5-5, to=3-1]
	\arrow["\sim", from=3-1, to=3-3]
	\arrow[from=3-3, to=3-5]
	\arrow[from=3-5, to=1-1]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMiw0LCJIXjEoUykiXSxbNCw0LCJIXjEoUywgU1xcc20gQik9MCIsWzAsNjAsNjAsMV1dLFswLDIsIkheMihTXFxzbSBcXEJCXjQpIl0sWzIsMiwiSF4yKFMpIl0sWzQsMiwiSF4yKFMsIFNcXHNtIEIpPTAiLFswLDYwLDYwLDFdXSxbMCwwLCJIXjMoUywgU1xcc20gQikiXSxbMCwxXSxbMSwyXSxbMiwzLCJcXHNpbSJdLFszLDRdLFs0LDVdXQ==)

We have $H^i(S, S\sm \BB^4) = H^i(T, T\sm \BB^4) = H^i( \BB^4, \bd)$, and by Poincaré-Lefschetz duality, this is isomorphic to $H_{4-i}(\BB^4)$.
This is equal to 0 if $i\neq 0$ or $4$.
Writing $\hat{S} = (S\sm \BB^4) \disjoint_{S^3} (\bar{\CP^2}\sm \BB^4)$ and applying Mayer-Vietoris yields

\begin{tikzcd}
	{H^2(\hat{S})} && {H^2(S\sm \BB^4) \oplus H^2(\bar{\CP^2} \sm \BB^4)} && \textcolor{rgb,255:red,214;green,92;blue,92}{H^2(S^3) =0} \\
	\\
	&& \cdots && \textcolor{rgb,255:red,214;green,92;blue,92}{H^1(S^3) =0 }
	\arrow[from=3-5, to=1-1]
	\arrow["\sim", from=1-1, to=1-3]
	\arrow[from=1-3, to=1-5]
	\arrow[from=3-3, to=3-5]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbNCwyLCJIXjEoU14zKSA9MCAiLFswLDYwLDYwLDFdXSxbMCwwLCJIXjIoXFxoYXR7U30pIl0sWzIsMCwiSF4yKFNcXHNtIFxcQkJeNCkgXFxvcGx1cyBIXjIoXFxiYXJ7XFxDUF4yfSBcXHNtIFxcQkJeNCkiXSxbNCwwLCJIXjIoU14zKSA9MCIsWzAsNjAsNjAsMV1dLFsyLDIsIlxcY2RvdHMiXSxbMCwxXSxbMSwyLCJcXHNpbSJdLFsyLDNdLFs0LDBdXQ==)

Combining this with the isomorphisms from earlier, we can write the direct sum as $H^2(S) \oplus H^2( \bar{\CP^2})$ where the latter is equal to $\ZZ \ell = [E]$ for $\ell$ a line class.
:::

:::{.question}
What is the intersection form on $H^2(\hat{S}; \ZZ)$?
:::

:::{.remark}
Using the proposition, along with the fact that 

1. its an orthogonal decomposition, 
2. $\pi^*$ is an isometry, and 
3. $[E]^2 = -1$, 

we know that the Gram matrix for $H^2(\hat{S})$ is the same as that for $H^1(S) \oplus [-1]$, i.e. it is of the form
\[
\begin{bmatrix}
A & 0 
\\
0 & -1
\end{bmatrix}
.\]

:::

:::{.proof title="of 2"}
Consider $[\Sigma_1], [\Sigma_2]\in H^2(S; \ZZ)$ where the \( \Sigma_i \) are real surfaces, and suppose \( \Sigma_1 \transverse \Sigma_2 \) and $p\not\in \Sigma_1, \Sigma_2$.
We then have 
\[
[ \pi\inv( \Sigma_i ) ] = \pi^* [ \Sigma_i ] 
.\]

\begin{tikzpicture}
\fontsize{35pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-19_14-07.pdf_tex} };
\end{tikzpicture}

The intersection number is preserved because $\pi$ is generically injective.
:::

:::{.proof title="of 1"}
It also follows that if $p\not\in \Sigma$, $\pi^*[\Sigma] = [\pi\inv \Sigma]$ where the latter is disjoint from $E$.
So $\pi^*[\Sigma] \cdot E = 0$.
:::

:::{.proof title="of 3"}
Since $[E] \sim [\text{line}] \in \bar{\CP^2}\sm \BB^4$, and $E^2 = [E] \cdot [E] = -1$ since the orientations disagree in $\bar{\CP^2}$.
:::

:::{.proposition title="Computing the pullback of a curve"}
Let $C \subset S$ be a curve on a surface and suppose $C$ is locally cut out by\[
f(x, y) = a_{m, 0} x^m + a_{n-1, 1} x^{m-1} y + \cdots + a_{0, m} y^m + O(x^{m+1}, y^{m+1})
,\]
near $p\in S$, so the lowest order terms in the Taylor expansion are degree $m$.
Then \[
\pi^* C = \hat{C} + mE
.\]
:::

:::{.proof title="?"}
On the blowup, take local coordinates $(x, \mu)$ where $y = x\mu$ and write
\[
V(\pi^* f) 
&= V( x^m \qty{a_{m, 0} + a_{m-1, 1} \mu + \cdots + a_{0, m} \mu^m + O(x^{m+1}, \mu^{m+1} )  } ) \\
&= m V(x) + V( a_{m, 0} + \cdots ) \\
&= E + \hat{C}
.\]
:::

:::{.example title="?"}
Take 
\[
C = \ts{ y^2 = x^3-x^2 } \subseteq \CC^2
,\] 
where $\Bl_0 \CC^2 \to C$.
Then $\pi^* C = \hat{C} + 2E$, so 
\[
C = V(x^2 + y^2 + O(\deg(3))
.\].
:::

:::{.corollary title="Computing the square of the strict transform"}
$\hat{C}^2 = C^2 - m^2$.
:::

:::{.proof title="?"}
Write $\pi^* C = \hat{C} + mE$, then $\hat{C} = \pi^* C - mE$ implies that $\hat{C}^2 = (\pi^* C - mE)^2$.
This equals\[
(\pi^* C)^2 - 2m \pi^* C\cdot E + m^2 E^2
&= C^2 - 0 - m^2 \\
&= C^2 - m^2
,\]
where we've used (2), (1), and (3) respectively to identity these terms.
:::

:::{.example title="?"}
Let 
\[
C\da \ts{ zy^2 = x^3 - x^2 z } \subset \CP^2
,\]
then $C^2 = (3\ell)^2 = 9$.
The multiplicity of $C$ at the point $[0:0:1]$ is 2.
Taking the coordinate chart \( \ts{ z=1 } \cong \CC^2 \), we recover the curve $y^2 = x^3 - x^2$ which has multiplicity 2 at $(0, 0)$.
We can conclude $\hat{C} = \Bl_{[0:0:1]} \CP^2$ has self-intersection number $\hat{C}^2 = 9-2^2 = 5$.
:::

:::{.theorem title="Castelnuovo Contractibility Criterion"}
Let $S$ be a complex surface and let $E \subset S$ be a holomorphically embedded $\CP^2$ such that $E^2 = -1$
Then there exists a smooth surface $\bar{S}$ and $p\in \bar{S}$ such that $S = \Bl_p \bar{S}$ with $E$ as the exceptional curve.
:::

:::{.definition title="Blowdown"}
This $\bar{S}$ is called the **blowdown** of $S$ along $E$.
:::

:::{.remark}
Note that this is the exact situation when we blow things up.
This is a converse: if we have something that looks like a blowup, we can find something that blows up to it.
:::

:::{.exercise title="?"}
Show that the category $\Mfd_\CC$ is not closed under blowdowns, i.e. there is no blowdown of a holomorphically embedded $\CP^1$, say $E$, with $E^2 = 1$.

> Hint: think about $\CP^2$.

:::

:::{.remark}
This is interesting because there does exist a blowdown in the smooth category $\Mfd(C^\infty(\RR))$.
This is because $S \to S\# \bar{\CP^2}$ and $S\to S\# \CP^2$ are indistinguishable here.
One can just reverse orientations.
:::

:::{.example title="?"}
A complex surface with a holomorphically embedded $\CP^1$ of self intersection $-1$.
Let $p, q\in \CP^2$ be distinct points, and let $\Bl_{p, q} \CP^2 \da \Bl_p \Bl_q \CP^2$.
Note that these two operations commute since these are distinct points and blowing up is a purely local operation.
Let $\ell \subset \CP^2$ be the unique line through $p$ and $q$.
Viewing $p, q$ as lines in $\CC^3$, they span a unique plane, which is a line in projective space, so this makes sense and we can write $\ell \approx \spanof \ts{ p, q }$.
Since $\ell$ is defined by a linear equation in local coordinates near $p, q$, we have $\mult_p \ell = \mult_q \ell = 1$.
We hve
\[
\hat{\ell} = \pi^* \ell - E_p - E_q \\
\hat{\ell}^2 = \ell^2 - 1^2 - 1^2 = 1-1-1 = - 1
.\]
Under $\pi: \Bl_{p, q} \CP^2 \to \CP^2$, we have $\hat{\ell} \mapsvia{\sim} \ell$.

\begin{tikzpicture}
\fontsize{35pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-19_14-40.pdf_tex} };
\end{tikzpicture}

Here since all of the lower order terms have degree 1, there is a well-defined tangent line.
Since $\ell \cong \CP^2$, we have $\hat\ell \cong \CP^2$.
Letting $\sigma$ be the blowdown of $\hat\ell$, we have

\begin{tikzcd}
	& {\Bl_{p, q}\CP^2} \\
	\\
	{\CP^1 \cross \CP^2} && {\CP^2}
	\arrow["\sigma"', from=1-2, to=3-1]
	\arrow["\pi", from=1-2, to=3-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMSwwLCJcXEJsX3twLCBxfVxcQ1BeMiJdLFsyLDIsIlxcQ1BeMiJdLFswLDIsIlxcQ1BeMSBcXGNyb3NzIFxcQ1BeMSJdLFswLDIsIlxcc2lnbWEiLDJdLFswLDEsIlxccGkiXV0=)

:::

:::{.remark}
There's a way to do this with Kirby Calculus.
:::