# Friday, February 26


:::{.remark}
Last time: Riemann-Roch for surfaces, today we'll discuss some examples.
Recall that if $S \in \Mfd_\CC^2$ is closed and compact (noting that $S\in \Mfd_\RR^4$) and $L\to S$ is a holomorphic line bundle then
\[
\chi(S, L) = \chi(\OO_S) + {1\over 2}(L^2 - L \cdot K)
\]
where $K = c_1(K_S)$ for $K_S \da \Omega_S^2$ the canonical bundle and $L = c_1(L)$.
We also saw
\[
\chi(\OO_S) = {1\over 12}(K^2 + \chi_{\Top}(S))
,\]
where $\chi_\Top$ is the Euler characteristic and is given by 
\[
\chi_\Top(S) = 2 h^0(S; \CC) - 2 h^1(S, \CC) + h^2(S; \CC)
.\]
:::


:::{.example title="?"}
Let $S = \CP^2$, which can be given in local coordinates by 
\[ 
\ts{ [x_0: x_1: x_2 ] \st (x_0, x_1, x_2) \in \CC^3\smz } 
\] 
where we only take equivalence classes of ratios \( [x,y,z] = [\lambda x, \lambda y, \lambda z] \) for any \( \lambda\in \CC\units \).
This decomposes as 
\[
\CP^2 \union \CC \union \ts{ \pt } = \ts{ [1: x_1: x_2] } \union \ts{ [0 : x_1: x_2] } \union \ts{ [0:0:1] }
,\]
i.e. we take $x_0 \neq 0$, then $x_0 = 0, x_1\neq 0$, then $x_0 = x_1 = 0$.
Note that 
\[
h^i(\CP^n; \ZZ) = 
\begin{cases}
\ZZ &  0 \leq i \leq 2n \text{ even} 
\\
0 & \text{else}.
\end{cases}
\]

We can use this to conclude that $\chi_\Top(\CP^n) = n+1$ and $\chi_\Top(\CP^2) = 3$.
Over $\CP^n$ we have a **tautological line bundle** $\OO(-1)$ given by sending each point to the corresponding line in $\CC^{n+1}$, i.e. $\OO(-1) \to \CP^n$ given by 
\[
\lambda (x_0, \cdots, x_n) \mapsto [x_0: \cdots: x_n]
.\]
Note that the total space is $\Bl_0(\CC^{n+1})$ is the **blowup** at zero, which separates the tangents at 0.
:::

:::{.remark}
Let $X$ be an algebraic variety, i.e. spaces cut out by polynomial equations, for example \( \ts{ xy = 0 } \subseteq \CC^2 \) which has a singularity at the origin.
A **divisor** is a $\ZZ\dash$linear combination of subvarieties of codimension 1.
Note that for a curve $X$, this recovers the definition involving points.
For $D$ a divisor on $X$, we associated a bundle $\OO_X(D)$ which had a meromorphic section with a zero/pole locus whose divisor was precisely $D$.

Recall the construction: we chose a point, then a trivializing neighborhood where the transition functions where $V$.

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For a higher dimensional algebraic variety or complex manifold, for $D$ a complex submanifold, pick a chart around a point that the nearby portion of $D$ to a coordinate axis in $\CC^n$, which e.g. can be given by \( \ts{ z_1 = 0 } \).

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As before there's a distinguished section $s_D \in H^0(X; \OO_X(D) )$ vanishing along $D$.
Note that a line bundle is a free rank 1 $\OO\dash$module, and analogously here the functions vanishing along $D$ are $\OO\dash$modules generated by (here) $z_1$.
:::

:::{.definition title="Hyperplane"}
A **hyperplane** in $\CP^n$ is any set of the form
\[
H = \ts{ [x_0: \cdots : x_1 ] \st \sum a_i x_i = 0 } \cong \CP^{n-1}
.\]
:::

:::{.example title="?"}
Take $\CP^{n-1} \subseteq \CP^n$, e.g. \( \ts{ x_0 = 0 } \).
This is an example of a **divisor** on $\CP^n$, i.e. a complex codimension 1 "submanifold".
We can take the line bundle constructed above to get $\OO_{\CP^n}(\CP^{n-1})$ which vanishes along $\CP^{n-1}$.
More generally, for any hyperplane $H$ we can take $\OO_{\CP^n}(H)$, and these are all isomorphic, so we'll denote them all by $\OO_{\CP^n}(1)$.
The implicit claim is that is the inverse line bundle of the tautological bundle, so $\OO(1) \tensor \OO(-1)$ is the trivial bundle since the transition functions are given by reciprocals and multiplying them yields 1.
We can classify complex line bundles on $\CP^n$ using the SES
\[
0 \to \constantsheaf{\ZZ} \to \OO \mapsvia{\exp} \OO\units \to 1
.\]

We know that $H^1(X; \OO\units)$ were precisely holomorphic line bundles, since they were functions agreeing on double overlaps with a cocycle condition.
We have a LES coming from sheaf cohomology:

\begin{tikzcd}
	&&&& \cdots \\
	\\
	{H^1(X; \OO)} && {H^1(X; \OO)} && {H^1(X; \OO\units)} \\
	\\
	{H^2(X; \OO)} && \cdots
	\arrow["{c_1}", from=3-5, to=5-1, out=0, in=180]
	\arrow[from=3-1, to=3-3]
	\arrow[from=3-3, to=3-5]
	\arrow[from=5-1, to=5-3]
	\arrow[from=1-5, to=3-1, out=0, in=180]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMiwyLCJIXjEoWDsgXFxPTykiXSxbNCwyLCJIXjEoWDsgXFxPT1xcdW5pdHMpIl0sWzAsNCwiSF4yKFg7IFxcT08pIl0sWzAsMiwiSF4xKFg7IFxcT08pIl0sWzQsMCwiXFxjZG90cyJdLFsyLDQsIlxcY2RvdHMiXSxbMSwyLCJjXzEiXSxbMywwXSxbMCwxXSxbMiw1XSxbNCwzXV0=)

Applying this to $X\da \CP^n$, we have $H^1(\OO) = H^2(\OO) = 0$.
This can be computed directly using that $\CP^n = \union_{n\geq 1} \CC^n$ by taking charts $x_i\neq 0$, and this yields an acyclic cover.
Thus $c_1$ is an isomorphism above, and $\Pic(\CP^n) \cong \ZZ$, where $\Pic$ denotes isomorphism classes of line bundles.
We can identify $\Pic(\CP^n) = \ts{ \OO_{\CP^n}(k) \st k\in \ZZ }$.
:::