# Tuesday, February 14

:::{.proposition title="Riemann-Roch for curves"}
\[
h^0(C, L) - h^0(C, K_C \tensor L\inv) = \deg L + 1-g
.\]
:::

:::{.remark}
Recall: 

- $h^i(X, F) \da \dim_k H^i(X; F)$.
- $\chi(X, F) \da \sum_{i\geq 0}h^i(X, F)$ for $F\in \Sh(X, \kmod)$, provided these numbers are finite.
- $H^0(\AA^1\slice k; \OO) = k[x] t^0$, and note $\dim_k k[x] = \infty$.
- For $X$ an irreducible Noetherian topological space with $\dim X = d$, $H^i(X, F) = 0$ for $i > d$.
- For $X\in \Proj\Var\slice k$ and $F$ a finitely presented $\OO_X\dash$module, i.e. there is an exact sequence $\OO_X\sumpower m\to \OO_X\sumpower n\surjects F$, we have $h^i(X, F) < \infty$.
- Finitely presented sheaves are coherent.
  An analytic coherent sheaf is defined in the same way with respect to $\OO_X^\an$ (the sheaf of holomorphic functions).
- $h^0(X, K_C\tensor L\inv) = h^1(C, L)$.

:::

:::{.theorem title="Serre duality"}
Let $X$ be a compact complex manifold and let $E\to X$ be a holomorphic vector bundle.
Then $H^i(X, E) \iso H^{\dim_\CC X - i}(X, E\dual \tensor K_X)\dual$ where $K_X = \det\Omega_X \da \Omega_X^{\dim X}$.
:::

:::{.proof title="?"}
Regard $s\in H^i(X, E)$ as an element in Dolbeault cohomology,
\[
H^i(X, E) \cong { \ker \qty{ E\tensor A^{0, i}(X) \mapsvia{\delbar} A^{0, i+1}(X) } \over \im\qty{E\tensor A^{0, i-1}(X) \mapsvia{\delbar} E\tensor A^{0, i}(X) } }
.\]
Note that $K \tensor_\OO C^\infty = A^{n, 0}$.
Let 

- $t\in H^{n-i}(X, E\dual \tensor K)$
- $\tilde s\in E\tensor A^{0, i}(X)$
- $\tilde t\in E\dual \tensor K \tensor A^{0, n-i}(X) \cong E\dual \tensor A^{n, n-i}(X)$.

One can then pair $\inp{\tilde s}{\tilde t}\in A^{n, n}(X)$, and $\int_X \inp{\tilde s}{\tilde t} \in \CC$ is a perfect pairing.
:::

:::{.remark}
Upshot: the LHS in RR is $\chi(C, L)$.
:::

:::{.proposition title="?"}
The following is an important exact sequence of sheaves: for any $D\in \CDiv(X)^\eff$, one has 
\[
\OO_X(-D) \injects \OO_X \surjects \OO_D \qquad \in \Sh(X)
\]
where $\OO_D \da \iota_* \OO_X$ for $\iota: D\injects X$ the inclusion.
:::

:::{.proof title="?"}
Note $\OO_X(D)(U) = \ts{f\in \OO_(U) \st \div f \geq D}$, so $\OO_X(D) = I_D$ is the ideal sheaf of $D$.
If $D$ is cut out by a single function on $U$, we have $I_D(U) = (f) \subset \OO_X(U)$.
This yields an inclusion $\OO_X(-D) = I_D \injects \OO_X$.
By definition, the quotient $\OO_X/I_D$ are functions defined on $D$, at least on affine opens $U$.
Since exactness of sheaves is local, this check suffices.
:::

:::{.remark}
In particular, on a curve one has 
\[
\OO_C(-p) \injects \OO_C \surjects \OO_p
\]
where $p\in C$ is a point.
This can be tensored with any vector bundle $L$ to get 
\[
L(-p) \da L\tensor \OO_C(-p) \injects L \surjects \ro{L}{p}
,\]
which is exact since $L$ is locally free.
For $s_p\in H^0(\OO_C(p))$, we have $V(s_p) = [p]$ as a divisor: 

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:::

:::{.proposition title="?"}
For $F_1\injects F_2 \surjects F_3$, 
\[
\chi(F_2) = \chi(F_1) + \chi(F_3)
.\]
:::

:::{.proof title="?"}
Take the LES in cohomology, where $H^n(F_i) = 0$ for large enough $n$.
Now for a LES of vector spaces $V_1\injects V_2\to \cdots \surjects V_n$, one has $\sum (-1)^i \dim_k V_i = 0$.
:::

:::{.claim}
$\ro L p \cong \OO_p$ satisfies $H^i(C, \OO_p) = \CC t^0$, which is more generally true for a skyscraper sheaf at a point.
:::

:::{.remark}
Take a fine enough open cover (e.g. an affine cover) so that $p$ appears in only one set, and use Čech cohomology.
That $\ro L p \cong \OO_p$ follows from the fact that this holds on a small enough open $U$ and both are identically zero away from $p$.
:::

:::{.remark}
Now use that $\chi(C, \OO_p) = 1$, we then claim that $\chi(L) = \chi(L(-p)) + 1$ and thus $L\cong \OO_C(\sum n_p [p])$ and $\chi(L) = \sum n_p + \chi(\OO_C)$ by repeatedly applying this fact.
Note $\sum n_p = \deg L$, so 
\[
\chi(L) = \deg L + \chi(\OO_C)
.\]
We have $\chi(\OO_C) = h^0(\OO_C) - h^1(\OO_C) = 1- h^0(K)$ by Serre duality.
Applying the above version of RR to $K$ yields $\chi(K) = \deg K + \chi(\OO) = 2g-2 + \chi(\OO)$.
On the other hand, this equals $h^0(K) - h^0(K\dual \tensor K) = h^0(K) - 1$.
Combining these yields 
\[
\chi(\OO) = -(2g-2 + \chi(\OO)) \implies 2\chi(\OO) = 2-2g \implies \chi(\OO) = 1-g
.\]
Plugging this back into the first equation yields 
\[
\chi(L) = \deg L + 1-g
.\]
:::

:::{.remark}
Note that $\chi^\Top(C) = g$ was defined as the index of a vector field, and this shows that also $g = h^1(\OO_X)$.
:::

:::{.remark}
An application of Serre duality: the Hodge diamond.
Let $n\da \dim_\CC X$ and recall $h^{p, q} \da \dim_\CC H^q(X, \Omega^p_X)$.
By duality, $h^{p, q} = \dim_\CC H^{n-q}(X, (\Omega_X^p)\dual \tensor \Omega^n)$.
We first claim 
\[
(\Omega^p_X)\dual \tensor \Omega^n \cong \Omega^{n-p}
.\]
Note that sections of $\Omega^p$ are of the form $\sum a_I \dz_{i_1}\wedge \cdots \wedge \dz_{i_p}$ with $a_I$ holomorphic functions on $U$, and sections of \( (\Omega^p)\dual \) look like $\sum a_I \dd{}{z_{i_1}} \wedge \cdots \wedge \dd{}{z_{i_p}}$ whose transition functions are the inverse of those for \( \Omega^p \).
Noting that $\Omega^n$ is a line bundle with local sections of the form $f \dz_1\wedge\cdots \dz_n$, one can contract forms (interior multiplication) to obtain 
\[
\qty{\sum a_I \dd{}{z_I}} \tensor \qty{f \dz_I} = f \sum_{j\in I^c} \dz_{j_1}\wedge\cdots \wedge \dz_{j_{n-p}} 
.\]
Thus $h^{p, q} = \dim H^{n-q}(X, \Omega^{n-p}) = h^{n-p, n-q}$.

\begin{tikzcd}
	&& {h^{n,n}} \\
	& {h^{n-q, n-p}} && {h^{n-p, n-q}} \\
	{h^{n, 0}} &&&& {h^{0, n}} \\
	& {h^{p, q}} \\
	&& {h^{0,0}} \\
	&& {}
	\arrow[dashed, no head, from=1-3, to=5-3]
	\arrow[curve={height=-12pt}, Rightarrow, no head, from=4-2, to=2-4]
	\arrow[dashed, no head, from=3-1, to=3-5]
	\arrow[Rightarrow, no head, from=2-4, to=2-2]
\end{tikzcd}
> [Link to Diagram](https://q.uiver.app/?q=WzAsOCxbMiwwLCJoXntuLG59Il0sWzIsNV0sWzIsNCwiaF57MCwwfSJdLFswLDIsImhee24sIDB9Il0sWzQsMiwiaF57MCwgbn0iXSxbMSwzLCJoXntwLCBxfSJdLFszLDEsImhee24tcCwgbi1xfSJdLFsxLDEsImhee24tcSwgbi1wfSJdLFswLDIsIiIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9LCJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzUsNiwiIiwwLHsiY3VydmUiOi0yLCJsZXZlbCI6Miwic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFszLDQsIiIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9LCJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzYsNywiIiwwLHsibGV2ZWwiOjIsInN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XV0=)

This yields $K_4$ symmetry, and we'll see that for a Calabi Yau there is a $D_4$ symmetry.
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