# Thursday, February 09

:::{.remark}
Last time: $\Pic(X) \cong H^1(X; \OO_X\units)$ and 
\[
c_1: \Pic(X) &\to H^2(X; \ZZ) \\
\OO(D) &\mapsto [D]
\]
with $\im c_1 = \NS(X)$ and $\ker c_1 = \Pic^0(X) \cong \CC^g/ \Lambda$ where $H^1(X; \OO) \cong \CC^g$.
Today: consider the cohomology of vector bundles on a complex manifold $X$.
:::

:::{.definition title="$(p,q)\dash$forms"}
A **smooth $(p, q)\dash$form** is locally of the form 
\[
\sum_{ \abs{I} = p, \abs{J} = q} a_{I, \bar J} \dz_{i_1}\wedge\cdots \dz_{i_p} \wedge \dzbar_{j_1}\wedge \cdots \wedge \dzbar_{j_q}
.\]
Let $A^{p, q}$ be the sheaf of smooth $(p, q)\dash$forms
:::

:::{.example title="?"}
Some examples: 

- $A^{0, 1}(\CC) \ni \omega \da z\bar{z} \dzbar$.
- $A^{1, 0}(\CC\units) \ni \alpha \da \log\abs{z} \dz$
- $A^{1,1}(\CC^2)\ni \alpha \da e^{z_1}\dz_1 \wedge \dzbar_2 + \bar{z}_2 \dzbar_1 \wedge \dz_2$.
- An example of differentiation: $d(e^{z_2} \dz_1 + \dzbar_1 \dz_2) = e^{z_2} \dz_2\wedge \dz_1 + \dzbar_1 \wedge \dz_2$.
:::

:::{.remark}
Let $\Omega^{p}$ be the holomorphic $(p, 0)$ forms, noting that differentiation $d$ on smooth forms is not a map of $C^\infty(X, \CC)\dash$modules since $d(f\alpha) = fd(\alpha) + df\wedge \alpha$ for $f \in C^\infty(U, \CC)$ and $\alpha A^k(U)$.
There is a decomposition 
\[
A^k(U) = \bigoplus _{p+q=k} A^{p, q}(U)
,\]
which leads to a decomposition of sheaves.
Define $\del = \pi_{p+1, q}(d)$ and $\delbar: \pi_{p, q+1}(d)$, this yields a complex
\[
0 \to \OO_X \to A^{0,0} \mapsvia{\delbar} A^{0, 1} \mapsvia{\delbar} A^{0, 2} \to \cdots \to A^{0, \dim X} \to 0
,\]
which is an exact sequence of sheaves.
Noting that $d^2 = 0$, one has 

- $\del^2 = 0$,
- $\delbar^2 = 0$,
- $\del\delbar + \delbar\del = 0$.

See the Poincaré $\delbar$ lemma.

:::

:::{.remark}
More generally, for a vector bundle $E \in \oxmods$, note that $\OO\injects A^{0, 0}$ yields $\OO\injects \CC^\infty$, so can form $E\tensor_{\OO} C^\infty$.
Locally, $E\cong \OO\sumpower{r}$ on $U$, so one has $E\tensor A^{0,0} \cong (C^\infty)\sumpower{r}$ on $U$.
This yields $0\to E\injects E \tensor A^{0,0}$, and the claim is that there is a well-defined map $E\tensor A^{0,0}\to E\tensor A^{0, 1}$.
Locally this is given by $\tv{f_1,\cdots, f_r} \mapsto \tv{\delbar f_1,\cdots, \delbar f_r}$.
In a different trivialization, $s_V = t_{UV}(f_1,\cdots, f_r)$ where $t_{UV}$ is a holomorphic function valued in $\GL_r(\CC)$.
One has $\delbar(t_{UV} \circ (f_1,\cdots, f_r)) = t_{UV} (\delbar f_1, \cdots, \delbar f_r)$ since $\delbar (t_{UV}) = 0$, noting that in the first expression one is carrying out matrix multiplication.
:::

:::{.definition title="Dolbeault complex"}
\[
0 \to E \mapsvia{i} E\tensor A^{0,0} \mapsvia{\delbar} E\tensor A^{0, 1} \mapsvia{\delbar} \cdots \mapsvia{\delbar} E\tensor A^{0, \dim X} \to 0
,\]
and $H^*(X; E)$ can be computed from the homology of this complex.
:::

:::{.fact}
For any smooth vector bundle $V\to M$ over a manifold $M$, $\Hc^{i \geq 1}(M; V) = 0$ since $M$ admits partitions of unity.
Moreover if $\mcf \injects \complex{I}$ with $\complex{I}$ acyclic, so $H^{i\geq 1}(\complex I) = 0$, then $H^k(X; F)$ is computed as the homology of $\complex I$.
:::

:::{.remark}
Since $\Omega^p$ is a holomorphic vector bundle on $X$, this yields the Dolbeault resolution 
\[
0\to \Omega^p \to A^{p, 0} \to A^{p, 1} \to \cdots
,\]
and $H^{p, q} \da H^q(X; \Omega^p)$ is the homology of this complex.
Define $h^{p, q} \da \dim_\CC H^{p, q}$ -- note that this forms a diamond since for $p,q\geq \dim X$ there are no $p\dash$forms or $q\dash$forms whatsoever.
:::

:::{.theorem title="Hodge decomposition and symmetry theorems"}
There is a decomposition
\[
H^k(X; \ul{\CC}) \cong \bigoplus _{p+q=k} H^{p, q}
,\]
and a symmetry
\[
H^{p, q}(X) \cong \overline{H^{q, p}(X)}
.\]
:::

:::{.remark}
Note that $\bar{V}$ doesn't necessarily make sense yet for $V = H^{p, q}(X)$, since we don't know that it is a subspace of some real vector space $W$ -- here we'll take $H^k(X; \ul{\CC}) = H^k(X; \ul{\RR}) \tensor \CC$.
E.g. writing $\CC^2 = \RR^2 \tensor_\RR \CC$, if $V = \gens{u, v}_\CC$ then $\bar{V} = \gens{\bar u, \bar v}_\CC$.
:::