# Monday, February 01


:::{.remark}
Last time \( \ul{\RR} \) on a manifold $M$ has a resolution by vector bundles:
\[
0 \to \ul{\RR} \injects \Omega^1 \mapsvia{d} \Omega^2 \mapsvia{d} \cdots
.\]
This is an exact sequence of sheaves of any smooth manifold, since locally $d \omega = 0 \implies \omega = d \alpha$ (by the *Poincaré $d \dash$lemma*).
We also want to know that \( \Omega^k \) is an acyclic sheaf on a smooth manifold.
:::

:::{.exercise title="?"}
Let $X\in Top$ and \( \mathcal{F}\in \Sh(\Ab)_{/X} \).
We say \( \mathcal{F}  \) is **flasque** if and only if for all $U \supseteq V$ the map \( \mathcal{F}(U) \mapsvia{\rho_{UV}} \mathcal{F}(V)   \) is surjective.
Show that \( \mathcal{F}  \) is acyclic, i.e. $H^i(X; \mathcal{F}) = 0$.
This can also be generalized with a POU.
:::

:::{.example title="?"}
The function $1/x\in \OO(\RR\smz)$, but doesn't extend to a continuous map on $\RR$.
So the restriction map is not surjective.
:::

:::{.remark}
Any vector bundle on a smooth manifold is acyclic.
Using the fact that \( \Omega^k \) is acyclic and the above resolution of $\ul{\RR}$, we can write $H^k(X; \RR) = \ker(d_k) / \im d_{k-1} \da H^k_{dR}(X; \RR)$.
:::

:::{.remark}
Now letting $X \in \Mfd_\CC$, recalling that \( \Omega^p \) was the sheaf of holomorphic \( p \dash \)forms.
Locally these are of the form $\sum_{\abs{I} = p} f_I(\vector{z}) dz^I$ where $f_I(\vector{z})$ is holomorphic.
There is a resolution
\[
0 \mapsvia{} \Omega^p \mapsvia{} A^{p, 0}
,\]
where in $A^{p, 0}$ we allowed also $f_I$ are *smooth*.
These are the same as bundles, but we view sections differently. 
The first allows only holomorphic sections, whereas the latter allows smooth sections.
What can you apply to a smooth $(p, 0)$ form to check if it's holomorphic?
:::

:::{.example title="?"}
For $p=0$, we have
\[
0 \to \OO \to A^{0, 0}
.\]
where we have the sheaf of holomorphic functions mapping to the sheaf of smooth functions.
We essentially want a version of checking the Cauchy-Riemann equations.
:::

:::{.definition title="$\del$ and $\delbar$ operators"}
Let \( \omega\in A^{p, q}(X) \) where 
\[
d \omega = \sum \dd{f_I}{z_j} dz^j \wedge dz^I \wedge d\bar{z}^J + \sum_j \dd{f_I}{\bar{z}_j} d\bar{z}^j \wedge dz^I d\bar{z}^J\da \del + \bar{\del} 
\] 
with $\abs I = p, \abs J = q$.
:::

:::{.example title="?"}
The function $f(z) = z\bar{z} \in A^{0, 0}(\CC)$ is smooth, and $df = \bar{z} dz + z d\bar{z}$.
This can be checked by writing $z^j = x^j + iy^j$ and $\bar z^j = x^j - iy_j$, and $\dd{}{\bar z} g = 0$ if and only if $g$ is holomorphic.
Here we get $\del \omega \in A^{p+1, q}(X)$ and $\bar{\del} \in A^{p, q+1}(X)$, and we can write $d(z \bar z) = \del(z\bar z) + \delbar(z\bar z)$.
:::

:::{.definition title="Cauchy-Riemann Equations"}
Recall the Cauchy-Riemann equations: \( \omega \) is a holomorphic $(p, 0) \dash$form on $\CC^n$ if and only if $\delbar \omega = 0$.
:::


:::{.remark}
Thus to extend the previous resolution, we should take
\[
0 \to \Omega^p \injects A^{p, 0} \mapsvia{\delbar} A^{p, 1} \mapsvia{\delbar} A^{p, 2} \to \cdots
.\]
The fact that this is exact is called the *Poincaré \( \delbar \dash \)lemma*.
:::


:::{.remark}
There are no bump functions in the holomorphic world, and since \( \Omega^p \) is a holomorphic bundle, it may not be acyclic.
However, the $A^{p, q}$ *are* acyclic (since they are smooth vector bundles and thus admit POUs), and we obtain 
\[
H^q(X; \Omega^p) = \ker( \delbar_q) / \im(\delbar_{q-1})
.\]
Note the similarity to $H_{\dR}$, using $\delbar$ instead of $d$.
This is called **Dolbeault cohomology**, and yields invariants of complex manifolds: the **Hodge numbers** $h^{p, q}(X) \da \dim_\CC H^q(X; \Omega^p)$.
These are analogies:

Smooth     | Complex |
-----------|---------|
$\ul{\RR}$ | $\Omega^p$ |
$\Omega^k$ | $A^{p, q}$ |
Betti numbers $\beta_k$ | Hodge numbers $h^{p, q}$ |

Note the slight overloading of terminology here!
:::


:::{.theorem title="Properties of Singular Cohomology"}
Let $X\in \Top$, then $H_{\sing}^i(X; \ZZ)$ satisfies the following properties:

- Functoriality: given $f\in \Hom_\Top(X, Y)$, there is a pullback $f^*: H^i(Y; \ZZ) \to H^i(X; \ZZ)$.
- The cap product: a pairing
\[
H^i(X; \ZZ) \tensor_\ZZ H_j(X; \ZZ) &\to H_{j-i}(X; \ZZ) \\
\varphi\tensor \sigma &\mapsto \varphi\qty{\ro{\sigma}{\Delta_{0, \cdots, j}}} \ro{ \sigma}{\Delta_{i, \cdots, j}}
.\]
  This makes $H_*$ a module over $H^*$.

- There is a ring structure induced by the cup product:
\[
H^i(X; \RR) \cross H^j(X; \RR)\to H^{i+j}(X; \RR) && \alpha\cup \beta &= (-1)^{ij} \beta \cup \alpha
.\]

- Poincaré Duality: If $X$ is an oriented manifold, there exists a fundamental class $[X] \in H_{n}(X; \ZZ) \cong \ZZ$ and $(\wait)\cap X: H^i \to H_{n-i}$ is an isomorphism.

:::


:::{.remark}
Let $M \subset X$ be a submanifold where $X$ is a smooth oriented $n\dash$manifold.
Then $M \injects X$ induces a pushforward $H_n(M; \ZZ) \mapsvia{\iota_*} H_n(X; \ZZ)$ where \( \sigma \mapsto \iota \circ \sigma \).
Using Poincaré duality, we'll identify $H_{\dim M}(X; \ZZ) \to H^{\codim M}(X; \ZZ)$ and identify $[M] = PD( \iota_*( [M]))$.
In this case, if $M\transverse N$ then $[M] \cap [N] = [M \cap N]$, i.e. the cap product is given by intersecting submanifolds.
:::


:::{.warnings}
This can't always be done!
There are counterexamples where homology classes can't be represented by submanifolds.
:::