# Lecture 4 (Friday, January 22)

## The Exponential Exact Sequence

:::{.remark}
Let $X = \CC$ and consider $\OO$ the sheaf of holomorphic functions and $\OO\units$ the sheaf of *nonvanishing* holomorphic functions.
The former is a vector bundle and the latter is a sheaf of abelian groups.
There is a map $\exp: \OO \to \OO\units$, the **exponential map**, which is the data $\exp(U): \OO(U) \to \OO\units(U)$ on every open $U$ given by $f\mapsto e^f$.
There is a kernel sheaf $2\pi i \ul{\ZZ}$, and we get an exact sequence
\[
0 \to 2\pi i \ul{\ZZ} \to \OO \mapsvia{\exp} \OO\units \to \coker(\exp) \to 0
.\]
:::

:::{.question}
What is the cokernel sheaf here?
:::

:::{.remark}
Let $U$ be a contractible open set, then we can identify $\OO\units(U) / \exp(\OO\units(U)) = 1$.

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Any $f\in \OO\units(U)$ has a logarithm, say by taking a branch cut, since $\pi_1(U) =0 \implies \log f$ has an analytic continuation.
Consider the annulus $U$ and the function $z\in \OO\units(U)$, then $z\not\in \exp(\OO(U))$ -- if $z=e^f$ then $f=\log(z)$, but $\log(z)$ has monodromy on $U$:

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Thus on any sufficiently small open set, $\cok(\exp) = 1$.
This is only a presheaf: there exists an open cover of the annulus for which $\ro{z}{U_i}$, and so the naive cokernel doesn't define a sheaf.
This is because we have a locally trivial section which glues to $z$, which is nontrivial.

:::

:::{.exercise title="Fixing the sheaf cokernel"}
Redefine the cokernel so that it is a sheaf.
Hint: look at sheafification, which has the defining property 
\[
\Hom_{\presh}(\mathcal{G}, \mathcal{F}^{\presh} ) =\Hom_{\Sh}( \mathcal{G}, \mathcal{F}^{\Sh})
\]
for any sheaf \( \mathcal{G}  \).
:::

## Global Sections

:::{.definition title="Global Sections Sheaf"}
The **global sections** sheaf of \( \mathcal{F}  \) on $X$ is given by \( H^0( X; \mathcal{F}) = \mathcal{F}(X)   \).
:::

:::{.example title="?"}
\envlist

- $C^ \infty (X) = H^0(X, C^ \infty )$ are the smooth functions on $X$
- $VF(X) = H^0(X; T)$ are the smooth vector fields on $X$ for $T$ the tangent bundle
- If $X$ is a complex manifold then $\OO(X) = H^0(X; \OO)$ are the globally holomorphic functions on $X$.
- $H^0(X; \ZZ) = \ul{\ZZ}(X)$ are ??
:::

:::{.remark}
Given vector bundles $V, W$, we have constructions $V \oplus W, V \otimes W, V\dual, \Hom(V, W) = V\dual \otimes W, \Sym^n V, \Wedge^p V$, and so on.
Some of these work directly for sheaves:

- \( \mathcal{F} \oplus \mathcal{G}(U) \da \mathcal{F}(U) \oplus \mathcal{G}(U)     \) 
- For tensors, duals, and homs $\HHom(V, W)$ we only get presheaves, so we need to sheafify.

:::

:::{.warnings}
$\Hom(V, W)$ will denote the *global* homomorphisms $\HHom(V, W)(X)$, which is a sheaf.
:::

:::{.example title="?"}
Let $X^n \in \Mfd_{\smooth}$ and let \( \Omega^p \) be the sheaf of smooth $p\dash$forms, i.e $\Wedge^p T\dual$, i.e. \( \Omega^p(U) \) are the smooth $p$ forms on $U$,
which are locally of the form \( \sum f_{i_1, \cdots, i_p} (x_1, \cdots, x_n) dx_{i_1} \wedge dx_{i_2} \wedge \cdots dx_{i_p} \)
where the $f_{i_1, \cdots, i_p}$ are smooth functions.

:::{.example title="Sub-example"}
Take $X= S^1$, writing this as $\RR/\ZZ$, we have \( \Omega^1(X) \ni dx \).
There are two coordinate charts which differ by a translation on their overlaps, and $dx(x + c) =dx$ for $c$ a constant:

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:::

:::{.exercise title="?"}
Check that on a torus, $dx_i$ is a well-defined 1-form.
:::

:::

:::{.remark}
Note that there is a map $d: \Omega^p \to \Omega^{p+1}$ where \( \omega\mapsto d \omega \).
:::

:::{.warnings}
$d$ is **not** a map of $\OO\dash$modules: $d(f\cdot \omega) = f\cdot \omega + {\color{red} df \wedge \omega}$, where the latter is a correction term.
In particular, it is not a map of vector bundles, but is a map of sheaves of abelian groups since \( d ( \omega_1 + \omega_2) = d( \omega_1 ) + d ( \omega_2) \), making $d$ a sheaf morphism.
:::

:::{.remark}
Let $X \in \Mfd_\CC$, we'll use the fact that $TX$ is complex-linear and thus a $\CC\dash$vector bundle.

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:::{.remark title="Subtlety 1"}
Note that \( \Omega^p \) for complex manifolds is \( \Wedge^p T\dual \), and so if we want to view $X \in \Mfd_\RR$ we'll write $X_{\RR}$.
$TX_\RR$ is then a real vector bundle of rank $2n$.
:::

:::{.remark title="Subtlety 2"}
\( \Omega^p \) will denote *holomorphic* $p\dash$forms, i.e. local expressions of the form
\[
\sum f_I(z_1, \cdots, z_n) \Wedge dz_I
.\]
For example, $e^zdz\in \Omega^1(\CC)$ but $z\bar z dz$ is not, where $dz = dx + idy$.
We'll use a different notation when we allow the $f_I$ to just be smooth: $A^{p, 0}$, the sheaf of $(p, 0)\dash$forms.
Then $z\bar z dz\in A^{1, 0}$.
:::

:::{.remark}
Note that $T\dual X_\RR \otimes _\CC = A^{1, 0} \oplus A^{0, 1}$ since there is a unique decomposition \( \omega = fdz + gd\bar z \) where $f,g$ are smooth.
Then \( \Omega^d X_\RR \otimes_\RR \CC = \bigoplus _{p+q=d} A^{p, q} \).
Note that \( \Omega_{\smooth}^p \neq A^{p, q} \) and these are really quite different: the former are more like holomorphic bundles, and the latter smooth.
Moreover \( \dim \Omega^p(X) < \infty \), whereas \( \Omega_{\smooth}^1 \) is infinite-dimensional.
:::