# 2025-01-29-15-22-42

:::{.remark}
Newlander-Nirenberg theorem: $(X, I)$ is a complex manifold iff $[\T_X^{1, 0}, \T_X^{1, 0}] \subseteq \T_X^{1, 0}$.

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:::{.lemma}
$(X, I)$ is integrable iff $\forall \alpha\in A_\CC^{1, 0}(X)$, one has $(d\alpha)^{0, 2}=0$.
:::

:::{.proof}
If $\alpha$ is a 1-form and $\chi,\psi$ are vector fields, 
\[
d\alpha(\chi, \psi) = \chi(\alpha(\psi)) - \psi(\alpha(\chi)) - \alpha([\chi, \psi])
.\]
Note that $(d\alpha)^{0, 2} = 0\iff d\alpha(\chi, \psi) = 0$ for all $\chi, \psi\in \T_X^{0, 1}$.
Since $\alpha$ is a $(1, 0)$ form, $\alpha(\psi) = 0$ and $\alpha(\chi) = 0$.
Here we write
\begin{align*}
\alpha &= \sum_i \alpha_i \dx_i \\
\chi &= \sum_i \chi_i \dd{}{x_i} \\
\psi &= \sum_i \psi_i \dd{}{x_i} \\
d\alpha(\chi, \psi) &= \sum_{i, j} \dd{\alpha_i}{x_j}\qty{\chi_j\psi_i - \chi_i \psi_j}
\end{align*}
We also have $(d\alpha)^{0, 1} = 0 \iff \alpha([\chi,\psi]) = 0 \iff [\chi,\psi]$ is a $(0, 1)$ form.
:::

:::{.remark}
Note that if $n\da \dim_\CC X = 1$, then every almost complex structure $(X, I)$ is integrable.
In any dimension, $(X, I)$ induces a canonical orientation on $X$ given by $\ts{e_1, Ie_1, e_2, Ie_2,\cdots, e_n, Ie_n}$.
:::

:::{.theorem}
If $X$ is an oriented 2-manifold, then there exists a complex structure on $X$ inducing its orientation.
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:::{.lemma}
For oriented 2-manifolds $X$, there is a bijection
\[
\ts{\text{Almost complex structures on } X} 
\mapstofrom \ts{\text{Conformal structures on } X}
= \ts{\text{Riemannian metrics on } X} / \sim
\]
where $m_1\sim m_2 \iff \exists f$ a positive function such that $m_1 = fm_2$.
:::

:::{.remark}
Why: let $I$ be rotation by an angle of $\pi/2$, and use the fact that $\U_1 = \SO_2$.
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:::{.theorem}
Let $X$ be a simply connected Riemann surface.
Then either

- $X = \CP^1 \cong S^2$,
- $X = \CC$, or
- $X = \DD \subseteq \CC = \ts{ z\in \CC \st \abs{z} < 1} \cong \HH = \ts{z\in \CC \st \Im(z) > 0 }$.

:::