# Wednesday, March 24

:::{.remark}
Last time: we defined a **Kähler manifold**: $X\in \Mfd(\CC)_\cpt$ and \( \omega \in \Omega^2(X_\RR) \) a closed real 2-form such that $g(x, y) \da \omega(x, Jy)$ is a metric.
By the Hodge theorem, we have a space \( \mch^k(X) \) of harmonic $k\dash$forms for $(X, g)$ which represents $H^k_{\dR}(X; \RR)$.
We can consider the $\CC\dash$valued harmonic forms \( \mch^k_\CC \da \mch^k(X) \tensor_\RR \CC \), which represents $H^k_\dR(X; \CC)$
:::

:::{.question}
How does this interact with the decomposition of the smooth $k\dash$forms
\[
\Omega^k(X_\RR)\tensor_\RR \CC = \bigoplus_{p+q=k}^K A^{p, q}(X)
,\]
where $\mch^k_\CC(X)$ is contained in this.
Note that this is a small finite dimensional space in an infinite dimensional space!
The following miracle occurs:
:::

:::{.theorem title="Kähler manifolds admit a Hodge decomposition?"}
If $X \in \Mfd(\Kahler)$,
\[
\mch^k_\CC = \bigoplus_{p+q = k} \mch^{p, q}(X)
,\]
where 
\[
\mch^{p, q}(X) \da 
\qty{ \mch^K(X) \tensor_\RR \CC } \intersect A^{p, q}(X) \subseteq \Omega^k(X_\RR)
.\]
:::

:::{.example title="?"}
Let $X = \CC/ \Lambda$ be an elliptic curve where \( \Lambda \) is a lattice.
The standard metric $dx^2 + dy^2$ on $\CC$ descends to a metric on $X$ since translation is an isometry on the metric space $(\CC, dx^2 + dy^2)$.
Let $z=x+iy$ be a complex coordinate on $\CC$ so $dz = dx + idy$ and $d\bar z = dx - idy$, then \( dx^2 + dy^2 = dz d\bar{z} \in \Sym^2(\T\CC) \).
The symplectic form is given by 
\[ 
\omega(v, w) = \pm g(v, Jw) = i \dz \dzbar (v, w) 
\] 
since $J$ is given by $i$ on $\CC$.
Then \( \omega(v, w) = i \dz(v) \dzbar(w)\), i.e. \( \omega = i\dz \wedge \dzbar \).
So 
\[ 
\bar{ \omega} = \bar i \dzbar \wedge \dz = -i \dzbar \wedge \dz = i\dz \dzbar = \omega
, \] 
and this determines the Kähler geometry on $X$.
What are the harmonic 1-forms on $X$, $\mch^1(X) \tensor_\RR \CC$?
Note that \( \omega= \dV \) is the volume form.
The smooth 1-forms are given by 
\[ 
\Omega^1(X_\RR) \tensor_\RR \CC = A^{1, 0}(X) \oplus A^{0, 1}(X) = \ts{ f(z, \bar{z} )\dz } \oplus \ts{ g(z, \bar z)\dzbar}
,\]
where $f,g$ are smooth and \( \Lambda\dash \)periodic on $\CC$ to make them well-defined.
We can find the Hodge star:
\[
\hodgestar: ? &\to ? \\
\dz & \mapsto i\dzbar \\
\dzbar &\mapsto -i\dz
.\]
Writing \( \alpha\da f(z, \bar z)\dz + g(z, \bar z) \dzbar \), this is harmonic if \( d \alpha = 0 \) and \( \stardstar \alpha = 0 \).
The first implies \( \partial_{\bar z} f - \partial_{z} g = 0\).
What does the second imply?
We can compute
\[
\hodgestar \alpha 
&= if(z, \barz) \dzbar - ig(z, \bar z) \dz \\
\implies \partial_z f + \partial_{\bar z} g 
&= 0
,\]
and so \( \partial_{\bar z} f = \partial_z g \) and \( \partial_{\bar z}^2 f = \partial_{\bar z} \partial_z g = - \partial_z^2 f \), so
\[
\qty{ \partial_{\bar z}^2 + \partial_z^2 }f &= 0 \\
\qty{ \partial_{\bar z}^2 + \partial_z^2 }g &= 0 
.\]

Note that this recovers the usual notion of harmonic functions on $\CC$, i.e. being in the kernel of the Laplacian.
The only biperiodic functions that satisfy these equations are constants, since there is a maximum modulus principle for harmonic functions.
Thus 
\[
\mch^1(X) \tensor_\RR \CC = \ts{ c_1 \dz + c_2 \dzbar } = \CC \dz \oplus \CC \dzbar = H^{1, 0}(X) \oplus H^{0, 1}(X) 
.\]
:::

:::{.remark}
There is a generalization to higher genus curves.
Recall the following theorem:
:::

:::{.theorem title="Uniformization"}
Let $C \in \Mfd^1(\CC)_\cpt$ of genus $g\geq 2$.
Then the universal cover admits a biholomorphism 
\[
\tilde C \cong \HH \da \ts{ z\in \CC \st \Im(z) > 0 }
.\]
:::

:::{.remark}
This essentially follows from the Riemann mapping principle.
:::

:::{.corollary title="Every curve of genus g>1 is the plane mod a subgroup of biholomorphisms"}
Any curve $C$ of genus $g\geq 2$ is of the form $C = \HH/ \Gamma$ where \( \Gamma \leq \BiHol(\HH) \) is a subgroup that acts freely.
By covering space theory, \( \Gamma = \pi_1(C) \), and it's known that \( \BiHol(\HH) \cong \PSL_2(\RR) \) by the map
\[
\begin{bmatrix}
a & b 
\\
c & d
\end{bmatrix}
z
\mapsto 
{az + b \over cz + d}
.\]
:::

:::{.proposition title="The upper half-plane admits a PSL-invariant hyperbolic metric"}
The upper half plane $\HH$ admits a **hyperbolic metric** which is invariant under \( \PSL_2(\RR) \) given by 
\[
g_{\hyp} = {dx^2 + dy^2 \over y^2 } = {\dz \dzbar \over \Im(z)^2 }
.\]
:::

:::{.proof title="?"}
This follows from a computation:
\[
d\qty{ az + b \over cz + d} 
&= {a\dz \over cz + d} - {c (az+b)\dz \over (dz+d)^2 } \\
&= {a (cz+d) - c(az+b) \dz \over (cz+d)^2} \\
&= { (ad-bc)\dz \over  (cz + d)^2 } \\
&= {\dz \over  (cz + d)^2 } \\
&= { d\qty{ az+b \over cz+d} d \qty{\bar{ az + b \over cz + d}} \over \Im\qty{az+b \over cz+d}^2 } \\
&= { \dz \dzbar \over (cz+d)^2(c \bar{z} + d)^2 \Im\qty{az+b \over cz+d} } \\
&= { \dz \dzbar \over \Im(z)^2 } 
.\]
:::

:::{.remark}
It's miraculous! 
The biholomorphisms of $\HH$ preserve a metric.
So $C$ has a canonical metric, $g_\hyp$, which descends along the quotient map $\HH \to \HH/\Gamma \cong \CC$.
:::

:::{.question}
What are the harmonic 1-forms on $(C, g_\hyp)$?
:::

:::{.remark}
By lifting we can write 
\[ 
\Omega^1(C_\RR) \tensor_\RR \CC = A^{1, 0}(C) \oplus A^{0, 1}(C) 
= 
\ts{ f(z, \bar z)\dz + g(z, \bar z) \dzbar \st z\in \HH, \, f,g\in C^{\infty }(\CC, \RR) }  
\]
But $\dz$ is *not* invariant under the map $z\mapsto {az+b \over cz+d}$, since \( \dz \mapsto {\dz \over (cz+d)^2 } \).
In order to descend $f(z)$ to $C$, we need 
\[
f\qty{az +b \over cz + d} = (cz+d)^2 f(z)
&&
\text{ for all } 
\begin{bmatrix}
a & b 
\\
c & d
\end{bmatrix}
\in \Gamma
\]
This says that \( f \) is a **modular form of weight 2**.
:::

:::{.exercise title="?"}
Check that this implies that $f$ must be holomorphic and $g$ must be antiholomorphic.
:::

:::{.fact}
There is a decomposition
\[
\mch^1(C_\RR) \tensor_\RR \CC = \mch^{1, 0}(C) \oplus \mch^{0, 1}(C)
,\]
and the first space will be the space of holomorphic 1-forms $H^0(K_C)$, and the second term will be $\bar{H^0(K_C)}$.
This shows the power of the Hodge decomposition theorem!
:::