# Friday, March 26th


:::{.remark}
Recall the Hodge decomposition theorem.
Let $(M, g) \in \Mfd_\RR^n(\Riem, \cpt)$, then choosing an orthonormal basis \( \ts{ v_j } \) for $T_p M$ yields a corresponding orthonormal basis in $T_p\dual M \da \Hom_\RR(T_p M, \RR)$ given by taking \( \ts{ e_i \st e_i(v_j) = \delta_{ij} } \).

\begin{tikzpicture}
\fontsize{44pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-15_23-21.pdf_tex} };
\end{tikzpicture}


There is a map
\[
\hodgestar: \Wedge^k T_p\dual M &\to \Wedge^{n-k} T_p\dual M \\
\Wedge_{j=1}^k e_{i_j} &\mapsto \pm \Wedge_{\ell = 1}^{n-k} e_{j_\ell}
\]
where the $e_{j}$ are defined such that $\Wedge_{j=1}^k e_{i_j} \wedge \Wedge_{\ell = 1}^{n-k} e_{j_\ell} \da \dV$, where $\dV$ is the volume form on $M$ at $p$.
Thus we have a map
\[
\hodgestar: \Omega^k &\to \Omega^{n-k} \\ 
1 &\mapsto \dV
.\]
 We defined $d^\dagger \da \stardstar$, and said a form \( \omega \) was *harmonic* iff $\Delta \omega=0$, where \( \Delta \da dd^\dagger + d^\dagger d \).
 The space of such forms was denoted \( \mathcal{H}^k(M) \subseteq \Omega^k(M)  \).

:::


:::{.theorem title="Hodge Theorem"}
\[
\mathcal{H}^k(M) \cong H^k_{\dR}(M; \RR) 
.\]
:::


:::{.question}
What kinds of extra structure can we put on a complex manifold?
:::


:::{.definition title="Kähler Form"}
A **Kähler form** is a closed 2-form \( \omega\in \Omega^2_\RR \) such that the following equation defines a metric on $T_p M$:
\[
g(u, v) \da \omega(u, iv)
.\]
I.e., this is a closed symplectic form that defines a metric.
:::


:::{.example title="?"}
Consider $M = \CC^n$ with holomorphic coordinates $\elts{z}{n}$, where $z_{j} \da x_j + iy_j$.
Then take
\[
\omega \da \sum_{j=1}^n \dx_j \wedge \dy_j
.\]

Note that multiplication by $i$ induces a map
\[
\cdot i: T_p \CC^n &\selfmap \\
\dd{}{x_j} &\mapsto \dd{}{y_j} \\
\dd{}{y_j} &\mapsto - \dd{}{x_j} \\
.\]
Moreover, \( \omega(u, iv) \) recovers the standard metric on $\CC^n$ given by 
\[
g_{\std} = \sum (\dx_j)^2 + (\dy_j)^2 \in \Sym^2 T\dual \CC^n
,\]
which is incidentally positive-definite, where $(\dx)^2(u, v) \da (\dd{}{x_j})u \cdot *(\dd{}{y_j}) v$.
Is this closed?
We need to check to see if $d\omega = 0$, but this is true: applying $d$ to all of the coefficients yields the constant 1.
:::


:::{.remark}
So for $M\in \Mfd(\CC)$ a complex manifold, we have a decomposition
\[
\Omega^k(M) &= \bigoplus_{p+q=k} A^{p, q}(M) \\ \\ 
A^{p, q} &\da \ts{ 
  \sum_{\substack{ \abs I = p \\ \abs J = q}} 
  \qty{\dz_{i_1} \wedge \cdots \dz_{i_p} } 
  \wedge 
  \qty{ \dz_{j_1} \wedge \cdots \dz_{j_q} } 
}
.\]

For $M$ a Kähler manifold, we have
\[
\mathcal{H}^k(M) = \bigoplus _{p+q = k} \mathcal{H}^{p, q}(M) 
\\  \\
\mathcal{H}^{p, q}(M) = \mathcal{H}^k(M) \intersect A^{p, q}(M)   
.\]

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\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-17_19-13.pdf_tex} };
\end{tikzpicture}

:::


:::{.remark}
Why is this true?
We have a map
\[
d: A^{p, q}(M) \to A^{p+1, q}(M) \oplus A^{p, q+1}(M)
,\]
where for example if $f(z) \da z \zbar \in A^{0, 0}(\CC)$, we have $df = \zbar \dz + z\dzbar$ where the first is a $(1, 0)$ form and the latter is a $(0, 1)$ form.
Write $d = \del + \delbar$ where $\del \da \sum \dz_j$ and $\delbar = \sum \dzbar_j$, as well as 
\[
d^\dagger: A^{p, q}(M) \to A^{p-1, q}(M) \oplus A^{p, q-1}(M)
.\]
Now $\hodgestar$ of a $(p, q)$ form is an $(n-p, n-q)$ form, and so 
\[
\hodgestar \qty{ \dz_{i_1} \wedge \cdots \wedge \dz_{i_r} \wedge \dzbar_{j_1} \wedge \cdots \wedge \dzbar_q } \da \hodgestar (\dz_I \wedge \dzbar_J) = \pm \dz_{I^c} \wedge \dzbar_{J^c}
,\]
and we have $d^\dagger = \del^\dagger + \delbar^\dagger$.
We can thus move around the bigraded group in several ways:

\begin{tikzcd}
	{A^{2, 0}} && {A^{2, 1}} && {A^{2, 2}} \\
	\\
	{A^{1, 0}} && {A^{1, 1}} && {A^{1, 2}} \\
	\\
	{A^{0, 0}} && {A^{0, 1}} && {A^{0, 2}}
	\arrow["\del", dashed, from=5-1, to=3-1]
	\arrow["\del", dashed, from=3-1, to=1-1]
	\arrow["\del", dashed, from=3-3, to=1-3]
	\arrow["\del", dashed, from=5-5, to=3-5]
	\arrow["\del", dashed, from=3-5, to=1-5]
	\arrow["\delbar", dashed, from=1-1, to=1-3]
	\arrow["\delbar", dashed, from=1-3, to=1-5]
	\arrow["\delbar", dashed, from=3-1, to=3-3]
	\arrow["\delbar", dashed, from=3-3, to=3-5]
	\arrow["\delbar", dashed, from=5-1, to=5-3]
	\arrow["\delbar", dashed, from=5-3, to=5-5]
	\arrow["\del", dashed, from=5-3, to=3-3]
	\arrow["{\delbar^\dagger}", color={rgb,255:red,214;green,92;blue,92}, curve={height=-18pt}, from=5-3, to=5-1]
	\arrow["{\delbar^\dagger}", color={rgb,255:red,214;green,92;blue,92}, curve={height=-18pt}, from=5-5, to=5-3]
	\arrow["{\del^\dagger}"', color={rgb,255:red,92;green,92;blue,214}, curve={height=24pt}, from=1-1, to=3-1]
	\arrow["{\del^\dagger}"', color={rgb,255:red,92;green,92;blue,214}, curve={height=24pt}, from=3-1, to=5-1]
\end{tikzcd}


> [Link to Diagram](https://q.uiver.app/?q=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)

:::


:::{.theorem title="Kähler Identities"}
Let 
\[
\Delta_{\delbar} &\da \delbar \delbar^\dagger + \delbar^\dagger\delbar \\
\Delta_\del &\da \del \del^\dagger + \del^\dagger \del \\
\Delta_d &\da dd^\dagger + d^\dagger d
.\]
Then
\[
{1\over 2} \Delta_d = \Delta_{\delbar} = \Delta_\del
.\]
:::


:::{.remark}
See Griffiths-Harris for details.
Note that this is a local statement, i.e. it can be checked in coordinate charts.
:::


:::{.remark}
The upshot: 
\[
\mathcal{H}^k(M) = \ker \Delta_d = \ker \Delta_{\delbar} 
,\]
and moreover
\[
\Delta_{\delbar}: A^{p, q}(M) \selfmap
\]
which implies that on \( \Omega^k(M) \),
\[
\ker \Delta_{\delbar} \circ \bigoplus _{p+q=k} \ker \qty{ A^{p, q}(M) \mapsvia{\Delta_{\delbar }} A^{p, q}(M) }
&= \bigoplus _{p+q = k} \ker \Delta_d
,\]
which yields the Hodge decomposition theorem 
\[
\mathcal{H}^k(M) = \bigoplus _{p+q=k} \mathcal{H}^{p, q}(M)  
.\]
:::


:::{.remark}
This is a strong restriction on what manifolds can admit a Kähler structure.
Moreover, since $\Delta_d$ is a real operator, we obtain 
\( \bar{ \mathcal{H}^{p, q}(M) } \cong \mathcal{H}^{p, q}(M)  \).
:::


:::{.remark}
Some consequences:

For $M$ a Kähler manifold, the odd Betti numbers $\beta_{2i+1}(M) \da \dim H_{\dR}^{2i+1}(M; \CC)$ are even.
This is because
\[
\bigoplus _{p+q=k} \mathcal{H}^{p, q} \cong \mathcal{H}^{2i+1}(M) \cong H_{\dR}^{2i+1}(M) 
.\]
If we define \( h^{p, q}(M) \da \dim_\CC \mathcal{H}^{p, q}(M)  \), we clearly have 
\[
\beta_{2i+1} = \sum_{p+q = 2i+1} h^{p, q}(M)
.\]
Now using that \( \bar{\mathcal{H} } \cong \mathcal{H}  \), we can rewrite this as
\[
\beta_{2i+1} 
&= \sum_{p+q = 2i+1} h^{p, q}(M) \\
&= 2 \sum_{\substack{ p+q = 2i+1 \\ p < q} } h^{p, q}(M)
.\]

:::


:::{.remark}
Is this just some fact about arbitrary complex manifolds, with no extra structure?
The answer is no, and the counterexample is the *Hopf surface*
\[
X \da \qty{ \CC^2 \sm \ts{\vector 0}} / (x,y)\sim (2x, 2y)
,\]
which we can roughly identify as $\RR^4$ "modulo doubling".
We can take a fundamental domain $1\leq \abs r \leq 3$, this yields an annulus-like sphere with the inner shell glued to the outer:

\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-17_19-49.pdf_tex} };
\end{tikzpicture}


This is homeomorphic to $S^1 \cross S^3$, but $\beta_1(M) = 1$, so this won't yield a Kähler structure.

:::