# Wednesday, March 10

:::{.warnings}
Missing some stuff from the first few minutes here!
:::

:::{.remark}
Can we always get a Hermitian metric?
Let $X \in \Mfd_{C^{\infty }(\RR)}$ and \( \bundle{E} \to X \in \VectBundle_{\CC} \) a smooth complex vector bundle.
Then any section $h\in \bundle{E}\dual \tensor \conj{\bundle{E}}\dual(X)$, we have
\[
h: \bundle{E} \tensor \conj{\bundle{E}} \to \OO \\
h( e\tensor f) 
.\]
for \( e, f \in \bundle{E}_p \) is a Hermitian form for all $p$.
In local trivializations, $\ro{\bundle{E}}{U} \cong \OO_U^{\oplus r}$, and one can take the standard Hermitian form here.
Then for $(f_1, \cdots, f_r) \in \OO^{\oplus r}(U)$, we have \( \sum f_i \bar f_i\in \OO(U) \).
This can be extended to all of $X$ using a partition of unity subordinate to the coordinate charts.

The thing to check here is that on $\CC^r$, for any collection $h_1, \cdots, h_n$, any positive linear combination \( \sum a_i h_i \) is again a Hermitian metric for any \( a_i \in \RR^+ \).
One can regard these as skew-symmetric matrices, which are closed under addition, and the positive-definite property ensures it's still a metric since $h(v, v) = \sum a_i h_i(v, v) > 0$ for $v\neq 0$.
:::

:::{.remark}
Recall that we start with a Riemannian manifold $(X, g)$ where $g: TX^{\otimes 2} \to \OO$ is a metric on the tangent bundle.
Locally choose $f_1,\cdots, f_n$ an orthogonal frame of $TX$, then setting $e_i \da f_i\dual$ yields an orthogonal frame of $T\dual X$ and thus an orthogonal frame $e_{i_1} \wedge \cdots e_{i_p}$ of $\Wedge^p T\dual X \da \Omega^p X$.
So we get a metric on the smooth $p\dash$forms \( \Omega^p X \).
We defined the Hodge star operator
\[
\hodgestar: \Omega^p &\to \Omega^{n-p} \\
e_{i_1} \wedge \cdots e_{i_p} &\mapsto \pm e_{j_1} \wedge \cdots \wedge e_{j_{n-p}}
.\]
where \( \ts{ i_1, \cdots, i_p, j_1, \cdots, j_{n-p} } = \ts{ e_1, \cdots, e_n } \).
We saw that 
\[
e_{i_1} \wedge \cdots \wedge e_{i_p} \hodgestar \qty{ e_1 \wedge \cdots e_{i_p}} &= e_1 \wedge \cdots \wedge e_n \\
\hodgestar \qty{ \sum_{\abs I = p } f_I e_I} &= \sum_{ \abs I = p} e_{I^c} (-1)^{\sign(I)}
.\]

Moreover, 
\[
\inner{ \alpha}{ \beta} = \int_X g( \alpha, \beta) dV = \int_X \alpha\wedge \qty{\hodgestar \beta}
,\]
and we showed that 
\[
\inner{ \alpha}{ d \beta} = \pm \inner{ d^\dagger \alpha}{ \beta}
&& 
d^\dagger \da \hodgestar d \hodgestar, \beta\in \Omega^{p-1}(X), \alpha\in \Omega^p(X)
,\]
yielding an adjoint operator
\[
d^\dagger: \Omega^p(X) \to \Omega^{p-1}(X)
.\]
:::

:::{.definition title="Laplacian"}
The **Laplacian** is the differential operator
\[
\Delta \da dd^\dagger + d^\dagger d: \Omega^p(X) \to \Omega^p(X)
.\]
:::

:::{.definition title="Harmonic Forms"}
A $p\dash$form \( \omega \) is **harmonic** if and only if \( \Delta \omega = 0 \).
We define \( \mathcal{H}^p(X)  \) as the space of harmonic $p\dash$forms.
:::

:::{.remark}
This operator is $\RR\dash$linear, so \( \mathcal{H}^p(X) \in \Vect_\RR  \).
Note that this whole construction can be made to work over $\CC$ by adding conjugates in appropriate places.
:::

:::{.proposition title="Characterization of when a smooth p-form is harmonic"}
A smooth $p\dash$form \( \omega \) is harmonic if and only if \( d \omega = d^\dagger \omega = 0 \).
:::

:::{.proof title="?"}
$\impliedby$:
This direct is easy, since $\Delta \omega \da (dd^\dagger + d^\dagger d) \omega = d(0) + d^\dagger 0 = 0$.

$\implies$:
A nice trick! 
Using the adjunction $d, d^\dagger$ we have
\[
\inner{ \Delta \omega}{ \omega}
&=
\inner{ d d^\dagger \omega}{ \omega} +
\inner{d^\dagger \omega}{ \omega}
\\
&=
\inner{ d^\dagger \omega}{ d^\dagger \omega} +
\inner{d \omega}{ d \omega}
.\]
We now use that since $g$ is positive definite, it is a non-negative smooth function, and 
\[
\inner{ \alpha}{ \alpha} \da \int_X g( \alpha, \alpha) \, dV \geq 0 \text{ with equality } \iff \alpha \equiv 0 \text{ on } X
.\]
So we can conclude that $d^\dagger \omega = d \omega = 0$.

:::

:::{.warnings}
Note that we've used that the inner product is symmetric over $\RR$.
Over $\CC$, there are bars introduced from conjugation when swapping the variables.
:::

:::{.proposition title="Orthogonal decomposition of p-forms"}
The following three subspaces of \( \Omega^p(X) \) are mutually orthogonal:
\[
d \Omega^{p-1}(X), \mathcal{H}^p(X), d^\dagger \Omega^{p+1}(X) 
.\]
:::

:::{.proof title="?"}
We can write
\[
\inner{ d \alpha}{ d^\dagger } = 
\inner{ d^2 \alpha}{ \beta} =
\inner{0}{ \beta}
,\]
showing that the 1st and 3rd spaces are orthogonal.
If \( \alpha\in \mathcal{H}^p(X)  \) then by the above proposition, \( d \alpha = d^\dagger \alpha = 0 \), and so
\[
\inner{ \alpha }{ d \beta} = \inner{d^\dagger \alpha}{ \beta} = 0 \\
\inner{ \alpha }{ d^\dagger \beta} = \inner{d \alpha}{ \beta} = 0
.\]
Thus the 2nd space is orthogonal to the 1st and 3rd.
:::

:::{.observation}
Suppose something false ($\danger$): that \( \Omega^p(X) \) is a *complete* vector space with respect to the inner product.
Remember that it is **not**!
But if it were, there would be a decomposition
\[
\Omega^p(X) = d \Omega^{p-1}(X) \oplus \mathcal{H}^p(X) \oplus d^\dagger \Omega^{p+1}(X) 
.\]
Let \( \alpha\in \qty{ d \Omega^{p-1}(X) \oplus  d^\dagger \Omega^{p+1}(X)}^\perp  \) where we take the orthogonal complement with respect to the inner product.
Then 
\[
\inner{ \alpha } { d \beta } &= 0 \forall \beta \\
\inner{ \alpha } { d^\dagger \gamma } &= 0 \forall \gamma\\ \\
\implies \inner{ d^\dagger \alpha}{ \beta} = 0 \forall \beta \\
\implies d^\dagger \alpha \equiv 0 && \text{setting} \beta\da d^\dagger \alpha
.\]
Similarly, \( d \alpha = 0 \) and so \( \alpha\in \mathcal{H}^p(X)  \).

The conclusion (which is true *without* the false assumption) is that 
\[
\qty{ d \Omega^{p-1}(X) \oplus d^\dagger \Omega^{p+1}(X)}^\perp = \mathcal{H}^p 
.\]
However, this doesn't yield the full direct sum decomposition: if $W \subseteq V$, then it's not necessarily true that \( V \cong W \oplus W^\perp \), which only holds if 

- $V$ is complete,

- $W$ is closed.

:::

:::{.fact}
For smooth $p\dash$forms, this decomposition **does** hold despite the false assumption:
\[
\Omega^p(X) = d \Omega^{p-1}(X) \oplus \mathcal{H}^p(X) \oplus d^\dagger \Omega^{p+1}(X) 
.\]

:::

:::{.corollary title="p-forms have harmonic representatives"}
Thus \( \mathcal{H}^p(X)  \) represents $H^p(X; \RR)$.
:::

:::{.remark}
We have 
\[
H^p(X; \RR) 
&= {\ker d \over \im d} \\
&= { d \Omega^{p-1}(X) \oplus \mathcal{H}^p(X) \over d \Omega^{p-1}(X) } \\
&= \mathcal{H}^p(X) 
.\]
Note that there is a map
\[
\mathcal{H}^p(X) \to H^p(X; \RR) 
\]
since \( \alpha\in \mathcal{H}^p(X)  \) satisfies \( d \alpha = 0 \) in addition to \( d^\dagger \alpha = 0 \).
:::

:::{.remark}
Note that one can complete these spaces using Sobolev spaces, but there are issues. 
Take $S^1$, then 
\[
L_2(S^1) \da \ts{ \sum a_n e^{2\pi i n z} \st \sum \abs a_i < \infty  }
,\] 
but for $f\in L_2(S^1)$ we have \( df = \sum 2\pi i n a_n e^{2\pi i n z} \) which may not converge.
:::