# Wednesday, March 17

## Inverting Bundles

:::{.remark}
Continuing review: let \( \bundle{E} \to X \in \Bun(\RR^n) \).
A **metric** on \( \bundle{E} \) is a smoothly varying positive definite inner product on the fibers.

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\todo[inline]{Fix this diagram! Need to remember what it was demonstrating.}


For $v, w\in \bundle{E}_p$, we want a pairing \( g_p(v, w): \bundle{E}_p^{\tensor 2} \to \RR \).
To think about this globally, this should be a map
\[
g: \bundle{E}^{\tensor 2} \to \OO
.\]
where \( g_p: \bundle{E}_p^{\tensor 2} \to \RR \).
Note that this map is \( \OO\dash \)linear, which follows from the fact that it's $\RR\dash$linear on each fiber, or equivalently it is a map of vector bundles.
We should also have that $g(s\tensor s) \in \OO(X)$ is a smooth function, and we require $g(s\tensor s) \geq 0$.
We also require $g(s\tensor s)(p) = 0 \iff s_0 = 0$ and $g(s\tensor t) = g(t\tensor s)$.
This implies that $g\in (\bundle{E}^{\tensor 2})\dual \tensor \OO = (\bundle{E}\dual)^{\tensor 2}(X)$.
The symmetric condition means that $g\in \Sym^2 \bundle{E}\dual(X)$.
:::

:::{.remark}
For Hermitian forms, we take
\[
h: (\CC^n)^{\tensor 2}\to \CC
\]
where $h$ is conjugate linear, so $h(cv, c'w) = \bar{c}c' h(v, w)$.
Note that we can write $h(v, w) = \conjugate{v}^t H w$ where $H$ is Hermitian, so $\conjugate{H}^t = H$.
This implies that $h(v,v) \in \RR^{\geq 0}$ and $h(v,v) = 0 \iff v=0$ with $h(v, w) = \conjugate{h(v, w)}$
The great thing about metrics: we can identify zero sections by self-pairing, multiplying by a volume form, and integrating.
For \( \bundle{E}\to X \in \Bun(\CC) \), there is another bundle \( \conjugate{\bundle{E}} \to X \in \Bun(\CC) \).
Supposing that 
\( \ro{ \bundle{E}}{U} \mapsvia{\varphi_U} \OO_U^{\oplus n} \) 
in a local trivialization, conjugating all of the transition functions gives the transition functions
\( \ro{ \conjugate{ \bundle{E}} }{U} \mapsvia{\mathrm{conj} \circ \varphi_U} \OO_U^{\oplus n} \).
This yields a map
\[
h: \conjugate{ \bundle{E} } \tensor_\CC \bundle{E} \to \OO \in ( \conjugate{\bundle{E}} \tensor \bundle{E} )\dual
.\]
In local trivializations we have \( \ro{ \bundle{E} }{U} = \OO_U^{\oplus n} = \CC^n \cross U \), and $h$ is described by $h_U \in (\conj{ \OO}^{\oplus n} \tensor \OO^{\oplus n})(U)$.
:::

:::{.remark}
When \( \rank \bundle{E} = 1 \) we abuse notation! 
For \( h\in (\conj{\bundle{E}}\dual \tensor \bundle{E}\dual)(X) \), this is locally a $1\times 1$ Hermitian matrix, thus of the form \( [a] \) for \( a\in \RR^{\geq 0} \).
So we write 
\[
h(s, t) = hs\conj{t} \da h\tensor s \tensor \conj{t} \in (\conj{\bundle{E}}\dual \tensor \bundle{E}\dual) \tensor \bundle{E} \tensor \conj{\bundle{E}} = \OO
\]
if \( \bundle{E} \) is a line bundle.
Why is \( V\tensor V\dual = \OO \) in this case? 
There is a pairing \( v\tensor \lambda \mapsto \lambda(v) \), or more generally a trace pairing.
:::


## Serre Duality Revisited

:::{.remark}
Let $X$ be a Riemann surface, so $X\in \Mfd^1(\CC)$.
Let \( L\to X \in \Bun^1(\Hol)\), then we have a resolution
\[
0 \to L \injects L\tensor A^{0, 0} \mapsvia{\delbar} L \tensor A^{0, 1} \to 0
,\]
where the first map is inclusion of smooth holomorphic sections into smooth sections.
What is this cut out by?
We had $s\mapsto \delbar s$ and thus $f \mapsto \dd{f}{\zbar} \dzbar$.
Note that $H_1(L) = \coker \delbar$.
:::

:::{.remark}
Serre duality said that 
\[
h^1(L) = \dim H^1(L) = h^0( L\dual \tensor K) && K = \Omega^1
,\]
where $\Omega^1$ is the sheaf of holomorphic 1-forms.
Choose a metric to identify $H^1(L)$ and $H^0(L\dual \tensor K)$.
Choose a hermitian metric on $L$ and take $s, t\in H^0(L\tensor A^{0, 0}) = C^\infty(L; \CC)$, then we get $h(s, t) \in C^{\infty }(X; \CC)$ a smooth complex function.
We abuse notation by writing this as $h(s, t) = hs\conj{t}$, viewing $h\in C^{\infty }(L\dual \tensor \conj{L}\dual)$ locally.
Note that we can't integrate a function on a manifold without a form, so choosing a volume for $dV$ we can define a pairing on sections
\[
\inner{s}{t} \da \int_X hs\conj{t} dV
.\]

Now for two sections \( \alpha, \beta\in H^0(L\tensor A^{0, 1}) \) we can write
\[
\int_X h \alpha \conj{ \beta} = \int_X \omega
,\]
where \( \omega \) is a smooth $(1, 1)\dash$form since $h\in \conj{L}\dual \tensor L\dual$, \( \alpha\in L\tensor A^{0, 1} \), and $\conj{ \beta} \in \bar{L} \tensor A^{1, 0}$.
We now have metric on both the source and target spaces here:
\[
H^0( L\tensor A^{0, 0}) \mapsvia{\delbar} H^0(L\tensor A^{0, 1})
,\]
where on the left-hand side we take \( (s, t) \mapsto \int_X hs\conj{t}dV \) and on the right-hand side we have \( (\alpha, \beta) \mapsto \int_X h \alpha\conj{\beta} \).
:::

:::{.remark}
Given a map of metric vector spaces \( V \mapsvia{\varphi} W \), the *adjoint* \( \varphi^\dagger \) satisfies
\[
\inner{ \varphi(v) }{w} = \inner{v}{ \varphi^\dagger(w)}
.\]
and $\coker( \varphi) = \ker( \varphi^\dagger)$.
So $H^1(L) = \coker \delbar = \ker \delbar^\dagger$, and after integrating by parts we have
\[
\inner{ \alpha}{ \delbar s} 
&\da \int_X \alpha\conj{ \delbar s } h \\
&= \int_X \alpha\del(\conj{s}) h \\
&= -\int_X \conj{s} \del( \alpha h) && \text{IBP} \\
&= -\int_X \conj{s} {\del(\alpha h) \over dV} dV \\
&= \inner{ - { \del( \alpha h ) \over dV}}{s}
.\]
So we could define
\[
\delbar^\dagger \alpha = \conj{- {\delbar(\conj{\alpha} h )} \over dV }
.\]
Note that \( \alpha\mapsto \conj{ \alpha} h \), so \( \alpha\in \ker \delbar^\dagger \iff \conj{ \alpha}h\in \ker \delbar\).
Then $\ker (\delbar^\dagger) = H^0(L\dual \tensor K)$.
:::