# Monday, March 08

:::{.remark}
Recall that given a differential complex \( (\complex{ \bundle{E} }, d) \) we had a symbol complex \( ( \pi^* \complex{\bundle{E}}, \sigma(d) ) \) where $\pi: T\dual X\to X$ and 
\[ \sigma\qty{  \sum_{\abs{I} \leq N} f_I \bd_I } \da \sum_{\abs{I} = N} f_I y^I 
,\] 
where we take the top-order differentials, $\dd{}{x_j} \mapsto y_j$ and 
\[
T\dual X &\to \RR \\
\alpha &\mapsto \alpha\qty{\dd{}{x_j} }
.\]
We say that \( ( \complex{\bundle{E} }, d ) \) is **elliptic** if the symbol complex is exact on $T\dual X \smz$ where we delete the zero section.
The Atiyah-Singer index theorem stated
\[
\chi( \complex{\bundle{E}}, d) = \int_X { \ch( \complex{ \bundle{E} }) \over \eul(X) } \td( TX\tensor_\RR \CC)
.\]
What's the connection to elliptic operators?
Given a 2-term complex 
\[
0 \to \bundle{E}^0 \mapsvia{D} \bundle{E}^1 \to 0
,\]
then $D$ is an **elliptic operator** if this is an elliptic complex.
This means the symbol complex is an isomorphism, i.e. 
\[
0 \to \pi^* \bundle{E}^0 \mapsvia{\sigma(D)} \pi^* \bundle{E}^1 \to 0
\]
where $\sigma(D)$ is an isomorphism away from the zero section.
:::

:::{.remark}
Every elliptic complex can be converted into a 2-term complex using a hermitian metric.
Given 
\[
\bundle{E}^0 \mapsvia{d^0} \bundle{E}^1 \mapsvia{d^1} \bundle{E}^2 \to \cdots
,\]
we map this to 
\[
0 \to \bundle{E}^{\text{even}} \da \bigoplus_{i \text{ even} } \bundle{E}^i 
\mapscorrespond{D^\text{even}}{D^{\text{odd} }} 
\bundle{E}^{\text{odd}} \da \bigoplus_{i \text{ odd}} \to 0
\]
where 
\[
D \da ((d^{2i-1})^{\dagger} , d^{2i} ) : \bundle E^{2i} \to \bundle{E}^{2i-1} \oplus \bundle{E}^{2i+2} \\
\]
and $(d^{2i-1})^{\dagger}$ is defined by the following property: for \( \alpha\in \bundle{E}^{2i-1} \) and \( \beta \in \bundle{E}^{2i}(X) \), 
\[
\inner{ d^{2i-1} \alpha}{\beta}_h = \inner{ \alpha } { ( (d^{2i-1})^{\dagger} \beta}_h
.\]
Here this pairing depends on a hermitian metric $h$, which is a hermitian form on each fiber:
\[
h_i: \bundle{E}^i \tensor \conj{ \bundle{E}^i} \to \CC
.\]
Using this, we can fix a volume form $dV$ on $X$ and define
\[
\inner{u}{v}_h \da \int_X h_i(u, \conj{v}) \, dV && u, v\in \bundle{E}^i(X)
.\]
This yields the desired two-term complex, and $( \complex{\bundle{E}}, d)$ is elliptic if and only if $D^e \circ D^o: \bundle{E}^o \selfmap$ and $D^o \circ D^e: \bundle{E}^e \selfmap$ are elliptic operators.
:::

:::{.example title="?"}
Taking the de Rham complex
\[
0 \to \OO \mapsvia{d} \Omega^1 \mapsvia{d} \Omega^2 \to \cdots
,\]
one can define 
\[
\Omega^{\even} \mapscorrespond{d + d^\dagger}{ d + d^{\dagger}} \Omega^{\odd}
.\]
Then using adjoint properties, we have
\[
\inner{\alpha}{ d^\dagger d^\dagger \beta} = 
\inner{ d \alpha}{ d^\dagger \beta} =
\inner{ d^2 \alpha}{ \beta } = 
0
,\]
using that $d^2 = 0$, and since this is true for all \( \alpha, \beta \) we have \( (d^\dagger)^2 \beta = 0 \)  for all \( \beta \).
Noting that $d d^\dagger + d^\dagger d: \Omega^i(X) \selfmap$, and this operator is **the Laplacian**.
Moreover \( \ker (d d^\dagger + d^\dagger d ) \) is the space of **harmonic $i\dash$forms**.
:::

:::{.remark}
Note that this space of harmonic forms depended on the Hermitian metrics on $\bundle{E}^i$ and the volume form $dV$.
In the case $\bundle{E}^i \da \Omega^i$, there is a natural metric determined by any Riemannian metric on $X$.
Recall that this is given by a metric
\[
g: TX \tensor TX \to \RR
.\]
This determines an isomorphism
\[
T_p X &\mapsvia{\sim} T_p\dual X\\
v &\mapsto g(v, \wait)
,\]
which we can invert to get a metric on the cotangent bundle $T\dual X$.
This induces a metric on $i\dash$forms using the identification \( \Omega^i \da \Wedge^i T\dual X \) and induces a volume form
\[
dV \da \sqrt{ \det g}: \Wedge^{\text{top}} TX \to \RR
.\]

In this case, $d d^\dagger + d^\dagger d$ on \( \Omega^i(X) \) is called the **metric Laplacian**.
:::

:::{.remark}
Let $(X, g)$ be a Riemannian manifold.
We thus have a symmetric bilinear form on \( \Omega^p(X) \) given by pairing sections:
\[
\inner{ \alpha}{ \beta} \da \int_X g( \alpha, \beta)
.\]
Note that we have orthonormal frames on \( \Omega^p(X) \) of the form \( e_{i_1} \wedge \cdots \wedge e_{i_p} \) where the \( \ts{ e_i } \)  are orthonormal frames on $T\dual X$.
:::

:::{.definition title="Hodge Star Operator"}
Let $n\da \dim(X)$.
The **Hodge star** operator is a map 
\[
\hodgestar: \Omega^p \to \Omega^{n-p}
.\]
defined by the property
\[
\alpha\wedge \hodgestar \beta= g( \alpha, \beta) dV
.\]
Concretely, we have 
\[
\hodgestar \qty{ \sum f_I dx_{i_1} \wedge \cdots \wedge dx_{i_p} } 
&= \hodgestar \qty{ \sum f_I e_{i_1} \wedge \cdots \wedge e_{i_p} } \\
&= (-1)^\ell \sum_{j_k \in \ts{ 1, \cdots, n } \sm I} f_I e_{j_1} \wedge \cdots \wedge e_{j_{n-p}}
\]
for some sign $\ell$.
:::

:::{.example title="?"}
Let $X\da \RR^4$ and $g$ the standard metric, i.e. $d = dx_1^2 + \cdots + dx_4^2$.
Take an orthonormal basis of $T\dual \RR^4$, say \( \ts{ e_1, e_2, e_3, e_4 } \) where $e_i \da dx_i$.
Then the induced volume form is $dV \da e_1 \wedge e_2 \wedge e_3 \wedge e_4$.
We can then compute
\(
\hodgestar(e_1 \wedge e_2)
\)
which is defined by the property
\[
\alpha\wedge \hodgestar( e_1 \wedge e_2) = g( \alpha, e_1 \wedge e_2) dV
.\]
On the right-hand side, $g( \alpha, e_1 \wedge e_2) = c_{12}(\alpha) e_1 \wedge e_2 \wedge e_3 \wedge e_4$ where $c_{12}$ is the coefficient of $e_1 \wedge e_2$.
To extract that coefficient, we can take \( \alpha( e_3 \wedge e_4 \), writing \( \alpha = \sum c_{ij} e_i \wedge e_j \).
Similarly,  \( \hodgestar)e_1 \wedge e_3) = -e_2 \wedge e_4 \).
This follows from writing 
\[
\alpha \wedge \hodgestar(e_1 \wedge e_3) =
c_{13}(\alpha) {\color{blue} e_1} \wedge e_2 \wedge {\color{blue} e_3 } \wedge e_4 =
(-1) c_{13}(\alpha) e_1 \wedge {\color{blue} e_3 \wedge e_2} \wedge e_4
.\]

From this, $\hodgestar: \Omega^p \to \Omega^{n-p}$ is defined fiber-wise as
\[
\inner{ \alpha}{ \beta} = \int_X \alpha\wedge \hodgestar \beta
.\]


:::

:::{.exercise title="?"}
Show that \( \hodgestar^2 = (-1)^{p(n-p)} \).
:::

:::{.proposition title="Formula for the adjoint of the Hodge star"}
Let $d^\dagger \da (-1)^{n(p-1) +1} \hodgestar d \hodgestar$.
Then 
\[
\inner{\alpha}{ d \beta} = \inner{d^\dagger \alpha}{ \beta} && \alpha\in \Omega^p(X), \beta\in \Omega^{p-1}(X)
.\]
:::

:::{.proof title="?"}
A slick application of Stokes' theorem!
Using that \( \hodgestar \) is an isometry, we have
\[
\inner{ \alpha}{ d \beta} 
&= \int_X \alpha\wedge \hodgestar d \beta \\
&= \int_X \hodgestar \alpha \wedge d \beta 
(-1)^{p(n-p)} 
&& \text{applying $\hodgestar$ to both} \\
&= -\int_X d( \hodgestar \alpha) \wedge \beta (-1)^{p(n-p)}
&& \text{Stokes/IBP} \\
&= (-1)^{p(n-p)+1} \int_X \hodgestar d \hodgestar \alpha \wedge \hodgestar \beta 
&& \text{isometry}\\
&= (-1)^{p(n-p)+1} \inner{\hodgestar d \hodgestar \alpha}{ \beta}
,\]
which shows that the term in the left-hand side of the inner product above is the adjoint of $d^\dagger$.

:::