# Spin Bundles and Dirac Operators (Monday, May 03)


:::{.remark}
Last time: we defined a Spin structure on an oriented manifold $M$ as a lift of the principal $\SO(n)$ bundle $P\to M$ (*unassociated* to $TM$) to a $\Spin(n)$ bundle $\tilde P$.
There was a **spin representation** $\Spin(n) \actson \SS$, which is irreducible for $\Cl(\RR^n)$ and splits as $\SS = \SS^+ \oplus \SS^-$, which are $\Spin(n)$ subrepresentations.
We defined **spinor bundles** 
\[
\tilde P \fiberprod{\Spin(n)} \SS = \SS_M = \SS_M^+ \oplus \SS_M^-
.\]
:::


:::{.example title="Dimension 4"}
If $\dim_\RR M = 4$, then $\SS_M^\pm \in \Vect_\CC^{\rk = 2}$, i.e. they are complex vector bundles of rank 2.
Consider the eigenspaces $-e_1e_2 e_3 e_4 \actson \SS$, then $e_i\cdot(\wait): \SS^{\pm} \to \SS^{\mp}$.
:::


:::{.remark}
Principal bundle: fibers are left $G\dash$torsors.
In the fiber product, the group sits in the middle and acts on each factor. 
So $\tilde P$ eats the right $G\dash$action, and $\SS$ eats the left action.
Remarkably, for Spin bundles, there is an action leftover.
:::


:::{.proposition title="The spin bundle is a Clifford module"}
The spin bundle $\SS_M$ naturally has the structure of a $\Cl(M)\dash$module.
:::


:::{.proof title="?"}
We have a Clifford action
\[
\Cl(\RR^n) \tensor \SS &\to \SS \\
x\tensor s &\mapsto x\cdot s
.\]

Recall that we have a natural conjugation action $\Spin(n) \actson \Cl(\RR^n)$ where $g \mapsto g(\wait)g\inv$, and similarly $\Spin(n) \actson \SS$ by $g\mapsto g\cdot(\wait)$.
Given any $V\to W$ of $G\dash$modules, any $P\in \Prinbun(G)$ yields an induced module 
\[
P \fiberprod{G} V \to P \fiberprod{G} W
,\]
and moreover $\tilde P \fiberprod{\Spin(n)} \Cl(\RR^n) = \Cl(M)$.
We then conclude that there is an action $\Cl(M) \tensor \SS_M \to \SS_M$, the **Clifford multiplication**.
:::


:::{.remark}
We have an isomorphism of bundles (not of algebras) $\Cl(M) \cong \Extalg T\dual M$, and any one form $\omega$ is an analogue of an element of $V^{\tensor 1}$, and $\omega \cdot (\SS^+, \SS^-) \in \SS_M^- \oplus \SS_M^+$.
:::


:::{.definition title="Clifford connection"}
A connection $\nabla$ on $\SS$ is a **Clifford connection** if 
\[
\nabla(x\cdot s) = x \cdot \nabla (s) + d(x) \cdot s 
&& x\in H^0 \Cl(M) = H^0\qty{ \Extalg^* T\dual M} ,\,\, s\in H^0 (\SS_M)
,\]
where $d$ is the de Rham differential.
:::


:::{.remark}
It is not obvious that a Clifford connection exists!
We have $\SS_M = \tilde P \fiberprod{\Spin(n)} \SS$, so it suffices to give a connection on $\tilde P$ which is $\Spin(n)$ invariant, since any associated bundle will inherit the connection.
Idea: we need a notion of parallel transport.
This is a principal $\Spin(n)$ bundle, so the fibers look like $\Spin(n)$, and we want to lift paths in $M$ to paths in $\tilde P$:

\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}{2021-05-03_14-16.pdf_tex} };
\end{tikzpicture}

It suffices to give a connection on $P$, and using that $\tilde P\to P$ is a 2 to 1 covering map, we can take a connecting on $P$ coming from $\OFrame(T\dual M, g\dual)$.
So it further suffices to produce a connection on $T\dual M$ preserving orthogonality of frames under parallel transport, which is essentially the definition of the Levi-Cevita connection $\nabla^\LC$.
Then the $\nabla$ associated to $\nabla^\LC$ on $P$ is a Clifford connection, yielding existence.
:::


:::{.remark}
The set of Clifford connections is a torsor over $\Omega^1(M)$.
The association is $\nabla \mapsto \nabla - \nabla^\LC$, and one can compute
\[
(\nabla - \nabla^\LC )(x\cdot s) = x \cdot( \nabla - \nabla^\LC )(s)
,\]
which exactly says that this is a $\Cl(M)\dash$linear map $\SS_M \to \SS_M \tensor \Omega^1$.
We can write $\Cl(M) \cong \Endo(\SS_M)$, and one can check that \( [\Endo \SS_M, \Endo \SS_M] \) consists only of scalars.
:::


:::{.definition title="Dirac Operator"}
Let \( \nabla \) be a Clifford connection on $\SS_M$ and $s\in H^0(\SS_M)$, so \( \nabla(s) \in \SS_M \tensor \Omega^1(M) \).
Then the **Dirac operator** is defined as 
\[
\dirac: H^0(\SS) &\to H^0(\SS) \\
s &\mapsto \sum _{e_i \in \Fr(T\dual M) } e_i \cdot \nabla_{e_i\dual}(s)
\]
where

- \( \nabla(s) = H^0(\SS_M \tensor \Omega^1) \)
- \( \nabla_{e_i\dual}(s) = \nabla(s)(e_i\dual) \in H^0(\SS_M) \) 

:::


:::{.remark}
This makes sense locally, and is well-defined independent of choice of frame.
Henceforth, we'll take $\nabla = \nabla^\LC$ -- in this case, if $s^+ \in H^0(\SS^\pm)$ then \( \nabla_v^{\LC}(s^\pm) \in H^0(\SS^\pm) \).
This is an order 1 differential operator:
\[
\dirac_{ \nabla^\LC} = \dirac: H^0(\SS^\pm) \to H^p(\SS^{\mp})
.\]

:::



:::{.proposition title="Relation between Dirac operator and Laplacian"}
\[
\dirac^2 = - \laplacian
.\]
:::


:::{.proof title="?"}
Given \( \psi \in H^0(\SS) \), write $\psi = \sum_{i=1}^4 \psi_i s_i$ with the $s_i$ forming a local frame of $\SS = \SS^+ \oplus \SS^-$.
We can write
\[
\dirac \psi = \sum e_i \partial_{x_i} \psi = \sum_{i=1}^4 \gamma_i \psi_{x_i}
.\]
where $\psi_{x_i} = \tv{ (\psi_1)_{x_i}, (\psi_2)_{x_i}, \cdots }$.
We then have
\[
\dirac^2 \psi
&= \sum_{i, j} \gamma_i \gamma_j \psi_{x_i x_j} \\
&= -\sum_{ij} 2 (e_i \cdot_g e_j) \psi_{x_i x_j} \\
&= -2 \sum_{ij} \delta_{ij} \psi_{x_i x_j} \\
&= -2 \sum_i \psi_{x_i x_i} \\
&= -2 \qty{ \sum_{i=1}^4 \partial^2_{x_i} }\psi \\
&= -2 \laplacian
.\]
where we sum over *all* $i, j$ and can pair terms,
and we use that \( \gamma_i \gamma_j + \gamma_j \gamma_i = -2 e_1 \cdot e_j \) 
\

Upshot: $\dirac \in \sqrt{ \laplacian}$, which is why the Dirac is an invariant in quantum mechanics.
This reduces the 2nd order Schrödinger operator a 1st order operator.
Note that $\dirac \psi = 0$ is the equation for a massless particle.
:::

> See maybe Lawson's spin geometry?
  Or Salamon.