# Wednesday, April 28

:::{.remark}
Last time: we defined $\Pin(n) \subseteq \Cl(\RR^n)$ which was generated by $S^1(\RR^n)$.
These were units because $v^2 = -\norm{v}^2 = -1$, so $v\inv = -v$, and formed a group contained in $\Cl(\RR^n)\units$.
There is a decomposition $\Cl(V) = \Cl_0(V) \oplus \Cl_1(V)$ with a $\ZZ/2\dash$grading, and we defined 
\[
\Spin(V) \da \Pin(V) \intersect \Cl_0(V) = \gens{ vw \st v,w\in S^1(\RR^n) } 
.\]
There is a map
\[
\Pin(n) &\surjects O(n) \\
v & \mapsto (u\mapsto vuv\inv) = -R_{v^\perp}
,\]
which preserves $V^{\tensor 1} \subset \Cl(V)$, and was reflection about the hyperplane $v^\perp$.
There is also a SES
\[
0 \to \ZZ/2 \to \Spin(n) \mapsvia{\pi}  \SO(n) \to 0
,\]
where we used the fact that $\ker \pi \subset Z\Cl(\RR^n)$.
It turns out that $\Spin(n) = \univcover{ \SO(n) }$, using that $\pi_1( \SO(n), \pt) = \ZZ/2$ and checking that $\pm 1\in \Spin(n)$, yielding a nontrivial kernel.
:::

:::{.remark}
This is local, at a single vector space, so we'll now try to globalise this to the tangent space of a manifold.
:::

:::{.definition title="Clifford Bundle"}
Let $(V, g)$ be an oriented smooth Riemannian manifold where $g$ is a metric on $TX$.
Define the **Clifford bundle** of $X$ by 
\[
\Cl(X) \da \Cl(T\dual X, g\dual)
,\]
where we've used the dual metric $g\dual$ on the cotangent bundle.
:::

:::{.remark}
We showed that $\gr \Cl(\RR^n) = \Extalg \RR^n$, and so there is a bundle isomorphism 
\[
\Cl(X) \mapsvia{\sim} \Extalg^* T\dual X
,\]
but the ring structure is different.
On the right, we have a way of multiplying sections, namely \( \omega_1 \wedge \omega_2 \), but on the left we have the Clifford multiplication \( \alpha_1 \cdot \alpha_2 \).
Note that \( \omega^{\wedge 2} = 0 \), but \( \alpha^{\cdot 2} \in \RR \) is some scalar.
We define \( \omega\cdot \omega= g^*( \omega, \omega) \), so we use the metric fiberwise to define a Clifford multiplication.
:::

:::{.definition title="The principal oriented frame bundle"}
Given an oriented bundle with a metric, there is a principal $\SO(n)$ bundle $P \da \OFrame$, the space of orthogonal oriented frames.
:::

:::{.remark}
This is principal since any two elements are related by a unique element of $\SO(n)$.
Recall that we had an *associated bundle* construction, so taking the standard representation $\rho: \SO(n) \to (\RR^n, g)$ where elements act by their transformations (?), there is an oriented bundle $P \fiberprod{\rho} \RR^n$.
If the bundle is $TX$ with a metric $g$, this yields a distinguished $\SO(n)$ bundle $P\to X$.
:::

:::{.definition title="Spin Structures"}
A **spin structure** is a lift $\tilde P$ of $P$ to a principal $\Spin(n)$ bundle.
:::

:::{.proposition title="Spin iff nontrivial $w_2$"}
$X$ admits a spin structure iff the second Stiefel--Whitney class $w_2(X) = 0$ in $H^2(X; \ZZ/2)$.
If $w_2(X) = 0$, then the spin structures are torsors over $H^1(X; \ZZ/2)$.
:::

:::{.remark}
Recall that a $G\dash$torsor is a set with a free transitive $G\dash$action.
For example, the fibers of a principal bundle are torsors.
Given any two torsors, we can compare them using elements of $G$, but there is no distinguished element.
For example, $\AA_n$ is a torsor over the vector space $k^n$.
:::

:::{.proof title="?"}
Consider transitions for $P\to X$:
\[
t_{ij}: U_i \intersect U_j \to \SO(n) \\
,\]
where $t_{ij} = t\inv_{ji}$ and the cocycle condition $t_{ij} t_{jk} t_{ki} = 1$ is satisfied.
We want a lift:

\begin{tikzcd}
	&& {\ZZ/2 \in Z\Spin(n)} \\
	\\
	{U_i \intersect U_j} && {\Spin(n)} \\
	\\
	{U_i \intersect U_j} && {\SO(n)}
	\arrow["\one", from=3-1, to=5-1]
	\arrow["{t_{ij}}", from=5-1, to=5-3]
	\arrow["{\tilde t_{ij}}", from=3-1, to=3-3]
	\arrow["{2:1}", from=3-3, to=5-3]
	\arrow[hook, from=1-3, to=3-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMCwyLCJVX2kgXFxpbnRlcnNlY3QgVV9qIl0sWzIsMiwiXFxTcGluKG4pIl0sWzIsNCwiXFxTTyhuKSJdLFsyLDAsIlxcWlovMiBcXGluIFpcXFNwaW4obikiXSxbMCw0LCJVX2kgXFxpbnRlcnNlY3QgVV9qIl0sWzAsNCwiXFxvbmUiXSxbNCwyLCJ0X3tpan0iXSxbMCwxLCJcXHRpbGRlIHRfe2lqfSJdLFsxLDIsIjI6MSJdLFszLDEsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV1d)

We can always lift to *some* $\tilde t_{ij}$ using the path-lifting property of covers if $U_i \intersect U_j$ is contractible, using that $\ZZ/2$ is discrete.
We can arrange $\tilde t_{ij} = \tilde t_{ji}\inv$ since $U_i \intersect U_j = U_j \intersect U_i$, but we may not have the cocycle condition on the lift.
We have $t_{ij} t_{jk} t_{ki} = 1$, so 
\[
\tilde t_{ij} \tilde t_{jk} \tilde t_{ki} \in \ker (\Spin(n) \to \SO(n)) = \ts{ \pm 1 }
,\]
using that everything in sight needs to be a group morphism.
So define
\[
\tilde t_{ijk} \da ( \tilde t_{ij} \tilde t_{jk} \tilde t_{ki} )_{i,j,k} \in \Cc^2_{\mathcal{U}}(X, \constantsheaf{\ZZ/2})
.\]
The claim is that $\bd^2( \tilde t_{ijk}) = 0$, but it turns out that regardless of choice of lift we obtain
\[
\bd^2(\tilde t_{ijk}) = \tilde t_{ijk} \tilde t_{ikl}\inv \tilde t_{ijl} \tilde t_{ijk}\inv = 0 \implies [ \tilde t_{ijk} ] \in \Hc^2(X, \constantsheaf{\ZZ/2})
.\]

Is this class well-defined?
Consider replacing $\tilde t_{ij}$ with $-\tilde t_{ij}$.
In general, we have
\[
i, j\in \ts{ a,b,c } \implies \tilde t_{abc} \mapsto -\tilde t_{abc}
,\]
and so this is a Čech coboundary in $\bd^1(1, \cdots, 1, -1, 1, \cdots, 1)$ where the $-1$ occurs in the $t_{ij}$ coordinate.
Thus $\tilde t_{ijk}$ is well-defined moduli $\bd^1 C_{\mathcal{U}}^1(X, \constantsheaf{\ZZ/2})$.
\

Note that $w_2(X)$ was produced from the pair $(X, g)$, but the space of metrics is connected and thus $w_2(X)$ depends only on $X$.
Suppose $w_2(X) = 0$, then $[ \tilde t_{ijk} ] = 0$ which implies that there is some $(s_{ij})$ with $\bd^1 (s_{ij}) = (\tilde t_{ijk})$.
So replace each $\tilde t_{ij}$ with $\tilde{ \tilde t}_{ij} \da s_{ij} \tilde t_{ij}$ is a new lift which satisfies the cocycle condition.
Thus they define the transition functions of a principal $\Spin(n)$ bundle lifting $P \to X$.

\
To see the claim about torsors, given any $\ell_{ij} \in \ker \bd^1$, note that any $\tilde{ \tilde t}_{ij} \ell_{ij}$ also satisfies the cocycle condition.
There is a map
\[
\ts{ \text{Spin structures} } &\from \ker \bd^1 \\
\tilde{\tilde t}_{ij} \ell_{ij} &\mapsfrom \ell_{ij}
,\]
which is a torsor because we needed to start with a given lift $\tilde{\tilde t}_{ij}$.
Then $\tilde P_1 \cong \tilde P_2$ iff there exists an $(m_i) \in \Cc^0_{\mathcal{U}}(X, \constantsheaf{\ZZ/2})$ such that $(\ell_{ij})_1 = (\ell_{ij})_2 + \bd^0 (m_i)$, which are different trivializations of the same bundle. 
:::

:::{.remark}
This is a nice example to get a hang of the use and importance of Čech cohomology.
We then use the isomorphism $\Hc \to H_{\sing}$.
:::

:::{.theorem title="Existence of spin representation of Clifford algebras in even dimension"}
Assume $n \da \dim V$ is even, then $\Cl(V)$ has a unique nontrivial irreducible finite dimensional complex representation $S$ of dimension $\dim S = 2^{n/2}$, the **spin representation**.
:::

:::{.remark}
It turns out that $\Cl(V) \tensor_\RR \CC \cong \Endo(S)$.
The left-hand side contains $\Spin(n)$, so given $\rho: \Cl(V) \to \Endo(S)$ a representation (i.e. a ring homomorphism) in matrices, we can restrict $\rho$ to $\Spin(n)$ to get \( \rho \ro{}{\Spin(n)}: \Spin(n) \to \GL(S) \).
Next time: spin representations.
Spinor bundle will be sections of associated bundle of the Clifford bundle.
:::