# Friday, April 30

:::{.remark}
Last time: we defined 
\[
\Cl(V, \cdot) &\da \bigoplus_n V^{\tensor n} / \gens{ v\tensor v = -\norm{v}^2 1 } \\
\Pin(V) &\da \gens{ v \st \norm{v} = 1 } \subseteq \Cl(V)
.\]
There is a $\ZZ/2$ grading $\Cl(V) = \Cl_0(V) \oplus \Cl_1(V)$ where $\Cl_0(V)$ is the image of even tensors and $\Cl_1(V)$ is the image of odd tensors.
We also had 
\[
\Spin(V) \da \Pin(V) \intersect \Cl_0(V) = \gens{ v\cdot w \st v,w \in V, \, \norm{v} = \norm{w} = 1 }
.\]
There was a map
\[
\Pin(V) &\to O(V) \\
v &\mapsto -R_v
,\]
where $R_v$ was reflection about $v^\perp$, where we identified this as an action on $V^{\tensor 1} \subset \Cl(V)$ where $u\to vuv\inv$.
For any Riemannian manifold $(X, g)$, we could define the Clifford bundle $\Cl(X) = \Cl(T\dual X, g\dual)$ to globalise this from vector spaces to bundles with metrics.
We defined a spin structure on $X$ as any lift of the principal $\SO(n)$ bundle over $(T\dual X, g)$ (namely $\Frame(X)$) to a $\Spin(n)$ bundle.
:::

:::{.warnings}
Each fiber is a metric space, so what happens if you just try to define \
\[
Y \da \Disjoint_{x\in X} \gens{ v \st \norm{v}^2 = 1,\, v\in T_x\dual X }
\,\,?\]
This seems to be isomorphic to a spin structure, but we do not have a distinguished action of any *fixed* group $\Spin(n)$.
We would have to choose isomorphisms to the standard spin group at each fiber, but the isomorphisms are not unique -- there is ambiguity up to the entire spin group.
So this does not define a spin structure.
:::

:::{.remark}
We showed that there exists a spin structure iff some cohomology class $w_2(K) \in H^2(X; \ZZ/2)$ vanishes.
:::

:::{.theorem title="Classification of complex representations of Clifford algebras"}
If $\dim_k V$ is even, there is a unique finite-dimensional complex irreducible $\Cl(V)$ representation of dimension $2^{n/2}$.
If $\dim_k V$ is odd, there are two complex conjugate representations of dimension $2^{\floor{n/2}}$.
:::

:::{.example title="?"}
Consider $\Cl(\RR^2) \cong \HH$.
There is an irreducible complex representation of dimension 2:
\[
1 &\mapsto \matt 1 0 0 1 \\
i &\mapsto \sigma_1 \da \matt 0 i i 0 \\
j &\mapsto \sigma_2 \da \matt i 0 0 {-i} \\
k &\mapsto \sigma_3 \da \matt 0 1 {-1} 0
.\]
:::


:::{.definition title="Pauli matrices"}
The \( \sigma_i \) defined above are referred to as the **Pauli matrices**.
:::


:::{.example title="?"}
Consider $\Cl(\RR^4)$.
By the theorem, there is a unique complex representation of $2^{4/2} = 2^2 = 4$, although the 4 here matching the dimension of $\RR^4$ is coincidental.
We'd like to find an isomorphism 
\[
\Cl(\RR^4) \mapsvia{ \sim} \Endo( (\CC^2)^{\tensor 2} ) \cong \Endo(\CC^4) = \Mat(4\times 4; \CC)
.\]
Note that $\Cl(\RR^4) \mapsvia{\sim} \Endo( (\CC^2)^{\tensor 3})$, which is why the dimensions multiply.
We can write 
\[
\Cl(\RR^4) = { \RR \gens{ e_1, e_2, e_3 ,e_4 } \over e_i e_j + e_j e_i = 2\delta_{ij} }
.\]
So define a map
\[
e_1 &\mapsto \gamma_1 \da 1\tensor \sigma_1 \\
e_2 &\mapsto \gamma_2 \da 1\tensor \sigma_2 \\
e_3 &\mapsto \gamma_3 \da \sigma_1 \tensor i \sigma_3 \\
e_4 &\mapsto \gamma_4 \da \sigma_2 \tensor i \sigma_3 
.\]


:::{.definition title="Dirac matrices"}
The matrices appearing above are called the **Dirac matrices**.
:::


:::{.exercise title="?"}
Determine a similar map for $\Cl(\RR^6)$ continuing this pattern.
:::

We can check that this is a representation.
Note that we can tensor matrices in a simple way:

\begin{tikzcd}
	& {e_1} & {e_2} && {} & {f_1} & {f_2} \\
	{e_1} & a & b && {f_1} & e & f \\
	{e_2} & c & d && {f_2} & g & h
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMTcsWzAsMSwiZV8xIl0sWzAsMiwiZV8yIl0sWzEsMCwiZV8xIl0sWzIsMCwiZV8yIl0sWzEsMSwiYSJdLFsyLDEsImIiXSxbMSwyLCJjIl0sWzIsMiwiZCJdLFs0LDBdLFs1LDAsImZfMSJdLFs2LDAsImZfMiJdLFs0LDEsImZfMSJdLFs0LDIsImZfMiJdLFs1LDEsImUiXSxbNiwxLCJmIl0sWzUsMiwiZyJdLFs2LDIsImgiXV0=)

Checking $e_2 \cdot e_2 = -1$, we have 

\[
(1 \tensor \sigma_2) \cdot (1 \tensor \sigma_2) &= ? \\
1_2 \tensor \sigma_2^2 &= -I_2 \oplus I_2 \\
\gamma_2 \gamma_3 &= - \gamma_3 \gamma_2
.\]

> Todo: messed up!

One can similarly check
\[
(1 \tensor \sigma_2 ) \cdot (\sigma_1 \tensor i \sigma_3) = - ( \sigma_1 \tensor i \sigma_2) (1\tensor \sigma_2)
.\]
:::

:::{.remark}
We thus have $\Cl(\RR^4) \actson \CC^4$ by sending $e_i \mapsto \delta_i$, the Dirac matrices. Using that $\Pin(4) \intersect \Cl(\RR^4) = \Spin(4) \subseteq \Cl(\RR^4)$, we can a spin representation, but this may no longer be irreducible.
In fact, as a $\Spin(4)$ representation this splits into two irreducible representations.
We know that $\Spin(4) \subseteq \Cl_0(\RR^4) = \Cl(\RR^3)$ which has two complex conjugate irreducible representations.
The key is to define an element $\omega_\CC \in \Cl(V) \tensor_\RR \CC$ with $\omega_\CC^2 = 1$, which yields a decomposition of $\SS = \SS^+ \oplus \SS^-$ as the $\pm 1$ eigenspaces of the action.
Here \( \omega_C \da -e_1 e_2 e_3 e_4 \mapsto \gamma_5 \).
One can define 
\[
\gamma_5 \da \im(\omega_\CC) = - \gamma_1 \gamma_2 \gamma_3 \gamma_4 = - \sigma_3 \tensor \sigma_3
\]
and one obtains the matrix
\[
-\sigma_3 \tensor \sigma_3 = 
\begin{bmatrix}
0 & 0 & 0 & -1 
\\
0 &  0& 1 & 0
\\
0 &  1& 0 & 0 
\\
-1 & 0 & 0 & 0
\end{bmatrix}
.\]
One can check that \( \gamma_5 \) anticommutes with the \(\delta_i \) for $1\leq i \leq 4$, and thus commutes with $\Cl_0(\RR^4)$.
We can write $\SS^+$, the positive 1 eigenspace of \( \gamma_5 \), as \( \CC(s_1 - s_4 ) \oplus \CC(s_1 + s_2) \).
So we have $\Spin(4) = \Cl(\RR^4) \actson \CC^2 \oplus \CC^2 = \SS$, which splits into $\gamma_5\dash$eigenspaces $\SS^+ \oplus \SS^-$, the **positive and negative spinors**.
This means that \( \gamma_5 \) commutes with the image of $\Spin(4) \injects \GL( \CC^2 \oplus \CC^2)$.
:::

:::{.fact}
If the action commutes with everything in the representation, the representation splits. (??? missed)
:::

:::{.remark}
Let $g\in \Spin(4)$, and $v^+ \in \SS^+ \subseteq \SS$.
Question: is it true that $g\cdot v^+ \in \SS^+$?
If so, this yields a subrepresentation.
If so, \( \gamma_5 v^+ = v^+ \) since we're in the $+1$ eigenspace, and on the other hand, $g\cdot v^+ = g \cdot \gamma_5 v^+ = g \omega_\CC \cdot v^+$ where the last identification comes from the map \( \gamma_5 \mapsto \omega_\CC \), and this is equal to \( \omega_\CC g \cdot v^+ \) using commutativity.
So $g\cdot v^+$ is in the $+1$ eigenspace of \( \gamma_5 \).
:::

:::{.remark}
Now take \( \gamma_i \).
This actually switches spinors: by anticommutativity of the \( \gamma_i \) with \( \gamma_5 \), we have
\[
\gamma_i \cdot v^+ = \gamma_i \gamma_5 v^+ = - \gamma_5 \gamma_i v^+
,\]
which means \( \gamma_i v^+ \) is in the $-1$ eigenspace for \( \gamma_5 \), i.e. \( \gamma_i v^+ \in \SS^- \).
:::

:::{.remark}
Suppose one has a spin structure and $\tilde P \to X$ is a principal $\Spin(n)$ bundle.
There are bundles over this of the form $\rho: \Spin(n) \to \GL(\SS^\pm)$, yielding the **spinor bundle** 
\[
\tilde P \fiberprod{\Spin(n)} \SS = \SS_x^+ \oplus \SS_x^-
.\]
:::

:::{.remark}
Let $G \mapsvia{\rho} \GL(V)$ be any representation.
If $\phi \in \Endo(V)$ commutes with $\rho(G)$, then the eigenspaces of $\phi$ are subrepresentations.
In other words, $G\actson V = \bigoplus _{i=1}^n V_{\lambda_i}$, then $G\actson V_{\lambda_i}$ is a subrepresentation, using that
\[
\phi(v) = \lambda v \implies gv = g\phi(\lambda \inv v) = \phi \rho(g) \lambda\inv v 
,\]
which says $\phi( \rho(g) \cdot v) = \lambda (\rho(g) \cdot v) \implies \rho(g) \cdot v \in V_{\lambda}$.
We used it here by
This rephrases Schur's lemma!
:::