# Wednesday, May 05

## Fun Physics Aside

:::{.remark}
Last time: we showed $\Cl(X) \da \Cl(\dualof{T} X, \dualof{g})$ acts on the spinor bundle $\SS_X \da \tilde P \fiberprod{\Spin(n)} \SS$ by Clifford multiplication.
For $\dim_\RR X = 4$, we have a splitting $\SS^+ \oplus \SS^-$ as complex rank 2 vector bundles.
If \( \omega\in H^0 \Cl(X) \)  is a one form, then \( \omega\SS_X^\pm \subset \SS^{\mp}\) .
:::

:::{.definition title="Clifford Connection"}
A **Clifford connection** is a map
\[
\nabla: \SS_X \to \SS_X \tensor \Omega^1 \\
.\]
where \( \alpha \cdot s \mapsto \alpha\cdot \nabla s + dx \cdot s \).
:::


:::{.remark}
There is a distinguished Clifford connection associated to \( \nabla^\LC \).
Also recall that we defined a Dirac operator $\dirac$ and showed $\dirac^2 = -2 \laplacian$.
:::


:::{.definition title="The Dirac Equation"}
The **Dirac equation** is defined on \( \psi\in H^0(X, \SS) \) as
\[
(i \dirac + m \omega)\psi = 0
.\]

Here $m$ denotes a mass, \( \omega = \omega_\CC = \prod_{i=1}^4 \gamma_i \).
:::


:::{.remark}
This describes fermions in a vacuum, e.g. an electron where \( \psi \) is its wave function.
Applying this to $\RR^4$ with $g = (dt)^2 - (dx)^2 - (dy)^2 - (dz)^2$, then this equation in $\psi$ is invariant under the Lorentz group $O(\RR^4, g)$.
:::



## Rohklin's Theorem


:::{.theorem title="Rohklin's Theorem"}
Let $X$ be a smooth closed oriented spin 4-manifold.
Then the signature $\sigma(X) \da b_2^+(X) - b_2^-(X)$ (the dimensions of positive/negative definite subspaces of $H^2(X; \RR)$ is divisible by 16.
:::


:::{.remark}
This restricts what topological manifolds can admit smooth structures.
Freedman constructed a topological manifold of dimension 4 with signature 8, which thus can not admit a smooth structure.
Recall that having a spin structure was having a lift
of a principal $\SO(n)$ bundle over $(\dualof{T} X, g)$ (namely $\Frame(X)$) to a $\Spin(n)$ bundle.

\begin{tikzcd}
	{\tilde P_{\SO(V)} \da P_{\Spin(V)}} \\
	\\
	{P_{\SO(V)}} & {} \\
	\\
	& X
	\arrow["\pi"', from=3-1, to=5-2]
	\arrow["{\tilde p}"', from=1-1, to=3-1]
	\arrow["\exists", dashed, from=1-1, to=5-2]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMSw0LCJYIl0sWzEsMl0sWzAsMiwiUF97XFxTTyhWKX0iXSxbMCwwLCJcXHRpbGRlIFBfe1xcU08oVil9IFxcZGEgUF97XFxTcGluKFYpfSJdLFsyLDAsIlxccGkiLDJdLFszLDIsIlxcdGlsZGUgcCIsMl0sWzMsMCwiXFxleGlzdHMiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=)

\todo[inline]{Diagram doesn't match definition, check Phil's notes!}

:::

### Proof
Consider $\SS_X \da \tilde P \fiberprod{\Spin(n)} \SS$, then define
\[
\dirac^\pm: H^0(\SS_X^\pm) \to H^0(\SS^{\mp})
.\]
Note that we can write $\dirac = \dirac^+ + \dirac^-$; 

- Step 1: Show $\ind\dirac^+ = -\sigma(X) / 8$,
- Step 2: Show $\ind \dirac^+$ is even.

### Step 1

What is the symbol $\Symb(\dirac)$?
By definition
\[
\Symb \dirac: \pi^* \SS \to \pi^* \SS
.\]
where $\pi:\dualof{T} X\to X$, and the symbol was defined by replacing $\dd{}{x_i}$ with a function $y_i: \dualof{T} X\to \RR$.
We can write
\[
\dirac \phi = \sum_{e_i \in \Fr} e_i \cdot \nabla_{e_i\dual} \psi
,\]
and so 
\[
\Symb \dirac(\psi) = \sum_i y_i e_i = \psi
.\]

We have a tautological form \( \alpha\in H^0(\dualof{T} X, \pi^* \Omega^1) \) where $(p, \alpha) \mapsto \alpha$, and 
so \( \Symb(\dirac)(\wait) = \alpha\cdot(\wait) \).


:::{.claim}
\[
\dirac: H^0(\SS) \selfmap && \text{is an elliptic operator}
.\]
:::

We need to check that the map \( \alpha\cdot(\wait) \) is exact if \( \alpha\neq 0 \).

We have \( \alpha\cdot(\wait): \SS\to \SS \)  and
\[
  (- \alpha)(\wait) \, \alpha(\wait) = (- \alpha \cdot \alpha) = \norm{ \alpha}^2 \neq 0
,\]
which makes the operator invertible away from zero.
Thus we can apply Atiyah-Singer.


:::{.lemma title="Formula for Chern characters"}
There is a nice formula for Chern characters:
\[
\ch \SS^+ - \ch \SS^- = \prod_{i=1}^n( e ^{x_i/2} - e^{-x_i/2})
.\]
where \( \ts{ \pm x_i } \) are the Chern roots of $\dualof{T} X$.
:::


:::{.proof title="?"}
Use the splitting principle to write
\[
\dualof{T} X \tensor_\RR \CC = \bigoplus _{i=1}^n L_i \tensor L_i\inv
.\]
Then $\SS^+$  is a sum of all tensor products of $L_i \tensor L_i\inv$ where the number of $-1$s appearing is even.
:::


:::{.remark}
Note there is ambiguity up to 2-torsion in the formula, but this gets moved into the choice of spin structure, which amounts to choice of a square root of each of these line bundles.
:::


Setting $2n\da \dim X$, we have
\[
\ind\dirac^+ 
&= (-1)^n \int_X { \ch\SS^+ - \ch \SS^- \over \eul X} \td(TX\tensor \CC) \\ \\
&= \int_X { \prod e^{x_i/2} - e^{-x_i /2} \over (-1)^n \prod x_i} \prod {x_i \over 1 - e^{x_i} } \prod {x_i \over 1 - e^{-x_i}} \\ \\
&= \int_X \prod { (e^{x_i/2} - e^{-x_i / 2 )} x_i \over (1-e^{x_i} ) (1 - e^{-x_i}) } \\ \\
&= (-1)^n \int_X \prod_I {x_i \over e^{x_i/2} - e^{-x_i/2}} \\ \\
&= \int_X \qty{ 1 - {x_1^2 \over 24} } \qty{ 1 - {x_2^2 \over 24}} \\ \\
&= -{1\over 24} \int_X x_1^2 + x_2^2 + (x_1 + x_2)^2 - 2x_1 x_2 \\ \\
&= -{1\over 24} \qty{c_1^2 - 2c_2}
.\]


:::{.remark}
See the $\hat{A}$ genus.
:::


:::{.claim}
\[
c_1^2 - 2c^2 = 3\cdot \sigma(X)
.\]
:::
This is another application of Atiyah-Singer, applied to a slightly different operator.
Recall the Hodge star operator,
\[
\hodgestar: \Omega^k(X) \to \Omega^{4-k}(X)
.\]

Defining $\tau \da i^{k(k-1) + 4 \over 2}$, we get $\tau^2 = 1$, so define an operator $\tau \hodgestar$.
This yields a splitting into $\pm 1$ eigenspaces:
\[
\Omega(X) = \Omega^+(X) \oplus \Omega^-(X)
.\]
Recalling that $d^\dagger$ was the adjoint of $d$, one can check that $d+d^\dagger: \Omega^{\pm}(X) \to \Omega^{\mp}(X)$ interchanges these.
It turns out that $\ind(d + d^\dagger) = \sigma(X)$, which by Atiyah-Singer and Hermite forms will equal $c_1^2 -2 c_2 \over 3$.
This yields the desired formula for step 1.


## Step 2

We now want to show $\ind \dirac^+$ is divisible by 2.
The key point is that $\ker \dirac^+$ and $\coker \dirac^+ = \ker \dirac^-$ admit a quaternionic vector space structure.
This comes from the fact that 
\[
\Spin(4) \cong \SU(2) \cross \SU(2) \cong S^1(\HH) \oplus S^1(\HH) \da \SS^+ \oplus \SS^-
,\] 
so we have a splitting into subspaces of unit quaternions.
It turns out that $\dirac$ is $\HH\dash$linear.
So we get an equality
\[
-\sigma(X) / 8 = \ind \dirac^+ = 2\lambda
\]
for some $\lambda$, yielding $8\divides \sigma(X)$.

## Remarks

:::{.remark}
If $H_1(X; \ZZ)$ has no 2-torsion, e.g. if $\pi_1X = 0$, then $w_2(X) = 0$ iff the intersection form on $H^2$ is even, where $w_2$ is the obstruction to existence of spin structures.
Note that this makes sense for topological manifolds and not just smooth manifolds, and in this case $\sigma(X)$ is divisible by 8.
This restriction comes from number theory: since we have a unimodular lattice, it breaks into sums of $E_8, -E_8$, and $H$ if indefinite, and any even unimodular lattice has signature divisible by 8.
So this can work as an obstruction to the existence of smooth structures.
:::


:::{.remark}
Note that $\CP^2$ has no spin structure, and $\sigma(\CP^2) = 1$.
There's a way to modify the invariant to set $\sigma(X)/8 = ? \mod 2$.
:::