# Tuesday, October 12

## More Euler Class

> Review how to construct the Euler and Thom classes.

:::{.remark}
Recall that the Euler class is the obstruction to finding a nowhere vanishing section on the $n\dash$skeleton.
Given $E\to X \in \VectBundle(X)^{\dim = n}$, we can form the sphere bundle $S^{n-1}\to \SS E \to X$.
Define a section $s: X\to \SS E$ over $X\skel{0}$, then inductively if $s$ is defined over $X\skel{i-1}$ for $i<n$, let $j: \Delta^i \to X\skel{i}$ be the attaching map for an $i\dash$cell.
Note that $\Delta^i$ is contractible, and we can form a pullback


\begin{tikzcd}
\Delta^i \cross S^{n-1} 
  \ar[r] 
  \ar[d]
& 
\SS E
  \ar[d] 
\\
\Delta^i
  \ar[r, "j"]
& 
X\skel{i} 
\end{tikzcd}

If $s$ is defined over $X\skel{i-1} \subseteq X\skel{i}$, we have $\ro{s}{\bd \Delta^i}: \bd \Delta^i \to S^{n-1}$.
But $\bd \Delta^{i} \cong S^{i-1}$, and if $i\leq n-1$, $\pi_i S^{n-1} = 0$ so $s$ extends over $\Delta^i$.
If $i=n$, note that $\pi_{n-1} S^{n-1} \cong \ZZ$, so we define $e_s(\Delta^n)$ to be the corresponding homotopy class of $s$, i.e. $e_s(\Delta^n)$ is the degree of this map between spheres.

We showed that

- $e_s$ doesn't depend on the trivialization chosen, and
- If $s'$ is another nowhere vanishing section on $X\skel{n-1}$, then $e_s - e_{s'} \in \im \delta$ for $\delta: C^{n-1}(X) \to C^n(X)$.
- If $\delta e_s = 0$, then $[e_s] \in H^n(X; \ZZ)$.
  The content here is that the cochain descends to a cocycle.

For $i:\Delta^{-1} \to X\skel{n-1}$ and $\SS E\to X\skel{n-1}$, form the pullback $i^* \SS E$.
Generally given two sections $s, s': \Delta^{n-1} \to i^* \SS E \cong \Delta^{n-1} \times S^{n-1}$, we can glue them to define a map $-\Delta^{n-1} \disjoint_{\bd} \Delta^{n-1} \mapsvia{-s \disjoint s'} S^{n-1}$.
The glued space is an $S^{n-1}$, so the degree of this map will be $\alpha^{\Delta^{n-1}}$
:::

:::{.lemma title="?"}
For any $\alpha\in C^{n-1}(X)$, there exists a section $s'$ on $X\skel{n-1}$ such that $e_{s'} - e_s = \delta \alpha$.
:::

:::{.exercise title="?"}
Prove this!
:::

:::{.proposition title="?"}
$[e_s] \in H^n(X)$ agrees with the Euler class.
:::

:::{.proof title="?"}
We can equivalently define $e_s$ as follows: extend $s$ to any section $\bar s$ of the disc bundle $\DD E \to X$.
Then for any $n\dash$cell $\Delta^n$, we can form the composition:

\begin{tikzcd}
	{(\Delta^n, \bd \Delta^n)} && {(\Delta^n \times \DD^n, (\bd \Delta^n)\times S^{n-1})} && {(\DD^n, S^{n-1})} \\
	\\
	{\bd \Delta^n} &&&& {S^{n-1}}
	\arrow["{\bar{s}}", from=1-1, to=1-3]
	\arrow["{p}", from=1-3, to=1-5]
	\arrow["\psi", curve={height=-30pt}, dashed, from=1-1, to=1-5]
	\arrow[hook, from=3-1, to=1-1]
	\arrow["{\psi'}"', from=3-1, to=3-5]
	\arrow[hook, from=3-5, to=1-5]
\end{tikzcd}

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

We can then define $e_s(\Delta^n) \da \deg \psi$, and we claim that this also equals $\deg \psi'$.
Look at the LES of a pair:

\begin{tikzcd}
	\textcolor{rgb,255:red,214;green,92;blue,92}{H_n(\DD^n)=0} &&& \textcolor{rgb,255:red,214;green,92;blue,92}{H_n(\DD^n) = 0} \\
	\\
	{H_n(\DD^n; S^{n-1})\cong \ZZ} &&& {H_n(\DD^n; S^{n-1})\cong \ZZ} \\
	\\
	{H_{n-1}(S^{n-1})\cong \ZZ} &&& {H_{n-1}(S^{n-1})\cong \ZZ} \\
	\\
	\textcolor{rgb,255:red,214;green,92;blue,92}{H_{n-1}(\DD^n) = 0} &&& \textcolor{rgb,255:red,214;green,92;blue,92}{H_{n-1}(\DD^n) = 0}
	\arrow["{\psi^*}", from=3-1, to=3-4]
	\arrow[from=1-1, to=3-1]
	\arrow[from=1-4, to=3-4]
	\arrow["\cong"', color={rgb,255:red,92;green,214;blue,92}, from=3-1, to=5-1]
	\arrow["\cong"', color={rgb,255:red,92;green,214;blue,92}, from=3-4, to=5-4]
	\arrow["{(\psi')^*}"{description}, from=5-1, to=5-4]
	\arrow[from=5-1, to=7-1]
	\arrow[from=5-4, to=7-4]
	\arrow[from=7-1, to=7-4]
	\arrow[from=1-1, to=1-4]
\end{tikzcd}

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

We may assume that $X$ is $n\dash$dimensional since for $(X, X\skel{n})$ we have
\[
H^n(X, X\skel{n}) = 0 \to H^n(X) \injects H^n(X^n)
,\]
so anything equal in $H^n(X\skel n)$ must be equal in $H^n(X)$.
Fix a positive generator $\gens{x} = H^n(\DD^n, S^{n-1})$ and $\gens{y} = H_n(\Delta^n, \bd \Delta^n)$ to be the fundamental class (positive generator).
Then 
\[
e_s(\Delta^n) = \ip{\bar{s}^* p^* x}{y}
.\]
Consider the map $(\Delta^n \times \DD^n, \bd \Delta^n \times S^{n-1}) \mapsvia{p} (\DD^n, S^{n-1})$, we have $p(\Delta^n\times S^{n-1}) = S^{n-1}$.
Then we claim that $p^*(x) = u$ will be the Thom class.
Using the attaching map $i: \Delta^n \to X\skel n$, we obtain 
\[
H^n(\DD E, \SS E) \mapsvia{i^*} H^n(\Delta^n \times \DD^n, \Delta^n \times S^{n-1})
.\]

Use that $\bar{s}: X\to \DD E$ induces an isomorphism $\bar{s}^*: H^n(\DD E)\to H^n(X)$, inducing the same isomorphism as the zero section.
So $(s')^* = \bar{s}^*$, i.e. any two sections of the disc bundle will be homotopic and thus induce equal maps in homology.
Now doing exactly what we did for the Euler class, we get a diagram:

\begin{tikzcd}
	&& {} & \textcolor{rgb,255:red,92;green,92;blue,214}{u} \\
	& {H^n(\DD E)} & {} & {H^n(\DD E, \SS E) } && {H^n(\Delta^n \times \DD^n, \Delta^n \times S^{n-1})} \\
	& {H^n(X)} \\
	\textcolor{rgb,255:red,92;green,92;blue,214}{e} &&& {H^n(X\skel n, X\skel{n-1})} && {H^n(\Delta^n, \bd \Delta^n)}
	\arrow["{i^*}", from=2-4, to=2-6]
	\arrow["{i_*}", from=4-4, to=4-6]
	\arrow["{\bar{s}^*}", from=2-6, to=4-6]
	\arrow["{\bar{s}^*}"', from=2-4, to=4-4]
	\arrow["LES"', from=2-4, to=2-2]
	\arrow["{\bar{s}^*}"', from=2-2, to=3-2]
	\arrow["LES"', from=4-4, to=3-2]
	\arrow[color={rgb,255:red,92;green,92;blue,214}, curve={height=18pt}, dashed, maps to, from=1-4, to=4-1]
\end{tikzcd}

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


More directly, we can map the LESs to get a commutative square:

\begin{tikzcd}
	{H^n(\DD E, \SS E)} && {H^n(X\skel n, X\skel{n-1})} \\
	\\
	{H^n(\DD E)} && {H^n(X\skel n)}
	\arrow["{\bar{s}^*}", from=1-1, to=1-3]
	\arrow["{\bar{s}^*}", from=3-1, to=3-3]
	\arrow[from=1-1, to=3-1]
	\arrow[from=1-3, to=3-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJIXm4oXFxERCBFLCBcXFNTIEUpIl0sWzAsMiwiSF5uKFxcREQgRSkiXSxbMiwwLCJIXm4oWFxcc2tlbCBuLCBYXFxza2Vse24tMX0pIl0sWzIsMiwiSF5uKFhcXHNrZWwgbikiXSxbMCwyLCJcXGJhcntzfV4qIl0sWzEsMywiXFxiYXJ7c31eKiJdLFswLDFdLFsyLDNdXQ==)


In other words, regard $e\in H^n_\cell(X)$, so $e$ corresponds to $[c]\in \Hom_{\mods{\ZZ}}(C^n(X), \ZZ)$.
We then claim that for $c\in H^n(X\skel n, X\skel{n-1})$, we have $e_s(\Delta^n) = i^* c(\Delta^n, \bd\Delta^n)$ using $i: (\Delta^n, \bd \Delta^n)\to (X\skel n, X\skel{n-1})$.
Final diagram:

\begin{tikzcd}
	&& {} & \textcolor{rgb,255:red,92;green,92;blue,214}{u} && \textcolor{rgb,255:red,92;green,92;blue,214}{p^*x = i^* u} \\
	& {H^n(\DD E)} & {} & {H^n(\DD E, \SS E) } && {H^n(\Delta^n \times \DD^n, \Delta^n \times S^{n-1})} \\
	\textcolor{rgb,255:red,92;green,92;blue,214}{e} & {H^n(X)} \\
	&&& {H^n(X\skel n, X\skel{n-1})} && {H^n(\Delta^n, \bd \Delta^n)} & \textcolor{rgb,255:red,92;green,92;blue,214}{\bar{s}^* p^* x = \bar{s}^* i^* u}
	\arrow["{i^*}", from=2-4, to=2-6]
	\arrow["{i_*}", from=4-4, to=4-6]
	\arrow["{\bar{s}^*}", from=2-6, to=4-6]
	\arrow["{\bar{s}^*}"', from=2-4, to=4-4]
	\arrow["LES"', from=2-4, to=2-2]
	\arrow["{\bar{s}^*}"', from=2-2, to=3-2]
	\arrow["LES"', two heads, from=4-4, to=3-2]
	\arrow[color={rgb,255:red,92;green,92;blue,214}, curve={height=30pt}, dashed, maps to, from=1-4, to=3-1]
	\arrow[color={rgb,255:red,92;green,92;blue,214}, dashed, maps to, from=1-4, to=1-6]
	\arrow[color={rgb,255:red,92;green,92;blue,214}, curve={height=-30pt}, dashed, maps to, from=1-6, to=4-7]
\end{tikzcd}

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

:::

## Computations on Smooth Manifolds

:::{.remark}
Recall that a smooth structure on a manifold $M$ is a collection $(U, \phi_U)$ where $U \subseteq M$ is open, $\phi_U: U\to \RR^n$ is a homeomorphism onto an open subset of $\RR^n$, and for all $U, V$ we have 
\[
\psi_{VU} \da \phi_V\inv \circ \phi_U: \phi_U(U \intersect V) \to \phi_V(U \intersect V) \in C^\infty
,\]
so there are all diffeomorphisms of open subsets of $\RR^n$.
A smooth atlas is maximal if not contained in any other smooth atlas, and two atlases are compatible if their union is again a smooth atlas.
We say two smooth structures are equivalent if they are compatible.
:::

:::{.exercise title="?"}
Show that any smooth manifold has a unique maximal smooth atlas.
:::

:::{.remark}
Recall that for $f: \RR^n\to \RR^m$, $df\in \Mat(m\times n; \RR)$ is given by $(df)_{ij} = \dd{f_i}{x_j}$.
For any $p\in U \intersect V$, we have $d_{\phi_U(p)} \psi_{VU} \in \GL_n(\RR)$, so we get a map $d\psi_{VU}: U \intersect V\to \GL_n(\RR)$.
By the chain rule they satisfying the cocycle definition, so these glue to a vector bundle $\T M\to M$.
:::

:::{.exercise title="?"}
Show that every other definition of $\T M$ coincides with this one.
:::