# Thursday, October 21

:::{.remark}
Recall that if $N^n \leq M^n$ is a submanifold, we have the following diagram:

\begin{tikzcd}
	{u_{\nu N}} & {H^{m-n}(\DD \nu N, \SS \nu N)} && {H^{m-n}(M, M\sm N)} && {H^{m-n}(M)} \\
	\\
	& {H_n(\DD \nu N)} && {H_n(N)} && {H_n(M)} \\
	&&& {[N]}
	\arrow["\PD", from=3-6, to=1-6]
	\arrow["\cong", from=3-2, to=3-4]
	\arrow["{i_*}", from=3-4, to=3-6]
	\arrow["{j^*}"', from=1-4, to=1-6]
	\arrow["{\cong, \text{ excision}}"', from=1-2, to=1-4]
	\arrow["\PD"', from=3-2, to=1-2]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsOCxbMSwwLCJIXnttLW59KFxcREQgXFxudSBOLCBcXFNTIFxcbnUgTikiXSxbMywwLCJIXnttLW59KE0sIE1cXHNtIE4pIl0sWzUsMCwiSF57bS1ufShNKSJdLFs1LDIsIkhfbihNKSJdLFsxLDIsIkhfbihcXEREIFxcbnUgTikiXSxbMywyLCJIX24oTikiXSxbMywzLCJbTl0iXSxbMCwwLCJ1X3tcXG51IE59Il0sWzMsMiwiXFxQRCJdLFs0LDUsIlxcY29uZyJdLFs1LDMsImlfKiJdLFsxLDIsImpeKiIsMl0sWzAsMSwiXFxjb25nLCBcXHRleHR7IGV4Y2lzaW9ufSIsMl0sWzQsMCwiXFxQRCIsMl1d)

Then the Thom class $u_{\nu N}\in H^{m-n}(\DD\nu N, \SS \nu N) \to H^{m-n}(\DD \nu N_x, \SS \nu N_x)$ is mapped to the generator specified by the orientations on fibers.
:::

:::{.theorem title="?"}
For $A^i \transverse B^j \leq X$ are smooth oriented submanifolds intersecting transversally, then
\[
\PD([A]) \cupprod \PD([B]) &= \PD([A \capprod  B]) \\
H^{n-i}(X) \times H^{n-j}(X) &\to H^{2n-i-j}(X)
.\]

:::

:::{.remark}
Then 
\[
[( \DD \nu N_x, \SS \nu N_x) \capprod N] = [\ts{x}]
,\]
which is the positive generator.
So
\[
\PD[ (\DD\nu N_x, \SS\nu N_x) ] \cupprod \PD[N] = \PD[x]
.\]
Now we can cap this to obtain
\[
[\DD \nu N, \SS \nu N] \capprod \qty{ \PD[ (\DD\nu N_x, \SS\nu N_x) ] \cupprod \PD[N]} 
&=
\qty{ [\DD\nu N, \SS\nu N] \capprod \PD [\DD\nu N_x, \SS \nu N_x] } \capprod \PD[N] \\
&= 
[\DD \nu N, \SS \nu N] \capprod \PD[x] \\
&= 1
,\]
where we've used that $\inner{\PD [x]}{ [\DD \nu N, \SS \nu N] }$.
So $\PD[N]$ is $j^*$ of the Thom class of $\nu N$.
:::

:::{.theorem title="?"}
Let $M \in \smooth\Mfd^n$ be closed and oriented and $E \mapsvia{\pi} M$ a $k\dash$dimensional oriented vector bundle.
Consider a generic section $s$ of $E$, so $\im(s) \transverse M$ in $E$.
Then
\[
e(E) = \PD[Z] && Z \da \im(s) \intersect M = s\inv(0)
.\]
:::

:::{.remark}
Recall that $s\inv(0) \leq M$ is a smooth submanifold, and $\ro{ds}{Z}: \nu Z \iso \ro{E}{Z}$, and since this orients $\nu Z$ this orients $Z$ as well.
:::

:::{.proof title="?"}
Let $N$ be a tubular neighborhood of $Z$ in $M$, such that $N \cong \DD \nu Z$.
By the lemma, we have two maps

\begin{tikzcd}
	{H^k(N, N\sm Z)} && {H^k(M, M\sm Z)} && {H^k(M)} \\
	{u_{\nu Z}} &&&& {\PD[Z]}
	\arrow["\cong", from=1-1, to=1-3]
	\arrow["{j^*}", from=1-3, to=1-5]
	\arrow["{\text{Lemma}}"', curve={height=30pt}, from=1-1, to=1-5]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMCwwLCJIXmsoTiwgTlxcc20gWikiXSxbMiwwLCJIXmsoTSwgTVxcc20gWikiXSxbNCwwLCJIXmsoTSkiXSxbMCwxLCJ1X3tcXG51IFp9Il0sWzQsMSwiXFxQRFtaXSJdLFswLDEsIlxcY29uZyJdLFsxLDIsImpeKiJdLFswLDIsIlxcdGV4dHtMZW1tYX0iLDIseyJjdXJ2ZSI6NX1dXQ==)

on the other hand, since $Z \injectsvia{i} M$, we get

\begin{tikzcd}
	{\ro{E}{Z}} && E \\
	\\
	& N
	\arrow["\iota", from=1-1, to=1-3]
	\arrow["{\ro{ds}{Z}}", from=3-2, to=1-1]
	\arrow["s"', from=3-2, to=1-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwwLCJcXHJve0V9e1p9Il0sWzIsMCwiRSJdLFsxLDIsIk4iXSxbMCwxLCJcXGlvdGEiXSxbMiwwLCJcXHJve2RzfXtafSJdLFsyLDEsInMiLDJdXQ==)


:::{.exercise title="?"}
Show $\ro{ds}{Z} \homotopic s$ are homotopic sections.
:::

Then $(\ro{ds}{Z})^* \circ i^* (u_E) = u_{\nu Z}$, so $u_{\nu Z} = s^* u_E$, and we have

\begin{tikzcd}
	\textcolor{rgb,255:red,92;green,92;blue,214}{u_E} &&&& \textcolor{rgb,255:red,92;green,92;blue,214}{u_{\nu Z}} \\
	& {H^k(E, E\sm M)} && {H^k(M, M\sm Z)} \\
	\\
	& {H^k(E)} && {H^k(M)} \\
	\textcolor{rgb,255:red,92;green,92;blue,214}{?} &&&& \textcolor{rgb,255:red,92;green,92;blue,214}{\PD[Z]} \\
	&&&& \textcolor{rgb,255:red,92;green,92;blue,214}{e(E)}
	\arrow["{\text{LES}}", from=2-2, to=4-2]
	\arrow["{s^*}", from=2-2, to=2-4]
	\arrow["{s^*, \cong}", from=4-2, to=4-4]
	\arrow[from=2-4, to=4-4]
	\arrow[color={rgb,255:red,92;green,92;blue,214}, dashed, from=1-1, to=1-5]
	\arrow[color={rgb,255:red,92;green,92;blue,214}, dashed, from=1-5, to=5-5]
	\arrow[color={rgb,255:red,92;green,92;blue,214}, dashed, from=5-1, to=6-5]
	\arrow[color={rgb,255:red,92;green,92;blue,214}, dashed, from=1-1, to=5-1]
\end{tikzcd}

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

Since the diagram commutes, we get $e(E) = \PD[Z]$.
:::

:::{.remark}
The Euler class for a bundle $E\to X\in \CW$ is the obstruction to finding a nowhere vanishing section on $X\skel{n}$ for $n\da \dim E$, and $e(E) = 0$ iff there is a nowhere vanishing section on $X\skel{n}$.
For a smooth manifold $M$ and $E \mapsvia{\pi} M$ with $\dim M = n, \dim E = k$ and $s:M\to E$ a section, then $\PD[s\inv(0)] = e[E]$ since $\dim s\inv(0) = n-k$.
If $E = \T M$, then $e(\T M) = 0$ implies $\chi(M) = 0$ and there exists a nowhere vanishing vector field.
:::