# Tuesday, October 19

:::{.remark}
Today: computations involving the Euler class.
:::

:::{.theorem title="?"}
Let $M\in \smooth\Mfd^{\dim =n}$ be closed and oriented, then noting that $e(\T M) \in H^n(M; \ZZ)$,
[^orient_coefs]

\[
\inner{ e(\T M)}{ [M] } = \chi_\Top(M)
,\]
the topological Euler characteristic of $M$.

[^orient_coefs]: 
$\ZZ$ can be replaced with $\ZZ/2$ if $M$ is not oriented.

:::

:::{.theorem title="?"}
Let $M\in \smooth \Mfd^{\dim = n}$ closed and $E \mapsvia{\pi} M \in \VectBundlerk{k}(X)$ oriented.
Suppose a generic section $s$ of $E$, i.e. $\im s \da \ts{s(x) \in E \st x\in M} \subseteq E$, intersects $M \subseteq E$ transversally.
Then

- $z\da s\inv(0)$ is a submanifold of $M$ is a submanifold of $M$
- $\ro{ds}{Z}: \nu Z \iso \ro{E}{Z}$

:::

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

:::{.corollary title="?"}
$e(E) = \PD([Z])$, where $[Z] \in H_{n-k}(M)$.
:::

:::{.claim}
The second theorem implies the first.
:::

:::{.proof title="?"}
Note that a section of $\T M$ is a vector field.
Use Morse functions $f: M\to \RR$:

- $\crit(f) = \ts{p\in M \st df_p = 0}$.
- $p\in \crit(f)$ is **nondegenerate** if the Hessian $H_p(f) \da \qty{ {\partial^2 \over \partial x_i \partial x_j } } (p)$ is nonsingular, i.e. $\det H_p\neq 0$.
- $f$ is **Morse** if all critical points are nondegenerate.
- If $p\in \crit(f)$ is nondegenerate, then $\ind_p(f)$ is the number of negative eigenvalues of $H_p$.
- $f$ Morse on $M$ induces a CW complex structure with exactly one $k\dash$cell corresponding to each index $k$ critical point $p\in \Crit(f)_k$.
  Thus we can compute $\chi(M) = \sum_k (-1)^k \dim C_k^\cell = \sum_k (-1)^k \# \Crit(f)_k$.

:::

:::{.proof}
$f$ Morse induces a gradient vector field on $M$: picking a Riemannian metric $g$ on $M$, define $df(\wait) \da g( \grad f, \wait)$ to get a section $\grad f: M\to \T M$.
Note that if $p\in \crit(f)$ then $d_p f = 0$ since $\grad_p f = 0$.
So the vector field $df$ vanishes at $\crit(f)$ and $(\grad f)\inv(0)$

:::{.exercise title="?"}
Show that the sign of a zero of the gradient vector field is $(-1)^{\ind_p(f)}$.
:::

So taking $Z = \crit(f)$, we can write
\[
e(\T M) = \PD \adjoin{ \sum_{p\in \crit(f) } (-1)^{\ind_p(f)} [x] }
,\]
where $[x]$ is the dual of a generator of $H_0(M)$
Then
\[
\inner{ e(\T M)}{ [M] } = \sum_{p\in \crit(f) } (-1)^{\ind_p(f)} = \# \crit_{\text{even}}(f) - \# \crit_{\text{odd}}(f) = \chi_\Top(M)
.\]
:::

:::{.remark}
Given $N^n \embeds M^m$ an oriented closed submanifold with $M$ closed, we can consider Thom class of the disc/sphere bundles.
We identify the disc bundle $\DD \nu N$ with a tubular neighborhood of $N$ in $M$ and apply excision to get the following:

\begin{tikzcd}
	{u_{\nu N} \in H^{m-n}(\DD \nu N, \SS \nu n)} && {H^{m-n}(M, M\sm N)} && {H^{m-n}(M)} \\
	\\
	&& {H^{m-n}(\DD \nu N, \DD \nu N \sm N)}
	\arrow["\cong", from=1-1, to=1-3]
	\arrow["\cong", from=1-1, to=3-3]
	\arrow["{\text{excision} \cong}", from=1-3, to=3-3]
	\arrow[from=1-3, to=1-5]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJ1X3tcXG51IE59IFxcaW4gSF57bS1ufShcXEREIFxcbnUgTiwgXFxTUyBcXG51IG4pIl0sWzIsMCwiSF57bS1ufShNLCBNXFxzbSBOKSJdLFsyLDIsIkhee20tbn0oXFxERCBcXG51IE4sIFxcREQgXFxudSBOIFxcc20gTikiXSxbNCwwLCJIXnttLW59KE0pIl0sWzAsMSwiXFxjb25nIl0sWzAsMiwiXFxjb25nIl0sWzEsMiwiXFx0ZXh0e2V4Y2lzaW9ufSBcXGNvbmciXSxbMSwzXV0=)

We can also consider the composition:

\begin{tikzcd}
	{[N]\in H_n(N)} && {H_n(M)} && {H^{m-n}(M)}
	\arrow["{i_*}", from=1-1, to=1-3]
	\arrow["\PD", from=1-3, to=1-5]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwwLCJbTl1cXGluIEhfbihOKSJdLFsyLDAsIkhfbihNKSJdLFs0LDAsIkhee20tbn0oTSkiXSxbMCwxLCJpXyoiXSxbMSwyLCJcXFBEIl1d)

:::

:::{.claim}
These two classes are equal in $H^{m-n}(M)$.
:::

:::{.proof title="?"}
Consider the following:

\begin{tikzcd}
	\textcolor{rgb,255:red,92;green,92;blue,214}{u_{\nu N}} \\
	{\gens{u_{\nu N}} = H^{m-n}(\DD \nu n, \SS \nu n)} && {H_n(\DD \nu N)} \\
	\\
	{H^{m-n}(M, M\sm N)} && {H_n(N) = \gens{[N]}} & \textcolor{rgb,255:red,92;green,92;blue,214}{\pm [N]} \\
	\\
	{H^{m-n}(M)} && {H_m(M)}
	\arrow["{\PD, \cong}", tail reversed, from=2-1, to=2-3]
	\arrow["{\PD(\wait) = [M]\capprod(\wait), \cong}", tail reversed, from=6-1, to=6-3]
	\arrow["\cong", from=2-3, to=4-3]
	\arrow["\cong"', from=2-1, to=4-1]
	\arrow[from=4-1, to=6-1]
	\arrow[from=4-3, to=6-3]
	\arrow[color={rgb,255:red,92;green,92;blue,214}, curve={height=-12pt}, dashed, from=1-1, to=4-4]
\end{tikzcd}

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

The claim is that $u_{\nu N}$ is mapped to the *positive* generator, so the both paths from $H^{m-n}(\DD \nu N, \SS \nu N)\to H^{m-n}(M)$ agree.
:::

:::{.remark}
An aside on cup/cap products.
The cap product is a map:
\[
\capprod: H_m(X) \tensor_\ZZ H^i(X) &\to H_{m-i}(X) \\
.\]
On chains, it's given by
\[
\capprod C_m(X) \tensor_\ZZ C^i(X) &\to C_{m-i}(X) \\
(\sigma, \phi) &\mapsto \phi\qty{ \ro{\sigma}{\tv{v_0,\cdots, v_i}} } \ro{\sigma}{v_{i+1}, \cdots, v_m}
.\]
Then $\PD(\wait) \da [X]\capprod (\wait)$.

The cup product is a map
\[
\cupprod: H^i(X) \tensor_\ZZ H^j(X) &\to H^{i+j}(X) \\
C^i(X) \tensor_\ZZ C^j(X) &\to C^{i+j}(X) \\
(\phi, \psi) &\mapsto \qty{\sigma \mapsto \phi\qty{\ro{\sigma}{[v_0,\cdots, v_i]}} \psi\qty{\ro{\sigma}{[v_{i+1}, \cdots, v_{i+j}}} }
.\]
Then fixing any element yields a map $\alpha \cupprod (\wait): C^j(X) \to C^{i+j}(X)$, which is induced by a map $\phi\capprod(\wait): C_{i+j}(X) \to C_i(X)$.
:::