# Thursday, October 07

## The Euler Class

:::{.remark}
Let $E\to X$ be an orientable bundle, then


\begin{tikzcd}
	{H^n(\DD E, \SS E; \ZZ)} &&& {H^n(\DD E; \ZZ) \cong H^n(X; \ZZ)} \\
	{u_{-E} = -u_E} &&& {e(E)}
	\arrow[maps to, from=2-1, to=2-4]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJIXm4oXFxERCBFLCBcXFNTIEU7IFxcWlopIl0sWzMsMCwiSF5uKFxcREQgRTsgXFxaWikgXFxjb25nIEhebihYOyBcXFpaKSJdLFswLDEsInVfey1FfSA9IC11X0UiXSxbMywxLCJlKEUpIl0sWzIsMywiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoibWFwcyB0byJ9fX1dXQ==)

:::

:::{.lemma title="?"}
$e(-E) = -e(C)$.
:::

:::{.proof title="?"}
Using naturality:


\begin{tikzcd}
	{E = f^*(\RR^n)} && {\RR^n} \\
	\\
	X && \pt
	\arrow["\pi", from=1-3, to=3-3]
	\arrow["f"', from=3-1, to=3-3]
	\arrow[from=1-1, to=3-1]
	\arrow[from=1-1, to=1-3]
	\arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJFID0gZl4qKFxcUlJebikiXSxbMiwwLCJcXFJSXm4iXSxbMiwyLCJcXHB0Il0sWzAsMiwiWCJdLFsxLDIsIlxccGkiXSxbMywyLCJmIiwyXSxbMCwzXSxbMCwxXSxbMCwyLCIiLDEseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XV0=)

Then $e(\RR^n\to \pt) = 0$, and $0 = f^*(0) = e(f^*(\RR^n\to \pt)) = e(E)$.

:::

:::{.lemma title="?"}
$e(E) = 0$ if $E$ is the trivial bundle.
:::

:::{.proof title="?"}
Using the exact sequence for the pair $(\DD E, \SS E)$:


\begin{tikzcd}
	{u_E} && e \\
	{H^n(\DD E, \SS E; \ZZ)} && {H^n(\DD E; \ZZ)} && {H^n(\SS E; \ZZ)} \\
	\\
	&& {\cong H^n(X\times \DD^n; \ZZ) \cong H^n(X) \tensor H^n(\DD; \ZZ)} && {\cong H^n(X\times S^{n-1}; \ZZ)} \\
	&& {H^n(X)} && {H^n(X)\oplus H^1(X)} \\
	&& \alpha && {(\alpha, 0)}
	\arrow[hook, from=5-3, to=5-5]
	\arrow[from=6-3, to=6-5]
	\arrow[from=2-1, to=2-3]
	\arrow[from=2-3, to=2-5]
	\arrow[maps to, from=1-1, to=1-3]
\end{tikzcd}

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

So the second map is the inclusion $\alpha \mapsto (\alpha, 0)$, which is thus injective.
The first map is 0, so $e(E) = 0$.
:::

:::{.example title="?"}
Compute $e( \T S^2)$.
:::

:::{.lemma title="?"}
If $E$ is an odd dimensional vector bundle, then $2e(E) = 0$
:::

:::{.remark}
Check that \( \alpha \cupprod \beta = (-1)^{ij} \beta \cupprod \alpha \).
:::

:::{.proof title="First way"}
The antipodal map will give an isomorphism $E\cong -E$, so $e(E) = e(-E) = -e(E)$ and thus $2e(E) = 0$.
:::

:::{.proof title="Second way"}
Use the map:


\begin{tikzcd}
	{H^n(\DD E, \SS E;\ZZ)} &&& {H^{2n}(\DD E, \SS E; \ZZ)} \\
	{u_E} &&& {u_E\cupprod u_E = - u_E \cupprod u_E}
	\arrow["{u_E \cupprod(\wait)}", from=1-1, to=1-4]
	\arrow[maps to, from=2-1, to=2-4]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJIXm4oXFxERCBFLCBcXFNTIEU7XFxaWikiXSxbMywwLCJIXnsybn0oXFxERCBFLCBcXFNTIEU7IFxcWlopIl0sWzAsMSwidV9FIl0sWzMsMSwidV9FXFxjdXBwcm9kIHVfRSA9IC0gdV9FIFxcY3VwcHJvZCB1X0UiXSxbMCwxLCJ1X0UgXFxjdXBwcm9kKFxcd2FpdCkiXSxbMiwzLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJtYXBzIHRvIn19fV1d)

Then $2u_E \cupprod u_E = 0$, making $2u_E = 0$ and thus $2e(E) = 0$.
:::

:::{.lemma title="?"}
Given $X\to X$ and $F\to Y$, then we can form the bundle $E\times F\to X\times Y$.
Writing $\pi_1:X\times Y\to X$ and $\pi_2:X\times Y\to Y$, we have
\[
e(E\times F) = \pi_1^*(e(E)) \cupprod \pi_2^*(e(F)) = e(E) \tensor e(F)
.\]
:::

:::{.exercise title="?"}
Prove this using the Kunneth formula.
:::

:::{.corollary title="?"}
Let $E,E'\to X$ be two vector bundles. 
Then $e(E \oplus E') = e(E) \cupprod e(E')$.
:::

:::{.proof title="?"}
Consider the diagonal $\Delta: X\to X\cartpower{2}$, then take the pullback:


\begin{tikzcd}
	{\Delta^*(E\times E')\cong E \oplus E'} && {E\times E'} \\
	\\
	{X\cartpower{2}} && X
	\arrow["\Delta", from=3-1, to=3-3]
	\arrow[from=1-1, to=3-1]
	\arrow[from=1-3, to=3-3]
	\arrow[from=1-1, to=1-3]
	\arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJcXERlbHRhXiooRVxcdGltZXMgRScpXFxjb25nIEUgXFxvcGx1cyBFJyJdLFsyLDAsIkVcXHRpbWVzIEUnIl0sWzIsMiwiWCJdLFswLDIsIlhcXGNhcnRwb3dlcnsyfSJdLFszLDIsIlxcRGVsdGEiXSxbMCwzXSxbMSwyXSxbMCwxXSxbMCwyLCIiLDEseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XV0=)

Then
\[
e(E \oplus E')
&= \Delta^*( e(E\times E'))\\
&= \Delta^*(\pi_1^*(e(E)) \cupprod \pi_2^*(e(E')) ) \\
&= \Delta^* \pi_1^* (e(E))\cupprod \Delta^* \pi_2^* e(E') \\
&= e(E) \cupprod e(E')
.\]

:::

:::{.corollary title="?"}
If $E' = L \oplus E$ for $L$ a trivial line bundle, $e(E') = 0$.
:::

:::{.proof title="?"}
\[
e(E') = e(L) \cupprod e(E) = 0
.\]
:::

## Obstruction Theory

:::{.remark}
The Euler class is the **obstruction** to finding a nonvanishing section over the $n\dash$skeleton of $X$.
Note that if we have a line bundle, we automatically have a section.
Conversely, a nonvanishing section will produce a line bundle summand.
:::

:::{.exercise title="?"}
Describe the correspondence between line subbundles and nonvanishing global section.
:::


:::{.remark}
Some review:
let $X\in \CW$ and write $H^*(X; \ZZ)$ for cellular homology.
Then $C^k_{\cell}(X) = \Hom(H_k(X^{(k)}, X^{(k-1)} ), \ZZ) = \Hom(C_k^{\cell}, \ZZ)$.
To produce the differential (green), use that the LES for the pairs (red and blue) are intertwined:


\begin{tikzcd}
	&&& \textcolor{rgb,255:red,92;green,92;blue,214}{\vdots} \\
	&&& \textcolor{rgb,255:red,92;green,92;blue,214}{H^k(X^{(k+1)}, X^{(k)}; \ZZ)} \\
	\\
	&&& \textcolor{rgb,255:red,92;green,92;blue,214}{H^k(X^{(k+1)}; \ZZ)} \\
	\\
	\textcolor{rgb,255:red,214;green,92;blue,92}{\cdots} & \textcolor{rgb,255:red,214;green,92;blue,92}{H^k(X^{(k-1)}; \ZZ)} && \textcolor{rgb,255:red,214;green,92;blue,92}{H^k(X^{(k)}; \ZZ)} && \textcolor{rgb,255:red,214;green,92;blue,92}{H^k(X^{(k)}, X^{(k-1)}; \ZZ)} & \textcolor{rgb,255:red,214;green,92;blue,92}{\cdots} \\
	\\
	&&& \textcolor{rgb,255:red,92;green,92;blue,214}{H^{k+1}(X^{(k+1)}, X^{(k)}; \ZZ)}
	\arrow[color={rgb,255:red,92;green,92;blue,214}, from=2-4, to=4-4]
	\arrow[color={rgb,255:red,92;green,92;blue,214}, from=4-4, to=6-4]
	\arrow[color={rgb,255:red,214;green,92;blue,92}, from=6-6, to=6-4]
	\arrow["\delta", color={rgb,255:red,92;green,214;blue,92}, from=6-6, to=8-4]
	\arrow["\delta", color={rgb,255:red,92;green,92;blue,214}, from=6-4, to=8-4]
	\arrow[color={rgb,255:red,214;green,92;blue,92}, from=6-4, to=6-2]
\end{tikzcd}

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

Thus we have a formula
\[
\inner{ \delta \alpha}{ [\Delta^{k+1}] } = \inner{ \alpha}{ \bd[\Delta^{k+1}] }
.\]
:::


:::{.remark}
Recall that if the fibers of a bundle were $n\dash$connected, we could construct sections on $X^{(n-1)}$. 
Given any section, we can use a Riemannian metric to project onto norm 1 elements to get a section for the sphere bundle, say $s: X^{(n-1)} \to \SS E$.
Note that $\pi_i(S^{n-1}) = 0$ for $i\leq n-2$, so we can construct such a section.
Let $i: \Delta^n\to X$ be the attaching map for some $n\dash$cell, then form the following pullback to get a trivial bundle, where we can pull back the section:


\begin{tikzcd}
	{\Delta^n \times S^{n-1}} && {i^* \SS E} && {\SS E} \\
	\\
	&& {\Delta^n} && X && {X^{(n-1)}}
	\arrow[from=3-3, to=3-5]
	\arrow[from=1-5, to=3-5]
	\arrow[from=1-3, to=3-3]
	\arrow[from=1-3, to=1-5]
	\arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-3, to=3-5]
	\arrow["\cong"{description}, no head, from=1-1, to=1-3]
	\arrow["{\pi_1}"', from=1-1, to=3-3]
	\arrow[hook', from=3-7, to=3-5]
	\arrow["s"', color={rgb,255:red,214;green,92;blue,92}, from=3-7, to=1-5]
	\arrow["{\tilde s}", color={rgb,255:red,214;green,92;blue,92}, curve={height=-30pt}, from=3-3, to=1-1]
\end{tikzcd}


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

So $s$ yields a map $\bd \Delta^n \cong S^{n-1} \to S^{n-1}$, so it has a degree.
Then a cocycle can be defined as $[\Delta^n] \mapsvia{e_s}  \deg( \bd \Delta^n \to S^{n-1})\in \ZZ$.
:::


:::{.lemma title="?"}
$e_S$ is independent of the choice of trivialization for the pullback.
:::


:::{.proof title="?"}
Since $\Delta^n$ is contractible, any two trivializations are homotopic.
Thus the sections $s_T, s_{T'}: \bd \Delta^n \to S^{n-1}$ corresponding to two trivializations are homotopic.
:::


:::{.lemma title="?"}
If $s'$ is a different section over $X^{(n-1)}$, then $e_{s'} - e_s = \delta(\alpha)$ for some $\alpha\in C^{n-1}_{\cell}(X)$. 
:::


:::{.proof title="?"}
See phone notes.
:::