# Thursday, November 04

:::{.remark}
Recall that $w(\T\RP^n) = (1+a)^{n+1}$ where $\gens{a} = H^1(\RP^n; C_2)$, and as an application, if $\RP^{2^r} \embeds \RR^N$, then $N \geq 2^{r+1} - 1$.
Similarly, $c(\T \CP^n) = (1+a)^{n+1}$ with $\gens{a} = H^2(\CP^\infty; \ZZ)$ the positive generator.
:::

:::{.theorem title="Leray-Hirsch"}
Fix a commutative ring $R$ and a fiber bundle $F\injects E \to B$ such that

1. $H^j(F; R) \in \rmod$ is finitely generated and free,
2. For any $j$, there exist $c_{j_k} \in H^j(E; R)$ such that the restrictions of $\ts{c_{j_k}}_{k\geq 0}$ is a basis for $H^j(E_x; R) = H^j(F; R)$.

Then there is an isomorphism
\[
\Phi: H^*(B; R)\tensor_R H^*(F; R)\iso H^*(E; R) \\
b_i \tensor i^*(c_{j_k}) \mapsto \pi^*(b_i) \cupprod c_{j_k}
.\]


:::

:::{.example title="?"}
Let $\PP(E)$ be the projectivization of $E$, so each vector space fiber $V$ is replaced with $\PP(V)$.
Let $\gamma \subseteq \pi^*(E)$ be the canonical over $\PP(E)$, so the fibers are $\gamma_{(\gamma, [\gamma])} = \ts{w\in E_x \st w\in [v]}$.
For $E\in \VectBundlerk{n}(\CC)\slice {X}$, pick a metric on $E$, and define $E' \da \gamma^\perp$.
Now take the pullback:

\begin{tikzcd}
	&& \textcolor{rgb,255:red,92;green,92;blue,214}{\CC^n} \\
	& { \pi^*E = \gamma \oplus E'} && E \\
	\textcolor{rgb,255:red,92;green,92;blue,214}{\CP^n} \\
	& E && X
	\arrow[color={rgb,255:red,92;green,92;blue,214}, from=3-1, to=4-2]
	\arrow[from=4-2, to=4-4]
	\arrow[color={rgb,255:red,92;green,92;blue,214}, from=1-3, to=2-4]
	\arrow[from=2-4, to=4-4]
	\arrow[from=2-2, to=4-2]
	\arrow[from=2-2, to=2-4]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMSwxLCIgXFxwaV4qRSA9IFxcZ2FtbWEgXFxvcGx1cyBFJyJdLFszLDEsIkUiXSxbMywzLCJYIl0sWzEsMywiRSJdLFsyLDAsIlxcQ0NebiIsWzI0MCw2MCw2MCwxXV0sWzAsMiwiXFxDUF5uIixbMjQwLDYwLDYwLDFdXSxbNSwzLCIiLDAseyJjb2xvdXIiOlsyNDAsNjAsNjBdfV0sWzMsMl0sWzQsMSwiIiwwLHsiY29sb3VyIjpbMjQwLDYwLDYwXX1dLFsxLDJdLFswLDNdLFswLDFdXQ==)

So we have a bundle $\CP^{n-1} \injects \PP(E) \to X$, we'll now try to apply the theorem.
Note that $H^*(\CP^{n-1}; \ZZ) = \ZZ[u]/\gens{u^n}$, so $H^i$ is generated by $u^i$, fulfilling condition 1.

Let $\iota_x: E_x \injects E$ be the inclusion of a fiber.
If we restrict $\PP(E)$ to a fiber $\PP(E)_x$, this yields the canonical over $\CP^{n-1}$:

\begin{tikzcd}
	{\gamma_x} && \gamma \\
	\\
	{\PP(E)_x = \CP^{n-1}} && {\PP(E)}
	\arrow[from=1-1, to=3-1]
	\arrow[from=1-3, to=3-3]
	\arrow["{\iota_x^*}"{description}, hook, from=3-1, to=3-3]
	\arrow[hook, from=1-1, to=1-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMiwwLCJcXGdhbW1hIl0sWzIsMiwiXFxQUChFKSJdLFswLDIsIlxcUFAoRSlfeCA9IFxcQ1Bee24tMX0iXSxbMCwwLCJcXGdhbW1hX3giXSxbMywyXSxbMCwxXSxbMiwxLCJpX3giLDEseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFszLDAsIiIsMSx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV1d)

Then $\iota_x^* c_1(\gamma) = c_1(\gamma_x) = e(\gamma_x) = u$, so we have condition 2.


:::{.exercise title="?"}
For $L \in \VectBundlerk{1}(\CC)$,
\[
c_1(L\dual) = c_1(L)
.\]
:::

So we have classes 
$1, c_1(\gamma), c_1(\gamma)^2,\cdots, c_1( \gamma)^{n-1}$,
and we can take duals to obtain
$1, c_1(\gamma\dual), c_1(\gamma\dual)^2,\cdots, c_1( \gamma\dual)^{n-1}$.
There exist $c_1(E), c_2(E), \cdots, c_n(E) \in H^*(X)$ such that we can write the former as a linear combination of the latter:
\[
c_1( \gamma\dual)^n + c_1(E) c_1( \gamma\dual)^{n-1} + \cdots + c_n(E) = 0
,\]
and the $c_i(E)$ are called the **Chern classes**.
:::

:::{.remark}
Write $c(E) = 1 + c_1(E) + \cdots + c_n(E)$, so
\[
\pi^* c(E) 
&= c(\gamma) c(E') = (1 +c_1(\gamma))(1 + c_1(E') + \cdots + c_{n-1}(E'))\\
&= 1 + (c_1(\gamma) + c_1(E'))
+ (c_1( \gamma)c_1(E') + c_2(E'))
+ \cdots 
+ (c_1(\gamma) c_{i-1}(E') + c_1(E') )
.\]

Plugging this into the LHS above yields
\[
c_1( \gamma\dual)^n + 
(c_1(\gamma) + c_1(E')) c_1(\gamma\dual)^{n-1} +
(c_1(\gamma)c_1(E') + c_2(E'))c_1(\gamma\dual)^{n-2}
+ \cdots +
c_1(\gamma)c_{n-1}(E') 
&=
(c_1 (\gamma) + c_1(\gamma\dual))( c_1( \gamma\dual)^{n-1} + c_1(E') c_1( \gamma\dual)^{n-2} + c_2(E') c_1( \gamma\dual)^{n-2} + \cdots + c_{n-1}(E') )
.\]
Since $\gamma\dual$ and $\gamma$ are dual, the first term is zero, so this entire expression is zero.
:::

:::{.exercise title="?"}
Show that $c_1(E\dual) = -c_1(E)$.

> Hint: consider an explicit description in terms of transition functions.

:::

:::{.exercise title="?"}
Show that given $b_1 = 1, b_2 = x+a_1, b_3 = xa_1 + a_2, \cdots$, then $b_1 (-x)^n + b_2 (-x)^{n-1} + \cdots = 0$.

:::

:::{.remark}
Consider the following pullbacks:

\begin{tikzcd}
	{f^* E} && {\gamma \oplus E'} && E \\
	\\
	{\SS E} && {\PP(E)} && X
	\arrow[from=1-1, to=1-3]
	\arrow[from=1-3, to=1-5]
	\arrow[from=1-5, to=3-5]
	\arrow[from=3-3, to=3-5]
	\arrow[from=1-3, to=3-3]
	\arrow[from=3-1, to=3-3]
	\arrow[from=1-1, to=3-1]
	\arrow["f"{description}, curve={height=30pt}, tail, from=3-1, to=3-5]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMiwwLCJcXGdhbW1hIFxcb3BsdXMgRSciXSxbNCwwLCJFIl0sWzQsMiwiWCJdLFsyLDIsIlxcUFAoRSkiXSxbMCwyLCJcXFNTIEUiXSxbMCwwLCJmXiogRSJdLFs1LDBdLFswLDFdLFsxLDJdLFszLDJdLFswLDNdLFs0LDNdLFs1LDRdLFs0LDIsImYiLDEseyJjdXJ2ZSI6NSwic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoibW9ubyJ9fX1dXQ==)

What is $f^* E$?
There is a pullback

\begin{tikzcd}
	{\varphi^* \gamma} && \gamma \\
	\\
	{S^{2n-1}} && {\CP^{n-1}}
	\arrow["\varphi", from=3-1, to=3-3]
	\arrow[from=1-3, to=3-3]
	\arrow[from=1-1, to=3-1]
	\arrow[from=1-1, to=1-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwyLCJTXnsybi0xfSJdLFsyLDIsIlxcQ1Bee24tMX0iXSxbMiwwLCJcXGdhbW1hIl0sWzAsMCwiXFx2YXJwaGleKiBcXGdhbW1hIl0sWzAsMSwiXFx2YXJwaGkiXSxbMiwxXSxbMywwXSxbMywyXV0=)

Here $\phi^* \gamma$ will be trivial: we have \( \gamma \subseteq \CP^{n-1} \times \CC^n \), so $\phi^* \gamma \subseteq S^{2n-1} \times \CC^n$ and $\dim_\RR \phi^* \gamma = 2$.
Then the fibers are $\phi^*\gamma_x = \ts{(x, v) \st v = \lambda x,\, \lambda \in \CC}$.
It turns out that $\phi^* E = \CC \oplus q^* E'$ where $\SS E \mapsvia{q} \PP(E)$.
Writing $\pi: \PP(E) \to X$, we have
\[
(\pi \circ q)^* c_i(E) = c_i(E) && i < n-1
,\]
and $c_n(E) = e(E)$.
To find this we'll need $\pi \circ q$ to be injective.

Consider


\begin{tikzcd}
S^{2n-1} 
  \ar[r] 
& 
\SS E
  \ar[d, "\pi \circ q"] 
\\
& 
X 
\end{tikzcd}

By the Gysin sequence, for $j<2n-1$ we'll have $H^j(X) \cong H^j(\SS E)$ since the red terms vanish:

\begin{tikzcd}
	{H^j(\SS E)} && \textcolor{rgb,255:red,214;green,92;blue,92}{H^{j-2n+1}(X)} && {H^{j+1}(X)} \\
	\\
	{H^{j-1}(\SS E)} && \textcolor{rgb,255:red,214;green,92;blue,92}{H^{j-2n}(X)} && {H^j(X)}
	\arrow[from=3-1, to=3-3]
	\arrow[from=3-3, to=3-5]
	\arrow[from=3-5, to=1-1]
	\arrow[from=1-1, to=1-3]
	\arrow[from=1-3, to=1-5]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMCwyLCJIXntqLTF9KFxcU1MgRSkiXSxbMCwwLCJIXmooXFxTUyBFKSJdLFsyLDIsIkhee2otMm59KFgpIixbMCw2MCw2MCwxXV0sWzQsMiwiSF5qKFgpIl0sWzIsMCwiSF57ai0ybisxfShYKSIsWzAsNjAsNjAsMV1dLFs0LDAsIkhee2orMX0oWCkiXSxbMCwyXSxbMiwzXSxbMywxXSxbMSw0XSxbNCw1XV0=)

This uses that for $j<2n$ that $H^{j-2n}(X) = 0$.
For $i\leq n-1$, $2i<2n-i<2n-1$, and so the Chern classes for $X$ and $\SS E$ are isomorphic via $(\pi \circ q)^*: H^j(X) \to H^j(\SS E)$ in the LES.
By induction, we can define $c_i(E)$ for $i\leq n-1$.
:::

:::{.remark}
Idea: pull back to split off a line bundle, pull back further to split off another line bundle, and continue.
This is why we only need $c_1$ to determine a line bundle.
:::