# Chern and Stiefel-Whitney classes (Thursday, November 11)

:::{.remark}
Recall that we were proving the splitting principle: given $E\to X$ of dimension $n$, there is a space $Y$ with $f:Y\to X$ where 

- $f^* E = \bigoplus L_i$ splits as a direct sum of line bundles.
- $f^*: H^*(X) \to H^*(Y)$ is injective.

:::

:::{.lemma title="?"}
Assuming the exercise from last time that $c_1(L\dual) = - c_1(L)$
\[
c_i(E\dual) = (-1)^i c_i(E)
.\]
:::

:::{.proof title="?"}
It suffices to show this for $E = \bigoplus L_i$.
Write $E\dual = \bigoplus L_i\dual$, so
\[
c(E\dual) = \prod c(L_i\dual) 
&= \prod (1- c_1(L_i)) \\
&= 1 - \qty{\sum c_1(L_i)} + \sigma_2(c_1 L_1,\cdots, c_1 L_n) - \cdots \pm \prod c_1(L_i) \\
&= 1 - c_1(E) + c_2(E) \cdots \pm c_1(L_n)
.\]
:::

:::{.theorem title="?"}
For $E\in \VectBundlerk{n}(\CC)\slice X$, the top Chern class equals the Euler class:
\[
c_n(E) = e(E)
.\]
Similarly, for $E\in \VectBundlerk{n}(\RR)\slice X$,
\[
w_n(E) \equiv e(E) \mod 2
.\]
:::

:::{.proof title="?"}
Both classes satisfy the Whitney sum formula, so
\[
E &= \bigoplus L_i 
\implies c(E) = \prod c(L_i) \\
\implies e(E) 
&= \prod e(L_i) \\
&= \prod (1 + c_1(L_i)) \\
&= 1 + \sigma_1(c_1 L_1, \cdots, c_1 L_n) + \cdots + \prod c_1(L_i) \\
&\da 1 + \sigma_2(c_1 L_1, \cdots, c_1 L_n)+ \cdots + c_n(E)
.\]
Then $e(E) = \prod e(L_i) = \prod c_1(L_i) = c_n(E)$.

Now the claim follows from the splitting principle and naturality of characteristic classes.
:::

:::{.exercise title="?"}
Show that for any $E\in \VectBundlerk{n}(\CC)\slice{E}$,
\[
c_i(E) \equiv w_{2i} (E) \mod 2
.\]
:::

:::{.lemma title="?"}
For $E\in\VectBundlerk{n}(\RR)\slice X$,  
Then
\[
c_1(E) = c_1(\Extalg^n E)
,\]
noting that $\Extalg^n E$ is a line bundle.
A similar claim holds for $c_1$.
:::

:::{.fact}
\[
c_1(L_1\tensor L_2) = c_1(L_1) + c_1(L_2)
.\]
:::

:::{.remark}
ETS this holds for $E = \bigoplus L_i$, in which case $\Extalg^n E = \Tensor_{i=1}^n L_i$.
By the above fact, $c_1(\Extalg^n E) = \sum c_1(L_i)$.
Since $c_i(E) = \sigma_i(c_1 L_1, \cdots, c_1 L_n)$, we have 
\[
c_1(E) = \sigma_1(c_1 L_1,\cdots, c_1 L_n) = \sum c_1 L_i
.\]
::

:::{.remark}
Recall that the Euler class is the obstruction to finding an obstruction to extending a section over the $n\dash$skeleton.
:::

## Obstruction Theory

:::{.lemma title="?"}
$w_1 E$ is the obstruction to orienting $E$ in the following sense:
for any $S^1 \mapsvia{f} E$, write $f_*[S]$ for the image of the fundamental class, then
\[
\inner{w_1(E)} {f_*[S^1] } = \chi_{f^* E \text{ is orientable}}  =
\begin{cases}
1 & f^*E \text{ is nonorientable} 
\\
0 & f^* E \text{ is orientable}.
\end{cases}
.\]
:::

:::{.remark}
Why? For example, consider $E\to S^1$.
Trivialize over $S^1\smts{\pt}$, then glue the ends by some element $A\in \GL_n(\RR)$.
If $\det(A) > 0$, this will be orientable, and $\det(A) < 0$ will be nonorientable.
:::

:::{.proof title="?"}
If $E$ is a line bundle, define $\tilde w_1(E) \in H^1(X)$ by the following
\[
\inner{\tilde w_1(E)}{ f_* [S^1] } = \chi_{f^* E \text{ nonorientable}}
.\]
This is natural under pullback, consider the following diagram:

\begin{tikzcd}
	{f^* h^* E} && {h^*E} && E \\
	\\
	{S^1} && Y && X
	\arrow[from=1-3, to=1-5]
	\arrow[from=1-5, to=3-5]
	\arrow[from=1-3, to=3-3]
	\arrow[from=1-1, to=1-3]
	\arrow["f", from=3-1, to=3-3]
	\arrow["h", from=3-3, to=3-5]
	\arrow[from=1-1, to=3-1]
	\arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-3]
	\arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-3, to=3-5]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMiwwLCJoXipFIl0sWzQsMCwiRSJdLFs0LDIsIlgiXSxbMCwyLCJTXjEiXSxbMiwyLCJZIl0sWzAsMCwiZl4qIGheKiBFIl0sWzAsMV0sWzEsMl0sWzAsNF0sWzUsMF0sWzMsNCwiZiJdLFs0LDIsImgiXSxbNSwzXSxbNSw0LCIiLDAseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XSxbMCwyLCIiLDAseyJzdHlsZSI6eyJuYW1lIjoiY29ybmVyIn19XV0=)

We have
\[
&\inner{ \tilde w_1 h^* E}{ f_* [S^1] }
&= \inner{ \tilde w_1 h^* E}{ (h\circ f)_* [S^1] } \\
&= \inner{ h^*\tilde w_1 h^* E}{ f_* [S^1] } \\
&= \tilde w_1 (h^* E) \\
&= h^* \tilde w_1(E)
.\]

We need to show $\tilde w_1(\gamma) = w_1(\gamma)$ where $\gamma\to \RP^\infty$ is the canonical.
Write $H^1(\RP^\infty; C_2) = C_2 = \gens{a}$, so $w_1( \gamma) = a$.
Take the pullback:

\begin{tikzcd}
	{i^*\gamma} && \gamma \\
	\\
	{\RP^1 \cong S^1} && {\RP^\infty}
	\arrow["i", 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]
	\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=WzAsNCxbMiwwLCJcXGdhbW1hIl0sWzAsMCwiaV4qXFxnYW1tYSJdLFsyLDIsIlxcUlBeXFxpbmZ0eSJdLFswLDIsIlxcUlBeMSBcXGNvbmcgU14xIl0sWzMsMiwiaSJdLFswLDJdLFsxLDNdLFsxLDBdLFsxLDIsIiIsMSx7InN0eWxlIjp7Im5hbWUiOiJjb3JuZXIifX1dXQ==)

Then $i^* \gamma$ is not orientable since locally this looks like a Mobius band, so we have
\[
\inner{\tilde w_1( \gamma)}{ i_* [\RP^1] } = 1
.\]

Now $w_1(E) = w_1(\Extalg^n E)$ and we have to bundles:

\begin{tikzcd}
	{f^*(\Extalg^n E)} && {f^* E} \\
	\\
	& {S^1}
	\arrow[from=1-1, to=3-2]
	\arrow[from=1-3, to=3-2]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwwLCJmXiooXFxFeHRhbGdebiBFKSJdLFsyLDAsImZeKiBFIl0sWzEsMiwiU14xIl0sWzAsMl0sWzEsMl1d)

These are either simultaneously orientable or simultaneously nonorientable, by considering $\det A_1, \det A_2$ for the gluing maps between trivializations.
So \( w_1(E) = \inner{\Extalg^n E}{f_* [S^1] } = \chi_{f^* \Extalg^n E \text { nonorientable}} \).
:::

:::{.theorem title="?"}
$w_k(E)$ is the mod 2 reduction of the obstruction to extending $n-k+1$ linearly independent sections of $E$ over $X\skel{k}$ when $\dim_\RR E = n$.
:::

:::{.remark}
For $k=n$, $w_n(E)$ corresponds to 1 linearly independent section of $X\skel{n}$ since $e(E) \equiv w_n(E) \mod 2$.
For $k=1$, $w_1(E)$ corresponds to $n$ linearly independent sections over $X\skel{1}$.
:::