# Stiefel-Whitney and Chern Classes (Tuesday, November 02)

:::{.remark}
The Stiefel-Whitney classes $w_i \in H^i(X; C_2)$ will be defined for $\VectBundle(X)\slice \RR$ while Chern classes $c_i \in H^{2i}(X; \ZZ)$ will be defined for $\VectBundle(X)\slice \CC$.
We setting $w(E) \da \sum_i w_i(E) \in H^*(X; C_2)$, we mentioned several properties:

 - $w_i(f^* E) = f^* w_i(E)$
 - $w(E_1 \oplus E_2) = w(E_1) \cupprod w(E_2)$
 - $w_{> \dim E} E = 0$
- $w_1(\gamma)$ is the generator of $H^1(\RP^\infty; C_2) = C_2$ where $\gamma \to \RP^\infty$ is the canonical line bundle.
  For complex bundles, $c_1(\gamma)$ is the positive generator of $H^2(\CP^\infty; \ZZ) = \ZZ$.

These properties characterize the $w_i$ and $c_i$ uniquely.
:::

:::{.remark}
On why we need $C_2$ coefficients: we're pulling back $\gamma \to \BO_1$ and $\BO_1 \homotopic \RP^\infty$ where $H^1(\RP^\infty; \ZZ) = 0$!
For line bundles, we'll automatically have $w_{\geq 2} = 0$ for any such class, so we only have $w_1$ the work with, and this we need to pull back something nonzero to get anything interesting.
:::

:::{.corollary title="?"}
Some immediate consequences:

- $w_i(E \oplus \RR) = w_i(E)$, this if $E$ is trivial then $w_i(E) = 0$ and $w(E) = 1$.
  Moreover $w(E \oplus \RR) = w(E) w(\RR) = w(E)$ since $w(\RR) = 1$.

- If $E$ has $k$ linearly independent nonvanishing sections, then there is a splitting $E = E' \oplus R\sumpower{k}$ where $\dim E' = n-k$.
  Thus $w_i(E) = 0$ for $i> n-k$.

- If $E_1 \oplus E_2$ is trivial, then $w(E_1) \cupprod w(E_2) = 1$.
:::

:::{.example title="?"}
If $M \subseteq \RR^N$, then $w(\T M) \cupprod w(\nu M) = 1$.
If $S^n \subseteq \RR^{n+1}$, then $w(\T S^n) = 1$ since $w(\nu S^n) = 1$.
So for example $w(\T S^2) = 1$, but $\T S^2$ has no nonvanishing sections.
:::

:::{.remark}
Consider inverting a formal power series $\sum_{i\geq 0} a_i$:
\[
(1 + a_1 + a_2 + a_3 + \cdots)( 1 - a_1 + (a_1^2 + a_2) + (a_1^3 + 2a_1 a_2 - a_3)  + \cdots) = 1
.\]

So if $w(E_1)w(E_2) = 1$, we can solve for $w(E_1)$ in terms of $w(E_2)$.
:::

:::{.proposition title="?"}
Note that $H^*(\RP^n; C_2) = \FF_2[a]/\gens{a^{n+1}}$ where $H^1(\RP^n; C_2) = \gens{a}$.
Claim: $w(\T \RP^n) = (1+a)^{n+1}$
:::

:::{.proof title="?"}
Let $\gamma\to \RP^n$ be the canonical line bundle, which is a pullback of $\gamma^\infty \to \RP^\infty$ and $\gens{ w_1(\gamma^\infty) } =   H^1(\RP^\infty; C_2)$.
By the naturality property, $w_1(\gamma) = \alpha$ in $H^1(\RP^n; C_2)$.


:::{.lemma title="?"}
\[
\T\RP^n = \Hom(\gamma, \gamma^\perp)
.\]
:::

Recall that $\gamma \subseteq \RP^n \times \RR^{n+1}$ as a subbundle over $\RP^n$, and $\gamma = \ts{ ([\vector x], \lambda \vector x) \st \lambda \in \RR } = \pi\inv[\vector x]$.
We can write $\T \RP^n = \T S^n/ (\vector x, \vector v)\sim (- \vector x, - \vector v)$, so 
\[
\T\RP^n 
= \ts{ [(x, v), (- \vector x, - \vector v)] \st v\in \T_x S^n }
= \ts{ [(x, v), (- \vector x, - \vector v)] \st v\in \T_x S^n \iff \inner{x}{v} = 0 }
.\]

So define a map
\[
\T\RP^n &\to \Hom( \gamma, \gamma^\perp) \\
[(\vector x, \vector v), (-\vector x, \vector v)] &\mapsto \qty{\lambda \vector x \mapsvia{\ell} \lambda \vector v }
,\]
and one can check that this is well-defined.

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We can thus write
\[
\T\RP^n \oplus \RR 
&= \Hom( \gamma, \gamma^\perp) \oplus \Hom( \gamma, \gamma) \\
&= \Hom( \gamma, \gamma^\perp \oplus \gamma) \\
&= \Hom(\gamma, \RR^{n+1}) \\
&= \Hom(\gamma, \RR)\sumpower{(n+1)} 
.\]

Note that $\Hom( \gamma, \RR) \cong \gamma$, so $w(\T\RP^n \oplus \RR) = w(\gamma)^{n+1} = (1+a)^{n+1}$ and $w(\T\RP^n \oplus \RR) = w(\T\RP^n)$.
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:::{.corollary title="?"}
$w(\T\RP^n) = 1$ iff $n+1=2^r$ for some $r$.
:::

:::{.proof title="?"}
$\impliedby$:
By induction, $(1+a)^{2^r} = 1 + a^{2^r}$, using that $(1+a)^2 = 1 + a^2 + 2a$ and $2a=0$ when we have $C_2$ coefficients.
Now write
\[
(1+a)^{2^r} = ( (1+a)^{2^{r-1}} )^2 \equalsbecause{IH} (1 + a^{2^{r-1}})^2 = 1 + a^{2^r}
.\]
Now use that $\dim \T\RP^n = n$, so $(1+a)^{n+1} = 1$ since $a^{2^r} = a^{n+1} = 0$.

$\implies$:
Suppose $n+1 = m2^r$ with $m$ odd, then
\[
(1 + a)^{ m2^{r}}
= ((1+a)^{2^r})^m \\
= (1+a^{2^r})^m \\
= 1 + ma^{2^r} + \cdots
,\]
where the first nontrivial term doesn't vanish since $a^{2^r}\neq 0$.
:::

:::{.lemma title="?"}
If $\RP^{2^r}$ admits an immersion into $\RR^N$, then $N\geq 2^{r+1} - 1$.
:::

:::{.proof title="?"}
Why?
\[
w(\T \RP^{2^r}) = (1+a)^{2^r+1} = (1+a)^{2^r}(1+a) \\
= (1+a^{2^r})(1+a) \\
= 1 +a + a^{2^r} + a^{2^r + 1} \\
= 1 +a + a^{2^r} 
.\]
Try to invert this:
let $n\da 2^r$, then 
\[
(1+a+a^n)(1 + a + a^2 + \cdots + a^{n-1}) = 1
.\]

Then
\[
w(\nu \RP^n) = \sum_{0\leq i \leq n-1} a^i
,\]
so $\dim \nu \RP^n \geq n-1$.
:::