# Pontryagin Classes (Tuesday, November 23)

## Complexification

:::{.question}
How to we get integral cohomology classes when we don't have a complex structure?
:::

:::{.definition title="Complexification"}
Given a real vector bundle $E$, its **complexification** is defined as
\[
E\tensor_\RR \CC
\da \Disjoint_{x\in X} E_x\tensor_\RR \CC
.\]
Then $\dim_\RR E\tensor_\RR \CC = 2\dim_\RR E$.
:::

:::{.remark}
In terms of transition functions (i.e. Čech cohomology data), this is the inclusion $\GL_n(\RR) \injects \GL_n(\CC)$.
We can now consider the Chern classes $c_i(E\tensor \CC)$.
:::

:::{.definition title="Conjugate bundle"}
Given a complex vector bundle $E$, there is a **conjugate bundle** $\bar{E}$ with the complex structure $i_{\bar E}\cdot V\da -i_{E}\cdot V$.
This corresponds to conjugating Čech cohomology data, i.e. replacing transition functions $f_{ij}$ with $\bar{f_{ij}}$.
:::

:::{.lemma title="?"}
For any real vector bundle $E$, 
\[
E\tensor \CC \cong \bar{\EE\tensor \CC}
.\]
:::

:::{.proof title="?"}
The transition functions for the former are in $\GL_n(\RR)$, which is fixed under conjugation.
:::

:::{.remark}
Note that $E\tensor \CC \cong E \oplus E$, and we can map
\[
\RR\sumpower{2}\tensor \CC &\iso \CC\sumpower{2} \\
(x\oplus y) \tensor \gens{1, i} &\mapsto (x, ix) \oplus (y, iy)
.\]
:::

:::{.exercise title="?"}
Show that if $E$ is a complex vector bundle,
\[
E\tensor \CC \cong E \oplus \bar{E}
,\]
and 
\[
c_i(\bar E) = (-1)^i c_i(E)
.\]
:::

## Pontryagin Classes

:::{.lemma title="?"}
For $i$ odd, $c_i(E\tensor \CC)$ is 2-torsion.
:::

:::{.proof title="?"}
On one hand, 
\[
E\tensor \CC \cong \bar{E\tensor \CC} \implies
c_i(E\tensor \CC) = c_i(\bar{E\tensor \CC}) = (-1)^i c_i(E\tensor \CC)
.\]
So if $i$ is odd, $c_i = -c_i$ and thus $2c_i = 0$.
:::

:::{.definition title="Pontryagin classes"}
The **$i$th Pontryagin class** is
\[
p_i(E) = (-1)^{i} c_{2i} (E\tensor \CC) \in H^{4i}(X)
.\]
:::

:::{.remark}
Suppose $R$ is a coefficient ring where 2 is invertible (or not a zero divisor).
Some properties of the $p_i$:

a. $p_i$ is natural, so $p_i (f^* E) = f^* p_i(E)$.
b. $p_{i> \dim(E)/2}$.
c. $p(E \bigoplus E') = p(E) \cupprod p(E')$.
  Why? Note $p(E) = 1 - c_2 + c_4 - c_6 + \cdots$, so multiplying two such things yields
  \[
  p(E) p(E') = 1 - (c_2 + c_2') + (c_4 + c_2c_2' + c_4') - (c_6 + c_4 c_2' + c_2 c_4' + c_6') - \cdots = p(E \oplus E')
  .\]
d. If $\dim_\RR E = 2n$ then $p_n(E) = e(E)^2$.
  Why? 
  \[
  p_n(E) = (-1)^n c_{2n}(E\tensor \CC) = (-1)^n e(E\tensor \CC) = (-1)^n (-1)^{2n(2n+1) \over 2} = (-1)^{n + n(2n+1)} e(E)^2 = (-1)^{n(2n+2)}e(E)^2 = e(E)^2
  .\]

e. If $E$ is a complex line bundle,
\[
c(E \oplus \bar E) = c(E\tensor \CC) = 1 - p_1(E) + p_2(E) - \cdots = c(E) c(\bar E) = (1 + c_1(E) + c_2(E) + \cdots)(1 - c_1(E) + c_2(E) - \cdots)
.\]

:::

:::{.example title="?"}
We can compute $p(\T \CP^5)$.
Recall that $c(\T \CP^n) = (1+a)^{n+1}$ where $\gens{a} = H^2(\CP^n; \ZZ)$ is the positive generator.
So $c(\T\CP^5) = (1+a)^6 = 1 + {6\choose 1}a + {6\choose 2}a + \cdots$.
Using (e) above, we have $c(E) c(\bar E)= (1+a)^6(1-a)^6 = (1-a^2)^6 = 1-6a^2 + 15a^4$.
So $p(\T \CP^5) = 1 + 6a^2 + 15a^4$.
:::

:::{.corollary title="?"}
If $\CP^5\injects \RR^N$ immerses, then $p(\nu \CP^5) = {1\over 1 + 6a^2 + 15a^4} = 1 - 6a^2 + 21a^4$.
So $p_2(\nu \CP^5) \neq 0$, so $\dim(\nu)/2 \geq 2 \implies \dim(\nu) \geq 4$ and this forces $n\geq 14$.
Note that here we used that $p_i(E) = 0$ for $i > \dim E/2$.
:::

:::{.theorem title="?"}
\[
H^*(\BSO_{2n+1}; R) &\iso R[p_1,\cdots, p_n] \\
H^*(\BSO_{2n}; R) &\iso R[p_1,\cdots, p_{n-1}, e] \\
H^*(\BO_{2n+1}; R) &\iso H^*(\BO_{2n}; R) \iso R[p_1,\cdots, p_{n}] \\
,\]
where $p_i$ are the Pontryagin classes for the respective canonical bundles.
:::