# Tuesday, August 31


:::{.remark}
We want to show the equivalence between (isomorphism classes) of fiber bundles with $G$ structures with fiber $F$ and principal $G\dash$bundles.
Recall that $\Prin\Bun\slice G$ are fiber bundles $P \mapsvia{\pi} B$ with a right fiberwise $G\dash$action which is free and transitive on each fiber.


To send fiber bundles to principal bundles, we used a *mixing* construction.
Since $G\actson F$, we get an identification $G \subseteq \Homeo(F, F)$.
We constructed 
\[
P \fiberprod{G} F \da (P\cross F)/(pg, f)\sim (p, gf)
.\]
A lemma was that $P\fiberprod{G} F \to B$ is a fiber bundle with fiber $F$ and projection $\pi(p, f) \da \pi(p)$.

Today we'll talk about the reverse direction.
Note the composition of sending $E$ to $\Prin\Bun\slice G$ and then mixing recovers $E$ when $E$ is a vector bundle, but not generally.
\[
\correspond{
  \text{Fiber bundles with }G\dash\text{structures} \\
  \text{ and fiber }F
}
\adjunction{\text{Clutching}}{\text{Mixing}}{}{}
\correspond{
  \text{Principal $G\dash$bundles}
}
\]
:::




:::{.example title="?"}
For $E \mapsvia{\pi} B$ a real vector bundle, we sent it to $\Frame(E)$, which is a principal $\GL_n(\RR)\dash$action
Using a left action $\GL_n \actson \RR^n$, we can form $\Frame(E) \fiberprod{\GL_n} \RR^n$, a fiber bundle with a $G \da \GL_n$ structure, i.e. exactly a vector bundle.

:::{.exercise title="?"}
Show that there is a homeomorphism 
\[
\Frame(E) \fiberprod{\GL_n} \RR^n \mapsvia{\sim}E
.\]
:::

For the reverse map, take a map $f$ defined by $(\vector e_1, \cdots, \vector e_n) \in \pi\inv(b) \subset \Frame(E)$ and $\tv{\elts{b}{n}}^t \in \RR^n$ to $\sum_{i=1}^n b_i \vector e_i$.
For this to be well-defined, one needs to show the following:
\[
f(( \vector e_1, \cdots, \vector e_n) A, \vector b) = f( (\vector e_1, \cdots, \vector e_n), A\vector b) && \forall A\in \GL_n(\RR)
.\]
The left hand side is
\[
b_1 (a_{1, 1} \vector e_1 + \cdots + a_{n, 1} \vector e_n ) + \cdots + b_n(a_{1, n} \vector e_1 + \cdots + a_{n, n} \vector e_n)
= \sum_{i=1}^n b_i \qty{ \sum_{j=1}^n a_{j, i} \vector e_j }
.\]

The right-hand side is
\[
(a_{1, 1}b_1 + \cdots + a_{1, n}b_n)\vector e_1 + \cdots + (a_{n, 1} b_1 + \cdots + a_{n,n} b_n)\vector e_n
= \sum_{i=1}^n \qty{\sum_{j=1}^n a_{i, j} b_i } \vector e_i
,\]
and one can check that these sums match term by term. 
:::

:::{.remark}
Note that if we choose a basis for the fibers, we can set $A' \da \tv{\vector e_1, \cdots, \vector e_n}^t$ to be the matrix with columns $\vector e_i$, the map $f$ is given by $f(A', \vector b) \da A'\vector b$, and we're showing that $(A'A)\vector b = A'(A\vector b)$.
However, this involves choosing an isomorphism between the abstract fibers and $\RR^n$.
:::


:::{.remark}
What are local charts for a principal bundle?
For $P\fiberprod{G} F$, pick charts $(U, \phi)$ for $P \mapsvia{\pi} B$:
\[
\phi: \pi\inv(U) &\to U \times G \\
x & \mapsto (\pi(x), \gamma(x))
.\]
Then a local chart for the principal bundle is of the form
\[
\pi\inv(U) \fiberprod{G} F &\mapsvia{\tilde \phi} U \cross F \\
(x, f) &\mapsto ( \pi(x), \gamma(x) f)
.\]
We also have
\[
(U \times G)\fiberprod{G} F &\to U \cross F \\
( (x, g), f) &\mapsto (x, gf)
.\]
One can invert $\tilde \phi$ using $(a, f) \mapsto (\phi\inv(a, 1), f)$.
This yields transition functions: writing \[
\phi_V: \pi\inv(V) &\to V \cross G \\ 
x &\mapsto (\pi(x), \psi(x) )
,\]
then
\[
\phi_{VU} = \phi_V \circ \phi_U\inv: 
(a, f) & \\
&\mapsvia{\phi_U\inv} ( \phi\inv_U(a, 1), \,\, f) \\
&\mapsvia{\phi_V} (\pi \phi_U\inv(a, 1), \,\, \psi( \phi_U\inv(a, 1))f ) \\
&= (a,\,\, \psi(\phi_U\inv(a, 1)) f)
.\]
This says that $(a, 1) \mapsto \psi( \phi\inv_U(a, 1))$.
:::



:::{.remark}
In general, for a bundle $E \mapsvia{\pi} B$, taking local trivializations $\phi_U, \phi_V$, we get $\phi_{VU}: (U \intersect V ) \cross F \selfmap$, or currying an argument, $\phi_{VU}: U \intersect V \to \Homeo(F, F)$.
If the bundle satisfies the cocycle condition $\phi_{UW} = \phi_{VW} \circ \phi_{UV}$.
Given a covering $\ts{U_i}_{i\in I}\covers B$, we get $\phi_{ij}: U_i \intersect U_j \to G$ and a topological space $F$ with $G \subseteq \Homeo(F, F)$ satisfying the cocycle condition $\phi_{ik} = \phi_{jk} \circ \phi_{ij}$, then we can build a fiber bundle with fiber $F$ and structure group $G$ by setting $E = \disjoint_{i\in I} (U_i \cross F)/\sim$
We then set for $b\in U_i \intersect U_j$ the equivalence 
\[
(U_i \cross F) \ni (b, f)\sim (b, \phi_{ij}(b) f) \in (U_j \cross F)
.\]
This is an equivalence relation precisely when the cocycle condition holds.
This is referred to as **clutching data**.
:::


:::{.example title="The Mobius band is clutch"}
Let $\ZZ/2\actson \RR$ by $t\mapsto -t$ with $U, V$ defined as follows:

\begin{tikzpicture}
\fontsize{41pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/CharacteristicClasses/sections/figures}{2021-08-31_13-53.pdf_tex} };
\end{tikzpicture}

Labeling the intersections as 1, 2, we set
\[
\phi_{VU}: (U \intersect V) = (U \intersect V)_1 \disjoint (U \intersect V)_2 &\to \ZZ/2 && \subseteq \Homeo(\RR) \\
x \disjoint y &\mapsto x \disjoint -y
.\]
This yields the open Mobius band.
:::


:::{.question}
Actually, several questions.
Assume $F$ is a fixed fiber common to all of the following constructions, since bundles with non-homeomorphic fibers can't be isomorphic.

1. Given clutching data $\ts{\phi_{ij}}$, when is the resulting fiber bundle trivial?
2. Given two sets of clutching data $\ts{\phi_{ij}}$ and $\ts{\psi_{ij}}$ with the same open cover $\ts{U_i} \covers X$, when are the corresponding bundles $G\dash$isomorphic?
3. Given two sets of clutching data $\ts{\phi_{ij}}$ and $\ts{\psi_{ij}}$ with the *different* open cover $\ts{U_i} \covers X$ and $\ts{V_i} \covers X$, when are the corresponding bundles $G\dash$isomorphic?

:::

:::{.lemma title="?"}
The fiber bundle obtained from $\phi_{ij}$ is trivial iff there exists a map $\gamma_i: U_i \to G$ such that $\phi_{ij} = \gamma_i \gamma_j\inv$.
:::


:::{.proof title="?"}
The trivial bundle is $B\times F\to B$, so if we have $E\to B$, we can take a map
\[
U_i \cross F &\to U_i \cross F \\
(b, f) &\mapsto (b, \gamma_i(b) f)
.\]
Use that $B\times F$ is a trivial bundle, so it is its own trivialization.

> To be continued next time.

:::