\classheader{2013-01-30}

To construct a vector bundle on a base $B$, we can take an open cover $B=\bigcup_{i\in I}U_i$ and appropriate transition functions $\varphi_{ij}:U_i\cap U_j\to\GL_k(\R)$ (appropriate just means that if we are doing topological manifolds, they should be continuous; $\Cinfty$ manifolds, they should be $\Cinfty$; etc.) that satisfy the cocycle condition $\varphi_{ij}(x)\varphi_{jk}(x)=\varphi_{ik}(x)$ for all $x\in U_i\cap U_j\cap U_k$.

We then construct the vector bundle as
\[V:=\coprod_{i\in I}(U_i\times\R^k)\bigg/{\sim}\]
where $\sim$ is the equivalence relation generated by
\[(x,y)\in U_i\times\R^k\sim (x,\varphi_{ji}(x))\in U_j\times\R^k \text{ for all }x\in U_i\cap U_j.\]
The bundle map $p:V\to B$ is the map induced by the projection maps $p_1:U_i\times\R^k\to U_i$.

Given vector bundles $p_1:V_1\to B$ and $p_2:V_2\to B$, and letting $V_i(x)=p_i^{-1}(x)$ for each $x\in B$, we can construct new vector bundles
\[V_1\oplus V_2,\quad V_1\otimes_\R V_2,\quad V^*,\quad \Lambda^m(V),\quad\text{etc.}\]
whose fibers at $x\in B$ are $V_1(x)\oplus V_2(x)$, $V_1(x)\otimes_\R V_2(x)$, etc. 

For example, we normally define $V_1\oplus V_2=V_1\times_B V_2$, but here is another way: if $V_1$ and $V_2$ are the vector bundles constructed by $\varphi_{ij}':U_i\cap U_j\to \GL_k(\R)$ and $\varphi_{ij}'':U_i\cap U_j\to \GL_\ell(\R)$ respectively, we compose with the group homomorphism $\rho:\GL_k(\R)\times\GL_\ell(\R)\to\GL_{k+\ell}(\R)$ sending
\[(A',A'')\;\;\longmapsto\;\;\begin{pmatrix}
A' & 0\\ 0 & A''
\end{pmatrix},\]
and we see that $\rho\circ(\varphi_{ij}',\varphi_{ij}'')$ is a map from $U_i\cap U_j$ to $\GL_{k+\ell}(\R)$ that also satisfies the cocycle condition. Then $V_1\oplus V_2$ is the associated vector bundle to this information.

For the tensor product, the construction is essentially the same; we now take the homomorphism $\rho:\GL_k(\R)\times\GL_\ell(\R)\to\GL(\R^k\otimes_\R\R^\ell)$ sending $(A',A'')$ to $A'\otimes A''$.

For $V^*$, we take $\rho:\GL_k(\R)\to\GL_k(\R)$ defined by $\rho(A)=(\transpose{A})^{-1}$.

For $\Lambda^m(V)$, we take $\rho:\GL(\R^k)\to\GL(\Lambda^m(V))$ defined by $\rho(T)=\Lambda^m(T)$.

\begin{definition}
We define the bundle of frames of a vector bundle as follows. Let $p:V\to B$ be a real vector bundle of rank $k$. We can construct the $k$-fold fiber product $V\times_B\cdots\times_B V$, and we define
\[P=\{(v_1,\ldots,v_k)\in V\times_B\cdots\times_B V\mid v_1,\ldots,v_k\text{ are linearly independent}\}.\]
This is an open subset of the $k$-fold fiber product. It consists of the smoothly consistent ways of taking isomorphisms $p^{-1}(x)\xleftarrow{\cong}\R^k$ for each $x$. Note that for $\psi\in P$ and $g\in \GL_k(\R)$, we get $\psi g\in P$. Thus, we have a fiber bundle $\pi:P\to B$ with a right $G$-action on $P$, under which each fiber is stable (in fact the action on each fiber is simply transitive); in other words, the diagram
\begin{center}
\begin{tikzcd}
P\times G\ar{r}{\text{action}} \ar{d}[swap]{p_1} & P\ar{d}{\pi}\\ P\ar{r}[swap]{\pi} & B
\end{tikzcd}
\end{center}
commutes. Such a fiber bundle is, by definition, a principal $G$-bundle on $B$. You can see some different takes on this in Kobayashi and Nomizu's \textit{Foundations of Differential Geometry}, and Steenrod's \textit{Topology of Fiber Bundles}.
\end{definition}


\begin{definition}
Given a topological group $G$, a principal $G$-bundle $P$ on $B$, and an action of $G$ on a topological space $F$, the associated fiber space $P\times_G F$ is defined to be $(P\times F)/{\sim}$ where $\sim$ is the equivalence relation generated by
\[(zg,y)\sim(z,gy)\text{ for all $z\in P$, $y\in F$, and $g\in G$.}\]
In fact, $P\times_G F$ is a fiber bundle on $B$ with fiber $F$. The following diagram commutes:
\begin{center}
\begin{tikzcd}[row sep=0.25in]
P\times F \ar{rd} \ar{d}\\
P \ar{d} & (P\times F)/{\sim}\ar{ld}\\
B
\end{tikzcd}
\end{center}
\end{definition}

Given a homomorphism $\rho:G\to\GL_n(\R)$, we obtain an action of $G$ on $\R^n$. Taking our fiber $F$ to be $\R^n$, we can construct the associated fiber space $P\times_G\R^n$, which is a vector bundle on $B$. This is an equivalent way of constructing vector bundles - all of the constructions mentioned above can be done in this way too. For example, given a vector bundle $V$, we can construct $\Sym^m(V)$ by taking the associated fiber space to the natural homomorphism $\rho:\GL(\R^k)\to\GL(\Sym^m(\R^k))$.


Let $X$ be a topological space that is decent enough to have a universal covering space $\pi:\widetilde{X}\to X$, and let the group of covering transformations $\Gamma$ be given the discrete topology. Then $\widetilde{X}$ is a principal $\Gamma$-bundle on $X$, and we can consider associated fiber spaces, vector bundles, etc.

We often have an action of $\pi_1(X,x_0)$ on things. For example, if $p:E\to B$ is a fiber bundle and $B$ is a manifold or simplicial complex, then $H^i(p^{-1}(x_0))$ is an abelian group with an action of $\pi_1(B,x_0)$. For any $\gamma\in\pi_1(X,x_0)$, the corresponding action arises from the commutative diagram
\begin{center}
\begin{tikzcd}
\gamma^*E:=I\times_BE \ar{r}{p_2}\ar{d}[swap]{p_1} & E\ar{d}{p}\\ I=[0,1] \ar{r}[swap]{\gamma} & B
\end{tikzcd}
\end{center}
Here are the details of the construction.
\begin{lemma}
Every fiber bundle on $I$ is trivial.
\end{lemma}
\begin{proof}
There is a finite open cover of $I$ where the fiber bundle is trivial on each element (because $I$ is compact). By induction on the number of elements on the cover, we are reduced to the case that there are two elements of the cover. WLOG, we'll assume that the fiber bundle is trivial on (open sets containing) $[0,\frac{1}{2}]$ and $[\frac{1}{2},1]$, say with trivializations

\begin{center}
\begin{tikzcd}[column sep=0.1in]
\pi^{-1}({[}0,\frac{1}{2}{]}) \ar{rr}{\varphi} \ar{rd}[swap]{\pi} & & {[}0,\frac{1}{2}{]}\times F\ar{ld}{p_1}\\ 
& {[}0,\frac{1}{2}{]}
\end{tikzcd}\qquad
\begin{tikzcd}[column sep=0.1in]
\pi^{-1}({[}\frac{1}{2},1{]}) \ar{rr}{\psi} \ar{rd}[swap]{\pi} & & {[}\frac{1}{2},1{]}\times F\ar{ld}{p_1}\\ 
& {[}\frac{1}{2},1{]}
\end{tikzcd}
\end{center}

You can find a $c$ such that
\begin{center}
\begin{tikzcd}[row sep=0.2in]
 & F\\
\pi^{-1}(\frac{1}{2}) \ar{ru}{\phi|}[swap]{\cong} \ar{rd}{\cong}[swap]{\psi|}\\
& F\ar{uu}[swap]{c}
\end{tikzcd}
\end{center}
where $\phi|$ and $\psi|$ denote $\phi$ and $\psi$ suitably restricted.


{\color{myred}{\rule{\textwidth}{0.02in}}\vspace{-0.1in}\bf
\begin{center}
$\blacktriangledown$\quad Why is $\alpha$ smooth?\quad $\blacktriangledown$
\end{center}}

There is then an isomorphism with the trivial bundle, $\alpha:E\to [0,1]\times F$, defined by
\[\alpha(e)=\begin{cases}
\varphi(e) & \text{ if }e\in\pi^{-1}([0,\frac{1}{2}]),\\
( \id_{[\frac{1}{2},1]}\times c)(\psi(e)) & \text{ if } e\in\pi^{-1}([\frac{1}{2},1]).
\end{cases}\qedhere\]
\end{proof}
\vspace{-0.1in}{\color{myred}{\rule{\textwidth}{0.02in}}\vspace{-0.1in}\bf
\begin{center}
$\blacktriangledown$\quad Are the isomorphisms independent of the trivialization of $\pi$ chosen?\quad $\blacktriangledown$
\end{center}}
\begin{corollary-N}
If $\pi:E\to I$ is a fiber bundle, then the inclusions $\pi^{-1}(0)\to E$ and $\pi^{-1}(1)\to E$ are homotopy equivalences, so we get induced isomorphisms
\begin{center}
\begin{tikzcd}
H^i(\pi^{-1}(0)) & H^i(E)\ar{l}[swap]{\cong} \ar{r}{\cong} & H^i(\pi^{-1}(1)).
\end{tikzcd}
\end{center}
\end{corollary-N}
\vspace{-0.1in}{\color{myred}{\rule{\textwidth}{0.02in}}\vspace{-0.1in}\bf
\begin{center}
$\blacktriangledown$\quad Aren't these homomorphisms in fact isomorphisms?\quad $\blacktriangledown$
\end{center}}
\begin{corollary-N}
Let $\pi:E\to B$ be a fiber bundle and let $\gamma:[0,1]\to B$ be a path from $\gamma(0)=x$ to $\gamma(1)=y$. We get an induced homomorphism $H^i(\pi^{-1}(x))\to H^i(\pi^{-1}(y))$, as well as an induced homomorphism $H_i(\pi^{-1}(y))\to H_i(\pi^{-1}(x))$.
\end{corollary-N}
\begin{proof}
This is Corollary 1 applied to $\gamma^*E$.
\end{proof}
\vspace{-0.1in}{\color{myred}{\rule{\textwidth}{0.02in}}}

We're out of time, so let me request that you read the definitions of direct limit, presheaf, stalk, sheaf, and sheafification. Spanier is a good reference for this.


