I'm currently self-studying through the book Differential forms in Algebraic Topology by Bott & Tu and got stuck on an exercise 6.10, which asks to compute $\textrm{Vect}_k(\mathbb S^1),$
the isomorphism classes of rank $k$ vector bundles over $\mathbb S^1$. I've gotten the result for $k=1$, which was simple since $O(1)\cong \mathbb{Z}/2\mathbb{Z}$, however, I'm not sure how to generalize to higher-rank vector bundles. Since the circle is a 1-manifold, it seems like the classification of higher vector bundles should be able to be deduced from the $k=1$ case. Is this true? If so, how? Any hints would be appreciated.
EDIT: I apologize, should've been more detailed about the $k=1$ argument. given a basic open cover $\{U,V\}$ of $S^1$ since $U\bigcap V$ is just two disconnected intervals, there will be four possibe cocycles (with values in $O(1)$). Similarly, there are very few possible transition functions between the cocycles so its easy to compute that these transition functions partition the four possible cocyles into two groups of two, one of which clearly corresponds to the product bundle and so the other must correspond to the isomorphism class of the Mobius bundle.
Best Answer
$\require{AMScd}$Consider the map $f:t\in[0,1]\mapsto\exp(2\pi it)\in S^1$. If $E$ is a vector bundle in $S^1$, then $f^*(E)$ is a vector bundle on $[0,1]$. It is not difficult to show that all vector bundles on $[0,1]$ are trivial. You can see from this that there is a continuous bundle map $F$ such that the diagram \begin{CD} [0,1]\times\mathbb R^n @>F>> E \\ @Vp_2VV @V\xi VV \\ [0,1] @>f>> S^1 \end{CD} commutes and $F$ is an isomorphism on fibers. In particular, we can construct a unique linear isomorphism $\phi_f:\mathbb R^n\to\mathbb R^n$ such that $F(1,\phi_f(v))=F(0,v)$ for all $v\in\mathbb R^n$.
Deduce from this that all vector bundles on $S^1$ are obtained from a trivial vector bundle $[0,1]\times\mathbb R^n\to[0,1]$ by identifying $(0,v)$ with $(1,\phi(v))$ for a fixed automorphism $\phi:\mathbb R^n\to\mathbb R^n$.
Next, show that if $\phi$ and $\psi$ are two automorphisms of $\mathbb R^n$ in the same path component of $GL(n,\mathbb R)$, then the correspinding bundles are isomorphic.