Find a nontrivial bundle of $S^1$ with fibre isomorphic to $\mathbb{R}^n$ Show that such a nontrivial bundle exists for every $n\in\mathbb{N}$.
I don't really have any useful ideas here. I'm not sure if there is a general approach I should be taking or if there is just a standalone specific bundle/method that works here. Any help is appreciated.
 A: We know that there are two line bundles (ie. fiber $=\mathbb R$) on $S^1$, namely the Mobius band $M\to S^1$ and the trivial bundle $\epsilon ^1\to S^1$.
Now if we take $E= M \oplus \mathbb \epsilon^{n-1}$, where $\epsilon ^{n-1}$ is the trivial $(n-1)$ - vector bundle, we have its Stiefel Whitney class $\omega(E)= \omega (M)\omega (\epsilon ^{n-1})= \omega (M) $ which is non-trivial, hence $E$ is non-trivial.
If you want to argue with orientability, going once along a generator of $\pi_1S^1$ will give you an orientation reversing base change of the fiber $M$. Hence it will also give you one for $E$, namely:
$$
\left( \begin{matrix}
-x & 0 & 0 &\cdots \\
0 & 1 & 0& \cdots \\
0& 0&1 &\cdots \\
\vdots &\vdots & \vdots& \ddots
\end{matrix} \right),
$$
where the identity block is $(n-1)$ in size and comes from a trivialization of $\epsilon ^{n-1}$. Can you see what's going on?
In the latter variant you conclude by saying that orientability is an isomorphism invariant of a vector bundle and that trivial bundles are orientable.
