This question was asked in my exercise of differential geometry and I am not able to prove what I was asked.
Question: Denote by Mob, the Mobius "strip" which is vector bundle over $S^{1}$ with typical fiber $\mathbb{R}$. Then show that there exists a global smooth vector field on Mob that is nowhere $0$.
Unfortunately, I am not able to construct such a vector field. Can you please give some leads?
Thank you!
I am not a differential geometer, and I am sure there are much more elegant ways to write this down. You should try to find one such way yourself.
Write the Möbius strip $M$ as the strip $I\times \mathbb{R}$, $I=[0,1]$ with the points $(0,x)$ and $(1,-x)$ glued together. Then giving a nonvanishing vector field on $M$ is equivalent to just giving a smooth (including being smooth at the glued points) function $(\xi,\eta):I\times \mathbb{R}\to \mathbb{R}^2$ such that $\xi(0,x)=\xi(1,-x),\ \eta(0,x)=-\xi(1,-x).$ Here $\xi$ and $\eta$ correspond to the horizontal and vertical direction.
And now you can just take your nonvanishing vector field on a circle to get $\xi$ and take $\eta$ as identically zero.
The geometric picture is this: you just take a vector field orthogonal to the fibers $\mathbb{R}$ of the strip.