Define the Mobius strip $E$ to be the quotient $\mathbb{R}^{2}/\sim$ where

$\left(x_{1},y_{1}\right)\sim\left(x_{2},y_{2}\right)$ if and only if $$\left(x_{2},y_{2}\right)=\left(x_{1}+2\pi k,\left(-1\right)^{k}y_{1}\right)$$ for some integer $k$. It is a smooth vector bundle over $\mathbb{S}^{1}$ whose fiber at every point $e^{i\theta}\in \mathbb{S}^{1}$ is $E_{\exp\left(i\theta\right)}=\left\{ \left[\left(\theta,y\right)\right]:y\in\mathbb{R}\right\}$ with the obvious vector space structure (here, $\left[\left(\theta,y\right)\right]$ is a typical element of $E$) . I want to show that this smooth vector bundle is nontrivial. This is equivalent to showing that every smooth global section $s:\mathbb{S}^{1}\rightarrow E$ must vanish somewhere. This is how I proceeded:

Suppose that there exists a smooth global nowhere vanishing section $s:\mathbb{S}^{1}\rightarrow E$. We seek to find a contradiction by showing that $s$ must vanish somewhere. Define a smooth map $F:\left[0,2\pi\right)\rightarrow\mathbb{R}$ as follows. Given a real number $x\in\left[0,2\pi\right)$ , define $F\left(x\right)$ to be the real number such that $s\left(\exp\left(ix\right)\right)=\left[\left(x,F\left(x\right)\right)\right]$. It remains to show that there exists a real number $x\in\left[0,2\pi\right)$ such that $F\left(x\right)=0$. This will complete the proof.

I am stuck at this point. I think that I might have to use the intermediate value theorem to show that $F\left(x\right)=0$, so it suffices to show that $F$ changes sign somewhere. I am unsure how to show this however.


1 Answer 1


You can enlarge the domain of $F$ to the closed interval $[0,2\pi]$ using the same formula (and you'll have to check that $F$ is continuous, which you can do by checking it separately on $[0,2\pi)$ and on $(0,2\pi]$).

Then use the fact that $F(0)=-F(2\pi)$.

  • $\begingroup$ I think you can only understand a section $s: S^1 \rightarrow E$ using a formula such as $s(e^{i \theta}) = (\theta, F(\theta))$ if the bundle is trivial, i.e. if $E= S^1 \times \mathbb{R}$ (I am writting $\mathbb{R}$ as the fibre). Otherwise it doesn't make sense defining sections, since it is much easier to define directly the function $F: S^1 \rightarrow \mathbb{R}$, right? Or you actually can define $s(e^{i \theta}) = (\theta, F(\theta))$, as long as you don't take the full $[0, 2 \pi]$ (and take for instance $[0, 2 \pi)$)? $\endgroup$
    – MBolin
    Jul 28, 2020 at 21:57

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