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I'm trying to solve the following question:

Consider the vector bundle

$$E:= [0,1] \times \mathbb{R}/\sim \quad \to S^1$$

where $\sim$ is the equivalence rleation $(0,t) \sim (1,-t)$ for all $t \in \mathbb{R}$. Does there exist a section of $E$ that is nowhere vanishing?

How can I visualise $E$? It looks like some kind of infinite Möbius strip. Also, how is the map $E \to S^1$ explicitely defined? I cannot see what map is meant here, but there should be an obvious choice.

  • Projection onto the first factor gives the mapping to $S^1$, noting that the fiber over $0$ has been glued to the factor over $1$, so you're identifying the endpoints of the interval downstairs. – Ted Shifrin Jan 01 '20 at 19:08
  • I don't quite get it. We don't come back in the same point when we get to $1$, starting in $0$? –  Jan 01 '20 at 19:11
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    Yes, $[0,1]/(0\sim 1)$ is homeomorphic to the circle. But, as you said, this is the Möbius bundle — you've glued the fibers by the map $t\rightsquigarrow -t$ when you identify $0$ and $1$. – Ted Shifrin Jan 01 '20 at 19:16

1 Answers1

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The map $E\to S^1$ is explicitly defined by $\overline{(x,t)}\mapsto e^{2i\pi x}.$ You can show that it is an homeomorphism when restricted to $[0,1]\times\{0\}/\sim$.

Then, you can show that the (continuous) sections of this bundle are in 1-1 correspondance with continuous maps $\sigma:[0,1]\to\mathbb{R}$ such that $\sigma(0)=-\sigma(1)$ by sending $\sigma$ to $\tilde{\sigma}:S^1\to E$ defined by $\tilde{\sigma}(e^{2i\pi x})=\overline{(x,\sigma(x))}$, with $0\leq x\leq 1$. Finally, the existence of a nowhere vanishing section would imply the existence of a nowhere vanishing function $\sigma:[0,1]\to\mathbb{R}$ with $\sigma(0)=-\sigma(1)$. Since $\sigma(0)\neq 0$, then $\sigma(0)$ and $\sigma(1)$ are of opposed signs. By continuity of $\sigma$, it contradicts the intermediate value theorem.

Edit to answer your questions:

Sounds good to me!

For $\sigma(0)=-\sigma(1)$, note that $\tilde{\sigma}:S^1\to E$ is a section, which means that for all $x\in[0,1]$ there is a $y_x$ such that $\tilde{\sigma}(e^{2i\pi x})=\overline{(x,y_x)}$. Note that $\tilde{\sigma}(1)=\tilde{\sigma}(e^{2i\pi 0})=\overline{(0,y_0)}$ and $\tilde{\sigma}(1)=\tilde{\sigma}(e^{2i\pi 1})=\overline{(1,y_1)}$, so by definition of the equivalence relation we get $y_0=-y_1$. Since with our definition we get $\sigma(0)=y_0$ and $\sigma(1)=y_1$, we get the wanted result.

A non-vanishing section $s:B\to E$ of a vector bundle is a section such that

$$\forall x\in B,s(x)\neq 0_{E_x}$$

where $0_{E_x}$ is the $0$ of the vector space $E_x=\pi^{-1}(x)$.

Balloon
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  • Thanks I'll go through it! –  Jan 02 '20 at 00:02
  • I am stuck showing it is bijective. How do you associate such a continuous map $\sigma: [0,1] \to \mathbb{R}$ with $\sigma(0) = - \sigma(1)$ with a continuous section? –  Jan 02 '20 at 09:48
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    Consider the maps $\varphi_0:E\to [0,1)\times\mathbb{R}$ and $\varphi_1:E\to(0,1]\times\mathbb{R}$ defined by $\varphi_0\big(\overline{(x,t)}\big)=\varphi_1\big(\overline{(x,t)}\big)=(x,t)$. These are continuous maps. Then, if $\tilde{\sigma}$ is a continuous section, define $\sigma(x)=(\mathrm{pr}_2\circ\varphi_0\circ\tilde{\sigma})(e^{2i\pi x})$ if $x\in[0,1)$ and the same with $\varphi_1$ if $x\in(0,1]$. Try to show that $\sigma$ is well-defined, continuous, and that $\sigma(0)=-\sigma(1)$. – Balloon Jan 02 '20 at 11:00
  • I will try! Thank you! –  Jan 02 '20 at 11:00
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    To see your map $\sigma$ is well-defined and continuous: well-defined is obvious because only the points with $x=0,1$ can possibly give an ambiguous value and there we have a unique image by construction. Continuity follows because we are continuous on $[0,1/2]$ and $[1/2,1]$ as composition of continuous functions and then we invoke the pasting lemma. Does that seem correct? –  Jan 18 '20 at 18:24
  • And why exactly is this map $\sigma$ that you defined in the comments non-vanishing? –  Jan 18 '20 at 19:15
  • I have added some edits to answer your questions. – Balloon Jan 18 '20 at 19:16
  • Thanks! I deleted these previous comments because I found them yourself, but still struggling with seeing that $\sigma$ is non-vanishing. I tried to show that if $\sigma$ vanishes somewhere, then $\tilde{\sigma}$ must also vanish somewhere, but I was unsuccesful. Can you explain why this is the case? –  Jan 18 '20 at 19:17
  • If $\sigma(x)=0$, then $\tilde{\sigma}(e^{2i\pi x})=\overline{(x,0)}=0_{E_{e^{2i\pi x}}}$, so the points $x\in[0,1]$ where $\sigma(x)=0$ (the $0$ of $\mathbb{R}$) give points $e^{2i\pi x}\in S^1$ where $\tilde{\sigma}(e^{2i\pi x})=0_{E_{e^{2i\pi x}}}$ (the $0$ of $E_{e^{2i\pi x}}=\pi^{-1}(e^{2i\pi x})$). – Balloon Jan 18 '20 at 19:23
  • I see that, but why is $(\overline{(x,0)}$ the zero vector of $E_{e^{2 \pi i x}}$? –  Jan 18 '20 at 19:27
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    Ah I see, we have $\pi^{-1}(e^{2\pi i x}) = {\overline{(x,t)}: t \in \mathbb{R}}$ and say we fix representatives with $0 \leq x < 1$, then we can just add/ scalar multiply these classes with the vector space structure inherited from $\mathbb{R}$, i.e.. $(x,t) + (x,s) := (x, t+s), \alpha (x,t) := (x, \alpha t)$. Looks correct? –  Jan 18 '20 at 19:36
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    Exactly, you formulated this quicker than me! – Balloon Jan 18 '20 at 19:38
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    Thanks for all the help. You helped me understand vector bundles much better. I wish you the best in life. –  Jan 18 '20 at 19:39
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    You are welcome, I wish you the best too. See you! – Balloon Jan 18 '20 at 19:39
  • I am facing similar issue, like when considering the charts for $\mathbb S^1$ how the computation alter? Here I post a question, could you please help me with that? @Balloon – N00BMaster Feb 15 '24 at 03:40