Timeline for For a fiber bundle $f:M\to S^1$, is there a point whose horizontal lift is a circle?
Current License: CC BY-SA 4.0
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Sep 24, 2020 at 22:35 | history | edited | YCor | CC BY-SA 4.0 |
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Sep 24, 2020 at 22:12 | answer | added | Gael Meigniez | timeline score: 1 | |
Sep 23, 2020 at 15:56 | comment | added | mathmetricgeometry | @archipelago: I think the above cylinder is not a counter example. Their Horizontal lifting are circles. Klein bottle to $S^1$ is a counter example, the fibers are $S^1$, the diff is reflexion, no fixed point. Need Horizontal lifting of base $S^1$ twice, then it's a closed curve. | |
Sep 23, 2020 at 13:07 | comment | added | mathmetricgeometry | @archipelago: Thank you! Just consider the cyliner $f:S^1\times (-\infty,+\infty)\to S^1$, the diffeomorphisms of $f^{-1}(x)$ is $(0,t)\to (1,t+1)$. Then the Horizontal lifting lines are spiral rising lines. | |
Sep 23, 2020 at 12:47 | comment | added | Ben McKay | @mathmetricgeometry: archipelago has explained the general case: a mapping torus. | |
Sep 23, 2020 at 12:42 | comment | added | mathmetricgeometry | @archipelago: For Mobius band, this surely holds. I ask the general case. | |
Sep 23, 2020 at 12:35 | review | Close votes | |||
Sep 28, 2020 at 3:05 | |||||
Sep 23, 2020 at 12:34 | comment | added | archipelago | Every such bundle is a mapping torus of a self-diffeomorphisms of $f^{-1}(x)$. The horizontal lift you ask for exists if and only if this diffeomorphism has a fixed point. For $f^{-1}(x)=(-1,1)$ the Mobius bundle is induced by $t\mapsto -t$ which indeed has a fixed point. | |
Sep 23, 2020 at 12:10 | history | asked | mathmetricgeometry | CC BY-SA 4.0 |