Timeline for Lifting a quadratic system to a non-vanishing vector field on $S^{3}$ or $T^{1} S^{2}$
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| Jul 6, 2019 at 18:14 | history | edited | YCor | CC BY-SA 4.0 |
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| Jul 6, 2019 at 13:09 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
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| Nov 16, 2014 at 15:11 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Nov 15, 2014 at 8:39 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Nov 15, 2014 at 8:13 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Nov 15, 2014 at 8:04 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Nov 7, 2014 at 21:00 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Nov 7, 2014 at 20:55 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Nov 7, 2014 at 20:49 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Nov 6, 2014 at 19:19 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Nov 6, 2014 at 19:12 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
I add anote at the final part of the question
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| Oct 1, 2014 at 15:13 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Sep 30, 2014 at 15:11 | comment | added | Robert Bryant | @AliTaghavi: Finally, the $S^1$-bundles over $M$ correspond one-to-one, to elements in $H^2(M,\mathbb{Z})$. I think that there is a proof in Milnor-Stasheff, but another place might be Atiyah's Lectures on $K$-theory. Check in those places for proofs. | |
| Sep 30, 2014 at 15:08 | comment | added | Robert Bryant | @AliTaghavi: In $S^3$, two different fibers are linked, but, of course, they are not knotted. In $\Sigma_n$ with $n>1$, I'm not sure what you might mean by `knotted'. | |
| Sep 30, 2014 at 15:05 | comment | added | Robert Bryant | @AliTaghavi: No. $S^3\to S^2$ is the only simply-connected $S^1$-fiber bundle over $S^2$: Each of the $\Sigma_n$ is also an $S^1$-fiber bundle over $S^2$. Also, you have to distinguish the two $S^1$-fiber bundles you get by reversing the $S^1$-action on any given $S^1$-fiber bundle (this makes sense because $S^1$ is abelian). Thus, there is actually a $\mathbb{Z}$-family of $S^1$-fiber bundles, with $\Sigma_n$ and $\Sigma_{-n}$ being topologically the same space but not equivalent bundles (when $n\not=0$). Check Milnor and Stasheff for example, for proofs. | |
| Sep 30, 2014 at 14:11 | comment | added | Ali Taghavi | @RobertBryant Prof. Bryant Thank you for your comment. So We conclude that the Hopf fibration is the only $S^{1}$-fibre bundle from $S^{3}$ to $S^{2}$?I have another question in this fibration, are there two disjoint fibres which are "linked" to each other?Is there a fibre which is knoted? And what is an elementary refrence for classification of $S^{1}$ fibre bundles, in particular the space $\Sigma_{n}$ which you mentioned? | |
| Sep 30, 2014 at 13:52 | comment | added | Robert Bryant | @Ali: If you just divide $S^3$ by the action of the $n$-element cyclic subgroup of $S^1$ (when $n\ge 1$), you'll get the bundle $\Sigma_n\to S^2$, and these are all of the examples. | |
| Sep 30, 2014 at 13:04 | comment | added | Ali Taghavi | @abx thanks for your answer to the last question. The total space correspond to generator is $S^{3}$ with Hopf map. What is the total space of multiple generator by 2? and what is the projecting map? | |
| Sep 30, 2014 at 10:26 | comment | added | abx | Answer to the last question : the isomorphism classes of $\mathbb{S}^1$-bundles on a space $X$ form a group, identified with $[X,B\mathbb{S}^1]$. When $X=\mathbb{S}^2$ this group is $\pi _2(B\mathbb{S}^1)\cong \pi _1(\mathbb{S}^1)=\mathbb{Z}$. The Hopf fibration is a generator, but you have also all its multiples. | |
| Sep 30, 2014 at 10:02 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Sep 30, 2014 at 9:53 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Sep 30, 2014 at 9:47 | history | asked | Ali Taghavi | CC BY-SA 3.0 |