Timeline for Why is $\Omega_k(C^\infty(M))\to\Omega^1(M)$ surjective?
Current License: CC BY-SA 3.0
17 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Mar 11, 2017 at 17:56 | comment | added | Sebastian Goette | @DenisNardin On an $n$-manifold, one can always get a finite atlas where every "chart" is defined on a disjoint union of contractible subsets. This should be good enough for the argument. To get such an atlas, start with the first barycentric subdivision of a triangulation and consider the open stars of all vertices coming from degree $k$ simplices as one chart. Then you have $(n+1)$ types of charts, and so you can always do with a sum over $n(n+1)$ forms of the type $f\,dg$. | |
| Nov 25, 2016 at 23:42 | comment | added | Fallen Apart | @DenisNardin Actually manifolds addmiting finite atlas would be enought for me, but I wrote it in this way because in mentioned question they assumed nothing about $M.$ Thank you one more time for your first comment. This assured me that I haven't missed something. | |
| Nov 25, 2016 at 23:34 | comment | added | Denis Nardin | @FallenApart Oh I see. I was thinking only of compact manifolds. Sorry, I guess I was distracted :). Good call on the Whitney embedding theorem, by pulling back along a tubular neighbourood that should reduce the problem to $\mathbb{R}^n$, where it is obvious (since the cotangent bundle is trivial). | |
| Nov 25, 2016 at 23:25 | comment | added | Fallen Apart | @DenisNardin I see problem, cause in order to apply $\gamma$ I need to write 1-form $\alpha$ as a finite sum. When manifold $M$ is not compact, it possibly do not admin cover with finite amout of charts. And hence gluing by partition of unity would give me infinite sum. If I remember correctly the key in Swan's proof to make vector bundles finitely generated is Whitney Embedding theorem. | |
| Nov 25, 2016 at 23:08 | comment | added | Denis Nardin | @FallenApart Note that you don't need the Serre-Swan theorem to get that formula: it is enough to notice that it holds locally and that you can patch those local expressions using partitions of unity (which, admittedly, is half of the proof of the Serre-Swan theorem anyway...) | |
| Nov 25, 2016 at 22:59 | comment | added | Fallen Apart | @DenisNardin Your impression is fine, but since there was no explenation of surjectivity I assume that it should be obvious as well. But I think you are right. My explenation is standard one and among good mathematician there is no need to write down such trivial proof. I also want to point out that David Speyer in mentioned question also gave answer. And in his post he reffered to this n-cat caffe discussion in which they claim that $\gamma$ is obviously surjective. | |
| Nov 25, 2016 at 22:40 | comment | added | Denis Nardin | My impression is that "because the relations displayed above are valid in the classical interpretation of the calculus" refers to the existence of the map, not to its surjectivity (which I would prove exactly as you do). | |
| Nov 25, 2016 at 21:34 | history | edited | Fallen Apart | CC BY-SA 3.0 |
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| Nov 25, 2016 at 21:05 | history | edited | Fallen Apart |
It is not question about kahler manifolds. Do not add this TAG!
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| S Nov 25, 2016 at 20:47 | history | suggested | Henry.L | CC BY-SA 3.0 |
This is not really a question about modules. And I correct some obvious typos.
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| Nov 25, 2016 at 20:16 | review | Suggested edits | |||
| S Nov 25, 2016 at 20:47 | |||||
| Nov 25, 2016 at 19:12 | history | edited | Fallen Apart | CC BY-SA 3.0 |
added 20 characters in body
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| Nov 25, 2016 at 19:03 | comment | added | Fallen Apart | @LaurentMoret-Bailly Module of smooth vector fields on $M.$ | |
| Nov 25, 2016 at 19:01 | comment | added | Laurent Moret-Bailly | What is $\mathfrak{X}(M)$? | |
| Nov 25, 2016 at 18:52 | history | edited | Fallen Apart | CC BY-SA 3.0 |
edited title
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| Nov 25, 2016 at 18:37 | history | asked | Fallen Apart | CC BY-SA 3.0 |