Timeline for When is a `1-form' with continuous coefficients exact?
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18 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Nov 16, 2014 at 11:16 | vote | accept | DCM | ||
| Nov 16, 2014 at 11:14 | vote | accept | DCM | ||
| Nov 16, 2014 at 11:15 | |||||
| Nov 15, 2014 at 23:32 | comment | added | Igor Khavkine | I should say that I was perhaps overly pessimistic in my previous comment. Rereading the answer by Pedro at the linked question, I see that the obstacles to proving the Poincaré lemma at low regularities actually appear further along in the de Rham sequence. | |
| Nov 15, 2014 at 20:05 | comment | added | Deane Yang | Doesn't the Poincare lemma hold for distributions? And if the weak derivatives of $u$ are continuous, it follows that $u$ is $C^1$. | |
| Nov 15, 2014 at 14:53 | comment | added | DCM | @RB: I'm not sure why the integral condition didn't appeal to me yesterday... (and sorry if you feel I dismissed your suggestion - this was certainly not my intention). It's certainly grown on me since I had a look at the alternative! Thanks again. | |
| Nov 15, 2014 at 14:30 | answer | added | TaQ | timeline score: 4 | |
| Nov 15, 2014 at 11:03 | comment | added | Robert Bryant | @DCM: As people have pointed out, differential criteria are not adequate, but I'm not sure why you seem to be ignoring the integral condition, i.e., that, for $f_i$ continuous in $\Omega$, the $1$-form $\phi= f_i(x)\ dx^i$ is of the form $\phi = du$ for some $u \in C^1(\Omega)$ if and only if the integral of $\phi$ around any closed, piecewise $C^1$ path in $\Omega$ vanishes. Is there some reason you don't like this criterion? | |
| Nov 14, 2014 at 16:45 | comment | added | DCM | I was rather hoping that I had missed some subtle difference between my question and the one there which meant that mine had an easy answer... I will, however, do as you've suggested. Thanks for your help. | |
| Nov 14, 2014 at 15:36 | review | Close votes | |||
| Nov 15, 2014 at 19:59 | |||||
| Nov 14, 2014 at 15:33 | comment | added | Igor Khavkine | @DCM, your question is essentially the same question as the one you linked to, except with one level of regularity less. As the discussion there shows, this is a rather non-trivial question and the best available information seems fairly recent. So, if this question as posed is really of importance to you, you should look into that literature as cited there by Jochen and Pedro. | |
| Nov 14, 2014 at 15:30 | comment | added | Igor Khavkine | @TaQ, note that Horváth's Prop.4.3.9 guarantees only that there exists a distribution $g$ such that $f_j = \partial_j g$. It does not guarantee that $g$ is $C^1$. | |
| Nov 14, 2014 at 13:47 | comment | added | DCM | I will certainly investigate Horvath's book. Thanks! | |
| Nov 14, 2014 at 13:44 | comment | added | TaQ | Proposition 4.3.9 on page 334 in Horváth's Topological Vector Spaces and Distributions gives $\partial_i f_j=\partial_j f_i$ as a necessary and sufficient condition in the case $\Omega=\mathbb R\kern.4mm^n$ . Maybe the proof can be adapted to the case of bounded convex $\Omega$ ? | |
| Nov 14, 2014 at 13:43 | comment | added | DCM | Ah. Thankyou for that! Can you suggest any relevant references? | |
| Nov 14, 2014 at 12:04 | comment | added | Robert Bryant | If I remember correctly, the answer is 'when the integral of the $1$-form $\phi$ along any two paths joining $p$ and $q$ has a value $F(p,q)=-F(q,p)$ that depends only on the endpoints $p$ and $q$'. I think that there is also an answer based on $d\phi=0$ where the exterior derivative $d$ is suitably extended to the appropriate Sobolev spaces. | |
| Nov 14, 2014 at 12:02 | review | First posts | |||
| Nov 14, 2014 at 12:42 | |||||
| Nov 14, 2014 at 11:52 | history | asked | DCM | CC BY-SA 3.0 |