Timeline for Existence of divergence-free unit vector field in conformally rescaled euclidean metric
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Apr 30, 2021 at 12:26 | vote | accept | Leo Moos | ||
| Apr 27, 2021 at 17:48 | history | edited | Leo Moos | CC BY-SA 4.0 |
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| Apr 27, 2021 at 17:44 | comment | added | Leo Moos | @RobertBryant [...] I still have trouble reconciling this with the degeneration of the metric at the boundary. I understand the point you raise in 3), but I'm not sure how to answer in the framework that motivates the problem. What I was hoping is that phrasing it as an existence problem for the vector field $X$ would provide some elements that explain why some metrics admit geodesic foliations and others do not. | |
| Apr 27, 2021 at 17:39 | comment | added | Leo Moos | @RobertBryant [...] What I had worked out is that if $\omega_g$ is the area form, then $\iota_X \omega$ is a calibration form if $\lvert X \rvert \leq 1$ and $X$ is divergence-free. I thought that because there is a whole foliation of geodesics, that in fact $\lvert X \rvert = 1$ everywhere, and moreover that $X$ should be the unit normal to every leaf. I got that mixed up in the initial question - what I had in mind is that $X$ is orthogonal to the leaves of the foliation, and therefore tangential to the boundary. [...] | |
| Apr 27, 2021 at 17:28 | comment | added | Leo Moos | @RobertBryant I apologise for taking so long to get back to you. I'll try to address your questions in order. 1) I misspoke - you're right, I'd be happy to restrict to the case where $g$ is a multiple of the Euclidean metric. 2) See the first point. 3) I've tried to fix this, but I couldn't quite figure out the right way. The question arose when studying metrics $g$ which do degenerate near the boundary (in the sense that the factor $e^{2 \varphi}$ goes to zero), and which additionally admit foliations by geodesics which intersect the boundary orthogonally. [...] | |
| Apr 27, 2021 at 17:22 | comment | added | Leo Moos | @DenisSerre To be honest I was hoping for 'classical solutions', without singular points. However if you were aware of general existence results of weak solutions that would of course also be interesting - sorry for only replying to your comment now. | |
| Apr 27, 2021 at 17:21 | history | edited | Leo Moos | CC BY-SA 4.0 |
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| Apr 26, 2021 at 10:52 | comment | added | Robert Bryant | @LeoMoos: Three questions: 1) Why specify 'conformally flat' in the question, as all metrics in dimension $2$ are conformally flat? Did you mean to impose the stronger condition that $g$ be a multiple of the Euclidean metric? 2) If $g$ isn't assumed to be a multiple of the Euclidean metric, then why not just let $\Omega$ be an abstract surface with a boundary that consists of smooth arcs that meet at 'corners' and let $g$ be a Riemannian metric on $\Omega$ minus the 'corner points'. 3) If $g$ doesn't extend to the smooth parts of the boundary, then what do your boundary conditions on $X$ mean? | |
| Apr 25, 2021 at 12:08 | comment | added | Robert Bryant | Do you have a typo in your last paragraph? The domain $\Omega$ that you list is the right half-plane, and $X = \partial_2$ is tangent to the boundary line $x_1=0$, not normal to it. | |
| Apr 24, 2021 at 21:25 | answer | added | Robert Bryant | timeline score: 4 | |
| Apr 24, 2021 at 14:24 | comment | added | Denis Serre | Do you allow an interior singularity ? I mean, if $\Omega$ is the unit disk with the standard metric, a natural solution is $\vec e_\theta$ in polar coordinates, which is singular at the origin. | |
| Apr 24, 2021 at 14:07 | history | asked | Leo Moos | CC BY-SA 4.0 |