Timeline for A differential operator associated with a vector field on the torus
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Mar 25, 2019 at 12:45 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
edited body; edited tags
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| Mar 24, 2019 at 20:55 | history | undeleted |
Ali Taghavi Carlo Beenakker Stefan Kohl♦ |
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| Mar 23, 2019 at 0:01 | history | deleted | CommunityBot | Scheduled: RemoveDeadQuestions | |
| Jul 2, 2017 at 10:35 | comment | added | Ali Taghavi | @SebastianGoette thanks for this reference. I read your comment just now. | |
| Jun 25, 2017 at 7:24 | comment | added | Sebastian Goette | In some cases, there are exotic indices, e.g. the parity of dim ker could be an invariant. Please check the book by Lawson-Michelsohn for details. | |
| Jun 24, 2017 at 9:57 | comment | added | Ali Taghavi | @SebastianGoette I think of fredholm index as you mentioned. So according to your comment and comment of Thomas Richard I conclude that the index is zero. But what other type of index on can associate to differential operators? | |
| Jun 24, 2017 at 9:04 | comment | added | Sebastian Goette | What kind of index are you thinking of? The operator described by @ThomasRichard has selfadjoint principal symbol, so the classical definition (dim ker - dim coker) does not give anything interesting. | |
| Jun 24, 2017 at 8:29 | comment | added | Ali Taghavi | @ThomasRichard What is that deformation?. Moreover do you deform an arbitrary metric a to a metric with constant $g_{ij}$? | |
| Jun 23, 2017 at 11:38 | comment | added | Thomas Richard | Also, when $g$ is a flat metric and $X=\partial_x$, $D$ is just the usual laplacian. One can always deform (conformally for instance) a Riemannian metric on $\mathbb{T}^2$ to a flat one. I wonder if the index moves during the deformation. That might help. | |
| Jun 23, 2017 at 11:29 | comment | added | Ali Taghavi | @ThomasRichard Thank you for your comment. | |
| Jun 22, 2017 at 8:36 | comment | added | Thomas Richard | Computations show that $Df=|X|_g^2\Delta_g f+\dots$ where the terms in the dots depend on $f$ and $df$. So ellipticity is ok since your vector field is non vanishing, as for the index I have no idea. | |
| Jun 22, 2017 at 8:18 | comment | added | Ali Taghavi | @ThomasRichard Yes it is a tangent vector. | |
| Jun 22, 2017 at 8:16 | comment | added | Thomas Richard | so $h$ is a vector field right ? | |
| Jun 22, 2017 at 7:52 | history | asked | Ali Taghavi | CC BY-SA 3.0 |