Timeline for Can the Laplace operator on $n-$ manifolds be represented as a sum of $n$ second order derivational operators
Current License: CC BY-SA 3.0
12 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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| Aug 8, 2015 at 6:49 | comment | added | Ali Taghavi | @Raziel Thanks a lot for this new version. | |
| Aug 7, 2015 at 8:02 | comment | added | Raziel | I added more details on the $N >n$ argument in my original answer. | |
| Aug 6, 2015 at 21:27 | comment | added | Ali Taghavi | In combination to this one: mathoverflow.net/questions/182139/… I realy search for some elliptic PDE associated to a vector field which its index(or some thing similar) has a nice dynamical interpretation. | |
| Aug 6, 2015 at 21:21 | comment | added | Ali Taghavi | @Raziel according to your comment on "application of this question": I do not know any application, But I would like to say; appart from the usual structure of R^n, my other motivation(However very indirect and weak motivation) was the following post: mathoverflow.net/questions/182415/… | |
| Aug 6, 2015 at 20:18 | comment | added | Ali Taghavi | @Raziel Can you more explain why for some N>n we have N global vector field which square sum is $\Delta$?(Assuming parallelizability) | |
| Aug 6, 2015 at 20:15 | comment | added | Ali Taghavi | @DavidSpeyer In my question I used n for both dim M and the number of fields, since I was motivated by usual structure of R^n. | |
| Aug 6, 2015 at 19:03 | comment | added | David E Speyer | Agreed. So, is there any obstacle to doing this with an overdetermined system of fields? I would guess not, by some sort of partition of unity argument, but the nonlinearity of those trailing first order operators makes this hard. | |
| Aug 6, 2015 at 18:59 | comment | added | Raziel | Indeed! Since I don't know precisely the application the OP had in mind, I thought it was worth pointing out this possibility (i.e. avoid global issues by choosing an over-determined system of fields) | |
| Aug 6, 2015 at 18:56 | comment | added | David E Speyer | Agreed. The original question wrote $n$ for both $\dim M$ and the number of vector fields. It is possible that was a typo. I, and most of the other answers, are assuming it was not. | |
| Aug 6, 2015 at 18:54 | comment | added | Raziel | It is possible to write the Laplace operator as a "sum of square" with an overdetermined set of vector fields $X_1,\ldots,X_N$, with $N \geq n = \dim(M)$. For the case of $M=\mathbb{R}^n$ Just take Any $n\times N$ matrix $A$ such that $AA^* = \mathbb{I}$. Then the $N$ vectors $X_j:=\sum_{i=1}^n A_{ij}\partial_{x_i}$ give an example. So, in more general situations, you can hope to write your Laplacian as a sum of $N$ GLOBAL vector fields that, still, have maximal rank (equal to the dimension of the manifold) at each point. In this way I suspect you can avoid the parallelizability hypothesis. | |
| Aug 6, 2015 at 18:25 | history | answered | David E Speyer | CC BY-SA 3.0 |