Timeline for Existence of diagonalizing coordinates for the metric tensor
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| May 5, 2014 at 1:37 | history | edited | Arthur Suvorov | CC BY-SA 3.0 |
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| May 4, 2014 at 22:54 | comment | added | Arthur Suvorov | Perhaps I have given this question an unfortunate title; I was asking more along the lines of the existence of a sum decomposition of the form $ g_{\mu \nu} = S_{\mu \nu} +N_{\mu \nu}$ where $S$ is diagonalizable and $N$ has some other properties. I am aware that some metrics do not admit such systems; thank you Robert for the Fubini example. I drew the comparison with the Jordan-Chevalley decomposition to see perhaps what Nμν should look like here. Since we have an analogous 'diagonalizable' construction, maybe Nilpotent means some derivative of $N$ vanishes eventually. | |
| May 4, 2014 at 22:32 | comment | added | Arthur Suvorov | Perhaps the Einstein equations from scratch aren't simpler in the diagonal 'gauge' because there may be highly non-trivial dependencies. Usuaully though one takes a solution already found and modifies it to allow for extra freedoms that should be permitted on physical grounds. Also, solving perturbation equations such as: Klein-Gordon (wave), Regge-Wheeler, Dirac... all contain many less cross terms in the diagonal system. Thank you for that reference Deane; much appreciated. This means using a 3+1 ADM split one can always find a triply orthogonal system for the spatial metric | |
| May 2, 2014 at 13:05 | comment | added | Deane Yang | DeTurck and I (projecteuclid.org/euclid.dmj/1077303804) proved that any 3-dimensional Riemannian metric is locally diagonalizable. We were told at the time that diagonal co-ordinates had already been studied and found useful in general relativity (maybe by Chandrasekhar?). But I've never understood why the Einstein equations might be easier to study in diagonal co-ordinates. | |
| May 2, 2014 at 11:06 | comment | added | Robert Bryant | I don't know useful results for computing the Ricci tensor of the sum of two quadratic forms, so I can't help you there. What we do know is that in dimension $4$ the diagonalizable metrics depend on 8 functions of 4 variables locally while all metrics depend on 10 functions of 4 variables locally, so there have to be at least two independent identities that diagonalizable metrics satisfy that aren't satisfied by the general metric. They will be high order, though. In any case, not all Einstein metrics are diagonalizable: The Fubini-Study metric on $\mathbb{CP}^2$ is not locally diagonalizable. | |
| May 2, 2014 at 8:51 | comment | added | Ben McKay | Local coordinates in which the metric tensor is diagonal are traditionally called orthogonal systems. Darboux wrote a book about triply orthogonal systems, i.e. such coordinates, just for Euclidean space of dimension 3. But he wasn't looking at general metrics or decompositions, so I guess that isn't really relevant. | |
| May 2, 2014 at 4:09 | review | First posts | |||
| May 2, 2014 at 5:59 | |||||
| May 2, 2014 at 3:52 | history | asked | Arthur Suvorov | CC BY-SA 3.0 |