Timeline for Riemannian vs Non-Riemannian curvature
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Jan 11, 2018 at 3:03 | review | Close votes | |||
| Jan 11, 2018 at 10:52 | |||||
| Jan 6, 2018 at 8:56 | comment | added | Aureliano Skirzewski | I am interested to know the conditions for the curvature of a connection to be Riemannian. I sketched how I was planning to know the answer (by showing the exist a metric) but if the answer doesn't involve a metric it's alright too. The problem I originally have is an action where the only field is an affine connection (no metric in the kinematics). I obtain field equations that look simple when torsion is zero $$\nabla_{[a}R_{b]c}=0,$$ and to make connection with Einstein equations we explore the Riemannian sector. | |
| Jan 5, 2018 at 20:48 | comment | added | Deane Yang | I'm still confused by what your question is asking. Are you asking, given a $(1,3)$ tensor on, say, a neighborhood of a point, whether there exists a Riemannian metric whose $(1,3)$ Riemann curvature tensor is equal to the given tensor on some possibly smaller neighborhood? Or are you asking whether there exists a Riemannian metric, whose $(1,3)$ Riemann curvature is equal to the given tensor only at a single point? Note that the condition that you state in your question is necessary but not sufficient for the $(1,3)$ tensor to be the $(1,3)$ Riemann curvature of a Riemannian metric $g$. | |
| Jan 5, 2018 at 2:40 | answer | added | Robert Bryant | timeline score: 13 | |
| Jan 5, 2018 at 2:11 | comment | added | Aureliano Skirzewski | Without metric, the curvature is by definition a (1,3) tensor. The upper index has no part of the symmetries the other indices share as it cannot be lowered. Whether it's a Riemann or not I don't know, how could I? That's my question. If you know the metric at one point you can lower all the indices and the symmetry is that of a Young tableu (2,2). Infinitesimal loop holonomies are then recognized as a subgroup of so(d) and preserve the metric. It also generates the parallel transport and you can determine the metric elsewhere | |
| Jan 4, 2018 at 22:44 | comment | added | Deane Yang | When you say "Riemann", do you mean only that the (2,2) tensor satisfies the symmetries of a Riemann curvature tensor? Or do you mean that the (2,2) tensor really is the Riemann curvature tensor for the metric solved for? | |
| Jan 4, 2018 at 20:07 | comment | added | Aureliano Skirzewski | It is not the same question, in that post @orbit assumes a tensor with Young tableu (2,2). It is interesting but it assumes wrongly the Riemann is a (0,4) instead of a (1,3) tensor. I rather ask if there is a metric that would turn my (1,3) tensor into a (0,4) with Young tableu (2,2) | |
| Jan 4, 2018 at 19:55 | comment | added | Pedro Lauridsen Ribeiro | Possible duplicate of Equations satisfied by the Riemann curvature tensor | |
| Jan 4, 2018 at 19:23 | comment | added | macbeth | Is your question different from that answered at mathoverflow.net/questions/202211/… ? | |
| Jan 4, 2018 at 19:22 | comment | added | Aureliano Skirzewski | @BenMcKay, you are right but, If the curvature is zero at a point, that would just tell me that any non-degenerated tensor can be parallel transported to be the metric elsewhere (if parallel transport of that tensor doesn't depend on the path). | |
| Jan 4, 2018 at 19:21 | review | Close votes | |||
| Jan 4, 2018 at 21:12 | |||||
| Jan 4, 2018 at 19:07 | comment | added | Aureliano Skirzewski | Not all curvatures are Riemannian, if you don't know the metric how can you tell? | |
| S Jan 4, 2018 at 19:06 | history | suggested | Amir Sagiv | CC BY-SA 3.0 |
Latex and minor edits
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| Jan 4, 2018 at 19:02 | comment | added | Anton Petrunin | Any tensor which satisfy the symmetries of curvature tensor appears as a curvature tensor for some metric. Is that what you are asking? | |
| Jan 4, 2018 at 18:59 | review | Suggested edits | |||
| S Jan 4, 2018 at 19:06 | |||||
| Jan 4, 2018 at 18:41 | comment | added | Ben McKay | If the curvature vanishes, these equations don't say anything, so clearly they don't determine the metric. | |
| Jan 4, 2018 at 18:41 | review | First posts | |||
| Jan 4, 2018 at 18:59 | |||||
| Jan 4, 2018 at 18:37 | history | asked | Aureliano Skirzewski | CC BY-SA 3.0 |