Timeline for Why torsion is only defined for linear connection on TM?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 29, 2022 at 8:24 | vote | accept | ychemama | ||
| Jun 27, 2022 at 23:43 | answer | added | Josh Lackman | timeline score: 2 | |
| Jul 23, 2018 at 15:41 | answer | added | Ben McKay | timeline score: 6 | |
| Jul 23, 2018 at 15:13 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jun 23, 2018 at 17:25 | comment | added | Peter Michor | See mathoverflow.net/a/116487/26935 and other answers there for some aspects. | |
| Jun 23, 2018 at 15:13 | answer | added | Alex Gavrilov | timeline score: 0 | |
| Jun 22, 2018 at 18:13 | comment | added | Ivan Izmestiev | Then a possible approach to this could be by "natural operations", see for example the book "Natural operations in differential geometry" by Kolar, Michor, and Slovak. I would guess that any tensor "constructed out of $\nabla$" depends only on the curvature of $\nabla$. | |
| Jun 22, 2018 at 11:58 | comment | added | ychemama | @IvanIzmestiev I know that the usual definition cannot be extended as such. My question is : why there is no more general notion of torsion which applies to any vector bundle? Or is there ? Or to put it anther way : is the curvature the only tensor field which can be constructed out a connection $\nabla$ ? | |
| Jun 22, 2018 at 11:28 | comment | added | Ivan Izmestiev | The definition of the torsion tensor $T(X,Y) = \nabla_X Y - \nabla_Y X - [X,Y]$ cannot be extended to arbitrary vector bundles: $\nabla_X Y$ subsumes $X \in \Gamma(TM)$ and $Y \in \Gamma(E)$, and then you exchange $X$ and $Y$... When you speak about "the twist of a moving frame along a curve", don't you mean a different notion of torsion, like in Frenet-Serret formulas? | |
| Jun 22, 2018 at 10:22 | history | asked | ychemama | CC BY-SA 4.0 |