Timeline for Why are isometries of Minkowski space necessarily linear?
Current License: CC BY-SA 4.0
25 events
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| Jun 26, 2020 at 3:15 | history | edited | Tim Campion | CC BY-SA 4.0 |
correction: the lorentz metric is nondegenerate; it's just not postive-definite.
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| Jun 26, 2020 at 0:08 | history | edited | Martin Sleziak |
removed the deprecated (geometry) tag
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| Jun 22, 2020 at 7:40 | comment | added | Ben McKay | @RyanBudney: in Lorentz signature, geodesics don't minimize length. They are degenerate critical points of the energy integral functional. | |
| Jun 22, 2020 at 5:33 | answer | added | KConrad | timeline score: 6 | |
| Jan 14, 2012 at 18:43 | comment | added | Renato G. Bettiol | @Boaz and @Vitali: An isometry of a semi-Riemannian (or pseudo-Riemannian) manifold $(M,g)$ is a diffeomorphism $\phi$ that preserves the metric tensor, i.e., the pull-back of $g$ by $\phi$ coincides with $g$: $\phi^*(g)=g$. I have never seen any other definitions in the literature. Although there is a notion of distance in Lorentzian manifolds (see e.g. the book of O'Neill or the one of Beem, Ehrlich and Easley), there are very few results analogous to the Riemannian case. | |
| Jan 2, 2012 at 0:08 | comment | added | Boaz Haberman | I guess I was thinking of the minimal proper time between two points. But maybe this is not a meaningful quantity. | |
| Jan 2, 2012 at 0:06 | vote | accept | Boaz Haberman | ||
| Jan 1, 2012 at 4:00 | comment | added | Vitali Kapovitch | @Boaz: What proper time would that be? you have a continuous map between pseudo-Riemannian manifolds. Proper time is defined for individual smooth timelike curves. Please give a full definition of isometry you want to use. | |
| Jan 1, 2012 at 2:28 | comment | added | Boaz Haberman | I guess I am thinking of a map which preserves "proper time" | |
| Dec 30, 2011 at 22:02 | comment | added | Vitali Kapovitch | @Boaz what's your definiton of an isometry of a Pseudo-Riemannian manifold? There is no notion of a distance or of arc-length of a continuous curve here. | |
| Dec 30, 2011 at 21:44 | answer | added | Deane Yang | timeline score: 3 | |
| Dec 30, 2011 at 18:48 | comment | added | Boaz Haberman | Now that I think of it, it is not so obvious to me that isometries of pseudo-Riemannian manifolds are continuous. | |
| Dec 30, 2011 at 18:47 | comment | added | Boaz Haberman | Sorry, I meant that usually in the context of Riemannian geometry isometries are defined to be smooth, but a priori an isometry (distance-preserving map) could be continuous but not smooth (in fact there are pathological examples). A sticky point is that an isometry is not necessarily a bijection, and a lot of these theorems are only available for surjective isometries. In particular, I'm not sure that it's obvious for an arbitrary isometry that it takes geodesics to geodesics. @Tom: I saw the Mazur-Ulam theorem come up in a proof of the Myers-Steenrod theorem . . . | |
| Dec 30, 2011 at 18:45 | vote | accept | Boaz Haberman | ||
| Jan 2, 2012 at 0:06 | |||||
| Dec 30, 2011 at 18:45 | vote | accept | Boaz Haberman | ||
| Dec 30, 2011 at 18:45 | |||||
| Dec 30, 2011 at 18:45 | vote | accept | Boaz Haberman | ||
| Dec 30, 2011 at 18:45 | |||||
| Dec 30, 2011 at 18:32 | comment | added | Ryan Budney | @Boaz: by definition isometries are continuous. | |
| Dec 30, 2011 at 18:20 | vote | accept | Boaz Haberman | ||
| Dec 30, 2011 at 18:45 | |||||
| Dec 30, 2011 at 18:10 | comment | added | Boaz Haberman | I guess what is confusing me is that usually isometries are defined to be smooth maps, but a priori they could just be continuous. Apparently the Myers-Steenrod theorem says that all isometries of Riemannian manifolds are smooth, however this theorem requires the isometry to be surjective. | |
| Dec 30, 2011 at 17:50 | vote | accept | Boaz Haberman | ||
| Dec 30, 2011 at 18:20 | |||||
| Dec 30, 2011 at 16:34 | answer | added | Pablo Lessa | timeline score: 11 | |
| Dec 30, 2011 at 5:59 | answer | added | Anton Petrunin | timeline score: 29 | |
| Dec 30, 2011 at 4:59 | comment | added | Tom LaGatta | Off-topic question: what are some applications of the Mazur-Ulam theorem? | |
| Dec 30, 2011 at 0:31 | comment | added | Ryan Budney | There are a lot of ways to prove these kinds of theorems. One strategy is to argue isometries carry geodesics to geodesics -- i.e. geodesics in the differential-geometric sense are rectifiable curves that locally minimize length. | |
| Dec 29, 2011 at 23:49 | history | asked | Boaz Haberman | CC BY-SA 3.0 |