Timeline for A question on long line
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Nov 7, 2016 at 10:35 | comment | added | Tom | Although this question is 3 years old, just a comment to one of the comments above: The Long Line IS path connected. Any two points are contained in a chart which is diffeomorphic to a unit interval. The 1-point compactification of the Long Line is NOT path connected but since this it not a manifold, it is not interesting. | |
| Dec 18, 2013 at 8:21 | comment | added | abx | Triviality of the cotangent bundle implies triviality of the tangent bundle (its dual), hence $T M\cong T^*M$ and you can apply my answer. | |
| Dec 18, 2013 at 7:38 | comment | added | Ali Taghavi | do you mean :triviality of cotangent bundle implies that M is metrizable? if so, why? | |
| Dec 18, 2013 at 6:16 | comment | added | abx | No, for the same reason. | |
| Dec 18, 2013 at 6:07 | comment | added | Ali Taghavi | Is $T^{*}(M)$ trivial? | |
| Dec 18, 2013 at 6:06 | vote | accept | Ali Taghavi | ||
| Dec 18, 2013 at 6:04 | vote | accept | Ali Taghavi | ||
| Dec 18, 2013 at 6:05 | |||||
| Dec 18, 2013 at 5:58 | comment | added | Ryan Reich | Ah, I did not realize that even a very large "manifold" with a Riemannian metric could be topologically metrized. I was thinking of the obvious definition where you measure the infimum length of piecewise-smooth paths between two points, as computed by arc length, and which obviously doesn't work for a non-path-connected space. | |
| Dec 18, 2013 at 5:54 | comment | added | abx | I don't get your point. It is well known that the long line does not admit a riemannian metric, see here. | |
| Dec 18, 2013 at 5:42 | comment | added | Ryan Reich | The metric gives a local distance, but since the long line is not path-connected, the local distance doesn't compare two arbitrary points. | |
| Dec 18, 2013 at 5:37 | history | answered | abx | CC BY-SA 3.0 |