Timeline for There is no arcwise isometry from a high dimensional manifold into a low dimensional manifold
Current License: CC BY-SA 3.0
11 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Dec 22, 2016 at 10:44 | vote | accept | Asaf Shachar | ||
| Dec 14, 2016 at 16:12 | comment | added | Asaf Shachar | @IgorRivin These length-preserving maps need not be metric isometries, take for instance $\alpha(t)=(\cos t,\sin t)$. Also, even when they are injective isometries, they are not always surjective ($(x \to (x,0)$). | |
| Dec 14, 2016 at 15:19 | comment | added | Igor Rivin | @RyanBudney I might be dense, but since $d(f(x), f(y)) = d(x, y),$ how can an isometry fail to be bijective? | |
| Dec 14, 2016 at 8:07 | answer | added | Asaf Shachar | timeline score: 2 | |
| Dec 14, 2016 at 7:19 | comment | added | Ryan Budney | @IgorRivin: Asaf's definition of isometry allows for maps to be not bijections -- he allows covering spaces, for example. It also allows for isometric embeddings. He's trying to make a calculus-ish argument that the domain's dimension needs to be less than or equal the target space's. | |
| Dec 14, 2016 at 4:29 | answer | added | Anton Petrunin | timeline score: 5 | |
| Dec 14, 2016 at 1:21 | comment | added | Igor Rivin | Doesn't this follow from the fact that an isometry is a homeomorphism? | |
| Dec 13, 2016 at 22:41 | history | edited | Asaf Shachar | CC BY-SA 3.0 |
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| Dec 13, 2016 at 22:35 | comment | added | Ryan Budney | Any metric-defined notion of dimension would work. For example Lebesgue covering dimension is defined for metric spaces. It behaves as expected for manifolds and metric isometries of this sort give you a lower bound on the dimension of the target. So it's an immediate contradiction. All the results you need are in Munkres's point-set topology textbook, for example. | |
| Dec 13, 2016 at 22:27 | history | asked | Asaf Shachar | CC BY-SA 3.0 |