Timeline for Least reasonable regularity on Riemannian metric tensor to define a metric [closed]
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Oct 12, 2016 at 3:02 | review | Reopen votes | |||
| Oct 12, 2016 at 10:49 | |||||
| Oct 5, 2016 at 6:40 | comment | added | Willie Wong | @shuhalo: you still didn't correct the part that reads "how low can the regularity of a smooth curve be so that we still have a well-defined metric in terms of path lengths". | |
| Oct 5, 2016 at 2:58 | review | Reopen votes | |||
| Oct 5, 2016 at 8:10 | |||||
| Oct 5, 2016 at 2:43 | comment | added | shuhalo | @NateEldrege: right, that was just an annoying typo. I corrected it. | |
| Oct 5, 2016 at 2:42 | history | edited | shuhalo | CC BY-SA 3.0 |
Corrected typo + clarified
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| Oct 4, 2016 at 17:32 | history | closed |
Nate Eldredge Ben McKay Willie Wong Misha Anton Petrunin |
Needs details or clarity | |
| Oct 4, 2016 at 8:24 | comment | added | Thomas Richard | Although tangentially relevant, you might be interested to know that there is a way to define a distance for metrics with very low regularity which doesn't require being able to compute the length of curves : $d_g(x,y)=\sup\{|f(y)-f(x)|,\,f:M\xrightarrow{C^1}\mathbb{R},\, |\nabla f|^2_g\leq 1\}$. This in particular makes sense for measurable metrics. However, it won't see the pathology Anton built. | |
| Oct 3, 2016 at 21:07 | history | edited | Willie Wong |
edited tags
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| Oct 3, 2016 at 19:43 | comment | added | Anton Petrunin | Consider the following example: your metric is standard on the plane, but along one line it is twice smaller. Note that $L^2$-norm on the vector fields is the same as in the standard plane, but you can make a shortcut using the spacial line --- depending on the application you have in mind, try to tell what would be right choice of the induced metric in this case? | |
| Oct 3, 2016 at 18:50 | review | Close votes | |||
| Oct 3, 2016 at 22:14 | |||||
| Oct 3, 2016 at 18:27 | comment | added | Nate Eldredge | Wait - are you talking about the regularity of the curve $c$ or the metric $g$? It seems you are switching back and forth. | |
| Oct 3, 2016 at 18:26 | comment | added | Nate Eldredge | Since the expression for $L(c)$ involves the derivative of $c$, how does it make sense if $c$ is only continuous but not differentiable? It could even be nowhere differentiable. "Almost everywhere differentiable" is plausible but not correct (see the Cantor staircase function). I suspect the condition you are really looking for is "absolutely continuous". | |
| Oct 3, 2016 at 18:03 | history | edited | Myshkin | CC BY-SA 3.0 |
+ top level tag (dg.)
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| Oct 3, 2016 at 16:12 | history | asked | shuhalo | CC BY-SA 3.0 |