Timeline for Are all $C^1$ arcs tame?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Sep 21, 2011 at 3:50 | vote | accept | CommunityBot | ||
| Sep 20, 2011 at 21:27 | comment | added | George Lowther | You can always define the parameterization $i$ by making it piecewise linear (approximating $p$ by a polygon), then applying $C^1$ bump on the linear segments to make it line up with $p$, then another $C^1$ bump on each segment to make the derivatives line up at the joins. | |
| Sep 20, 2011 at 21:26 | comment | added | Sergei Ivanov | You need $C^1$ functions $v_1,\dots,v_{n-1}:(-\varepsilon,1+\varepsilon)\to\mathbb R^n$ such that $p'(t),v_1(t),\dots,v_{n-1}(t)$ are linearly independent for every $t$; then you define $i(t,x_1,\dots,x_{n-1}) = p(t)+x_1v_1(t)+\dots+x_{n-1}v_{n-1}$ and by the Inverse Function Theorem this is a diffeomorphism in a neighborhood. To construct smooth $v_i(t)$ first construct continuous ones (e.g. let them be a basis of the orthogonal complement of $p'(t)$ constructed locally by orthogonalization of constant family), then smoothen. | |
| Sep 20, 2011 at 21:07 | comment | added | user5810 | How do you show the existence of $U$ and such a parametrization? | |
| Sep 20, 2011 at 20:13 | history | answered | Sergei Ivanov | CC BY-SA 3.0 |