Timeline for Checking that the image of a curve is not contained in a hyperplane
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 9, 2022 at 10:37 | comment | added | Robert Bryant | Unless you tell us more about how your curve $\gamma:[0,1]\to\mathbb{R}^n$ is specified and what sorts of information about it can be easily computed, it's hard to give advice about an 'easy-to-check condition'. For example, it's a simple matter to construct examples of smooth $\gamma$ that don't lie in a proper subspace of $\mathbb{R}^n$ but for which uniform sampling from $[0,1]^m$ to test $m$ points $\gamma(t_1),\ldots,\gamma(t_m)$ for linear independence will have a 99.99999% probability of 'yes', but still $\gamma([0,1])$ does not lie in a proper subspace. | |
| Dec 6, 2022 at 14:25 | answer | added | Anton Petrunin | timeline score: 4 | |
| Dec 6, 2022 at 11:13 | comment | added | Mark Wildon | If the image of $\gamma$ is contained in a proper subspace then so is its derivative. You could use any linear algebra method to test if the tangent vectors are in a plane. | |
| Dec 6, 2022 at 5:57 | history | became hot network question | |||
| Dec 6, 2022 at 1:18 | answer | added | Piotr Hajlasz | timeline score: 13 | |
| Dec 5, 2022 at 22:03 | comment | added | user44143 | Consider $\gamma$ at $n$ random points in $[0,1]$, and calculate the determinant of those points: $\det\neq 0$ means that $\gamma$ definitely is not contained in an $n-1$-dimensional subspace; $\det =0$ means that all those points lie in some lower-dimensional subspace, and it may well contain the whole curve. | |
| Dec 5, 2022 at 21:57 | history | asked | J. Swail | CC BY-SA 4.0 |