Timeline for A necessary and sufficient condition for a space curve to lie on a ellipsoid
Current License: CC BY-SA 3.0
7 events
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| Apr 3, 2018 at 8:40 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Made a slight improvement of the statement for spherical curves.
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| Mar 27, 2018 at 19:37 | comment | added | Willie Wong | ...the five dimensional vector space $a|x|^2 + b\cdot x + c$, which linear structure is used in arguing that the condition on the top level derivative gives an integrability condition. Thanks! (Your parenthetical on allowing any quadric surface is what made it click for me.) | |
| Mar 27, 2018 at 19:35 | comment | added | Willie Wong | Okay, I think I get it now. The idea is that by your definition if $q$ solves $(d/dt)^k q\circ \gamma(t_0) = 0$, so does $\lambda q$. So the one dimensionality is natural. The $Q$-nondegeneracy condition states that based on the data at $t_0$, there is only one quadric hypersurface that can be compatible (depending on the point $t_0$). The condition on $Q^{10}_\gamma$ is an integrability condition saying that the hypersurfaces can be pieced together. // What I did wrong in my previous comment was that I looked at only the 4 dimensional set of polynomials of the form $|x-a|^2 - r^2$, and not... | |
| Mar 27, 2018 at 17:51 | comment | added | Robert Bryant | @WillieWong: It's 'non-simultaneous vanishing', i.e., I don't want all of them to vanish at once, so, for example, it's OK if all but one of them vanish for the $Q$-nondegeneracy, and the $Q$-nondegeneracy condition only involves the derivatives of $\kappa$ up to order $6$ and $\tau$ up to order $5$. It is somewhat like the sphere case, in that the appropriate nondegeneracy condition there is the nonvanishing of $\kappa\tau$, while the equation $P=0$ that must be satisfied also involves $\kappa''$ and $\tau'$. (I allow any quadric surface for $Q$-curves, not just ellipsoids.) | |
| Mar 27, 2018 at 16:56 | comment | added | Willie Wong | Is the condition being $Q$-nondegenerate the condition of "non-simultaneous vanishing" of some polynomials, or is it the condition of "simultaneous non-vanishing"? Also, why is the number of derivatives up to order 6? In analogy with the sphere case (which has $Q$ being dimension 4), don't we expect the non-degeneracy condition to be up to $\mathrm{dim} Q - 3 = 7$ derivatives in $\kappa$ and $6$ in $\tau$? | |
| Mar 27, 2018 at 11:17 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Fixed an error at the beginning about the right notion of nondegeneracy
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| Mar 27, 2018 at 9:21 | history | answered | Robert Bryant | CC BY-SA 3.0 |