Timeline for Smooth curve in $\mathbb{R}^3$ not contained in real analytic surface?
Current License: CC BY-SA 4.0
11 events
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| Feb 8, 2023 at 19:24 | comment | added | Robert Bryant | @CrashBandicoot: Oh...OK. I think that it is probably true that the 'generic' smooth curve in $\mathbb{R}^n$ is not contained in any analytic submanifold of $\mathbb{R}^n$, but I don't know any literature about this. | |
| Feb 8, 2023 at 18:03 | comment | added | Robert Bryant | @CrashBandicoot: I don't. I'm not even sure what 'meager' means. | |
| Sep 17, 2022 at 22:15 | comment | added | Robert Bryant | @IanAgol: Here's an interesting result: It turns out that a (smooth, but I think that $C^3$ would suffice) space curve $\gamma$ of constant torsion equal to $1$ lies in a surface $S$ of constant Gauss curvature $-1$ in such a way that $\gamma$ is an asymptotic curve in $S$ if and only if the integral of its curvature $\kappa$ with respect to arc length $\mathrm{d}s$ between any two points of $\gamma$ is less than $\pi$. This follows from the solvability of a doubly characteristic IVP for the Sine-Gordon Equation, plus the standard structure equations. | |
| Sep 5, 2022 at 17:09 | comment | added | Robert Bryant | @IanAgol: I meant to come back to your comment about curves of constant torsion. A theorem of Beltrami and Enneper says that the square of the torsion of an asymptotic curve on a surface is the negative of the Gauss curvature of the surface along the curve. In particular, for a pseudospherical surface, the asymptotic curves have constant torsion. A natural question is whether every curve of (nonzero) constant torsion $\tau$ is an asymptotic curve on a surface of constant Gauss curvature $-\tau^2$ (which may not be complete). That seems plausible, but I don't know a proof. | |
| Aug 19, 2022 at 18:36 | comment | added | Ian Agol | @RobertBryant okay, thanks for the clarification, I didn’t realize that they aren’t necessarily analytic. | |
| Aug 19, 2022 at 17:09 | comment | added | Robert Bryant | @IanAgol: Pseudospherical surfaces, by which people usually mean surfaces in $3$-space with constant negative Gauss curvature (though some sources specifically require Gauss curvature equal to $-1$), do not have to be analytic. I believe it's true that any smooth space curve is locally contained in a (local) smooth surface of Gauss curvature $-1$. In particular, the example that I gave above must lie in some smooth surface of Gauss curvature $-1$. Meanwhile, smooth surfaces with Gauss curvature equal to $+1$ in $3$-space do have to be real-analytic. | |
| Aug 19, 2022 at 16:27 | comment | added | Ian Agol | If I recall correctly, any curve of constant torsion can be embedded locally in a pseudospherical surface (which I think is analytic?). | |
| Aug 19, 2022 at 14:16 | comment | added | Robert Bryant | @OtisChodosh: You're welcome. It's an intresting question. I guess a related question is whether every smooth germ of a curve in $\mathbb{R}^3$ lies in some $2$-dimensional (locally) real-analytic subset of $\mathbb{R}^3$ (which may be singular). I don't know how to answer that, but I do expect the answer to be 'no', even if you assume the curve to be smoothly embedded. | |
| Aug 19, 2022 at 13:05 | comment | added | Otis Chodosh | This a very elegant solution, thank you Robert! | |
| Aug 19, 2022 at 13:03 | vote | accept | Otis Chodosh | ||
| Aug 19, 2022 at 11:03 | history | answered | Robert Bryant | CC BY-SA 4.0 |