Is there a $C^\infty$-smooth embedding $\gamma : I \to \mathbb{R}^3$ so that there is no real analytic $2$-dimensional submanifold $M \subset \mathbb{R}^3$ with $\gamma(I)\subset M$?
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asked Aug 18, 2022 at 23:12
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2$\begingroup$ Presumably you want $\gamma$ to be an embedding? Otherwise you can just arrange $\gamma(I)$ to have a triple self-intersection from three linearly independent directions. $\endgroup$– Willie WongCommented Aug 19, 2022 at 2:22
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$\begingroup$ @Willie Wong, Thanks, I fixed this! $\endgroup$– Otis ChodoshCommented Aug 19, 2022 at 2:44
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$\begingroup$ Is $M$ allowed to have boundary, and does $M$ have to be embedded as well? $\endgroup$– A. Thomas YergerCommented Aug 19, 2022 at 5:20
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2$\begingroup$ An obvious guess is to take a non-analytic curve in the plane e.g. $y = e^{-1/x}$ and "wiggle" it in the z-direction -- maybe have $z = e^{-1/x} \sin(1/x)$ or something. No idea how you'd prove it though. $\endgroup$– Kevin CastoCommented Aug 19, 2022 at 6:36
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1 Answer
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Here is an example: Let $\gamma:\mathbb{R}\to\mathbb{R}^3$ be defined by $\gamma(t) = \bigl(t,\exp(-1/t^2),0\bigr)$ for $t<0$, $\gamma(0) = (0,0,0)$ and $\gamma(t) = \bigl(t,0,\exp(-1/t^2)\bigr)$ for $t>0$. Then I claim that there is no nonsingular real-analytic surface $M\subset\mathbb{R}^3$ that contains the image of $\gamma$.
To see this, suppose that such an $M$ does exist and let $u\in S^2$ be a unit normal to $M$ at the origin. If $u\not=(0,0,\pm1)$, then $M$ and the plane $z=0$ intersect transversely at $(0,0,0)$ along a real-analytic curve that contains $\gamma\bigl((-\epsilon,0]\bigr)$ for some $\epsilon>0$, which is clearly impossible. Similarly, we reach a contradiction if $u\not=(0,\pm1,0)$. Thus, $M$ does not exist.
answered Aug 19, 2022 at 11:03
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1$\begingroup$ This a very elegant solution, thank you Robert! $\endgroup$ Commented Aug 19, 2022 at 13:05
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2$\begingroup$ @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. $\endgroup$ Commented Aug 19, 2022 at 14:16
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1$\begingroup$ @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. $\endgroup$ Commented Aug 19, 2022 at 17:09
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1$\begingroup$ @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. $\endgroup$ Commented Sep 5, 2022 at 17:09
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2$\begingroup$ @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. $\endgroup$ Commented Sep 17, 2022 at 22:15