Let $S$ be a surface in $\mathbb{R}^{3}$ with the following property:

There is a uniform constant $M$ such that for every Frenet curve $\gamma(t)$, contained in $S$, $| \tau(t) | \leq M$, for all t, where $\tau$ is the torsion. Does this imply that $S$ is a part of a plane?

And what would be a possible generalization of the above for a submanifold of $\mathbb{R}^{n}$?

The motivation and comment:

We do not assume that $S$ is compact. Even in the compact case $S^{2}$, there is a Frenet curve $\gamma :(0, \infty) \rightarrow S^{2}$ such that $\tau(t)$ is unbounded. The reason: a necessary and sufficient condition for that a regular curve $\gamma$ lies on $S^{2}$ is that the pair $(\kappa, \tau)$, curvature and torsion, satisfies in certain differential equation. see Elements of diff geometry by A. Pressley. It can be easily shown that this differential equation has a solution which $\tau$ is unbounded. On the other hand for every smooth functions $\kappa(t), \tau(t)$, there is a Frenet curve for which the curvature and torsion are $\kappa (t), \tau (t) $, respectively . So there is a Frenet curve in $S^{2}$ which torsion is unbounded.


closed as off-topic by Ryan Budney, Daniel Moskovich, Stefan Kohl, Carlo Beenakker, Jack Huizenga Dec 23 '13 at 14:06

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  • $\begingroup$ This isn't a research question. The answer is no, and there are elementary examples. $\endgroup$ – Ryan Budney Dec 20 '13 at 9:20
  • $\begingroup$ could you please give an example $\endgroup$ – Ali Taghavi Dec 20 '13 at 11:42
  • $\begingroup$ What is a "Frennet curve''? (By the way, you probably mean "Frenet"). $\endgroup$ – Igor Belegradek Dec 20 '13 at 16:50
  • $\begingroup$ Yes Frenet curve, a curve with nonzero curvature $\endgroup$ – Ali Taghavi Dec 20 '13 at 16:57
  • 5
    $\begingroup$ Actually, the correct answer is 'Yes, such a (connected) surface $S$ must be part of a plane'. This follows from the classical formula (easily derivable using moving frames) that expresses the torsion $\tau$ of a curve $\gamma$ on a surface $S\subset\mathbb{R}^3$ in terms of the second fundamental form and its first covariant derivative, $\rho$, and $\dot\rho$, where $\rho$ is the geodesic curvature of the curve on the surface. From this formula, one sees that, unless the second fundamental form vanishes identically, there cannot be such a bound on $\tau$. $\endgroup$ – Robert Bryant Dec 21 '13 at 12:05