3
$\begingroup$

I'm really stuck on this problem.

Let $\alpha:[a,b]\subset \mathbb{R}\to \mathbb{R}^3$ be a smooth arc-length parametrized curve and let $\kappa:[a,b]\to \mathbb{R}$ be its curvature.

I know from the "Fundamental theorem of the local theory of curves" that, roughly speaking, associated to each smooth non-zero curvature function and smooth torsion function there is a unique regular parametrized curve, up to rigid motions. In particular, defining a smooth non-zero curvature function, there is a unique plane curve associated.

Let $\beta:[a,b]\to\mathbb{R}^2$ be the plane curve endowed with the curvature $\kappa$ and suppose that $\alpha(I)$ is a closed curve.

Is it possible that $\beta(I)$ be a non-closed curve?

I tried to read a paper called "A differential-geometric criterion for a space curve to be closed", the author is Hwang Cheng-Chung. But I don't know how to apply it to my problem.

$\endgroup$
2
$\begingroup$

Even if $\alpha$ is a plane curve I think you could come up with examples when $\beta$ is not closed.

Say $\alpha$ may contain some straight line segments each followed by a turn. A straight line segment could be followed by a left turn or by a right turn. Let $\beta$ "copy" $\alpha$, except at just one turn, where say $\alpha$ makes a right turn, and $\beta$ makes a left turn, with the same curvature as $\alpha$. I assume your curvature is always non-negative (so it is not a signed curvature).

So, since $\beta$ made the wrong turn, it does not end where $\alpha$ ends, so it does not close, see picture.

$\endgroup$
  • 1
    $\begingroup$ you are right! It is enough what you wrote it, the curvature is invariant under changes of orientation, so I can choose any simple closed curve, I parametrize it until the middle and then goes back to the origin. The new curve is not closed. Thank you very much!!! $\endgroup$ – DonQuixote Mar 23 '15 at 2:44
  • $\begingroup$ Be careful. The Fundamental Theorem of Curve Theory does require $\kappa>0$ (and remember that for space curves there is no "signed" curvature). (If $\kappa$ vanishes at a point, there is likely no Frenet frame at that point.) So, if you have $\kappa$ nowhere zero, these counterexamples can't occur. $\endgroup$ – Ted Shifrin Mar 23 '15 at 13:16
  • $\begingroup$ @TedShifrin I thought I was careful, by adding straight line segments (where the curvature is 0) before a turn is made. For the case when the curvature does not vanish, perhaps you could add a separate answer explaining it? Also the OP seems to drop the torsion from the picture, so perhaps one could make an example by just twisting the curve. Hmm, I read the questions again, and it does say "non-zero curvature", somehow didn't see that before... $\endgroup$ – Mirko Mar 23 '15 at 13:27
  • $\begingroup$ @MirkoSwirko: The point is that if you do not allow points where $\kappa=0$ you will not be able to make up such examples!! :) $\endgroup$ – Ted Shifrin Mar 23 '15 at 13:32
  • $\begingroup$ @TedShifrin The point is that I barely passed my diff geom course, so please don't take me too seriously. I feel you are correct when we talk about plane curves, but I am not in a position to follow your comment for space curves (where we could conceivably play with the torsion alone, preserving the curvature?) unless you either provide more details (in another answer) or a reference (though I guess the Fundamental Theorem qualifies as a reference if I really want to find it and read about it). At least I may get partial credit for illustrating that the condition $\kappa>0$ is essential :) $\endgroup$ – Mirko Mar 23 '15 at 13:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.