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.

$\endgroup$

1 Answer 1

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.

$\endgroup$
6
  • 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, 2015 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. $\endgroup$ Mar 23, 2015 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, 2015 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!! :) $\endgroup$ Mar 23, 2015 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, 2015 at 13:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .