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Hopf's Umlaufsatz (better known as?) says:

Let $\gamma$ be a simple closed di fferentiable positively oriented curve in the plane. Then for its curvature $\kappa$ it holds:

$$\int_{\gamma}\kappa\ \text{d}s = 2\pi$$

I wonder if (and cannot see why not) the inversion holds, too:

Let $\kappa$ be a continuous function $\kappa: [0,1] \rightarrow \mathbb{R}$ with $\kappa(0) = \kappa(1)$ and $\int_0^{1}\kappa\ \text{d}s = 2\pi$. Then there is a simple closed di fferentiable positively oriented curve $\gamma$ (of length 1) with curvature $\kappa$.

(If this holds, $\gamma$ would be unique upto congruency via Euclidean motions.)

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    $\begingroup$ Sometimes this is called the 2-dimensional Gauss-Bonnet theorem, as in this article. // Yes, one can reconstruct a curve from its curvature, as on second page here. $\endgroup$
    – user53153
    Commented Jan 11, 2013 at 20:49
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    $\begingroup$ @5PM: A planar curve can be reconstructed from it's curvature, but the question is if the curve is closed or not. $\endgroup$ Commented Jan 16, 2013 at 20:51

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There is another necessary condition on $\kappa$. The so-called Four vertex theorem:

Theorem: Suppose $\kappa$ is the curvature of a simple closed curve. Then, either $\kappa$ is constant or has at least 2 minimums and 2 maximums.

The ellipse shows that there is no reason to expect more than 2 minimums or maximums of curvature.

It turns out that together with your integral condition and this condition together are necessary and sufficient for $\kappa$ to correspond to a closed simple curve. A write up, due to DeTurck, Gluck, Pomerleano, and Vela-Vick, (who say the result is due to Dahlberg) can be found here. See also a more informal (but similar) write up by Vela-Vick here.

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  • $\begingroup$ Thanks, now I remember reading about the theorem long time ago... Now I also see that @Hans wanted length $1$, which is not guaranteed by the construction as far as I can tell... Or maybe Hans even wanted an arclength parametrization? I hope he'll clarify. $\endgroup$
    – user53153
    Commented Jan 16, 2013 at 21:15
  • $\begingroup$ @5PM: I missed the unit length part. In that case, I'm not sure what more needs to be added. $\endgroup$ Commented Jan 16, 2013 at 23:22

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