Suppose $\alpha: \mathbb{R} \rightarrow \mathbb{R}^3$ is a $C^\infty$ curve, parameterized by arc length ($\left\|\alpha'(t)\right\| = 1$), and with $\alpha(0) = \alpha(\ell)$. Show that there exists a $t_0 \in [0,\ell]$ such that $ \left\| \alpha''(t_0)\right\| \geq \frac{2\pi}{\ell}. $


If we had that $\alpha'(0) = \alpha'(\ell)$, then the result would follow from Fenchel's theorem, but we don't. Actually there is a nice extension to Fenchel's theorem described in Toponogov's book ``Differential Geometry of Curves and Surfaces: A Concise Guide'', under the name Fenchel-Reshetnyak, which gives lower bounds on the integral curvature in an open curve. This can be adapted to our situation to show that $\left\| \alpha''(t_0)\right\| \geq \frac{\pi}{\ell}$, but that bound is only half as good as it presumably could be. It makes sense that the integral curvature bound will be only half as good if we don't constrain $\alpha'(0)$ w.r.t. $\alpha'(\ell)$ (in particular, one may be the negative of the other), but nonetheless it seems that the original bound on the maximum curvature in the loop should still stand. How can we recover it?


1 Answer 1


This smells like a problem in the Calculus of Variations and can be solved as such. We therefore make a slight reformulation: among all $C^3$ curves of length $l$ originating and terminating at the same point, find the one(s) for which the maximum curvature is smallest.

Let's begin by developing some intuition. In general, a minimax problem like this has a solution where the objective (the curvature in this case) is as constant as possible. A circle is an obvious candidate for a solution and indeed a circle of length $l$ has constant radius $l / (2 \pi)$ and constant curvature of $2 \pi / l$.

(As a tiny simplification, by choosing our linear units of measurement let's henceforth assume $l=1$.)

The technique of the Calculus of Variations is to assume we have a solution and perturb it a little, showing that any perturbation, no matter how small, cannot decrease the value of the objective. To motivate this in the present case, though, I find it more appealing to consider how one might go about reducing the maximum curvature of any unit-length loop. A point of maximum curvature is part of a relatively sharp "bump" on the curve. If we push that bump inwards a little, we should be able to decrease its sharpness and at the same time we slightly decrease the total length of the curve. To make up for the latter, uniformly dilate the entire curve relative to its point of origin until the dilated curve again has unit length. This uniformly decreases all curvatures along the loop. In this fashion the maximum curvature has strictly decreased. The only way this operation can fail is when there is no "bump": the curvature everywhere is constant.

To make this go through rigorously, use an adapted orthonormal frame for the curve: $T(t) = \alpha'(t)$ is the tangent, $N(t)$ is the inward-pointing unit normal with $T'(t) = \kappa(t)N(t)$, and $B(t)$ is the unit binormal, with $N'(t) = -\kappa(t)T(t) + \tau(t)B(t)$ and $B'(t) = -\tau(t)N(t)$. ($\tau$ is the torsion. These Serret-Frenet formulae generalize readily to higher dimensions and in two dimensions just forget about $B$.) Notice that $\kappa \ge 0$. Let $\epsilon > 0$ be small and let $\delta(t)$ be a smooth non-negative function in an open neighborhood of $[0,1]$, vanishing at $0$ and $1$ (in order to keep the endpoints of the new curve fixed). Set the "pushed" curve to be

$$\psi(t) = \alpha(t) + \epsilon \delta(t) N(t).$$

Compute the curvature of $\psi$ to first order in $\epsilon$ using the Serret-Frenet formulae. I find that its square becomes

$$\kappa^2 + 2 \kappa \left( \delta'' - \delta \tau^2 \right) \epsilon + O(\epsilon^2).$$

Suppose the curvature is not constant. Then there is a closed interval of maximum curvatures (perhaps reducing to a point) and within some neighborhood of that interval all curvatures are strictly less than the maximum. We can make $\delta$ zero outside this larger neighborhood, let it increase slowly enough (and choose $\epsilon$ sufficiently small) so that $\delta''$ keeps the new curvature less than the maximum within the outer neighborhood, and make $\delta''$ strictly negative at all points within the interval of maximum curvature. This procedure strictly decreases the maximum curvature within the outer interval. You can also check, again to second order in $\epsilon$, that the length of $\psi'$ decreases by $\epsilon \delta \kappa \ge 0$, so we will be able to apply our dilation trick in order to keep the total loop length constant after this perturbation.

A little more formally (and using less work, actually), the preceding equation shows that when $\kappa$ is constant and $\tau = 0$, we would need $\delta'' \lt \delta \tau^2 = 0$ everywhere in order to decrease the curvature, which implies $\delta$ is identically zero in the unit interval. Otherwise (namely, if $\kappa$ is not constant or $\tau \gt 0$), it will be possible to decrease the curvature subject to maintaining the unit length of the curve. Curves of constant curvature and zero torsion are circular arcs, and the only circular arc that returns to its starting point is a full circle. Therefore the maximum curvature in any sufficiently smooth unit-length loop can be no less than $2 \pi$ and is strictly greater than that if the loop is not a circle, QED.

I believe the same technique works in higher dimensions, too, although I have not done the calculations. (There are additional vectors in the adapted frame, and therefore additional torsion terms, but this doesn't seem to present any obstacle to carrying out the same program.)

  • $\begingroup$ Very nice, thank you. I haven't fully digested the details, but I see that this is a good approach. When you say "within some neighborhood of that interval all curvatures are strictly less than the maximum", do you mean 'in the complement of that interval ...'? $\endgroup$
    – yasmar
    Commented Sep 16, 2010 at 17:43
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    $\begingroup$ @yasmar: Sorry about being vague. A local maximum doesn't have to occur just at a point; there can be a closed interval in which curvature is constant and locally max. Around that interval, though, there must be a neighborhood in which the curvature is strictly less than the local max: that's what I'm talking about. (There's also a delicate bit about what happens at the endpoints of the loop, but there's no problem there because $\delta$ is zero at the endpoints.) $\endgroup$
    – whuber
    Commented Sep 16, 2010 at 18:35
  • $\begingroup$ Okay, I understand that your third to last paragraph implies that we could reduce the maximal curvature in alpha with that trick, unless the interval of maximal curvature is the entire interval. Since any curve purporting to have a lower maximal curvature than 2pi is not a circle, we can lower its maximal curvature more. Thus we'd have a contradiction if a curve (other than a circle) claimed to have the lowest maximal curvature. Do I need to demonstrate compactness in a space of admissible curves, or am I making things needlessly complicated? $\endgroup$
    – yasmar
    Commented Sep 16, 2010 at 20:32
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    $\begingroup$ (1) You don't have to cover all "admissible" curves; you just need to deal with a dense subset of them within a neighborhood of the zero perturbation. No (relative) compactness arguments about the loop space are needed. (2) That thought about a torus in R^4 is creative, but the curvature of a surface in which a loop might lie does not affect the loop's own curvature within R^4. The torus can be "flat" in the same sense that a cylinder in R^4 is: at every point it has a principal direction of zero curvature. In the perpendicular direction it usually curves! $\endgroup$
    – whuber
    Commented Sep 16, 2010 at 23:46
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    $\begingroup$ (Continuation) I should have mentioned in the penultimate paragraph that within its context $\delta$ is arbitrary, subject only to the smoothness requirements and that it vanish at the loop's endpoints. The import of the calculation is to show that any attempt at perturbing a circle-with-basepoint while keeping its length constant cannot decrease its maximum curvature. $\endgroup$
    – whuber
    Commented Sep 16, 2010 at 23:51

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