I'm interested in generalizations the following well-known theorem of Fáry and Milnor.

Theorem. (Fáry-Milnor) If a simple closed curve $\gamma \subset \mathbb{R}^3$ is knotted, then the total curvature of $\gamma$ is greater than $4\pi$.

I have found some sources that prove essentially the same result when $\mathbb{R}^3$ is equipped with a non-positively curved metric (e.g. the usual hyperbolic metric), maybe satisfying some extra hypotheses (for example, see [1,2]). I'm interested in the opposite situation.

Question. Is there a version of the Fáry-Milnor theorem for positively curved metrics on $S^3$?

It seems that one could derive some inequality from the usual one using contraction maps to and from Euclidean space and estimating the resulting effect on the total curvature along $\gamma$. However, this would depend strongly on the metrics/contraction maps involved. I would really prefer something metric independent if possible.

[1] F. Brickell and C. C. Hsiung. The total absolute curvature of closed curves in Riemannian manifolds.

[2] C. Schmitz. The Theorem of Fáry and Milnor for Hadamard Manifolds

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    $\begingroup$ Maybe the integral of the absolute value of the curvature is not the "right" integrand. The Fary-Milnor Theorem and generalizations are usually obtained by lifting the knot to some other homogeneous space by using the Frenet frame and then integrating some integrand (over the lift) that has an integral geometric representation. For instance, on the two sphere you can consider the integral of the absolute value of the curvature of a curve, but this is not too interesting. More interesting is to lift to SO(3) or S^3 by using the Frenet frame and measure the length in SO(3). $\endgroup$ Jul 9, 2021 at 16:20
  • $\begingroup$ Perhaps lifting the knot to some frame or Grassmannian bundle and using the geometry of the bundle induced from that of the manifold can yield something interesting. $\endgroup$ Jul 9, 2021 at 16:47

2 Answers 2


In fact, a knot in a positively curved metric can have zero curvature.

Consider a 2-bridge knot, then one can make it into a $\pi$-orbifold (or bifold in the terminology of Kronheimer-Mrowka) so that there is a cone metric with angle $\pi$ around the knot and constant positive sectional curvature. This is a quotient of $S^3$ by a dihedral group. The knot is geodesic in this metric, but the metric is not Riemannian. One can modify the metric in a small neighborhood of the knot in an $S^1\times S^1$ equivariant way (using cylindrical coordinates in a tubular neighborhood of the knot) to make it into a positively curved Riemannian metric. Then the knot is totally geodesic and has zero curvature. The modification of the metric can be done similarly to section 2 of this paper (I learned this technique from Thurston, but I think it may be originally due to Gromov).

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    $\begingroup$ It should also be possible to realize every $(p,q)$ torus knot with arbitrarilly small total curvature in the round sphere, by perturbing a geodesic run over $p$ times. $\endgroup$ Jul 9, 2021 at 12:21
  • $\begingroup$ More examples might come from perturbing Euclidean cone metrics on knots, but I don’t know any examples that don’t admit a spherical cone metric. However, I don’t see any reason that every knot (and maybe even link) couldn’t be realized as a geodesic in a positively curved metric. mathoverflow.net/q/391234/1345 $\endgroup$
    – Ian Agol
    Jul 9, 2021 at 13:53
  • $\begingroup$ @BenoîtKloeckner looks like reznikov did this. arxiv.org/abs/dg-ga/9410006 $\endgroup$
    – Ian Agol
    Jul 9, 2021 at 13:55
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    $\begingroup$ @BenoîtKloeckner actually, your suggestion works for any link: put it in braid position of braid index b, and perturb a geodesic running b times around to the braid with arbitrarily small curvature. I wonder if reznikov’s argument can be modified to make a positively curved metric with the link geodesic? In any case, you should post this as an answer. $\endgroup$
    – Ian Agol
    Jul 9, 2021 at 14:09

I expand a little on a comment made below Ian Agol's answer, including an improvement proposed by him.

Claim: every knot can be realized on the round sphere $\mathbb{S}^3$ with arbitrarily small (but positive) curvature.

Indeed, a knot can be presented as the closure of a braid; you an realize the closed braid inside a small tubular neighborhood of a closed geodesic, with the braid $C^2$-close to this core geodesic. This ensures the total curvature is as close to zero as you wish.

This is especially easy to represent oneself with torus knots, but actually works for every link, as mentionned by Ian Agol.


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