Fáry-Milnor theorem for positively curved metrics on $S^3$? 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
 A: 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).
A: 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.
