Converse of Scherk–Segre theorem on the number of vertices of a convex space curve

Question

It is well known that any smooth simple closed convex curve $\gamma$ in $\mathbb{R}^{3}$ that meets no plane in more than 4 points has exactly 4 vertices, i.e., points of vanishing torsion; here "convex" means that the curve lies on the boundary of its convex hull. This result, as mentioned for example by Pohl in On a theorem related to the four-vertex theorem, goes back to the work of Scherk and Segre in the 1930s.

I was wondering about the other direction:

Question. If $\gamma$ is a simple closed convex curve with (nonvanishing curvature and) exactly 4 vertices, does it then follow that no plane intersects it in more than 4 points?

Note that if $\gamma$ lies on $\mathbb{S}^{2}$, then points of vanishing torsion correspond to critical points of the geodesic curvature. So to find a counterexample it would suffice to construct a spherical curve with exactly 4 critical points of geodesic curvature that does not satisfy the given intersection condition.

Answer 1

This is not an answer but an extended comment to indicate that the answer would be no without the simplicity or convexity assumptions. First observe that there are curves in $\textbf{R}^2$ with exactly $4$ critical points of curvature which self-intersect an unlimited number of times, as shown for instance in the picture below. Pull this curve to $\mathbf{S^2}$ via stereographic projection, which will preserve critical points of curvature. So we obtain a spherical curve whose torsion changes sign exactly $4$ times and intersects some plane more than $4$ times. This example is convex but not simple. A perturbation of it yields a simple closed curve in $\mathbf{R}^3$ which still has exactly $4$ vertices and intersects some plane more than $4$ times, but this perturbation will not preserve convexity.