Smoothly homotoping a sphere in $\mathbb{R}^3$

Question

Start with the standard sphere $S^2$ and consider another (diffeomorphic) sphere $S$ such that there is a family of deformations of $S^2$ in $\mathbb{R}^3$ that ends in $S$. If $S$ is positively curved, I strongly suspect that there can be a family of deformations (if not the given one) which starts with $S^2$, ends with $S$ and every member of the family in between is positively curved. How can I make this argument solid? Or is it not true at all?

Answer 1

Yes, this is correct. There are two steps to the proof:

1. (Easy and classical, going back to Gauss.) Positive curvature is equivalent to positive 2nd fundamental form, is equivalent to strict convexity of $S$ in $R^3$.

2. (Considerably harder, although, in $R^3$ this was probably known earlier): The normalized mean curvature flow of (strictly) convex hypersurfaces in $R^n$ deforms a hypersurface to the unit round sphere in $R^n$. This was first proved by G.Huisken "Flow by mean curvature of convex surfaces into spheres" in Journal of Diff. Geometry, 1984. An alternative proof is to argue intrincsically, that the set of Riemannian metrics of positive curvature on $S^2$ is connected. This can be achieved via Ricci flow (still, nonelementary). Then use Alexandrov-Pogorelov theorem that all positively curved spheres embed isometrically in $R^3$ as convex hypersurfaces and the embedding can be made continuous with respect to the metric (again, nonelementary).