Curves on potatoes

If we assume surfaces are rectifiable, then the answer is "yes"; the proof requires some geometric measure theory.

Let $S_1,S_2\subset\mathbb R^3$ be topological spheres, countably $2$-rectifiable, with finite $\mathcal H^2$ measure. Consider $F:S_1\times S_2\to\mathbb R^3$ defined by $F(x,y)=x-y$. By the coarea formula, $\mathcal H^1(F^{-1}(v))<\infty$ for almost every $v\in\mathbb R^3$. Hence $\mathcal H^1\bigl(S_1\cap(S_2+v)\bigr)$ is finite for almost every $v$.

Choose $v$ so that $S_1$ has points on both sides of $S_2+v$; this condition is open, so we may also assume the above finiteness.

Note that $K=S_1\cap(S_2+v)$ separates $S_1$. A finite-length compact separator of $S^2$ contains a Jordan curve. (Take a separating component; a finite-length continuum is locally connected, and a locally connected planar separator contains a Jordan curve. The needed statements are collected in "Coarea Inequality for..." §2.5 by Behnam Esmayli, Toni Ikonen, and Kai Rajala.)

Thus $S_1$ and $S_2$ contain congruent simple closed curves; in fact one is a translate of the other.

A version of this argument should work for embedded spheres of finite area. On the other hand, by Levin's theorem level sets of a generic function from the 2-sphere to $\mathbb R$ do not contain an arc; this suggests that the answer might be "no" in general.

The argument was developed in discussion with ChatGPT.