Given a compact closed surface $M$ (2-dim topological manifold) isometrically embedded in $\mathbb{R}^3$, are there 8 points $x\_i\in M(i=1,\dots,8)$ such that they are the vertices of a cube $C\subset\mathbb{R}^3$?


We may assume that (1)$M$ is smooth and homeomorphic to the 2-sphere $S^2$; (2)$M$ is piecewise-smooth; (3)$M$ is $C^2$-manifold.
The case (1) is actually what I'm mostly curious about.



Is there an inscribed cube for an arbitrary compact closed surface?
One can't inscribe cubes in generic surfaces by dimension reason. Indeed the space of cubes in $\mathbb R^3$ is $7=3+3+1$-dimensional, while a cube has $8$ vertices, and so a surface imposes $8$ conditions on the vertices of the cube.


To make this dimension reasoning rigorous one can do the following. Take the space of polynomials of degree $\le d$ on $\mathbb R^3$ that vanish at $8$ vertices of a (non-zero) cube. For $n$ large enough (probably $d\ge 3$ will suffice) this space has codimension $8$ in the space of all polynomials of degree $\le d$. So if we take the poly $x^2+y^2+z^2-1$ and add to it $\varepsilon F$, where $F$ is a generic poly of degree $d$ then the surface $\Sigma:=\{x^2+y^2+z^2-1+\varepsilon F=0\}$ doesn't contain a cube. And $\Sigma$ has a connected component diffeomorphic to a sphere.


However it is not so easy to construct a concrete example of such surface by hands, because it should be quite asymmetric.


