TITLE: Is there an inscribed cube for an arbitrary compact closed surface?
QUESTION [10 upvotes]: 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.

REPLY [7 votes]: 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.