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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.

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  • $\begingroup$ In which sense is meant "isometrically embedded"? $\endgroup$
    – YCor
    Commented Jun 7, 2021 at 14:12
  • $\begingroup$ @YCor: For this I follow the expression of Nash embedding theorem, meaning that the Riemann metrics of $M$ inherits from the ambient space $\mathbb{R}^3$(so as to let cube make sense). Or maybe i should just say that $M$ is a 2-dimensional compact surface of $\mathbb{R}^3$. $\endgroup$
    – user178596
    Commented Jun 7, 2021 at 17:21
  • $\begingroup$ Then every surface inherits the induced Riemannian metric, so indeed "a (piecewise?) smoothly/$C^2$ embedded surface/sphere" seems to be the assumption. $\endgroup$
    – YCor
    Commented Jun 7, 2021 at 17:27
  • $\begingroup$ @YCor: Yes, for piecewise smooth I follow Inscribed square problem suggesting that $M$ could admit some 'singular' points of codim at least 1 (e.g. polyhedrons). But what I really care about is the smooth surface case. $\endgroup$
    – user178596
    Commented Jun 7, 2021 at 17:40
  • $\begingroup$ Just for the analogy, there is an open question due to Toeplitz, whether every simple closed plane curve contains the vertices of a square. It's open in general, but known to hold in the piecewise $C^1$ case. See this Images des Mathématiques article by Ghys (French) for context and references. $\endgroup$
    – YCor
    Commented Jun 7, 2021 at 18:16

1 Answer 1

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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.

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    $\begingroup$ As I calculate it, the polynomials of degree $\le3$ on $\mathbb{R}^3$ have dimension $20$: one generator for each of three variants of $x^3, x^2, x, xy$, six variants of $x^2y$, and one variant each of $xyz$ and $1$. Meanwhile the subspace of polynomials vanishing on the vertices of the cube $(\pm1,\pm1,\pm1)$ has dimension $12$, being generated by $(x^2-1)(ax+by+cz+d)$, $(y^2-1)(ax+by+cz+d)$, $(z^2-1)(ax+by+cz+d)$. So indeed the vanishing polynomials have codimension $8$. $\endgroup$
    – user44143
    Commented Jun 8, 2021 at 1:34
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    $\begingroup$ @Lucellia Kassel, Yes, that's correct. By the way, I wonder if in any dimension (or at least in some dimension $>3$) one can always inscribe a polytope that is a convex hull of midpoints of edges of a regular tetrahedron. In dimension 3 this polytope is an Octahedron. In dimension $n$ it has $(n+1)n/2$ vertices, and the dimension of such polytopes is $n+1+n(n-1)/2$, which is larger $\endgroup$
    – Dmitri Panov
    Commented Jun 8, 2021 at 16:37
  • $\begingroup$ Yes, by a tetrahedron I mean a regular $n$-simplex. $\endgroup$
    – Dmitri Panov
    Commented Jun 9, 2021 at 11:45
  • $\begingroup$ @DmitriPanov: why this simplex edge midpoint polyptope rather than the crosspolytope, which has 2n vertices and dimension quadratic in n? I'm asking because the cross-polytope would generalize not only the octahedron but also the square in the original question by Toeplitz, where your question would be about a triangle. Is the cross-polytope "too easy" in large dimension? $\endgroup$
    – Stefan Witzel
    Commented Jun 13, 2021 at 20:38
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    $\begingroup$ @StefanWitzel I think the paper cited in the wiki article: link.springer.com/article/10.1007%2FBF02760036 solves the question for cross-polytopes in all dimensions. Though I haven't read the proof. $\endgroup$
    – Dmitri Panov
    Commented Jun 13, 2021 at 21:40

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