4
$\begingroup$

Let $\mathcal{A}_\Gamma$ be the space of convex (non-degenerate) Euclidean polyhedra with $1$-skeleton a certain polyhedral graph $\Gamma$. This space can be seen as a subset of $\mathcal{Gr}_2(\mathbb{R}^3)^F$ where $F$ is the number of faces of $\Gamma$.

It is a well known fact (for example in Proposition 17 of Deformations of hyperbolic convex polyhedra and cone-$3$-manifolds by Montcouquiol) that $\mathcal{A}_\Gamma$ is a manifold.

Question: Is $\mathcal{A}_\Gamma$ connected?

My gut tells me the answer is yes but I couldn't find an easy proof nor a reference.

Bonus questions: Is it simply connected? Contractible?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
5
$\begingroup$

I believe your question asks about isotopy:

Isotopy property: A combinatorial structure (such as a combinatorial type of polytope) has the isotopy property if any two realizations with the same orientation can be deformed into each other by a continuous deformation that maintains the combinatorial type. Equivalently, the isotopy property holds for a combinatorial structure if and only if its realization space is connected.

This definition is quoted from "Basic Properties of Convex Polytopes" by Martin Henk, Jürgen Richter-Gebert, and Günter M. Ziegler, in Handbook of Discrete and Computational Geometry, 2017. CRC link.

Although the isotopy property holds for $3$-polytopes (Steinitz's Theorem), it was proved by Richter-Gebert that it fails (quite badly) in dimension $4$. This is his "Universality Theorem."

Improve this answer
$\endgroup$
5
  • $\begingroup$ Thank you for your answer! I have a follow-up: I have seen Steinitz's Theorem stated roughly as "A graph is the $1$-skeleton of a $3$-polytope if and only if it is planar and $3$-connected". How does this relate to the Isotopy property? $\endgroup$
    – Giulio Belletti
    Commented Jan 29, 2021 at 12:31
  • $\begingroup$ @GiulioBelletti: I would have to look up the original Satz to be certain, but I believe Steinitz also proved the realization space is $\mathbb{R}^k$ where $k$ depends on the number of edges. (I realize this does not fully answer your followup question.) $\endgroup$
    – Joseph O'Rourke
    Commented Jan 29, 2021 at 12:46
  • 1
    $\begingroup$ @GiulioBelletti: 2nd try on your followup. One proof approach for Steinitz's theorem is via framework equilibrium stresses. Planar graphs with a self-stress with positive weights are Schlegel diagrams, which when lifted form a polyhedron. Varying the weights results in different realizations. The space of realizations is $\mathbb{R}^{E-6}$ where $E$ is the number of edges. $\endgroup$
    – Joseph O'Rourke
    Commented Jan 29, 2021 at 13:48
  • $\begingroup$ I'm still confused about the dimension. If the space of realizations of a graph is $\mathbb{R}^{E-6}$ then for the tetrahedron it would be a single point; however I expect there to be 12 degrees of freedom for tetrahedra in $\mathbb{R}^3$ (6 if we consider them up to isometry) $\endgroup$
    – Giulio Belletti
    Commented Jan 29, 2021 at 15:15
  • $\begingroup$ @GiulioBelletti: I think the discrepancy is that all is "up to projective equivalence," and all tetrahedra are projectively equivalent. $\endgroup$
    – Joseph O'Rourke
    Commented Jan 29, 2021 at 16:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .