Is there a way to parametrize the configuration space of all convex polyhedra of a given combinatorial type as a convex set?

Question

I'm sure this is easy/known, but I'm not hitting an appropriate search term for finding the answer and the coffee hasn't kicked in enough to come up with it myself:

Let $T$ be a simplicial 2-complex homeomorphic to the sphere with $n$ vertices. $\mathbb{R}^{3n}$ gives us a configuration space for $T$ of all possible realizations of $T$ as a polyhedron in $\mathbb{R}^3$. Take $C\subset \mathbb{R}^{3n}$ to be the configuration space of all realizations of $T$ as a convex polyhedron.

We might ask, is $C$ a convex set? Which is akin to asking if we have two realizations $P_0\in C$ and $P_1\in C$, both convex, and we linearly interpolate between the two in $R^{3n}$, giving us a continuous family of polyhedra $P_t$ for $t\in[0,1]$, is $P_t\in C$ for all $t\in[0,1]$? It seems to me that the answer here is no, because we can do something like simply rotate $P_0$ around some axis through the point at the average of its vertices by $\pi$ and then do this interpolation and it seems like the polyhedron will admit of some self-intersections at intermediate values of $t$ for lots of examples (though my intuition may be off here, I haven't actually constructed an example where this happens).

If I'm wrong about my intuition above, then I'm interested in why, and if I'm right, then I'm further interested in the following questions.

Question 1: What if we select a face $i_0j_0k_0$ of the polyhedron $P_0$ and require that in $P_1$, vertex $i_1 = i_0$ (so that $i_t$ is constant in $P_t$), $j_1$ lies on the ray $i_0 j_0$ (so that the segment $i_t j_t$ is just stretching in $P_t$ with one side fixed), and $k_1$ lies in the plane supporting $i_0 j_0 k_0$ such that the orientation of triangle $i_1 j_1 k_1$ is the same as $i_0 j_0 k_0$ (this way the triangle $i_t j_t k_t$ remains coplanar throughout the motion and doesn't flip orientation). Let $C_{i_0j_0k_0}\subset C$ be the configuration space with these additional requirements. It feels like $C_{i_0j_0k_0}$ is convex. Is it? Does this appear anywhere in the literature? Update: The answer here is no. I can construct examples.

Question 2: if the answer to question 1 is still no, it (again) feels like there ought to be a way to interpolate between two convex polyhedra with the same combinatorics in such a way that all intermediate polyhedra are convex. If there is, is there a different way of parametrizing the polyhedra such that the appropriately defined configuration space is convex, meaning we can interpolate linearly the parameters between two realizations while all intermediate polyhedra are also convex realizations?

Answer 1

The answer to the question in the title is yes. The realization space of polyhedra in 3-space is an open ball, and so you can choose to parametrize it as a convex set. This does not mean that the parametrization is easy to describe. Nor does it mean that the set $C\subset\Bbb R^{3n}$ from your question is convex. As far as I know there is no canonical convex parametrization.

Let me describe to you one way to parametrize the realization space (modulo some trivial operations) as the strictly positive orthant $\Bbb R^{m-3}_{>0}$, which is homeomorphic to the $(m-3)$-dimensional open ball (where $m$ is the number of edges of the 2-complex).

1. fix a (triangular) face $\Delta$ of your complex and an embedding of $\Delta$ into $\Bbb R^2$. Given one positive number $\omega_e>0$ per edge $e \not\in \Delta$ (this make $m-3$ positive numbers) there exists a unique so-called Tutte embedding of your complex into $\Bbb R^2$ which is in equilibrium w.r.t. the stresses $\omega_e$.

2. Theren exists a one-to-one correspondence between such stressed planar representation of your complex (with a positive stress on all edges $e\not\in \Delta$), and convex polyhedral liftings over the face $\Delta$ (aka convex polyhedra). This is known as the Maxwell-Cremona correspondence.

Both taken together give you a very implicitly defined bijection between the strictly positive orthant and the realization space of your simplicial sphere modulo certain affine and projective transformations.

Answer 2

Here’s a move towards an answer to questions 2 and 3.

First, a remark about triangles parametrized by their edge lengths—the configuration space of a single triangle parametrized by its edge lengths $x$, $y$, and $z$ is given by the intersection of three half-planes in $\mathbb{R}^3$, $x + y < z$, $x + z < y$, and $y + z < x$. This means that given two triangles $T_1$ and $T_2$, we can linearly interpolate between them by linearly interpolating their edge lengths.

Now, if $P_1$ and $P_2$ are two polyhedra with the same combinatorics, we can interpolate between their polyhedral metrics by a linear interpolation of their edge lengths. It’s currently unclear to me if $P_1$ and $P_2$ being convex metrics is sufficient to guarantee that the interpolation stays convex at all points between the two. But it does seem clear that the total discrete curvature is conserved by this interpolation. So assuming that the polyhedral metric $(1-t)P_1 + t P_2$ is convex, by Alexandrov’s theorem, there is a unique convex polyhedron realizing the metric. The problems with this are (a) I haven’t shown here that $(1-t)P_1 + t P_2$ is convex (indeed, I’m not sure if it is), and (b) Alexandrov’s theorem doesn’t necessarily give you back the same combinatorics, so it may be that for intermediate values of $t\in(0,1)$, edge flips have to occur.

Any ideas on (a) or (b)? I assume this is well known, but I do not know it…

Answer 3

I just wanted to add a picture that confirms that the linear interpolation between coordinates does not work. Let us have a look at the two black copies of ABCD. Connecting each of the corresponding pairs and walking at unit speed along the connecting lines gives at the half-time point the red configuration which is not an embedding of the given combinatorial type (e.g. a 4-gon with vertices labeled cyclically by ABCD).