# If we know the combinatorics of a polyhedron, and all but one of its dihedral angles, does that uniquely determine the remaining dihedral angle?

If we know the combinatorics of a polyhedron, and all but one of its dihedral angles, does that uniquely determine the remaining dihedral angle?

I’m happy to assume the polyhedron is simply connected, if that helps. I do not want to assume that it is convex, because the examples I’m interested in are non-convex.

The only partial result along these lines that I know is that the claim does hold for tetrahedra, by the argument outlined here.

On Mastodon, David Eppstein asked whether there is a fixed bound on the number of edges of a polyhedron that realizes a given Dehn invariant, depending only on the rank of the invariant. I strongly suspect there is no such upper bound.

I have been thinking about a simplified version of this question, where the polyhedra are restricted to polyhedra whose dihedral angles are multiples of $$\pi/2$$ on all but a single edge. Dmitrii Korshunov's answer below seems to settle this question.

Unfortunately I had somehow deluded myself into thinking that this special case implies the general case, which it does not, or not obviously.

• Everything you’re talking about is three dimensional, right? Commented Jul 15 at 20:20
• Yes, everything in this question is Euclidean and three-dimensional. Commented Jul 15 at 20:21
• Incidentally, the question whether the minimum number of edges to realize a Dehn invariant can be upper-bounded by a function of the rank comes from my recent paper "Orthogonal dissection into few rectangles" in Discrete Comput. Geom. (doi.org/10.1007/s00454-023-00614-w), where I observe that the rank lower-bounds this number of edges. Commented Jul 16 at 11:16
• Related: a few months ago I asked about the slightly more general possibility of a polyhedron whose dihedral angles can all weakly increase while preserving combinatorial equivalence, and noted the same connection to the Schläfli formula. Commented Jul 16 at 22:40

By Schläfli formula for a smooth family of polyhedra in $$\mathbb R^3$$ we have $$\sum l_i(t) \frac{\partial \theta_i(t)}{dt}=0$$
where $$\theta_i$$ is the angle corresponding to the $$i$$-th edge, and $$l_i$$ its length. Suppose now we have a one-parametric deformation of our polyhedron such that all $$\theta_i$$ except $$\theta_0$$ are constant. Then $$l_0(t) \frac{\partial \theta_0(t)}{dt}=0$$ hence $$\theta_0$$ is also constant. Thus $$\theta_0$$ is locally constant on the space $$P$$ of polyhedra with a given combinatorial type and fixed angles $$\theta_1,\dots \theta_n$$. Since there are only finitely many connected components of this space ( $$\sin$$ of the angle is algebraic in cordinates of vertices) we know that we can at least reconstruct $$\theta_0$$ up to finite choice. If we are lucky and $$P$$ is connected then we win.
• Interesting. I had a similar idea: I wanted to show the connectedness of $P$ by interpolating between two combinatorially-equivalent polyhedra in such a way that unchanged dihedral angles would be preserved. But I failed to find such an interpolation process. Commented Jul 15 at 21:35
• @RobinHouston Consider the space of all polyhedra of a given comb. type as a semi-algebraic subset of $\mathbb{R}^{3n}$, where $n$ is the number of vertices. the number of connected components of a semi-algebraic set is bounded in terms of the numb. of equations/inequalities and their degrees. Then take the map from $\mathbb{R}^{3n}$ to $\Pi_E S^1$. I claim that it is rational, in particular semi-algebraic hence maps semi-algebraic sets to s.a. sets. Fixing angles also imposes algebraic conditions. Also note that there are finitely many comb. types of polyhedra with given number of vertices. Commented Jul 16 at 1:34
• The map from $\mathbb{R}^{3n}$ to $\sum_E S^1$ is given as follows: an edge is shared by two simplices, each gives a pair a vectors -- take cross product of each pair. The angle between these cross products is our dihedral angle -- its sine and cosine are expressed rationally in terms the coordinates. Commented Jul 16 at 1:47