# What is Kept Fixed for Flexible Spheres

For background to this question much recent exciting related things, see this videotaped lecture by Alexander Gaifullin.

Consider a triangulation $K$ of a two-dimensional sphere and consider maps from the vertices of $K$ to $R^3$. A one-parameter family $\phi_t$ of such maps is called a flexing if the distance $d(\phi_t(v),\phi_t(u))$ is constant whenever $v$ and $u$ are adjacent vertices. A flexing is trivial if it keep the distances between every two vertices.

We can extend the maps of the vertices linearly to a map from $K$ considered as a geometric simplicial complex to $R^3$. When $\phi (K)$ is an embedding we can consider the solid body $L$ inside it.

The following fundamental facts are known:

1) (Bricard) If $K$ is the boundary complex of the Octahedron then there are maps of the vertices of $K$ with non-trivial flexes. (However when you extends linearly these maps to $K$ this is no longer an embedding.)

2) (Gluck) based on Alexandrov and Stainitz) A generic map of $K$ to $R^3$ is rigid.

3) (Connelly) There are embedded flexible spheres

4) (Sabitov) The volume of $L$ along a flexing of a sphere is constant for every flexing. (This extends to the case where we don't have an embedding by regarding the volume as a signed sum of volumes of bounded complements of the image of $K$.)

I would like to know what else might be fixed for flexing of 2-spheres into $R^3$.

Q1: (Perhaps this was asked by Bob Connelly) Are Dehn-invariants fixed for a flexing. In other words, are two embedded spheres along the flexing scissor-congruent?

Q2: Are the eigenvalues of the Laplacian of the (silid 3D-dimensional) body constant along the flexing?

Q3: Are lengths of closed geodesics in $L$ constant for a flexing?

Explanation (Added later, see Joe's comment): Here we consider $L$ as a 3-dimensional Billiard "table" and by "geodesics" we refer to billiard paths.

(Of course, positive answers to Q1 and Q2 would be a far-reaching if not far-fetched extension of Sabitov's theorem, and Q3 is also rather far-fetched.)

I don't know the answers to Q1, Q2, Q3 even for the known examples of flexible spheres. I dont know if Q2,Q3 can be extended to the non-embeddable case but I suspect Q3 could.

• The Schläfli's formula $$\sum_i\ell_i\theta_i=\mathrm{Const}$$ should be also on your list; here $\ell_i$ denote the length of the edges and $theta_i$ the corresponding dihedral angles. – Anton Petrunin Sep 6 '16 at 19:44
• A remark on Q3: Geodesics on the 2D surface of $K$ are unaltered by flexing---So those are kept fixed. But this bears no evident relation to billiard paths in the 3D solid $L$, which is the focus of Q3 – Joseph O'Rourke Sep 7 '16 at 11:40
• The result Anton mentions is due to R. Alexander (dx.doi.org/10.1090/S0002-9947-1985-0776397-6) and your Q1 is in the literature called the Strong Bellows Conjecture. – Yoav Kallus Sep 7 '16 at 18:50

## 2 Answers

Victor Alexandrov and Robert Connelly (http://arxiv.org/abs/0905.3683) constructed a counterexample to the Strong Bellows Conjecture i.e. a flexible polyhedral surface with non-constant Dehn invariant. It has a combinatorial type of the suspension over a hexagon. However, this surface is not embedded. As far as I know, no embedded counterexample is known up to now.

For several simplest types of flexible polyhedra, namely, for Bricard's octahedra and Steffen's 9-vertex flexor their Dehn invariants are constant during the flexion (Victor Alexandrov, http://arxiv.org/abs/0901.2989).

About 6 months ago, Victor Alexandrov discussed Q2 with me and told me that he is trying to make some progress towards it. Nevertheless, I have never heard about any result in this direction.

• Welcome to MO;)! – Ilya Bogdanov Sep 8 '16 at 8:44
• An update: Gaifullin and Ignashchenko have proved the strong bellows conjecture in arxiv.org/abs/1710.11247. So the answer to Q1 is positive. – Ivan Izmestiev Mar 7 at 12:31

I think the answer to Q3 is negative.

Take a flexible polyhedron and choose two adjacent faces $F_1$, $F_2$ such that the angle between them changes during the deformation. Now attach to $F_2$ a small tetrahedron so that one of its faces is parallel to $F_1$. If this is done near the edge $F_1 \cap F_2$, then there appears a short closed billiard trajectory (of period $2$). This trajectory is destroyed during the deformation.

The answer to Q2 seems to be negative as well (the idea came from a discussion with Daniel Grieser).

The Laplacian eigenvalues determine the simple part of the length spectrum (that is those lengths that are realized by a unique billiard trajectory). In the above example there are infinitely many geodesics between the parallel planes, therefore it doesn't work. Instead of adding a tetrahedron one should create a chamber that has a unique short closed billiard trajectory (with one of the collision points in the face $F_1$), and such that after rotating $F_1$ around the edge $F_1 \cap F_2$ the length of this trajectory changes. Since the simple length spectrum changes, the Laplacian eigenvalues change as well.