One can for discrete billiards ask questions which are classical in the continuum. One of them is Guillemin’s conjugacy problem, which is an inverse problem. It is problem 6 in my personal favorite list of open problems in Hamiltonian dynamics from 25 years ago. This is a problem which in the discrete can be asked. Given two discrete planar regions with the same area A, then the phase space of the billiard has n=6A elements and the geodesic flow permutation T defines a cycle structure. Is it possible to get the same cycle structure for different tables?
Update of March 31, 2025. I did some experiments and looked for smaller spheres which are ergodic meaning that there are exactly two orbits (orientation is preserved by the flow so that the phase space is covered by positively oriented and negatively oriented orbits). Wow, there are many even if we insist the area to be the same. Therefore: the Guillemin conjugacy question is almost trivial in the discrete. What does this indicate for the continuum? We do not know. For smooth geodesic flows on a 2-sphere, we expect a mixture of hyperbolic and stable motion. The hyperbolic part is expected to be governed by Pesin’s theory (non-uniform hyperbolicity is expected but hard to prove!) and near periodic elliptic orbits of the flow in the corresponding three dimensional phase space, we expect stable motion by KAM. There is a lot of structure coming in in the continuum which we do not see in a single discrete billiard. As usual in mathematics, we can approximate stuff with finite stuff but a single approximation does not say much. The approximation 22/7 of Pi does not say much pi. The empirical fact that we see ergodicity quite often could indicate however that we should see also ergodicity often in the continuum at least if the metric has no smoothness, like a generic polyhedron. Here is the code showing two examples (use the code from last week’s post) . We verify that both graphs lead to an ergodic flow (one orbit on each orientation class) and that the vertex degrees are different.
e1={1 -> 2, 1 -> 3, 1 -> 4, 1 -> 5, 1 -> 6, 2 -> 4, 2 -> 6, 2 -> 7, 2 -> 8, 2 -> 9, 2 -> 10, 3 -> 5, 3 -> 6, 3 -> 10,
3 -> 11, 3 -> 12, 4 -> 5, 4 -> 9, 4 -> 13, 4 -> 14, 4 -> 15, 4 -> 16, 4 -> 17, 4 -> 18, 5 -> 12, 5 -> 18, 5 -> 19, 5 -> 20,
5 -> 21, 6 -> 10, 7 -> 8, 7 -> 9, 7 -> 11, 7 -> 13, 7 -> 22, 7 -> 24, 7 -> 34, 8 -> 10, 8 -> 11, 9 -> 13, 10 -> 11,
11 -> 12, 11 -> 20, 11 -> 22, 11 -> 23, 11 -> 26, 12 -> 20, 13 -> 16, 13 -> 19, 13 -> 34, 13 -> 25, 13 -> 27, 13 -> 28,
13 -> 29, 13 -> 30, 13 -> 31, 13 -> 32, 14 -> 15, 14 -> 17, 14 -> 19, 15 -> 16, 15 -> 19, 15 -> 27, 15 -> 29, 15 -> 33,
16 -> 29, 17 -> 18, 17 -> 19, 17 -> 21, 18 -> 21, 19 -> 20, 19 -> 21, 19 -> 23, 19 -> 26, 19 -> 28, 19 -> 31, 19 -> 33,
20 -> 26, 22 -> 24, 22 -> 23, 22 -> 25, 24 -> 34, 24 -> 25, 34 -> 25, 23 -> 26, 23 -> 25, 23 -> 30, 23 -> 31, 23 -> 32,
25 -> 32, 27 -> 28, 27 -> 29, 27 -> 33, 28 -> 33, 30 -> 31, 30 -> 32};s1=UndirectedGraph[Graph[e1]];
e2={1 -> 2, 1 -> 3, 1 -> 4, 1 -> 5, 1 -> 6, 1 -> 7, 2 -> 4, 2 -> 7, 2 -> 8, 2 -> 9, 2 -> 10, 2 -> 11, 2 -> 12, 2 -> 13,
2 -> 14, 2 -> 15, 3 -> 6, 3 -> 7, 3 -> 12, 3 -> 14, 3 -> 16, 3 -> 17, 3 -> 18, 3 -> 19, 3 -> 20, 3 -> 21, 4 -> 5, 4 -> 11,
4 -> 16, 4 -> 22, 4 -> 23, 4 -> 24, 4 -> 25, 5 -> 6, 5 -> 16, 6 -> 16, 7 -> 14, 8 -> 13, 8 -> 15, 8 -> 17, 8 -> 18,
8 -> 20, 8 -> 22, 8 -> 29, 8 -> 30, 8 -> 34, 9 -> 10, 9 -> 15, 9 -> 22, 10 -> 11, 10 -> 22, 10 -> 24, 10 -> 25, 11 -> 25,
12 -> 13, 12 -> 14, 12 -> 18, 12 -> 30, 13 -> 30, 15 -> 22, 16 -> 19, 16 -> 23, 16 -> 26, 17 -> 19, 17 -> 20, 17 -> 29,
17 -> 27, 17 -> 28, 18 -> 21, 18 -> 30, 18 -> 34, 18 -> 32, 18 -> 33, 19 -> 26, 19 -> 28, 20 -> 21, 20 -> 34, 20 -> 32,
20 -> 33, 21 -> 33, 22 -> 23, 22 -> 24, 22 -> 29, 22 -> 27, 22 -> 28, 22 -> 31, 23 -> 26, 23 -> 31, 24 -> 25, 29 -> 27,
34 -> 32, 26 -> 28, 26 -> 31, 27 -> 28, 28 -> 31, 32 -> 33}; s2=UndirectedGraph[Graph[e2]];
Union[VertexDegree[s1] ] == {4, 5, 6, 7, 8, 9, 10, 12, 13}
Union[VertexDegree[s2]] == {4, 5, 6, 7, 8, 9, 10, 11}
Map[Length,T[Whitney[s1]]] =={192,192}
Map[Length,T[Whitney[s2]]] =={192,192}
