Getting a bit closer to our problem with incidence matrices for
k-
id.3335 :
- Consider a 6-coloring of the vertices of x3o3o3o5o to exist.
- Then, by restriction to the neighbouring vertices of any single vertex thereof, there will have to be a 5-coloring of the vertex figure of that tiling, i.e. of x3o3o5o or ex. (That one we already know to exist. This is just the tau-scaled inscribed chiral compound of 5 icoes.)
- Similarily, by restriction to the neighbouring vertices of any single vertex of ex, there will be a 4-coloring of its vertex figure, i.e. of x3o5o or ike. (That one too is known to exist. Just consider a tet being face-center inscribed. Any tet-vertex then induces exactly 3 (mutually disjoined!) tips of the adjacent triangles to that centered one of consideration.)
- Again likewise, by restriction onto the neighbouring vertices of any single vertex of ike, there will be a 3-coloring of its vertex figure, i.e. of the pentagon x5o.
At least at this point this seems to become freaky: what? a 3-coloring of a 5-gon?
But sure, yes, this is possible. And the sole possibility, which uses no adjacent identical colors, clearly is
a-b-c-a-b- . This obviously breaks down any symmetry of the pentagon: That single
c clearly fixes that point. And the remaining vertices (locally) would allow a rotation only - contradicting to that fixing. I.e. the only symmetry here is the identity!
Considering next the ike. Consider that already described 4-coloring. It becomes clear that any symmetry, which would interchange any 3 vertices of the same color, would be a symmetry of the corresponding tip of the inscribed tetrahedron. But any such action, applied onto the total tetrahedron, thereby would interchange the remaining vertices of that tetrahedron. Thus the other colors of the 4-colored ike thus would interchange as well. - This shows that here too the only symmetry of the so colored ike would be the identity.
Now coming onto incidence matrices. Incidence matrices always are based on symmetry. In fact, those are essentially based on 0-1-matrices, which describe the incidences of any vertex, any edge, any face, etc. If there is a symmetry of the described thingy, then the elements of the same track, can be grouped. This is how numbers larger than one occure. Well, as this coloring induces the identity to be the only symmetry at all, the according matrices would be 0-1-matrices only. And those, for sure, do exist - at least in the finite cases.
OTOH, we are not considering multicoloring of those vertex sets, we are effectively considering bi-colorings only: those vertices, which have to be chopped off, and those, which aren't. But chopping here is related to that so far considered multi-color scheme. We select the vertices of
k (arbitrary) colors for those
k-
id-operations. Thus both, these
k and the remaining
n-k colors too will be considered alike (in these 2 groups) then first. Here some non-identity transformation of the bi-coloring might take place again. And by the action of this symmetry that single class of non-diminished vertices gets re-divided again. - This is why we still have numbers larger than one in the matrices of teddi, sadi, bidex, et al.
E.g. teddi is derived from that 4-coloring of ike by chopping off the vertices of one color, while considering the other 3 colors alike. Just as the inscribed tet mentioned in the coloring-description, this bi-coloring will show up axial 3fold symmetry. Accordingly those remaining vertices will fall into different orbits according to this 3fold axial symmetry. In fact those are 3 cycles of 3 vertices.
For x3o3o3o5o we then seem to have 6 colors. Those define Wendy's proposed hyperbolic structures:
We could divide into 1 : 5, defining 1-id.3335. Cells ought to be pen + ex (= 0-id.335), vertex figure ought to be sadi (= 1-id.335).
We could use 2 : 4, defining 2-id.3335. Cells ought to be sadi (= 1-id.335), vertex figure ought to be bidex (= 2-id.335).
We could use 3 : 3, defining 3-id.3335. Cells ought to be bidex (= 2-id.335), vertex figure ought to be tridex (= 3-id.335).
We could use 4 : 2, defining 4-id.3335. Cells ought to be tridex (= 3-id.335), vertex figure ought to be quidex (= 4-id.335).
And we could use 5 : 1, defining 5-id.3335. Cells ought to be quidex (= 4-id.335), while here 2 types of vertex figures should occure.
Btw., even so this seems to be a big number, we already found some relations:
1-id.35 (teddi) is dual to 3-id.35
2-id.35 (Weimholt's hexahedron) is self-dual
1-id.335 (sadi) is dual to 4-id.335
2-id.335 (bidex) is dual to 3-id.335
So seemingly
1-id.3335 ought to be dual to 5-id.3335,
2-id.3335 ought to be dual to 4-id.3335,
3-id.3335 ought to be self-dual.
Now coming back onto our incidence matrix problems. Main difficulty here is: not to know that acting symmetry of the diminished structure. As stated above this most probably would result in dealing with every vertex on its own. Dealing with a (hyperbolic) tiling, we have an infinite amount of (remaining) vertices.
One could start with the counts of the provided by the asked for vertex figure. But it is not clear (and, from our so-far troubles, seemingly not true), whether the full symmetry of that would be supported.
--- rk