Klitzing wrote:[...] Displaying hex within 2D as vertex based tower of sections we have pt||hex||pt. Thus the only possible luna here is using 2 adjacent vertices, which results in squippypy (K-4.4).
Yes, because past adjacent vertices you get antipodes, which results in subdimensional cutting.
Displaying ico as such we'd get pt||cube||co||cube||pt. For 2 adjacent vertices we'd then get as luna the oct||tricu (K-4.30). For to orthogonal vertices we'd get as luna the {3}||gyro tricu (K-4.27 - which, btw., is nothing but {6}||oct !). Wonder whether 2 meta vertices would give something too or whether we here get non-CRF cell
intersections.
Be careful of using the vertex-distance approach, because for ico, bisecting with a hyperplane orthogonal to a vertex produces a non-CRF cutting with a rhombic dodecahedron cell!
To produce CRF luna, you need to cut with a hyperplane
parallel to an octahedral cell.
For my derivation, I made a model of bisected 24-cell, cutting with a hyperplane parallel to an oct, which is cuboctahedron||octahedron. Now since cuboctahedron = triangular gyrobicupola, it means it should be possible to bisect cuboct||oct with another hyperplane such that it bisects the cuboctahedron (since any other cutting will cut the cuboctahedron in non-CRF ways). This means the second hyperplane must make a hexagonal cross-section with cuboctahedron, and this hexagon defines a 2D plane. In 4D, fixing a 2D plane constrains the orientation of the hyperplanes that passes through it to a single degree of freedom. So the second hyperplane can be uniquely defined by the angle made with the first hyperplane.
Now, since the second cutting hyperplane, taken alone, must also make a (different) cuboctahedron cross-section with ico, the only way for the result of cutting with both hyperplanes to be CRF is that the result must have two 3-cupolae, joined at their hexagonal face. The angle between these 3-cupolae will be equal to the angle between the cutting hyperplane. So it is sufficient to start from the bisected 24-cell, pick a hexagonal cross-section of the bottom 3-cupola cell, and then inspect all possibilities of inserting a 2nd cupola that deletes some vertices and connects the rest, while keeping CRF cuttings of the lateral cells.
For me, this approach is easy to work with, because I use a convex hull algorithm with an input set of vertices; so starting from the bisected 24-cell, I try to find the next luna by finding the smallest set of vertices to delete to make another 3-cupola cell joined to the bottom cuboctahedron by the hexagonal cross-section. The simplest way is to delete a triangular face (i.e. remove 3 vertices) from the the cuboctahedron. This is the minimum, since otherwise you can't make two 3-cupola cells in the result joined by their hexagonal face. Then I run the convex hull algorithm, and use my program to verify that all edges are unit length, and use projections to determine the overall shape of the result.
It turns out that the result is oct||3cup. So this is the first luna of ico. Now, using the above argument that the second hyperplane can only vary by angle, and the result must have two 3cups joined at the 6gon, it follows that the next luna must be obtained by deleting the 3 vertices of the triangular face of one of the 3cups, opposite the hexagonal face. (Any other cutting in between will produce non-CRFs.) Looking at the projections, I can see that deleting this face will result in trigon||3cup (or, equivalently, 6gon||oct, as you mentioned). So this is the second luna.
Are any others possible? Well, note that in 6gon||oct, the distance between the top triangular faces of both 3cups (i.e. the 3gon opposite the 6gon) is 1 edge length. Again using the observation above that the second hyperplane can only vary in angle and the result must have two 3cups joined at the 6gon, this means that no more CRF lunae are possible, because the next cutting would require deleting the triangular face of one of the 3cups, which results in a subdimensional polytope with two coincident 3cup cells!
So the only possible lunae for ico are oct||3cup and oct||6gon (==3gon||3cup).
Displaying ex as such we'd get pt||ike||doe||f-ike||id||f-ike||doe||ike||pt. You already described the first 4 lunae (wedges) up to 2 orthogonal vertices, I suppose.
Actually, the 4 lunae come in increments of 36°, so there are no lunae produced by orthogonal vertices. The 3rd and 4th lunae are actually already past the orthogonal point. The 4th luna I listed is the narrowest one, with dichoral angle between the pentagonal rotundae at 36°, with a single 5-antiprism connecting their top faces.
Again, the situation here is that when you bisect the 600-cell (and replace bisected edges with pentagonal pyramids, since otherwise it's non-CRF), you get a o5x3o, which can be cut into two pentagonal rotundae. When you introduce a second cutting hyperplane, it must bisect the o5x3o with a 10-gon cross section; any other cuttings will cut the o5x3o into non-CRF fragments. So the second hyperplane is constrained by the plane of the 10-gon, which means not all pairs of vertices can be used for determining CRF lunae. It also means that there is only 1 degree of freedom between the two hyperplanes, so they can be defined solely by angle.
Furthermore, the second hyperplane bisecting the 600-cell must make a cross-section that can be CRF-bisected into something with a 10-gon face. But the only 600-cell cross-sections that are both CRF and allows bisection at a 10-gon is o5x3o, which means that the second hyperplane must also bisect the 600-cell at a o5x3o. Therefore, any lunae must consist of two pentagonal rotundae joined at their 10-gon faces.
The 4 lunae that I enumerated were derived in a similar way to the ico lunae; I start with the bisected 600-cell, then incrementally looked for the smallest set of vertices to delete such that it produces a CRF with two pentagonal rotunda cells. It turns out that given a pair of 600-cell vertices that lie on the same 10-gon great circle, you can generate a CRF luna, except for antipodes which produce a subdimensional result with two coincident o5x3o cells, and choosing the same vertex twice, which gives you the bisected 600-cell.
Because of the symmetry of the 10-gon, half of the lunae are identical to the other half, so that leaves only 4 unique possibilities besides the bisected 600-cell itself. Unfortunately, the two cutting hyperplanes are constrained such that the vertices must be selected from a 10-gon great circle (otherwise they produce non-CRFs), and this does not include orthogonal vertices.
