Alright, I'm having some trouble proving some things about lace prisms with n-simplex symmetry, which is why I haven't posted anything yet. But since I haven't been able to make any headway in that direction, I thought I might as well post my results for lace prisms with n-cube symmetry first.
Fundamental lace prismsTo make the problem of classifying lace prisms simpler, I make use of Klitzing's observation that if a lace prism A||B exists and is CRF, then if both A and B have an unringed node in their CD diagram in position i, then adding a ring to the respective nodes will produce a CRF with the same height. IOW, this is just the Stott expansion of the lacing cells of A||B. Therefore, given any lace prism A||B, we may
remove rings that occur in the same position in both A and B, until A's and B's CD diagram have no more ringed nodes in common. A lace prism of this form is called a
fundamental lace prism.
Given a fundamental lace prism, we can easily derive the corresponding non-fundamentals by adding rings to the same nodes in both A and B.
A fundamental lace prism also has the property that if the i'th node of A is ringed, then the i'th node of B must be unringed, and vice versa.
Note that it is legal for A or B to be a point (all nodes in CD diagram are unringed); this produces a B-pyramid (resp. A-pyramid). Also note that ooo...||ooo..., which is the line segment, is the fundamental lace prism for all uniform polytope prisms.
Fundamental lace prisms with n-cube symmetryLet A||B be a fundamental lace prism where A and B are both uniform polytopes with n-cube symmetry. Then we may derive coordinates for A||B by using my method of reading off the coordinates from the CD diagram:
- Start with the node at the end of the diagram connected to the edge labelled 4. If this node is ringed, then construct an n-vector whose first coordinate is 1; otherwise, the first coordinate of the vector is 0.
- For each subsequent node: if the node is ringed, then the corresponding coordinate in the vector is the previous coordinate plus sqrt(2), otherwise it is a copy of the previous coordinate.
- Take all permutations of coordinate and changes of sign of the vector: this generates the vertices of the corresponding polytope.
Example: x4o3x --> apacs(1,1,1+sqrt(2)). This is a rhombicuboctahedron of edge length 2.
Using this method of deriving coordinates for A and B, we may thus write the coordinates of A||B as:
(x1,x2,x3,...,0)
(y1,y2,y3,...,H)
where (x1,x2,...) and (y1,y2,...) are the coordinates generated by the above method (together with all permutations of coordinate and changes of sign thereof), and H is the height of the lace prism.
Since A and B are vertex-transitive, we may take the vectors generated for A and B (without the permutations/sign changes) as representative points on A and B, respectively. Then there must be an edge between (x1,x2,...) and (y1,y2,...). Which means that if A||B is CRF, then:
||(x1,x2,...,0) - (y1,y2,...,H)|| = 2
since the edge lengths of A and B are 2. Squaring both sides of this equation and solving for H, we get:
H^2 = 4 - ((x1-y1)^2 + (x2-y2)^2 + ...)
Since H>0 (otherwise A||B is either subdimensional or not CRF), this implies that:
((x1-y1)^2 + (x2-y2)^2 + ...) < 4
This inequality gives us a simple way of determining if a given lace prism with n-cube symmetry is CRF or not. I'll call this
Inequality_1, as I'll be referring to it repeatedly below.
Now observe that if the first node of either A or B (without loss of generality, say it's A) is ringed, then the (xi-yi) terms will all be of the form 1+k*sqrt(2) for some integer k. (Proof: the first nodes of A and B cannot both be ringed simultaneously, since otherwise A||B is not fundamental; so the first node of B is unringed. Then by the above method of coordinate derivation, all B's coordinates are multiples of sqrt(2), but A's coordinates are of the form 1+k*sqrt(2). So their difference will also be of the form 1+k*sqrt(2).) Let's call this the
First Category of lace prisms with n-cube symmetry.
OTOH, if both A and B have their first nodes unringed, then the (xi-yi) terms will all be multiples of sqrt(2). Let's call this the
Second Category of lace prisms with n-cube symmetry.
First Category: x4...||o4...Let's consider the first category first. Since the (xi-yi) terms are of the form 1+k*sqrt(2), then Inequality_1 can be written:
1 + (1+k1*sqrt(2))^2 + (1+k2*sqrt(2))^2 + ... < 4
for some integers k1, k2, .... (The first term in the sum is 1 because the first node of A is ringed, and the second node of B is unringed, so the difference in coordinate is precisely 1.)
Now, since sqrt(2) is irrational, 1+k*sqrt(2) is never zero, and so (1+k*sqrt(2))^2 is always positive. Furthermore, the smallest number of this form is (1-sqrt(2))^2. Therefore, by the
Archimedean property of the real numbers, the number of terms in the sum cannot exceed some constant number K. Since the number of terms corresponds with the dimension of A||B, this means that there is an upper limit K such that if the dimension n of A||B is greater than K, then A||B cannot be CRF.
What's the value of K? It's basically the maximum number of terms in the sum. Since the sum must be less than 4, maximizing the number of terms is equivalent to minimizing each term. As mentioned above, the smallest number of the form (1+k*sqrt(2))^2 is (1-sqrt(2))^2. So in the ideal case, we'd like to choose A and B such that every term (except the first) is (1-sqrt(2))^2. This ideal case is in fact attained by the lace prism x4o3o3o...||o4x3o3o.... As you can check for yourself, the coordinates for A and B are respectively (1,1,1,...) and (0,sqrt(2),sqrt(2),...), so the squared magnitude of their difference is exactly 1+(1-sqrt(2))^2+(1-sqrt(2))^2+... .
How many terms can we have in this sum before it exceeds 4? Let A||B be n-dimensional. Then A and B are (n-1)-dimensional, and so will have (n-1) coordinates. Therefore:
1+(n-2)*(1-sqrt(2))^2 < 4
(n-2)*(1-2*sqrt(2)+2) < 3
(n-2)*(3-2*sqrt(2)) < 3
Since (3-2*sqrt(2)) > 0, we can divide both sides without changing the sense of the inequality:
n-2 < 3 / (3-2*sqrt(2))
n-2 < (3 * (3+2*sqrt(2))) / ((3-2*sqrt(2)) * (3+2*sqrt(2))
n-2 < (9 + 6*sqrt(2)) / (9-8)
n-2 < 9 + 6*sqrt(2)
n < 11 + 6*sqrt(2)
11+6*sqrt(2) is approximately 19.485. Since n must be an integer, this means n≤19. Therefore K=19. IOW,
lace prisms with n-cube symmetry in the First Category are only CRF up to 19D; from 20D onwards, they are no longer CRF.Second Category: o4...||o4...Now we consider the 2nd category of lace prisms with n-cube symmetry. Since the first nodes of both A and B are unringed, the coordinates of A and B are all multiples of sqrt(2). Therefore, we may write Inequality_1 as:
(k1*sqrt(2))^2 + (k2*sqrt(2))^2 + ... < 4
for some integers k1, k2, .... But this is simply equivalent to:
2*k1^2 + 2*k2^2 + ... < 4
k1^2 + k2^2 + ... < 2
But since k1, k2, ... are all integers, this means that there cannot be more than 1 non-zero term in the left-hand side sum, since otherwise it will be ≥2. This, in turn, means that A and B can differ in not more than 1 coordinate.
Case 1: both A and B have no ringed nodes, so A||B is the line-segment. This is the fundamental lace prism that gives rise to all the uniform polytope prisms.
Case 2: A has a single ringed node, and B has no ringed nodes. In this case, A's ringed node can only occur in the last node; i.e., A's coordinates are (0,0,...,0,sqrt(2)). If the ringed node occurs anywhere else, the difference in coordinates between A and B will be 2*sqrt(2), and the squared sum of coordinates will be 8, which violates Inequality_1. So A||B in this case is o4o...o3x||o4o...o3o, that is, it is just the (n-1)-cross pyramid (or bisected (n-1)-cross).
Case 3: Both A and B have a single ringed node. In this case, if A's ringed node occurs at position i, then without loss of generality B's ringed node must occur at position (i+1). In this case, A's coordinates will be (0,0,...,0,sqrt(2),sqrt(2),...,sqrt(2)), and B's coordinates will be (0,0,...,0,0,sqrt(2),...,sqrt(2)), so the difference is exactly (0,0,...,0,sqrt(2),0,...0). Any other relative position of ringed nodes will cause there to be more than 1 difference in coordinate, which will violate Inequality_1.
There are no other CRF cases. Why? Because if, for example, A has two or more ringed nodes, then since A||B is fundamental, B cannot also have ringed nodes in the same positions, and so this introduces at least two differences in coordinate, again violating Inequality_1.
For case 1, the height of A||B is 2 (since A=B); for cases 2 and 3, the height of A||B is sqrt(2), since H^2 = 4 - (sqrt(2))^2 = 2, so H=sqrt(2). Furthermore, all 3 cases are independent of dimension, so they exist in all dimensions.
SummaryIn summary, CRF fundamental lace prisms A||B with n-cube symmetry are of one of the following forms:
1) x4...||o4...: these fundamental lace prisms only go up to 19D.
2) o4o
*x||o4o
+ (using Keiji's notation based on regular expressions[**]): these fundamental lace prisms exist in all dimensions, and are just the bisected (n-1)-cross. They have height sqrt(2).
3) o4o
*||o4o
*: this is just the line segment (of length 2); it exists in all dimensions, and give rise to the uniform polytope prisms (with n-cube symmetry).
4) o4o
ixo
+||o4o
ioxo
* (for some integer i): these fundamental lace prisms exist in all dimensions, and have height sqrt(2).
[**] Superscript * means "zero or more instances of the preceding node", and superscript + means "one or more instances of the preceding node". Superscript i means "exactly i instances of the preceding node".All other lace prisms with n-cube symmetry are derived from these fundamentals by adding rings to nodes in the same positions in A and B (equivalently, by Stott expansions of these fundamentals).
Further WorkI didn't look into the details of the fundamentals of the form x4...||o4.... There may be other interesting trends within that category, even if they are only CRF up to 19D.
, but that was based on analogy only. but I do agree that a new word is needed to remove confusion 
But in any case, my silence is due to having discovered something very interesting about lace prisms, but not having the full results yet. But maybe a teaser is due.
