a+1 b+1 a+1 b-1
[a-------b] (a------b) <a-----b> 1 2
a-1 b-1 a-1 b+1 | /
| /
|/
a=b mirror-edge digon mirror-margin o-----3
|\
a=b+1 miracle polygon ,, | \
| \
a<b+1 miracle wander ,, 5 4
[...] Apart from the cell of {5,4}, you can split a hexagon of {6,4} from mid-edge to opposite mid-edge, and quarter a cell of {8,4} in a similar mid-edge to opposite, twice. They all have the same group, and size, but do not fall on the same symmetry.
[...]
For hyperbolic linear Dynkin diagrams xPoQo we have for length formulas:
phi = hypdist(P0,P1) = arcosh[cos(pi/P)/sin(pi/Q)]
chi = hypdist(P0,P2) = arcosh[cot(pi/P)*cot(pi/Q)]
psi = hypdist(P1,P2) = arcosh[cos(pi/Q)/sin(pi/P)]
x5o4o
all edges alike: 2*phi = 2*arcosh[cos(pi/5)/sin(pi/4)] = 1.061275
regular pentagons a-a-a-a-a-
x6o4o (halving each hexagon gives a topological variant of x5o4o):
P0 o-+ P1
P1 / |
P0 o | P2
\ |
o-+ P1
edges a: hypdist(P0,P0) = 2*phi = 2*arcosh[cos(pi/6)/sin(pi/4)] = 1.316958
edges b: hypdist(P0,P1) = phi = arcosh[cos(pi/6)/sin(pi/4)] = 0.658479
edges c: hypdist(P1,P1) = 2*psi = 2*arcosh[cos(pi/4)/sin(pi/6)] = 1.762747
deformed pentagons a-a-b-c-b-
x8o4o (quartering each octagon gives a topological variant of x5o4o):
P0 o-+ P1
P1 / |
P0 o |
P1 +-----+ P2
edges a: hypdist(P0,P0) = 2*phi = 2*arcosh[cos(pi/8)/sin(pi/4)] = 1.528571
edges b: hypdist(P0,P1) = phi = arcosh[cos(pi/8)/sin(pi/4)] = 0.764285
edges c: hypdist(P1,P2) = psi = arcosh[cos(pi/4)/sin(pi/8)] = 1.224226
deformed pentagons c-c-b-a-b-
wendy wrote:[...] Any particular combination of faces, with assorted edge-reversals, etc, around a vertex, will make a uniform polyhedron, with some limitations. These limitations are in part handled by 'orbifolds', and 'archifolds', and in part by 'active elements'.
For higher dimensional groups, setting a set of angles between faces, is sufficient to define an exact polyhedron. Most of these reflective groups do not produce uniform polytopes, and the best way is at the moment is to randomly fish for them. The vast colony of 'lamina-apiculates' and 'lamina-truncates', derive from a p-gonal prism as symmetry group, but only one figure derived of this set is uniform: the tiling of truncated cubes, in x4x3o8o. [...]
x4x3o8o (N,M->oo)
. . . . | 6NM | 1 8 | 8 8 | 8 1
--------+-----+----------+----------+-------
x . . . | 2 | 3NM * | 8 0 | 8 0
. x . . | 2 | * 24NM | 1 2 | 2 1
--------+-----+----------+----------+-------
x4x . . | 8 | 4 4 | 6NM * | 2 0
. x3o . | 3 | 0 3 | * 16NM | 1 1
--------+-----+----------+----------+-------
x4x3o . | 24 | 12 24 | 6 8 | 2NM *
. x3o8o | 3M | 0 12M | 0 8M | * 2N
wendy wrote:[... @3] So if you imagine the rotation-cone as a 120-degree searchlight, it can only see one third of the triangle, usually including the right-angle between two mirrors (which is what the *2 symbolises.
When we drop a vertex in there, edges are fromed perpendicular to the mirrors, or in a similar way (to form a 60-degree angle), in the direction of the search-light. So if you're not standing at the @3, then you get a triangle, and only ever a triangle. This is different to the mirrors, where you can get a p or a 2p by dropping perpendiculars to one or two of the mirrors on either side of the angle. [...]
\
\ o
\ . :
.\ :
o. @3--------
/_ |
/ -_ |
/ * (seed)
/
\ o.
\ : .
\ : . o
\: .
@3--------
. /|
o. / |
/_ |
/ * (seed)
\ .o
.\ :
o. \:
.\
.\ . .o
.@3----.----:
./ . o
./
o. /|
/_|
/ * (seed)
--------------------------- Orbifold: 2 2 * *
\ : \
2 ^> 2 <v 2 Revised @2 * @2 *
\ : \
--------------------------- 2 = digonal rotation
/ : / ---- = mirror
2 v> 2 <^ 2 /,: = searchlight limits
/ : / ^> = decoration, with
-------------------------- ordered NS and EW
/
\
/ \
3 \
\ p
\ /
\/
\
wendy wrote:In both Coxeter-Dynkin and Thurston-Conway, all of the allowable constructions on a symmetry lead to a uniform figure. In CD, this is because the fundemental domain is a simplex, and the solution for all possible combinations of 0 and 1 (except all 0's), leads to a solution on the same symmetry.
"Group" is mathematics-talk for a thing you do with symbols, like ABAB = BABA etc. A "symmetry" has a "group", but several diverse symmetries can resolve to the same group, such as [3,3,3]+ amd [3,5]+ (the rotation groups of these figures), both resolve to the alternating group of order 60, A_5, but are evidently different symmetries.
In TC Orbifolds, the solutions work by distorting the symmetry. So a group like 2 2 2 2, a reflective rectangle, is a subgroup of {4,4} and {3,6}, but the ratios of the rectangles are different (1:1) vs (1:sqrt(3). The printer's term for the ratio of sides of a rectangle is 'aspect', i use this to denote that a symmetry description might lead to more than one thing on the ground, such as the 2 2 2 2 2 detailed earlier. I am not necessarily sure if 2 2 2 2 2 is topologically equal, even if they have the same group and same area.
No, you have not 7 but 14 wythoff constructs: 7 for o3o6o and 2 for oUo (U = infinity), i.e. 7 x 2 = 14.For example, the prismatic layers of the seven wythoff-constructs on o3o6o & oUo for the same base * 2 3 6 = [3,6] ,
have seven different heights, and you can't just assume here that because the same symbol denotes the symmetries, that they are exactly the same.
In fact, looking at o3o6o & oUo, there are five 'o', but only four are needed to set size. The fifth one sets an aspect, or height relative to base, which works here. If the fifth one did not resolve to aspect, then generally one could not create uniform figures here.
Another big difference between TC and CD is that TC allows you to have separate chains of unconnected mirrors.
[...]When one sets up a polyhedron by setting just its edge-angles, according to Conway, this leads to an singular shape. So, for example, an octagonal prism, whose corner angles are 2pi/3, and whose base of 2pi/4, will tile space (by Conway, if all angles divide 2pi, it tiles space). If all the angles divide pi, then you can use it as a reflective cell.
You can even go to the extent of dividing the octagonal prism into 32 symmetry cells (which are prisms based on o3o8o). But you can't assume that because it's a proper reflective group that uniform polytopes come from combinations of edges dropped to various walls (mirror-edge construction). The figures have no aspects, which means that there is only one figure that answers to a triangular prism, with those angles.
So having a method of generating reflective or other symmetry groups, is not going to generate masses of guarranteed uniforms in the way that CD and TC does.
This is why it's no great loss that CD or TC don't extend into higher hyperbolic dimensions.
The 'lamina-apiculates' and 'lamina-truncates' both use a common symmetry group of the style oPoQoRo, where PQ is a platonic solid, like 33, 34, 43, 35, 53, and QR is a bollohedron (not a horohedron like 44, 36). What happens is that you then replace the QR by a flat mirror,
so you get a generalised triangular prism or fustrum for the symmetry - ie a tetrahedron with a tip cut off it. The two figures in question are then xPxQoRo, which have cells xPxQo and xQoRo. You can adjust the edge ratio, so that xQoRo has the same edge as the 2d tiling, which means that you can replace xQoRo by a mirror,
and get a tiling of xPxQo. These are lamina-truncates.
The duals are lamina-apiculates. Here, the vertices of mPmQoRo would lie past the horizon, the cells have tips that fan out into an mQoRo = oQoRx. You can cut these tips off with a plane, and cause these tips (which are nominally pyramids), to become a fustrum, which by reflection on the little base, become a prism.
However, it is quite easy to demonstrate that the only instance of xPxQoRo that this works with both edges already equal, is x4x3o8o. It is a matter of looking at the vertical plane having the same edge as the horizontal one, that is, xPx, four at a corner = xQoRo.
So we seek x2Po4o = xQoRo, which has one solution in 4, 3, 8.
(In general, bollohedra have different edgelengths for the same curvature,
with these exceptions: x3oPo = xPoP/2o (including proper stars), and the following isolate equalities: x3o6o = x4o4o = x6o3o, x3o8o = x8o4o, x4o8o = x8o6o, x3o10o = x10o50 = x5o6o, and x5o12o = x12o10o.)
Since what we seek does not fall in this list, we can be sure that the truncated cubes, with a vf of oxo8ooo&#kt,
is the only example of this kind. This edge relation has been tested for P as high as 1800, using a much less fancy algorithm for deriving edges than you used.
wendy wrote:Regardless of where p falls, if it's not on the '3' itself, it will be rotated to make three images, and these form the triangle faces by "@3/" = x3o. If the vertex p actually falls on the '3', then evidently all the images are identical, and the triangle becomes "@3." = o3o.
wendy wrote:You are better off calculating the real edgelength as e=i/r. The reason for this, is that if you get ratios like 1:sqrt(2), then it might be possible to find the edge lengths of one in the other. The name of the measure is a 'sextant', since the sextant is the same chord as the radius of a sphere.
For example, E(x6o4o) = E(o4xUo) = sqrt(2) and o3xUo = 1.
I'm not really sure if (A ) we should be using dynkin symbols at all here, since these do not call down to lie groups,
and (B ) it is generally taken in orbifolds that if there is no explicit connection between symbols, they are unconnected. This is different to CD, where all mirrors are explicitly connected.
Of course, we're dealing with an entirely different field of things, so we need to be careful here. These are CT (conway-thurston symbols), not CD ones. They behave utterly differently, although all allowable CT markings lead to uniform tilings, there are restrictions on CT markings, which are different to the CD ones. For example, CT allows one to drop two perpendiculars to the same mirror from a vertex.
Regarding infinity: I rarely use the concept myself. A horogon is inscribed in something with Euclidean symmetry, and always refers to W4. An aperigon is a tiling, without an 'interior'. The greek means 'without a perimeter (fence)', and in my notion, is like laying solid cells across a space. aperigons can be bounded. A planotope is an apeirotope with an interior, such as the hexagonal tiles on the floor and the ground below it. A bollogon is Wx, where x>4. We can usually pick different bollogons apart, because the values of x can be measured directly, even if the edge-count can not. Generally, a bollogon has log(U) sides, where the horogon has U sides.
Gnerelized Dynkin diagrams
Usual Dynkin diagrams for such more general hyperbolics are clearly not defined, neither for the symmetry groups themselves, nor for the therefrom derived tesselations. Moreover, as the Wythoff kaleidoscopic construction essentially is based on those diagrams (resp. on the information decoded therein), this "more general case" thus also could be classically considered non-Wythoffian.
In view of the above cited theorem however, one might extend that diagrammal description by using a virtual simplex description, i.e. the number of nodes being again the number of facets of the (Coxeter) domain (which are orthogonal to the edges of the "omnitruncated" version), and the links are marked according to those m_i,j (i≠j) whenever such domain facets intersect. Additionally we introduce a non-intersection symbol: Ø.
E.g. the diagram of the provided picture then accordingly can be given as x3xØx4xØ*a5*c (*b2*d). (In fact, the nodes at positions b,a,c,d in this sequence cyclically provide the sides of the fundamental tetragon, or equivalently the orthogonal edges of the tiling incident to an arbitrary vertex. And the links provide the spanned faces: hexagon (m_b,a = 3), decagon (m_a,c = 5), octagon (m_c,d = 4), and (closing back) square (m_d,b = 2). I.e. node a represents the mirrors midway orthogonal to the hexagon/decagon edges, b correspondingly to the square/hexagon edges, c to the octagon/decagon edges, and d to the square/octagon edges. – But neither the domain sides b and c nor the sides a and d do intersect. Accordingly there are no tiles (polygons) to be spanned by those pairs of mirrors (m_b,c = m_d,a = Ø). At most they could be seen as pseudo bollogons, i.e. not really contained, faceting apeirogons with hyperbolic curvature.
As the number of facets of Coxeter domains exceeds those of simplices, the number of nodes in the symbol would be accordingly larger. Especially the usual relation, that the number of nodes equals 1 plus the dimension of the tesselation space (resp. of the embedded manifold), no longer holds in this generalization.
The symbol Ø in here was introduced ad hoc. Accordingly its usage clearly was designed by the current needs. Nonetheless there can be provided an according consistent complete theory!
o---Ø---o
\ /
p q
\ /
o
/ \
s r
/ \
o---Ø---o
To get an understanding what happens in case of 3D Coxeter domains, we consider first explicitely the general case of the to the right being shown symmetry, with has the linearised notation o-p-o-Ø-o-q-*a-r-o-Ø-o-s-*a. The domain here obviously is a (not necessarily straight) square pyramid: the base edges are the 4 not shown connections of pairs of nodes, representing links with mark numbers of 2; the diagonals of that base square then are those links with the non-incidence symbols Ø; the lacing edges finally are those links marked p,q,r,s.
First of all we have to ask: what are the subsymmetries? – For simplicial domains the answer was easy: just omit any single node together with its incident links; the remainder diagram would be the facet symmetry. Here we have to be a bit more careful! The facet symmetries clearly are that ones, which are provided by the facets of the domain. So, in our case of a pyramid, those are the base (still omitting a single node: the tip), and any of the lacing triangles, which each do omit more than just a single node! So we get here the subsymmetry o-Ø-o o-Ø-o (which happens to have euclidean curvature and therefore the ominous pseudo bollogons here become pseudo horogons, that is we could replace the Ø symbol here also by the ∞ symbol). Obviously the at the right displayed general symmetry therefore needs to be at least paracompact. The other subsymmetries here are given by o-p-o-r-o, o-q-o-s-o, o-q-o-r-o, and o-p-o-s-o (which, depending on the numbers p,q,r,s might make the total symmetry even a hypercompact one).
Next we have to ask for possible decorations of such generalized Dynkin diagrams. Sure, the omnitruncation always is possible, i.e. applying the x node symbol everywhere. But already in 2D we had seen, that not all applications of the o node symbols where allowed. – Considering first the base of the at the right displayed examplifying pyramidal domain, we do know the answer already: this is because of o-Ø-o o-Ø-o = o-∞-o o-∞-o (as described above). Thus we are back to A1×A1 symmetry, i.e. a reducible symmetry group (with disconnected graph). There we have to use at least one x node per component. – And the rules of the theorem of 2D effectively likewise ask, that at least one end of those links, marked by Ø signs, have to bear an x decoration. In other words: at most one end may carry an o decoration!
The facets of a polytope with Coxeter domain then can be read from the generalized Dynkin diagram in the very same way, as for non-generalized ones: they are the corresponding polytopes according to the provided decorated subgraphs, restricted as in the afore mentioned subsymmetries.
o o
p 2 / \2 / \ antiprism
o-----o o o o-p-o
| | | | | | prism
* 4 * *===* *===*
p/ */ 2/ 2/ */ 2/2 ø/ */ øa
...and that is in part, due to your misplaced use of ø.
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