Hi.gher. Space

by **Klitzing** » Mon Aug 19, 2013 2:30 pm

In euclidean and hyperbolic space the restriction to simplexial foundamental domains of reflection groups no longer holds. Such more general fundamental domains commonly are called Coxeter domains.

E.g. we have the general result, that any count of even numbered faces could be incident to a vertex, and a corresponding uniform hyperbolic tiling does exist. (This generalizes the omnitruncated figures of simplex domains.) The vertex configuration then would be [2p,2q,2r,..,2s,2t]. Esp. the vertex figure here is no longer restricted to a triangle, here it can be any n-gon!

Next we could un-truncate these tilings by choosing any edge and reducing that to zero. This results in the vertex configuration [p,2q,2r,..,2s,t,2s,..,2r,2q]. (This generalizes the rhombates and truncates of simplexial domains.)

Then we could choose a further edge - but it has to be neighbouring to the former - which also could be reduced to zero. This results in the vertex configuration [(q,2r,..,2s,t,2s,..,2r)^p]. (This generalizes the regulars and quasiregulars of the simplexial domains.)

And finally, as having started with all even numbered faces, we could consider the vertex alternations (i.e. snubs) here too. If we used n faces per vertex in the omnitruncated case, then the resulting vertex configuration here is [p,n,q,n,r,n,...,s,n,t,n].

As the Dynkin symbol clearly is based on the fundamental simplex (in fact it describes the dual graph), there are no Dynkin smbols for those more general figures. And further, as the kaleidoscopical construction of Wythoff effectively is based on the subdimensional decoding of that symbol, we well could consider all these more general figures as non-Wythoffian either.

This kind of seemed unfair to me. So I rethought all that in order to find a way out.

The key idea here was to consider such a more general domain still as a simplex, but no a degenerate one, i.e. having more than d+1 vertices in some subspaces of dimension d. Considering these "simplices", all the previous known Dynkin symbol and Wythoff construction would apply again. We just have to encounter an additional, new element here: non-intersecting domain facets.

Thus nodes of these generalized Dynkin symbols still represent domain facets, or dually the edges of those omnitruncates. Any 2 intersecting domain facets have to use a dihedral angle which is a submultiple of pi, say pi/m_ij. Accordingly the respective nodes have to be linked and given as link mark that number m_ij.

We clearly could consider to represent pairs of non-incident domain facets by nodes which are not linked. But this would run in troubles, as so far this very coding is in use for those links which should bear the number 2 (orthogonal facets). - Those were quite often used, and their corresponding withdrawel makes the symbols faster to be drawn and moreover easier to be recognized. Therefore we should not touch that. Esp. in order to not get into some authors using that, some the other intend, and finally no symbol can be read any longer.

Therefore I tried to introduce some special, additional "conection" character for generalized Dynkin symbols, which marks pairs of non-intersecting domain facets. For that special reason I use the empty set symbol, Ø.

Now it would be time to provide examples, in order to show its use in action. Let us consider the uniform tiling [10,8,4,6]. It then could be given the (linearized) generalized Dynkin symbol x3xØx4xØ*a5*c (*b2*d). An first un-truncation could then result in the uniform tiling [5,8,4,3, 4,8]. Its generalized symbol then would be o3xØx4xØ*a5*c. A second un-truncation could result e.g. in the uniform tiling [(4,4,3,4)^5] with generalized symbol o3xØo4xØ*a5*c. And the snub of the first one has vertex configuration [5,4,4,4,(2,)4,3,4] and symbol s3sØs4sØ*a5*c.

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.

--- rk

by **wendy** » Tue Aug 20, 2013 6:50 am

Most of the work here is done with 'decorated conway-thurston' symbols. In essence, it behaves as dynkin, but there are symmetry groups that do not occur in dynkin. Conway made some progress with the archiform notation, but this is not a process, but an outcome: many archifolds lead nowhere.

The actual CT symbol presents a number of problems i am still trying to overcome, specifically, it supposes all of the rotation-nodes appear before the mirror stripes, and that the wanders and miricales are insular, while they appear to be two separate entities each.

Dynkin-style graphs do not work, not much past a loop, which generally supposes the CT *p,q,r,s mirror-loop.

There are other kinds of edge to contend with, such as '%' which implements the 'over-grown snub' feature.

I've covered a fair bit of the groups like p * q, * p q r... and p q r ..., already. For example, the pyritohedral icosahedron is "3/ * / 2".

The root form of the pyritohedral group is "3/ * /2/" . This is a rhombo-cuboctahedron, where the cube-faces become rectangles b:c, and the triangles are a, and the squares that adjoin the squares and triangles are trapezia, of the type "bc&#at".

% is a dependent edge, that is, its size is derived from other conditions, including merging of faces.

3/ * /2/ rCO
3/ * /2% tC % is set so that it merges the attached faces into a single face
3/ * /2 I
3/ * %2 CO % appear as a diagonal on the square face.
3 * /2 C
3/ * 2 O

There are likewise seven figures on * 2 3 p, although i have not fully got down to sorting out the details.

If the number is followed by a % (eg * 2/ 3% 5/ ), this gives the equal of putting 'x' around the number with the %, ie x3x5o.

If the number is followed by nothing, there is only one kind of edge, and the other exists but is dependent, so 2 3/ 5% becomes o3x5o.

by **Klitzing** » Tue Aug 20, 2013 11:41 am

Suppose the Conway-Thurston notation was invented for mere groups, while the decorations are your invention. Suppose moreover that your comment "i have not fully got down to sorting out the details" also restricts to that decorational stuff only.

So, is there a freely accessible reference for those mere undecorated symbols, an internet link for example? - I still have not managed to wrap my mind around that notation. The namings of "miracles & wonders" do not contribute much clarity right by their choice. I'd like to see elaborated examples too, in order to see all these introduced notations in action.

Most probably this would be the first step, i.e. to get used to those undecorated symbols.

Thereafter we could struggle with your decorations, if still necessary.
In any case you should outline your inventions in a similar manner, i.e. definitions plus examples. The more the better.

--- rk

by **wendy** » Wed Aug 21, 2013 8:25 am

You could look at http://en.wikipedia.org/wiki/Orbifold_notation or its links. It is mostly by Tom Ruen, which probably means he got me to have a look over it. Heidi Bruegel is there on the talk page too. She is co-author of one of the books there. Google 'conway orbifold' without the quotes brings up plenty of references.

A 'miracle' is a mirrorless mirror, such as a rotary reflection. Basically, the thing becomes mirror-reflected, but there is no line of reflection.

A 'wander' is a rotationless rotation, such as a glide. Wander means here to go for a walk. It's no mirror image, but just a direct image placed by 'translation' or sliding.

John Conway also makes of a form of notation i called 'archifold' (he used orbifold for it, but archifold suggests the archimedean polyhedra it is designed to construct). It's a kind of edge-notation, where the edges arrive at a vertex-sector in numerical order (1,2,3,4...). The sectors can be reversed or run cyclically at a vertex, ie 1,2,3,4,1,2,3,4,1,2... or 1,2,3,4,4,3,2,1,1,2,... (or the pairs of '4' and '1' can be merged into single edge types.

The edges are of type [] mirror-like or () rotation-type, or <> mirror-margin. One can tell what sort of symmetries are involved from this table

Code: Select all: 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

The edge leaves a vertex through arm 'a' and arrives at arm 'b', and so has one or two symbols in the brackets. A single letter means it arrives and leaves in the same arm.

The Wythoff-mirror polytopes are of the form * [1]p [2]q [3]r, with the orbifold as *pqr, and the archifold at [1][2][3].

The snubs are of the form (1,2)p (3,4)q (5)2 or as antiprisms (1,2)p (3)2 (4)2.

You can see that the ordinary polygonal prism is (1,2)p * [3] and the antiprism is (1,2)(3)(4)

The frieze groups match up to the polygons, with p replaced by \infty. The effect of this is that any polygon could unwrap into a frieze.

So, the strip of squares becomes (1,2)u *[3] and the strip of triangles becomes (1,2)u (3)2 (4)2. If we were to join these freizes so that the horogon disappears we get the alternating band of triangles and squares, as (1)2 ,(2)2 (3,5) [4]*, that is 2 2 * (1) (2) (3,5) [4].

Unlike the orbifolds, the archifolds have a lot of duplication, in that many symbols lead to the same result. This is a recurring problem, even in my experiments, for which the cure is only partial until laws of symmetry are worked out.

Miracles and Wanders arise in what i describe as 'laminate polytopes' . In hyperbolic geometry, a lamina can be a yickle (see PG). Basically, any polygon can be opened up to make way for an infinitude, and then some other polyhedron similarly opened up glued into the holes. You can even have crossing opened-up polygons.

So, zB (1) (2,3) (4,5) is a classic snub 2 p q. You can open up a polygon, by adding symbols between its arrival and departure edge, eg

(1), (2,4), (5,6) (3) adds an edge from 'inside' the polygon (2,3). This means that we have converted the polygon into a frieze, and hooked it up to a similar rotated version. Putting (1), (2,4), (5,6), [3] makes it into a reflection, like a marek snub.

You can have as many snub faces as you want, so eg (1) (2,3), (4,5) (6,7) would be snub 2,p,q,r,s, and the cycle of polygons around the vertex is S2SpSqSrSs, and because S is repeated five times, S=5: 5,2,5,p,5,q,5,r,5,s. In my notation, this is 2/ p/ q/ r/ s/.

You can replace / with two active elements (nothing or %).

So, replacing / with "" causes the edge to disappear, and the pentagon becomes a quadralactral, and the polygon to which the / has been removed, becomes a wrapping around the vertex. So 2 p/ q/ r/ s/ gives a cycle around the vertex of 2(SpSqSrSs). S now becomes '4', but there are 16 polygons at a vertex.

Replacing a / with % causes the edge to be dependent, so that its length is such that it becomes an internal chord of a polygon. In short, a polygon swallows the snub faces, and becomes a p(S-1). So, eg 2% p/ q/ r/ s/ would create octagons at the '2', where the order of polygons around it run p,q,r,s,p,q,r,s. Likewise the vertex-around runs SpSqSrSs, but here S=2*4 = 8.

Note you can have only one %, or at most two zero-edges (on adjacent mirrors). You can't have a zero on a mirror and not on a mirror. These are 'active objects', that one needs to watch. You can have any number of 'passive objects'.

One problem with the Thurston-Conway orbifold is that it's designed to reflect groups, and not to lay out nodes for polytopes. For example, it supposes all of the rotation groups appear before the mirrors etc, while there are indications that these might appear in various order. For example the group p q *r * might actually need to be presented as @p * @q *r , where @ is a rotation-node, and * pqr is a string of mirror-angles. Likewise, the symbols x and o (the miracles and wanders), are actually bi-nodal, and these nodes can appear some distance apart: ie there can be a good deal of activity between the crossings supposed by x and o nodes.

Also, one can take a group like (1,3) (2,4), which is 'o', and present it in other ways. But this is a quartering of the space, and activity can occur in the various parts of the quarters. For example, '(1,4) (2) (3,5)'.

by **Klitzing** » Wed Aug 21, 2013 11:23 am

It looks to me that the orbifold notation rather is some mere listing of properties of the considered group, arranged in some predefined order of orbifold notational definition. And just that does not lend for needed rearrangement of characters, when you try to decorate those figures in some similar way as Coxeter-Dynkin diagrams. This very predefined order of property listing also makes it quite difficult to me to dechiffre those symbols: each character just describes a more or less uncorrelated single property. To me no larger structure springs into the eye.

Esp. the archifold then is even less reader friendly. The numbering of edges at the first vertex clearly is arbitrary. But then, being chosen, has to be remembered all over, in order to be compared with those of the next vertex. And, moreover, if those show up some partial symmetry, the interrelation even is not unique, how one numbering has to be mapped upon that of the next vertex! Additionally, even so the omission of the former orbifold characters within reduced archifolds might be redundand information, the recognition of those various notations is completely lost.

But from what I get out of the wiki description on orbifolds (nothing is said there about archifolds, neither by name, nor by content), the main problem for orbifolds to be "better than" coxeter-dynkin diagrams is that orbifolds are only defined for groups acting on 2D manifolds (with some additions allowing for 3D ones - using bold face instead of normal weight).

This seems to be a quite severe restriction, esp. as I know you usually dwell within 6D, or so.

--- rk

by **wendy** » Thu Aug 22, 2013 9:12 am

It's worth reviewing the scope of orbifolds.

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.

For two dimensions, though, the groups have 'aspect', which means that the same symbol (like 2 2 2 2 2), can describe a large number of disjoint shapes. 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.

Orbifolds differ from the coxeter-dynkin group, in that one can not set edges etc randomly to 0 or 1. It's true of CD as well, but because the only excluded case is when all edges are set to 0, gives a point, does not work in hyperbolic or euclidean tiling. But typically, in an orbifold, you might have m nodes over 2 dimensions, so you can only set 1 or 2 nodes to 0. Most have to stay at '1'.

The orbifold might be considered a symmetry-cell, which is variously rotated, reflected and slid, inverted-glided to its copies. These carry the edges onto the next copy of the vertex.

Since the vertex is a thing, it can only be tethered to one place in the cell. That means, that you can only set one or two (in a mirror-corner) edges to zero. You can do other things, like remove walls so that a cone swallows the medium area (over-grown snub, for example).

Conway's archifold and my active objects are complementry, and all nodes must be occupied by one or the other.

@r / This is new: it tells us that the following is a rotation group of order 'r', This allows us to move rotation groups along the symbol.
* / p / q / r / This is a chain of mirrors, with angles pi/p, etc. Unlike the dynkin symbols, unconnected mirrors don't cross, and '2' must be explicitly written
x A miracle - the distribution of this symbol and its nodes is unknown
o A wander - ditto.

So at the moment, instead of writing 2 2 * * (a group of rectangular shape, opposite sides are mirrors, and there's a digon rotation at each end), we write this in the order they might appear as you go around a fence: @2/ */ @2/ */

Currently several decorations are supported.

/ A passive node, representing a unit-length edge.
. An active node, representing a zero-length edge. Only one is allowed with an @, and one can be on each side of a number in one mirror-chain, eg *.2.
% A non-zero edge, designed to be removed that a polygon can swallow its neighbours, (overgrown snub), also a pre-set edge.

So, we could write the groups of the pyritohedrals as:

@3/ * /2/ rCO Note this drops two different edges to the same mirror.
@3/ * /2% tC % is set so that it merges the attached faces into a single face
@3/ * /2. I
@3/ * %2. CO % appear as a diagonal on the square face.
@3. * /2% C Note the cube faces are 1*1, not 1*0. The trapezium face is reduced to "/%"&#ot, so to speak.
@3/ * .2. O The vertex appears on each mirror around the '2' angle.

For the general CD group aPeQi one gets * 2 a P e Q i (since the mirrors fall between the angles in both cases)

For the general snub, one gets @P/ @Q/ @R/ .

If one sets the edge of @N/ to be the short-chord of a 2P, then the snub triangles are joined to N, and the outcome is to have a 2P, Q, R. When one of these is 2, you get a digon, but the edge remains significant. So @P% @Q/ @R/ is xPxQoRz (but rotational symmetry only.

When an edge is let go to zero, you get the P-gon disappear as a vertex-wrap, and you get alternating Q and R at a vertex, so the figure becomes @P. @Q/ @R%. Note that if there were more figures in the snub, then the last would not be dependent, You could have a quadralateral 0,1,1,1 for example.

As for the wanders and miricales, little has been done, because of the generally large size needed to produce the desired symmetry, but the early indications are that 'o' is probably two different edges, which needed to be split across the symbol, eg ( and ) for example. Likewise 'x' might need to be partitioned. In either case, they are listed as passive devices.

by **Klitzing** » Fri Aug 23, 2013 9:15 am

[...] 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.
[...]

Quite interesting indeed!

By "same group" you propably mean the same topological structure.

"[Same] size" might refer to hyperbolic areas of those various pentagons - I did not calculate those so far. (Most probably those would be related to a similar corner sum, as were the spherical ones - then those would be the same areas for sure, because we have 5 x 90° in each case.) - But clearly the various edge sizes will differ, cf. the respective calculations below!

Code: Select all: 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)]

Code: Select all: x5o4o all edges alike: 2*phi = 2*arcosh[cos(pi/5)/sin(pi/4)] = 1.061275 regular pentagons a-a-a-a-a-

Code: Select all: 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-

Code: Select all: 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-

--- rk

by **wendy** » Fri Aug 23, 2013 10:22 am

It's probably useful to look at how some of these CT diagrams build up. For this, we look at the pyritohedral group.

The simple group here is @3 *2 This is made up of a right-angle triangle, like an octahedron-face on a sphere, with a rotation-point inside of it. 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.

There is a further kind of edge one does not see in CD, which i represent by '%'. These are 'derived' edges, usually implemented for faces on either side to merge. The edge then becomes a chord of a larger polygon.

The only figure with the vertex on the searchlight is @3. */2%. The dot after the three represents a zero-length edge, an active object. Of course, we drop edges to each mirror, but only one is visible, and the second is derived. In effect /2% becomes a square. The outcome of this is a cube.

The rest all involve @3/, which means that they have triangles.

The polytope @3/ *.2. has triangles and a rectangular face of 0 * 0, that is, the vertex falls at the joins of the mirrors, as in CD. We get here an octahedron.

The polytope @3/ *%2. has triangles still, but now the rectangle has become a derived chord by 0. There is a "general" face, that lies between the triangle and the rectangle, which usually manifests itself in the style of a lace prism, "xa xb & # xc t" where the edges come from @3/c */a 2 /b. (a, b, c are subscript like). This face is then a o% &# xt, which is joined to a o2% rectangle, to another triangle, to make a square: % corresponds to 'q', the chord disappears, and the outcome is an cuboctahedron

The polytope @3/ * /2. has triangles, @3/, and the rectangles have become edges /2. = x2o. We see now that the trapezium now becomes a triangle, in the guise of xo &# xt . This has 20 triangles, gives an icosahedron. Note there is a reversed icosahedron on the same symmetry: @3 * .2/

The polyhedron @3/ */2/ has squares at the cube-sides, and triangles at the cube-corners. The cube-edges become xx&#xt, ie squares too. The outcome here is a rCO. Note here we have two edges dropping to the same mirror, which does not happen in CD. This is because the mirrors can see three vertices in the same cell.

A polyhedron @3/ */2% has a rectangle /2%, appended by two trapezia. The outcome here is x%%x &# xt, where one sees that %=k (r2+1), which join the six cube-faces and the twelve cube-edges into six octagons. The triangles are unaffected by this process.

by **wendy** » Fri Aug 23, 2013 10:40 am

All groups in even dimensions, can be calculated from the Euler excess or defect, which corresponds to the angle defect or excess in 2d. This is the basis of John Conway's orbifold cost.

So all of the groups '2 2 2 2 2' have an angle sum of 1;30 = 450 degrees, while the regular euclidean pentagon has angles of ;36, = 108 degrees five make 1;60 = 540 degrees. So the defect is 1;60 - 1;30 = 0;30, or 540 - 450 = 90 degrees.

The group subgroup to * 2 4 5, we find this angle as ;30 + ;15 + ;12 - ;60 gives 0;03 = 9 degrees defect. So the right pentagon is made of ten of these triangles, something that is pretty easy to see, since here, 2 2 2 2 2 is a cell of {5,4}.

The subgroup to *2 4 6, we find the defect as ;30 + ;15 + ;10 - 0;60, gives 0;05 = 15 degrees defect. Now there are 90/15 = 6 cells of {4,6} in a half-hexagon.

The subgroup to *2 4 8, we find the defect as ;30 + ;15 + ;07.60 - 0;60 gives 0;07.60, = 22 1/2 degrees defect, which is 90/4. One can see that this area divides into four triangles of {4,8}, and less obviously, into 12 sectors of {3,8} *2,3,8 gives ;30+;20+;07.60 -;60, is 0;02.60, which is 7.5 degrees or 90/12.

While it is very hard to find the size of reflective groups in 3d, in 4d, the Euler excess works again, and groups like {3,3,3,5}[1], {4,3,3,5}[8.5], {E,3,3,5}[17] and {5,3,3,5}[26] all reveal their symmetry sizes (in [], relative to (3,3,3,5), without any deeper investigation.

Coxeter used this exact approach to find that the 6D and 8D gosset polytopes 2_21 and 4_21 have 27 and 240 vertices respectively.

by **Klitzing** » Fri Aug 23, 2013 11:34 am

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. [...]

Sorry, what is your refered entity here at these highlighted words?
(Esp. what is the special property of x4x3o8o?)

At least, here is the IncMat of that hypercompact honeycomb (or bollochoron):

Code: Select all: 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

Its verf is the octagonal pyramid ox8oo&#k (where k=x(8)). Its cells are tics and x3o8o. (The latter should here better be seen as bollohedra rather than as hyperbolic tilings.)

--- rk

by **Klitzing** » Fri Aug 23, 2013 12:11 pm

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. [...]

Cannot follow that one! You'd get 3 cases as follows:

Code: Select all: \ \ o \ . : .\ : o. @3-------- /_ | / -_ | / * (seed) /

(single triangle, as described)

Code: Select all: \ o. \ : . \ : . o \: . @3-------- . /| o. / | /_ | / * (seed)

(two triangles - or bowtie?)

Code: Select all: \ .o .\ : o. \: .\ .\ . .o .@3----.----: ./ . o ./ o. /| /_| / * (seed)

(three triangles - or funny hexagon?)

--- rk

by **wendy** » Fri Aug 23, 2013 1:09 pm

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.

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. Here is a subgroup of {4,4}, consisting of stripes of non-intersecting mirrors (lamina), with digonal rotations. The different forms of the orbifold (TC and WK) are given, my version supposes you are standing in the middle of the field, and counting off what you see. So except for strips of mirrors (* a b c), you can move front to back cyclically.

When you "fence off" the various rotational corners, there is still room for one or more cells. In the snub group, these are the triangles that fall between the P,Q,R faces. In general, these are 'active' faces, not set by one or two edges in the symbol, but relying on many of the / marks to make an edge. These can be 'attracted' to a cone-point.

Code: Select all: --------------------------- 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

In the figure above, i have shown the 'searchlights' for the 2 digonal symbols at unrelated angles. The visible field is in the Upper-left, is carried by rotation and reflection into all of the other places. The decoration, an ordered NS and EW arrows, are variously reversed and flipped by the actions of the '2' and 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.

by **wendy** » Fri Aug 23, 2013 1:25 pm

With the searchlight idiom. The search-light is always mounted in the centre of the polygon it's creating. In euclidean geometry, the angle formed by the search light is the supplement (pi- angle) of the angle. There's no requirement that the edges drop onto the limits of the searchlight at right angles, but the total of visible edges is always 1e.

The chain of mirrors is in 3 * 2, a complete triangle, likewise centred on the '3' in the diagram, one sees only the length of one wall of that. But it's further out than the vertex p, so one can drop perpendiculars from p to these mirrors.

The difference between CT searchlights (cones in their talk), is that the limits are not restricted to specific points. You can see this by looking at the symmetry cells of the icosahedral, [3,5] vs octahedral [3,4], where five of the former and two of the latter make a 3*2, but they strike the octahedron edge at the ratio of 1:phi vs 1:1 respectively.

Code: Select all: / \ / \ 3 \ \ p \ / \/ \

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. Where this differs from the CD system, is that you can't generate hexagons by this process, because you would need to have two vertices in the spotlight at all times, which we don't allow.

Your images suppose that the limits of the searchlight are reflections, but the TC group has pure rotations as well. For example, in your bow-tie example, the edge with the colons is completely wrong - the only thing that exists between 11 and 3 on the clock there is a copy of the seed at 2 oclock, and between 7 and 11 oclock, there is a seed at 10 oclock.

by **Klitzing** » Fri Aug 23, 2013 3:29 pm

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.

Okay, by group you mean an algebraic object, and by symmetry the geometric realization.

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.

My fault. Meant to say combinatorically.

For example, the prismatic layers of the seven wythoff-constructs on o3o6o & oUo for the same base * 2 3 6 = [3,6] ,

No, you have not 7 but 14 wythoff constructs: 7 for o3o6o and 2 for oUo (U = infinity), i.e. 7 x 2 = 14.

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.

What? the length of edge symbol x is defined to be unity (for uniform figures). Therefore the height of those prismatic layers is always 1! In fact, you are dwelling here within 3D euclidean space, a mere prism-layer stack. (Nothing about calculating hyperbolic distances, that is: implied by curvature.)

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.

Rather unclear what you are on here.

Uniformity asks for unit lengths in either direction. It is just, as you have orthogonal components here, that you might use different scaling systems in those components, that is the up-unit != the horizontal unit.

Whether you use xUo or xUx in fact does not change the shape. It would only change the coloring.

Another big difference between TC and CD is that TC allows you to have separate chains of unconnected mirrors.

That exactly was the purpose of my initial mail within this thread: to allow for such unconnected (= non-intersecting) mirrors right within an appropriately extended Dynkin notation.

[...]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.

Not clear what you here on. - Suppose that the uniformity restriction would come in somehow?

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.

Are you speaking with respect to such decompositions here?

This is why it's no great loss that CD or TC don't extend into higher hyperbolic dimensions.

What? CD for sure extends into any dimension! Consider e.g. the E8 figure fy = 4_12 = x3o3o3o3o3o3o *e3o.

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,

Hmmm, does that work in general, or doe you have the additional request, that this surface manifold would have to intersect the sphere of infinity orthogonally, i.e. that this tiling (surface of bollohedron) shall have the same curvature as the embedding hyperbolic honeycomb? - Or would that just be again such a question of 4D embedding euclidean space, just as we dealt with at my partial Stott expansions?

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,

Ah, here you are stating it indeed!

and get a tiling of xPxQo. These are lamina-truncates.

Okay, understood.

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.

Hmmm, did not follow all this Catalan stuff. - But speaking of duals, you'd just replace any cell of x4x3o8o (or its "lamina-truncate", i.e. its mirror-reflected hyperbolic mere x4x3o filling) by a vertex, faces by edges, and vice versa edges by faces, and vertices by new cells? That one would seem much more direct for me. - Ah okay, there are those pyramids, which get truncated by those reflection surfaces, and by this very reflections become hyperbolic prisms. - Yep, understood as well.

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.

How does that relate? - So far I got that the curvature of that subsurface has to be the same as for the whole honeycomb. - And edges are all unity by definition of uniformity...

So we seek x2Po4o = xQoRo, which has one solution in 4, 3, 8.

What? That equation would have the general solution 2P=Q and R=4. Together with your assumption of oPoQo being spherical, for integral numbers this would require P=2 and Q=4. This in contradiction to your assumption that oQoRo has to be hyperbolic...

(In general, bollohedra have different edgelengths for the same curvature,

Again, uniformity requires unit edges throughout. It is the other way round, that this uniformity thing would ask for different curvatures in general. - But you seem to go the other way round: assuming the curvature to be equal, then yes, the accordingly scaled structures become differntly edged.

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.)

Impressive listing! How'd you derive their respective curvatures? Or did you just get those "for free" by occasional dissections / rejoins of pieces?

Since what we seek does not fall in this list, we can be sure that the truncated cubes, with a vf of oxo8ooo&#kt,

Your verf (dipyramid) tells, that you are refering here to the lamina-truncate version, not to the original x4x3o8o.

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.

Eeeh, which one are you refering here?

--- rk

by **Klitzing** » Fri Aug 23, 2013 3:42 pm

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.

I see, those cones (searchlight regions) get rotated, instead of being mirrored at their boundaries. Accordingly for @3 you'd get 1 seed point plus 2=3-1 images, or 3 vertices in total, in fact a triangle.

That is, I've now understood that "@P/" and that "@P." bit...
(Not really much, I fear.)

--- rk

by **wendy** » Sat Aug 24, 2013 10:27 am

The '1800' refers to Q, in the statement we tested every P,Q up to Q=1800, looking for equal lengths of edges. All values with the larger less than 12 are listed, and then the sole examples are those derived from the relation that xP/2xPo = xPo3o, which means that x3oPo and xPoP/2o (including the stars), are equal.

4,8 and 8,4 are the only hyperbolic examples of duals occuring in the list.

3,4 = 4,2
3,6 = 6,3 = 4,4
3,8 = 8,4
4,8 = 8,6
3,10 = 5,6 = 10,5
3,12 = 12,6
5,12 = 12,10
3,p = p, p/2

by **Klitzing** » Sun Aug 25, 2013 7:56 am

Fiddled a bit around with hyperbolic curvatures. Or curvatures in general. Found that "curvature" is more a unifying concept, but that its mathematical implementation is rather dependent on (mathematical) context. (Things like intrinsic and extrinsic ones e.g.) Moreover even the Gauß curvature is mainly defined for 2D manifolds only, there being the product of two orthogonal 1D curvatures, which in turn are the inverse of a local smoothing radius. Thus for uniformly curved spaces we there (2D) well have K = 1/r², which in hyperbolic spaces clearly becomes negative. But that concept does not lend for application to other dimensions alike. As then we should get something like K(d) = 1/r^d. And this would not allow for comparisions of curvatures of global one (e.g. honeycombs) to those of submanifolds (e.g. therein implemented tilings - as in your comparision of x4x3o8o vs. x3o8o).

This is why I took refuge to the good old circumradius itself. The formulae implemented in your spreadsheet (e.g. online accessible from my website at the download page) still apply for hyperbolics too, just that those would result in purely imaginary values then.

Based on unit edge lengths those 2 tesselations both evaluate to sqrt[ - sqrt(2) ]/2 = 0.594604 i, and thus proving indeed that those would have the same "curvature" (intuitive sense): I.e. x3o8o would cut the sphere of infinity of x4x3o8o indeed orthogonally. And therefore can be used as (an infinitude of) mirror planes for applications of external blends: I.e. omitting those bollohedral cells (tilings) and filling all space by mirrored copies of the compact cells (x4x3o) only (which you accordingly called a lamina-truncate of the former).

--- rk

by **wendy** » Sun Aug 25, 2013 9:05 am

Most of this curvatures are being dealt with in a different thread 'circle drawing'.

Curvature works in all dimensions. The model i use is the concept of the 'radius of the girthing polygon'. But keep an eye on the parallel thread viewtopic.php?f=3&t=1801 "The art of drawing circles". The approach is more an engineer's one than a mathematican's one, because i am not really good at trignometry or the calculus or other 'black arts'. It's more of what you see in Euclid, but with modern concepts.

There's even coming a 'hypercomplex plane' and the 'hyper-argand diagram', which leads to the back door into Coxeter Isomorphisms (which i have a weighty tome on, but still have no clue what that book is about).

The measures I use correspond to 'cho' and 'choh', which behave in a Euclidean fashion. The matricies I give work quite well in hyperbolic geometry, but because we see polytopes in terms of R/E (ie radius, given an edge), but hyperbolic tilings in terms of E/R (edge-lengths, given a radius), makes some interest.

Using chord-squares (the normal measure of my calculations), then the right-angle triangle becomes C = A+B-AB/2R (where Cap = low² throughout).

Curvature always remain 1/r^2 in all dimensions, because it's a kind of pythagerous triangle thing. The radius of a polychoron is done by the square-root of squares, not the fourth-root of biquadtates. Curvature is some kind of remainder in this thing.

One can then calculate the 'straight-line triangle, of the form E:E:C, by putting E=A=B, and then doubling the RHS, to get D = 4E - E^2/R, or 4 = D/E + E/R.

Since C/E is determined entirely locally (as the diameter over the radius of a circle), one then can see that E/R can be determined locally. For example,

In the tiling {12,4}, of edge E, a radius is sought by finding C. We need to use known chords of intersecting circles (where the chord is the intersect), to connect radius R to diameter D. So the circle that goes around the cell, tells us that if the edge is E, then the shortchord is 3.7320508 E. The second circle is drawn around the vertex, so that the dodecagon-shortchord now becomes an edge of a vertex-figure, the diameter D = 2 E' here, so we get D = 7.4641016 E.

Since we know that D/E = 7.4641016, and that this is 4-E/R, the edge comes out as 3.4641016, when R=-1. The unit here is not a 'radian', but a 'sextant'. We're measuring along the horocycle ruler, not against the straight one.

The spreadsheet works, because it calculates horocyclic lengths, which are directly convertable into straight lengths, if you know E/R. The main difference between the horocyclic lengths and the straight lengths, is that horocycles work on equidistants without change, so the edge rations given for E/R applies equally to the {12,4} that forms the cells of {12,4,3}, whereas the straight ruler does not. The edge of {12,4,3} is 7.156152, which is 2.077350 times the flat example.

by **Klitzing** » Mon Aug 26, 2013 1:59 pm

Started to calculate some of those purely imaginary circumradii of hyperbolic tesselations. (Those of the recently provided ones x4x3o8o and x3o8o had been the first ones.) Thereby found some nice coincidences!

R(o4o6x) = R(o4xUo) = 1/sqrt(-2)
R(o4x6o) = R(o3xUo) = sqrt(-1)

Moreover all 4 have tetravalent vertex figures with point symmetry. That is, we could cut those accordingly apart and past them wrongly together again, without getting out of the global curvature. - Sure the usage of triangles in the second case would prohibite from doing so uniformely (a run around that triangle would close in the wrong parity), but those of the first line truely allow for an uniform tiling: one with vertex configuration [4,6,6,U]. ("U" throughout this mail would represent "infinity", i.e. imply an horo- or apeirogon to be used.)

Btw., in view of my initial mail of this thread, we could even provide an extended Dynkin symbol for that one too, which here could be given in a two-loop form (as one - then omitted - link mark would be 2): x3xØxUxØ*a3*c.

We then further could consider several un-truncations thereof:
o3xØxUxØ*a3*c with vertex configuration [3,4,U,3,U,4],
x3oØxUxØ*a3*c with vertex configuration [3,6,U,U,6],
x3xØoUxØ*a3*c with vertex configuration [3,6,4,U,4,6],
x3xØxUoØ*a3*c with vertex configuration [6,6,6,6,U].

Or doubled un-truncations (which are not available at both sides of an Ø sign):
o3oØxUxØ*a3*c with vertex configuration [(3,U,U)^3],
o3xØoUxØ*a3*c with vertex configuration [(4,3,4,U)^3],
x3oØxUoØ*a3*c with vertex configuration [(6,3,6,U)^2],
x3xØoUoØ*a3*c with vertex configuration [(3,6,6)^U].

And for sure the snub of x3xØxUxØ*a3*c:
s3sØsUsØ*a3*c with vertex configuration [4,3,4,3,4,4,U].

--- rk

by **wendy** » Tue Aug 27, 2013 8:26 am

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.

by **Klitzing** » Tue Aug 27, 2013 11:32 am

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.

Well, this seems to be a choice of taste. You'd easily convert between those descriptions. But as hyperbolics - at least with respect to Dynkin symbol descriptions, just come out as an extension of sphercal ones, one clearly is used to use radii in dependance to edge lengths, and not the other way round...

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 so what? I explicitely stated that those usages are generalized ones.

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.

Indeed, CT != CD. And so rules for CT have nothing to do with rules for CD (up to the implications of dealt with groups). - Yes, usually all nodes of CD are regarded to be connected. And even then, where links are omitted (90°), those are considered to be given implicitely. But this exactly was the issue of generalization: to introduce a non-incidence symbol (Ø) exactly for this purpose. And thing of most interest here then is, that most of the technical stuff in dealing with those diagrams, just goes on to be true beyond. (Sure one should use some care and investigate each issue in detail, but most often the outcome is just as described!)

E.g. that funny restriction to un-truncations, that only neighboured edges could be reduced, just translates into: xØx is allowed for sure, likewise would be xØo. But not something like oØo.
Further, none of those describes a boundary polygon (at most it could be considered as a pseudo face).

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.

I do know that you differ between apeirogon (no interior) and horogon (with interior). But your term horogon is not as widely spread as the term apeirogon is. I.e. most people, concerned with geometrical mathematics, would understand the latter. And, being used within hyperbolical tilings, those would be then clearly be understood to have interior. (Thus the use of both was merely intended in a dictionarical sense. Not in equating your differenciation.)

--- rk

by **wendy** » Fri Aug 30, 2013 11:27 am

I have been playing around with the orbifold notation a bit.

It seems that there really is a need for a '\' symbol, to fill the role of Conways '<>' notation.

\ is already in use as a 'mirror-margin', that is, where the edge lies completely in the margin. This kind of edge is needed in a number of cases, so i have decided to adopt it throughout.

Most of the groups i have been working with are the euclidean groups, particularly of the families 2 2 2 2 and 2 2 * *.

In the first, @2/ @2/ @2/ @2/ is the general quadralateral (of any shape) as a tiler, based on rotations on the edge-midpoints, the uniform example is the square, but can, for example, be a trapezium. Replacing any of the @2/ with an @2. gives a tiling of triangles, six at a corner. In fact, the generalised triangle tiling is @2. @2/ @2/ @2/, where there is a digonal rotation on each of the three sides, and at the vertex.

The @2/ @2/ @2/ @2% is a tiling of generalised parallepeid hexagons, the opposite sides of which are parallel. The regular example is the {6,3}, the edges formed by the @2% are parallel to each other in every hexagon.

The final example is @2. @2/ @2/ @2%, formed by a series of quadralaterals, with the % forming parallel diagonals.

The groups formed on 2 2 * * are based on a parallel set of mirrors, with two digonal points between them. From the inside, the order around the viewer is @2 * @2 * . I was able to construct a herring-bone tiling (eg like chevrons), by lining up two sides of a parallel-pied on the mirrors, and placing one digon rotation at the centre and one at the edge-centre. But in order to make the symbol for this to work (as @2 *. @2% *\ ), the standard mirror-edge of wythoff's construction does not work, and one needs a "mirror-margin" (or edge in the plane of the mirror).

Then i started to think about the rules that might govern the placing of %.

% can be placed only once inside a kind of region governed by @ and *. Putting it twice would create infinite cells.

If % is placed at a cone-point @, then the vertex must fall on the end of a set of mirrors, and the \ edge must terminate the series, or all the rest of the symmetries must be @. So, it seems that "@3% *.2\ " is a valid representation of the octahedron, but putting .2/ is not.

If % falls on a mirror, then it must be at the end of a chain.

Having several mirrors may work, but i need to draw more pictures to get this to make me sense.

by **Klitzing** » Mon Sep 02, 2013 7:39 am

Interesting coincidence:

xØxPxØx = [4,4,4,2P] = x4o2Px
and also
xØoPoØx = [4^2P] = x4o2Po
as well as
sØsPsØs = [4,4,4,4,P] = s4o2Ps

but then
xØoPxØx = [4,4,4,4,P] (again), oØxPxØx = [4,4,2P,2P], oØxPoØx = [(4,4,P)^2]
while
o4o2Px = [2P^4], s4o2Px = [2P,4P,4P], x4o2Ps = [4,8,P,8]

--- rk

by **Klitzing** » Thu Oct 10, 2013 11:24 am

Hy Wendy (or others, which would like to pop in),

meanwhile I've worked out that Ø symbol extension of the Dynkin diagram logics a bit deeper, considered its implications, its to be applied "techniques", the difference to the already in use ∞ symbol. I'll provide below how far I did come.

But I got also further questions with respect to such "techniques". And I do know that you are quite genial at that point. So perhaps you might provide some ideas? - So please read the following outline carefully. Then you might provide an appropriate Extension of your "technical" advice to

... derive the vertex figure from the mere (full) generalized Dynkin symbol,
... to calculate the circumradius (okay, it would be purely imaginary) from the mere generalized Dynkin symbol (i.e. how to generalize that excel spread sheet calculation advice I once got from you?)

So here is how I do introduce that Ø sign in my current private copy of the future update of my Webpage on hyperbolics with Coxeter domains, dealing with 2D cases at that point:

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!

Here that word "theory" then will refer to a further outline of its usage, which will follow a bit down on that page, then concerning 3D cases, using for first examplifying toy the general symmetry

Code: Select all: 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.

This outline now cries for a continuation, providing a similar simple deduction of the vertex figure, of its purely imaginary circumradius, etc. - given nothing but the generalized Dynkin symbol!

Any help in sight?

--- rk

by **wendy** » Thu Oct 10, 2013 11:52 am

One of the enormous benefits to split 'wythoffian' from 'non-wythoffian', is that wythoff groups are simplex based, and that many interesting tricks work because it is a simplex base group. So you have for example, that you can always have uniforms from any configuration, and the dynkin matrix and stott matrix (what's inside the spreadsheet i sent richard), works.

These are conway-thurston symbols. In 4d, they work because you can wobble the group around (eg you can have a long skinny rectangle for (2222).

Still not fond of Ø though. It's better to acknowledge that you have a loop, and not worry about the criss-crosses over the field, because there ain't any.

I think if you hunt around in the polyhedron list, you will come up with the sort of thing that Don Hatch and Marek Ctranak and I did, in making a loop. In essence, regardless of what you end up with, it's going to be a field with a fence around it. Unlike the dynkin symbol, there are some fancy works around the fence, some of which i am still grappling with (CT's wanders and miracles), although we did breifly touch on some of the more exotica like mirrored snubs.

For example, one has eg " 2/ 3/ 3% 9/" You could feed your Ø rune at the front, eg Ø 2/ 3/ 3% 9/, but in the main, you're sitting in a field looking around the edge. There's only one of each at the vertex, ie 2, 3, 3, 9, but because you are not permitted to put cone-figures adjacent to each other, then you put a polygon p, whose sides are found like 2 p 3 p 3 p 9 p. Now the p represents a square or a tetragon in general.

The % is an active element, which means that it can only occur once in an environment. It's effect is to make an edge of a polygon into a chord, and that the polygon against the % "swallows" the surrounding 'p', so that you go from p to 3(p-1). The -1 is because one side of the p has been swallowed up by the 3, and the 3 means that its a triangle swallowing the stuff.

So the vertex figure goes to 2 P 3 P3P 9 P, where P and P3P are the same thing (a 9), and at P3P, we see the original 3 against the %, with two of its original edges as chords, and the P is now a trapezium that has become three sides and a chord of a nonagon.

So the vertex figure is 2 9 9 3 9 9 (and digons drop out) gives 9 9 3 9 9.

One has to tickle the magic of tyler to see what Melinda Green did, but i think she just used iterations.

The vertex-figure is pretty straight forward. John Conway worked a bit of it out with orbifolds, but in general it's a matter of figuring out what the active faces are. Unlike dynkin symbols, were we are nearly free to say a node is marked or unmarked, the CT diagrams, all nodes must be marked, and it is the exception that 'o' nodes occur. So a mirror nearly always correspond to x3x etc, and a cone to a polygon p.

The thing to look for is 'active objects'. These are things like vertices and snub-faces, which no not have a full internal symmetry.

As to the generalised edge (or radius), Tyler is the magic here.

by **wendy** » Thu Oct 10, 2013 12:13 pm

The other thing is that i don't think tossing around the dynkin/coxeter lie-group thing is a good idea. It's a different construction, a different countryside.

A group like '2 3 3 9' can have many different subgroups, but they're not by 'removing nodes', but 'applying the laws of symmetry'. It's the sort of thing which you get when you say, eg x3x3x4o = x3x4o3o etc (LS), not that x3x3x4o has x3x3x faces (node-removal).

A CT group can have say 11 things on its fence, but the only valid connections is where you poke the fence up. So you really don't need to show the 55 links, just the 11 around the edge: leave out the 44 occurances of Ø. We did this with the 'hatch loop', for example.

Conway did things like 'you can sort all the cones to the left, and have chains of mirrors and wanders and miracles following it, but i am not overly convinced that that's a really good thing. The abstract group might let that, but if you're standing at a point somewhere in the field, you can't just move things around like that.

That's why i have a cone-marker @. It kind of looks like a whirlpool. You need to break out of a chain of mirrors.

Figures mounted on cones can't be adjacent to each other, or anything else, really. A field of cones produces a new face inside the symmetry cell, rather like the snub triangles in sPsQs. In fact, they're examples of them. Miller's monster is a CT snub, but because it's hooked up to four cones, rather than three, the active faces are squares, not triangles.

Chains of mirrors work the same way as in the dynkin graphs, but generally, all sides are marked, and you can have two perpendiculars from the vertex to the same mirror (at the ends). So, for example, a series of mirrors like 3 * 2 (is actually a cone, rotating a mirror-right angle to three images, gives a triangle of mirrors). When you do something like 3/ */2/ (a rCO), you get 3 faces, the triangle is a cone, which means it can't be connected to any figure which has a full derivation. The */2/ is a rectangle, but each mirror has both kinds of edge at it.

The active face is formed between the 3/ and */2/ has three different kinds of edge, and a mirror running up the middle of it. It is the active face. It can also be 'swallowed' by the /2/ to give an octagon.

In any case, only some very simple cases have been bandied about.

It's known that if you replace a cone by a point, eg 2/ 3/ 5o, then the other figures repeat around the vertex n times (here 5), so this is an icosahedron, formed by repeating the 2,3, for 5 times.

It's known about %, but some of the early sketches on the pyritohedral group suggest that some more rigour is called for,

I have not played around with groups with 2 or 3 mirrors in a chain yet, or several chains of mirrors.

W

by **Klitzing** » Thu Oct 10, 2013 2:11 pm

I will have a look whether I'll do find such things that you adviced.

To me CT symbols still are kind of "miracles and wonders". And all what I've understood so far rather makes me dislike them. OTOH, Dynkin diagrams I'm heavily used to, and so any half-hooked generalization of that one serves much better to me than any unknown new theory. If both then will come out to be equivalent, much the better.

I wonder why you dislike my Ø symbol, tend to neglect non-connected (inner) stuff, and then complain that Coxeter domains would not allow for free decorations (o-nodes are restrictedly to be applied only). I had explained how that is related, that no mystics is in there! Just do(!) consider those Ø links, and then apply the corresponding (new) rule that those links can have at most one o-node at its ends. That's all.

Next to your favorisation of simplex domains based on the observation, that there the facets are derived simply by omitting one node. Yes, this is a true fact. But not a purpose, rather a consequence! In fact, what you realy have, you would have to use the sides of the domain, which define the individual subgroups. In case of simplices, this would result in omitting the opposite vertex. But that is not the to be generalized rule here. Rather use the domain facets instead in order to define those subelemental groupes, then you could freely use any other shaped domain too!.

You say the vertex figure is pretty straight forward. - I too managed to work that one out in any of my considered cases so far. But that always was by mere application of basic definitions, not by (generalized) Dynkin symbol derived techniques.

--- rk

by **wendy** » Fri Oct 11, 2013 7:51 am

Dear Richard.

You really should remember that the 'coxeter-dynkin' symbol has been going on for something like 150 years, and they still have not got it right. The main inspiration is due to neither of C or D, but to Stott, with Wythoff working it into the current mirror-notations. Stott seems to have got lost in the shuffle, though. None the same, even some modern authors have not grasped that it's really a coordinate system, and that the spreadsheet i made you really amounts to doing a dot-product in a system with oblique axies.

Still, the coxeter-dynkin symbol is totally inappropriate here, because we're not dealing with established constructions onto mirror-kaleidoscopes. We're actually dealing with something that people have bandied around 'possible' operators that might work on a group that contains four different kinds of symmetry operator (rather than just mirrors). Conway has done some interesting things here, but his process is rather loose: it allows constructions that lead nowhere.

Decorating the CT diagram may or may not be the way to go. In any case, some of the symmetry operators are not directly interactable, and might never appear in a dCT (decorated CT diagram). In essence, we have decorations in the box that no one has ever heard of, and does not sit on the CT nodes in the way that mirror-edges sit on the CD, but sit on new nodes between the CT diagram nodes.

Your claim that we need to look inside the polygon is in fact correct. Your particular reason for going in there (to fill it up with ø branches), is not. CT is already recieved as something that you arrange in some sort of circle, where the interactions are between adjacents on the circle. There is no point to drive this home. On the other hand, we were wrong to suppose that the active components are completely derived things, ie a group 2 2 * has only one possible arrangement of active components.

I spent a good deal of today, playing with sketches of 2 2 *, and its relationship with alternating bands of triangles and squares. In terms of the archiform, this resolves to "(1) (2) (3,5) [4]". However, by itself, 2 2 * gives at most 2/ 2/ */. This is edges (1) (2) [4], which produces a hexagon, the mirror bisecting opposite sides of it. There simply is no room for the edge [3,5]. It needs a new no-symmetry thing to create this edge, ie 2/ 2/ ø/ */øa.

It works like this. Unlike CD, the CT allows one to 'join' polygons, by replacing a p in each, by the complement. So the alternating band of triangles and squares comes to be a 'p=prism + p-antiprism', joined by connecting at the p's. However, all of the faces of both the prism and antiprism (except the p-gon) are active elements. So when you write out the passive elements of 2/ 2/ ø/ /* /øa, the sole list of passive elements gives "2 2 *" which gives nothing really. The passive elements consist of the hexagon of 2/ 2/ */, divided to split off two opposite triangles and a rectangle, which become the relevant faces of this polytope.

Code: Select all: o o p 2 / \2 / \ antiprism o-----o o o o-p-o | | | | | | prism * 4 * *===* *===* p/ */ 2/ 2/ */ 2/2 ø/ */ øa

The first two are instances of 'marek snubs'. The snub face is like an ordinary snub face, except one of the edges is opened up, and ran perpendicularly into mirrors. So the thing has twice the indicated edges. The first ought have 2 edges, it has four, the second gives a hexagon.

In the third case, we see the hexagon is split by an active edge: ie one that does not occur in the mirror-group, but consists of 'radial elements'. In any case, we see that there's a pretty ordinary 2/ 2/ p/ p-antiprism, and the bottom consists of a p/ */, which gives a tetragon.

The reason that we need to show these internal structures, is because they divide the central region into different active outcomes, which in turn, produces different polyhedra.

I then did some sketches of the 3 5 8 * orbifold. While you can shuffle the orbifold up, you can't do the same to the derived polytopes. 3/ 5/ 8/ */ is a different creature to 3/ 8/ 5/ */. Moreover, as long as an internal chord can be placed in several different ways, to give several different vertex-figires. For example

3/ 8/ 5/ */ gives a cycle (3 8 8 8 5 8 5 8 8 8 3 8 )

If we span it with a chord from 1 to 3, we get

ø/ 3/ 8/ øa 5/ */ gives (3 3 3 8 3 + 6 5 6 6 6 5 6)

A slightly different array gives ø/ 3/ 8/ 5/ øa */, runs ( 4 3 4 8 4 5 4 + 4 4)

Then you have to deal with constructs like [13] [24], an orbifold of crossing red and blue lines of {4,4}. You can 'add' an extra cone, along the lines of [13] [24] 3?, which gives the similar lines of {6,4}. These are what CT calls 'mirrorless mirrors', or miricales, which reverse a vertex like a mirror, but without help of one. A 'rotary reflection' is an example of one. Likewise, there's 'o', which corresponds to a generalised translation, a cone-less rotation.

But we starting to see that it's not just a matter of sticking things beside the symbols in the CT diagram like we can in the CD diagram. This makes it rather hard to simply pretend that CT is just a funny kind of CD. The active regions are not derived, but one can do things with it, which greatly makes things more complex.

But you can see that we can make 'active edges' cross, which means that while the miricales and wanders might end up in the CT, they don't necessarily need to end up in the dCT. On the other hand, i have not firmly convinced myself that there is just one kind of chordal edge. I think there are two.

Marek and JHC and a few others chimed in on the discussion on orbifolds, each adding their little etwases, but i don't think that any one of them had fully scoped the problem. Even I have not got my head completely around all of the intricies of the more complex chordal crossings.

So it's little use in making a definitive statement about how to decorate the CT when we really don't know what's going on in the CT. Even some of the simpler cases are causing me problems. But chordal segments to divide active regions helps, and that is in part, due to your misplaced use of ø.

by **Klitzing** » Fri Oct 11, 2013 11:07 am

...and that is in part, due to your misplaced use of ø.

Ahem! I was the one who introduced that Ø symbol. And this was done for that non-intersecting mirror stuff only. So it is definitely not misplaced in my usage! If you like to abuse that in a slightly different sense for your purposes, this is one thing. I don't weight that. But if you then tell me, I'd use that symbol in a (from your view) misplaced way, then this is surely wrong.

On the other hand, I will have to understand, what you are after now with your usage of that symbol. Might be I then come to understand, what you would like to call a "better" sense...

--- rk

by **wendy** » Fri Oct 11, 2013 11:55 am

I do quite agree with you in looking inside the orbifold for interesting things. I just don't agree with you over using a symbol to state that two unconnected symbols have no common interaction. Orbifolds are not CD diagrams. They're way more flexiable. Using something like ø might not actually reflect a zero-angle in any case, and it's inappropriate to connect such a thing to say, a cone.

I suppose my remark was a little harsh, though.

It seems that the interior of an orbifold belongs to a different kind of animal, such as might produce active edges, and with a little luck, the CT symbols of 'x' and 'o' (miracles and wonders, or reflections without mirrors, and movements without rotations). There is certianly a case for an active edge in one of the figures, which does not fall with or against any symmetry.

I've got a reasonable grasp on CT as such, and the conway archifolds, which do not produce a neat list of 'what works', but includes everything that works.

But conway was not really fussed over what Marek and I did, which is to create an operator-like system, and i have not fully tested if we have 'all the operators' necessary to start decorating the CT diagrams. This is why we are still plugging away at creating 'polyhedra that must be described in symmetry x'. Doing this is part and parcel of writing the 'laws of symmetry' necessary to reduce one figure to a different figure of a higher or lower order.

At the moment, the group 2 2 * is producing some interesting results. You just have to 'colour the edges' and pretend that 2 is really 32 or something.

A recent figure showed up in the previous discussion, but i suppose we already have % to handle this. Something like [13] [24] can described the rulled pages up and down in red, and across in blue, but there are similar things in {6,4} u, {8,4}, so it supposes that the real form is [13][24] 2% or 3% or 4%, where this wrap is entirely unexpected.

There is much to do here. It's nowhere near prime time like CD is.

Hi.gher. Space

Dynkin symbols for generalized Coxeter domains

Dynkin symbols for generalized Coxeter domains

Re: Dynkin symbols for generalized Coxeter domains

Re: Dynkin symbols for generalized Coxeter domains

Re: Dynkin symbols for generalized Coxeter domains

Re: Dynkin symbols for generalized Coxeter domains

Re: Dynkin symbols for generalized Coxeter domains

Re: Dynkin symbols for generalized Coxeter domains

Re: Dynkin symbols for generalized Coxeter domains

Re: Dynkin symbols for generalized Coxeter domains

Re: Dynkin symbols for generalized Coxeter domains

Re: Dynkin symbols for generalized Coxeter domains

Re: Dynkin symbols for generalized Coxeter domains

Re: Dynkin symbols for generalized Coxeter domains

Re: Dynkin symbols for generalized Coxeter domains

Re: Dynkin symbols for generalized Coxeter domains

Re: Dynkin symbols for generalized Coxeter domains

Re: Dynkin symbols for generalized Coxeter domains

Re: Dynkin symbols for generalized Coxeter domains

Re: Dynkin symbols for generalized Coxeter domains

Re: Dynkin symbols for generalized Coxeter domains

Re: Dynkin symbols for generalized Coxeter domains

Re: Dynkin symbols for generalized Coxeter domains

Re: Dynkin symbols for generalized Coxeter domains

Re: Dynkin symbols for generalized Coxeter domains

Re: Dynkin symbols for generalized Coxeter domains

Re: Dynkin symbols for generalized Coxeter domains

Re: Dynkin symbols for generalized Coxeter domains

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