## Development of the Coxeter-Dynkin diagram.

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### Development of the Coxeter-Dynkin diagram.

As you might know, I'm working on a school project about 4-dimensional CRF's.
Currently I'm stuck at the part about convex uniform 3-dimensional polytopes.
I would like to give a short overview of the development of the Coxeter-Dynkin diagram, starting with Archimedes who first described them, going to the Stott-operators that derive them out of the platonics (who invented those, was it Stott, Kepler, Conway, someone else?), and then going to the kaleidoscope construction (should I credit wythoff for it or someone else?) and finaly to the CD-diagram.

Now my question is: who was the first to describe how all the uniforms could be derived out of the platonics?
student91
Last edited by student91 on Wed Feb 12, 2014 10:37 am, edited 1 time in total.
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### Re: Development of the Coxeter-Dynkin diagram.

The Platonics etc were first described by J. Kepler. He divided the figures into Platonic (regular) and Archemedian (vertex-transitive)

Cayley demonstrates that the largest of the uniform figures represents the symmetry of the group. That is, the vertices of the tCO represents the symmetry of the cubic group. Cayley is more famous for vector analysis.

Schläfli is the first of many to observe the symbol that now bears his name. It is independently discovered over the next forty years by a number of different people, including Coxeter.

Thorald Gosset describes the semiregular figures: that is, those figures whose faces are regular. This includes the o3x3o3o, o3x3o5o, s3s4o3o, and a family that descends by using the first of these as a vertex-figure for the next.

Alicia Boole Stott (daughter of the computer engineer Charles Boole of 'boolean arithmetic' fame), devised the process of expansion. Here one expands a cube, by supposing that its faces are drawn outwards while retaining the same size. New faces form along the edges (generally rectangles, but can be squares), and at the vertices (triangles). Stott's paper describes the 15 figures of symmetry {3,3,5}, in the form e_1 e_2 C_600 for what we write o3x3x5o

Seven of the eight figures (except snub cube), can be derived from a point cube, by imagining that the vertices, edges, or faces are drawn outwards while keeping the same size. The icosahedron gives also seven+snub dodecahedron, and the remaining ones come from the tetrahedron.

[at this point, you could demonstrate canonical coordinates, eg (vertex -> cube = 1,1,1, edge gives CO, (r2, r2, 0)and octahedron = r2,0,0. So rco = vf = cube + octahedron = 1+r2, 1, 1, truncated cube = cube + CO = ve r2+1, r2+1, 1. and trunc octahedron = oct + CO = 2r2, r2, 0 and the tCO gives v+e+f = 2r2+1, r2+1, 1. ]

If you take a tCO, you can label the edges like this: a 'v' edge goes between the octahedron and square, an 'e' edge goes between the octahedron and the hexagon, and the 'f' edge goes between the hexagon and the square. By shrinking any combination of v, e, and f edges to 0, you reduce the tCO to the other six mirror-octahedrals.

1904: Wythoff showed that Stott's construction can be done in a kaleidoscope. One sets three mirrors to match the symmetry of a cube (vertex, edge-centre, face-centre). By dropping a point anywhere in this area, one can drop perpendiculars to the three mirrors, and make the seven different mirror-edge figures.

1933: Coxeter found a description of a "Lie group" that corresponds to mirrors, whereby a dot represents a single transform, eg A or B or C. These represent a reflection, so a reflection of an image into itself gives AA = 1. (back to the start). So the three mirrors give AA = BB = CC = 1 simply means that the image of an image is the thing itself.

The branches represent a relation like o------o means ABA = BAB, and when marked 4 or 5, gives ABAB = BABA, and ABABA = BABAB. If you look in the sorts of tube kaleidoscopes, the mirrors are a equalateral triangle, and you can count six copies of the centre around each corner. The one directly opposite the real thing can be reached by ABA or BAB.

In terms of the cayley diagram, if A=====B------C (where === is a '4' branch), then A is the octagon-square edge, B is the octagon-hexagon edge, and C is the hexagon-square edge. You can see then that something like ABAB = BABA because these walk around an octagon face in two different ways to the opposite point. AA brings you back to where you started from, etc.

Dynkin and deWitt independently discovered the notation during the 1940's, and while their construction is pretty much identical, the later researches use double-lines for '4' branches, and triple lines for '6'. The dynkin/deWitt form also allows for arrows on the branches. It is from these that we inherit the symbols like A_n, B_n, C_n, etc.

1935. Coxeter reads Wythoff's paper (which has no real notation for the kaleidoscope), and realises that you could indicate a mirror has a perpendicular edge, by marking the node: eg @-----o triangle, o-----@ inverted triangle. Coxeter wrote a 'wow' paper on the subject, with the idea that here we now have a map of all things and everyone's previous efforts seem somewhat feeble against it. (Galileo wrote a 'wow' paper when he pointed his spyglass at the sky. For his efforts the Pope had him on the stake.)

1938? Stott suggests to Coxeter, that one could represent the snubs, by showing the ring, but removing the mirror. This gives rise to the hollow node. ( )----( )====( ) snub cube.

1958: Coxeter, Miller and ?? wrote a book describing all the uniform 3d polyhedra, including the face-crossing ones, using a form of decorated schwarz triangle which they called a 'Wythoff Symbol'. The four-dimensional project is largely in the hands of Jonathan Bowers. Wythoff has nothing to do with either the symbol or the project, and this name simply throws a lot of people off-scent.

1967: John Horton Conway finds by extensive computer search, the remaining uniform polychora. There is just one: the grand antiprism.

Much of the more recent progress in this field has been done on the "Polyhedron" mailing list, and more lately at the 'higher dimensions' BBS (here), the work to be found there is considerably in lead of the field of published stuff. The current CRF project describes the 'johnson' figures in 4D.

Wendy Krieger devised the 'pseudo-regular trace', which allows nodes to be placed in clear sequence, the connecting branches are read as supports to the nodes. This is a major idiom shift, because previously it is the branches which describe the notation. Her efforts run to allowing multiple layers (lace prisms etc), and sizes (x, f, q, ...) to describe 'varieties' or non-equal edges. It should be noted that this notation is as much a 'wow' as Coxeter's 1935 paper, or Stott's 1904 paper. 3 wows a century is pretty good going. Its first appearence was in 2000, on the mailing list.

The future of the uniforms, lies in the description of the vast numbers of laminate and other tilings, which do not have a C.D. diagram. Instead, the plan for attack for the 2d hyperbolic tilings (where the number is legion), lies in the Conway-Thurston diagram, and trying to place 'decorations' (like x, o nodes) onto it.

The higher dimensions, where the uniforms are non-wythoffian (ie do not derive from a Coxeter-Dynkin diagram), then these are at the moment just maintained in a catalogue, but eventually, someone who is intensely clever will find rhyme in these too.
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### Re: Development of the Coxeter-Dynkin diagram.

That's an excellent explanation, Wendy, mind if I put it on the wiki verbatim?

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### Re: Development of the Coxeter-Dynkin diagram.

We could.

Ideally, we could do something along the mechanics of going from a tCO with coloured edges, to the various processes i describe here. Eg stott shrinks the edges, and the Coxeter-Lie symbol is a walk around the edges, and the two join together that way.

You could even use it to demonstrate how one counts faces, by restricting the walks to a lesser number of edges (eg all C edges are closed). That would help explain how we can use the CD diagram to count things.
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### Re: Development of the Coxeter-Dynkin diagram.

Thanks for the explanation, it really helped me out .
I still have a question though:
Who was the first to observe (just merely observe) that x3x4o, o3x4o, x3o4x, o3x4x, x3x4x and s3s4s can be placed in a "set/group", and x3x5o, o3x5o, x3o5x, o3x5x, x3x5x and s3s5s in a similar group, and x3x3o is left alone? I'm now adressing this to Kepler, but I really want to know whether I'm addressing it to the right person.
How easily one gives his confidence to persons who know how to give themselves the appearance of more knowledge, when this knowledge has been drawn from a foreign source.
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### Re: Development of the Coxeter-Dynkin diagram.

Despite of what Wendy said, the convex semiregular polyhedra already date back way beyond Kepler into the time of the old greeks.

The convex regular ones e.g. usually are attributed to Platon (Platonic solids). He also mapped them to the four old greek elemnts (i.e. tetrahedron <=> fire (pointing up), cube <=> earth (sitting down), octahedron <=> air (pointin into all 6 directions), icosahedron <=> water (the most drop like one)) and he thereby considered the dodecahedron as kind of outstandish, so he decided to map it to a fifth type, to heaven. That fifth type later either became ether, or, as the dodecahedron is deeply involved with the golden ratio, to the "gold" of the alchemists. - Whether he was the first who considered those figures seems doubtable, as cubes and even dodecahedra occured in caves of ice age. Square pyramids surely were known by the old egypts, so a consideration of pyramids ontop of other convex polygons at least would be straight forward. It rather seems that he counts as the first known written reference to the complete set.

The other 13 convex semiregular solids (except of the 2 infinite series of prismatic forms) usually are called the Archimedean solids.

From the mere listing of these 13 solids towards a division into subsets of solids having parallel face planes, it seems to be a rather small step. But I do not know a reference for that. Might be that already Archimedes did that. - Note that duality, even so investigated as a true inversion mapping only much later, in a mere combinatorical sense was known to the greeks as well. Thus the subdivision into the fire-, earth- and gold-series was a well-known fact at least within the middle ages (even so greek science in comparision to catholic science then rather led a shadowy existence).

--- rk
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### Re: Development of the Coxeter-Dynkin diagram.

The uniforms were certianly known before Kepler, but it is kepler's tradition that we follow in the naming of.

The question on whether Plato and Archemedes actually did discover them or were victims of 'honour-names' is still open. I tend to side with the latter. The greeks certianly did know of these figures, since most of them have constructions in Euclid's elements. and the idiom 'quintessence' (fifth element) comes from here too.

Kepler's naming of them seems to indicate that he certianly understood the relations between the two sets of eight and two sundries (x3o3o, x3x3o). But his construction is a hack and paste job.
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### Re: Development of the Coxeter-Dynkin diagram.

wendy wrote:The uniforms were certianly known before Kepler, but it is kepler's tradition that we follow in the naming of.

The question on whether Plato and Archemedes actually did discover them or were victims of 'honour-names' is still open. I tend to side with the latter. The greeks certianly did know of these figures, since most of them have constructions in Euclid's elements. and the idiom 'quintessence' (fifth element) comes from here too.

There is evidence that regular dodecahedra were known long before Plato, so at least that particular case must be an honour name. Don't know about the others.

Kepler's naming of them seems to indicate that he certianly understood the relations between the two sets of eight and two sundries (x3o3o, x3x3o). But his construction is a hack and paste job.

Not to mention that the names "truncated icosidodecahedron" and "truncated cuboctahedron" are inaccurate, requiring an unspecified amount of distortion to make them uniform, whereas Stott expansion of the truncated dodecahedron / truncated icosahedron (resp. truncated cube / truncated octahedron) yields uniform figures. But Stott came much later, of course.
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### Re: Development of the Coxeter-Dynkin diagram.

I just realized an interesting fact about CD diagrams and coordinates. For example, take a cube and an octahedron, both of unit edge length. If you compute their Minkowski sum, you'll see that the result is a rhombicuboctahedron! Similarly, the Minkowski sum of the cube and the cuboctahedron is the truncated cube, and the Minkowski sum of the cuboctahedron and the octahedron is the truncated octahedron. What has this to do with CD diagrams? Take a look:

x4o3o + o4o3x = x4o3x
x4o3o + o4x3o = x4x3o
o4x3o + o4o3x = o4x3x

Do you see the pattern? The CD diagram of the Minkowski sum of two convex polyhedra (of the same symmetry group) is simply the CD diagram where the respective nodes of the original are added together (o=0, x=1, so o+x=1).

This gives us a very nice way to compute coordinates for unevenly-scaled CD diagrams. For example, the phi-scaled tetrahedron f3o3o is just the original tetrahedron x3o3o with its coordinates multiplied by phi; so to compute the coordinates of, say, f3o3x, we simply compute the Minkowski sum of the phi-scaled tetrahedron f3o3o and the unit dual tetrahedron o3o3x. In other words:

x3o3f = x3o3o + f * (o3o3x)

where + denotes Minkowski summation. This works not just for polyhedra of a particular symmetry group; it works for all polytopes of any dimension of any symmetry group (as long as the operands of the sum are of compatible symmetry group).

Now, how to compute the Minkowski sum? That's easy: just compute the vector sums of each vertex of the first polytope with every vertex of the second polytope, and discard the vectors that lie inside the convex hull of the result. If the operand polytopes are origin-centered and orbiform, the second part becomes even easier: just discard all vectors whose norms are smaller than the maximum vector norm in the set.

This is very good news for me, because this process is easily automated, which means I can implement it in my polytope viewer.
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### Re: Development of the Coxeter-Dynkin diagram.

quickfur: You shoud check out, eg http://hddb.teamikaria.com/wiki/Wythoff_polytope where the definition of 'wythoff polytope' is exactly the 'position polytope' of a point in a kaleidoscope, and that you can compute the metric properties (like R_0 etc), directly from the coordinates.

I wrote a spreadsheet based on the theory given on that page, you can fetch Richard's more recent version from his web-page. Basically, you can write, eg o3o5o as a matrix, and then take its inverse.

Code: Select all
Dynkin                    Stott                     Product
(   2   -1    0  )        (  3-f     2    f )       4-2f     0      0
(  -1    2   -f  )    *   (   2      4   2f )    =   0     4-2f     0
(   0   -f    2  )        (   f     2f    3 )        0       0    4-2f
=========             =================        ==================
1/  2                   2/(4-2f)                   1/(4-2f)

Schläfli function for [3,5] is  4-2f

3    5                   5     3
1    f+1                f+1    1       Shortchord^2
1    2    3   4-2f        1   2   3-f   4-2f     Schläfli value

The Dynkin matrix is simply the dynkin symbol in matrix form. The main diagonal is all '2', the value d_ij is the cosine between the normals to the plane (ie of the supplement). So (2, 0), (5/2, -v), (3 , -1), (4, -q), (5, -f), (6, -h), (7, -1.801937736) &c.

The Stott-matrix corresponds to S_ij = v_i ̇· v_j, where v_i is an axial vector (like x3o5o or o3x5o, or o3o5x) To preform the dot product over a dynkin graph, you write the vector (eg x3x5o -> (1,1,0). Multiply this by S_ij and then dot it with the second vector, So the norm of (1,1,0) is

S_ij (1,1,0)_i (1,1,0)_j = (5-f, 6, 0)_j (1,1,0)_j gives (11-f). Because there is a factor 2/(4-2f), must be taken to account too.

Let's see. 11 = 111.11 base F. 11-f is 111.11 - f = 10f.ff1 Divide by 2-f = 0.1 gives 10ff.f1 base phi-square.

And that's the radius-square of the x3x5o.

In the spreadsheet, when we calculate the height for the lacing, eg for a polytope xo3oo5ox, the process is to calculate the norm of (1,0,-1), which gives the "drift" across the base symmetry. Because the implied edge-length is 4, the height is H² = 4L² - D², the matrix dot product gives the drift's square.

In short, the notation i designed was specifically to allow vector calculus direct. This is a deep and magical thing.

In fact, i spent a few months 'spotting' stott matrices for groups like o3o3o3o3o4z This involves, for example, calculating the vertex figures of say x3o3ox3o3o4z, and then calculating its diameter. From this, one gets a_ii + a_jj + 2 a_ij. Since we get a_ii and a_jj from the schlafli function, the a_ij come directly.

Schläfli Value

If you suppose that p, q, r, &c stand for the square of the shortchords of P, Q, R, then eg (ok i don't know the trig, just the values. Most of this work was done when the fashion was a four-function calculator (+, -, * and /). The values are simply memorised.

2->0, 3-> 1, 4-> 2, 5-> 2.61803398875, 6->3, 7-> 3.2467976037, 8-> 3.41421356238, 9-> 3.5320888806, 10-> 3.61803398875, 12 -> 3.73205080757

The diameter of a circum-circle around a triangle 1:1:A is D = 4/(4-A), where all values are squares of lengths. (recall, no sqrt key)

The diameter of a polygon from its 'p' value is then 4/(4-p).

The diameter of a polyhedron-vertex figure for p,q is 4p/(4-q) ie a polygon of this diameter. From this one finds

D(p,q) = 4 / (4- 4p/4-q)) = (4-q) / (4-p-q)

Repeating for a polychoron, *by moving p,q to q,r, and adding in a new p, we get

D(p,q,r) = 4(4-q-r)/(16-4p-4q-r4+pr)

If one has one's eyes open, the denominator is the same for (p,q,r) as it is for (r,q,p).

D(p,q,r,s) = 2. F(q,r,s) / F(p,q,r,s)

Where: F() = 2; F(p) = 4-p ; F(p,q) = 8-2p-2q), F(p,q,r) = 16-4p-4q-4r+pr

and F(p,q) = q.F(p) - 2 F(); F(p,q,r) = r F(p,q)-2F(p), F(p,q,r,s) = s F(p,q,r)- 2 (p,q) &c.

In short, it's kind of like a continued fraction thingie, where one iterates t(n+1) = p t(n) - 2t(n-1).

This also works for the removal of any node. For example, the D for something like oPxQoRo is 2 F(o) F(oRo) / F(oPoQoRo)
= 2 F() F(r) / F(p,q,r).

So you really don't need to derive the matrix to find these figures. It's only when you do something fancier.
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### Re: Development of the Coxeter-Dynkin diagram.

Actually, with o3o5o, you just need the vectors for the five symmetries, and apply EPAC to it.

[twelfty follows.... (some decimal translation in brackets)]
With o3o3o5o, the common relation between this and APAC 3.24 (384) is also EPAC, of order 1.72 (192). This means you would need to hold an array of 75 possible cells, to which EPAC is applied. You can reduce this in part by using that the highest shared symmetry is 4.96 (576), which gives just 25.

Other groups can be quite problematic. The group 4B (ie 2_21), has an order of 3 72 00 (51840), but this has different coordinates. With the 5D group, which is 16.00 (ie EPAC directly), one then would have to monitor 27 separate symmetries. It is also closely related to a group of order 196, being the tri-triangular group, so the sub-symmetry is 2,00 additional vertices (80 if one supposes that there is a cyclic order of hexagonal axies (ie A,B,C -> B,A,C &c). It is handled by a rather interesting complex group that equates to EPTC (trinary change of sign).

The vertices of 2_21 come to (1,-1,0), EPTC, where these are complex numbers, and TC means multiplying by cis(c/3).

The vertices of 3_21 are closely allied with a body-centred A_n (ie a 60-deg rhomboecton. That is, suppose the seven vectors are at 60 deg (c20), then you have all integers or all integer-halfs (in BCC form). However, the order of the 7-simplex is 2.96.00 (8!), while the figure in question gives an order of 56*3.72.00 = 1.81.72.00, which is 72 times as large. There is a coordinate system based on an elongated semicubic (all 1, except 1 at sqrt(2)), which allows one to create all of the figures of this symmetry.

The vertex of 4_21 can be derived into a quarter-cubic, or a thrice-body-centred rhomb. (ie at x, or x 1/3 or x 2/3). In terms of the bcc, the order of the total figure is 3.43.24.00.00 = 192.10! the symmetry of the half-cube is 128.8!, giving 2.E8.48.00, which differs from the first by a factor of 1.15 (135). You would need to maintain 135 separate simplexes, and then apply EPAC to the result.
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