Can anyone help with common notation?

Discussion of tapertopes, uniform polytopes, and other shapes with flat hypercells.

Can anyone help with common notation?

Postby ubersketch » Sat Dec 29, 2018 10:56 pm

I've seen a lot of people talk about polytopes with notations. Among which, I'll list lace cities, incidence matrices, and coxeter diagrams. Now although his pages are helpful, they still haven't quite drilled into my head how they actually work. Can someone help explain?
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Re: Can anyone help with common notation?

Postby Klitzing » Sun Dec 30, 2018 8:23 am

Dear übersketch,
I'd recommend to you the main explanatory pages of my website :nod:
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Re: Can anyone help with common notation?

Postby wendy » Sun Dec 30, 2018 9:50 am

Coxeter Diagrams

These come from group theory, and are not easily related to the models. Coxeter found these in the 1930s but it was some ten years before he joined them up to what we use now.

A dot in the graph is a node. This represents a mirror. A line in the graph is a branch, representing a set of mirrors.

A graph containing a single node, represents a single mirror. The 'identity' or I is one side of the mirror, and A might be the other side. So 'AA = I' means a reflection of a reflection is the self. You could then have a second mirror B, so that I = BB. Because there is no branch given, the two mirrors are at right angles, so you would see reflections A and B, which are reversed, and AB = BA which is not reversed.

If you draw two rays at 60 degrees, marked A and B, you could reflect the rays in each other, so that you have six rays pointing outwards, alternately marked A and B. Each of these cells then would have a name, based on what rays you cross. In order, I, A, AB, ABA, and from the other side I, B, BA, BAB. We have then that ABA=BAB, and the branch is drawn, but not numbered.

As the angles get sharper, eg 45° gives ABAB = BABA, marked '4', 36° gives ABABA = BABAB marked 5, and so forth.

If you now put a vertex in the sector between rays A and B, you can then draw lines perpendicular to the rays. The image is then a polygon of alternating edges A and B. This is the 'Cayley Diagram' for the group, or the omnitruncate. You can make the edges A and B at any length, independent of each other. If an edge B is 0, then the figure will only have edges crossing the 'A' mirrors,

so a3b is a hexagon a--------b
a3o is a triangle a-------o pointing up.
o3b is a triangle o-------b pointing down

When you add a third mirror, it makes three sides of a triangle. Usually one of these is a right-angle, which we do not draw the edges for. The cube is squares, three at a corner, looks like x4o3o, or x--4--o------o.

If you put your finger over a node, you will make a 'room' bounded by mirrors of that name. This will give here a polygon or whatever, that is entirely in that room.

So . - 4 - o-------o , by covering the first node, gives a hexagon whose alternating edges are b=0 and c=0. This makes the vertex.
x - 4 - . - - - o , covers the second node. It is a rectangle, of sides a=1 and c=0. This appears as a line.
x--4--o - - - . gives a twice-square, of alternating edges a=1, b=0. This becomes a simple square face.

Stott Addition

If we go back to the diagram of rays A and B, we could put lines parallel to ray A, (which is A=0), lines a=1, a=2, ... and the same for B. If we need to construct a polygon with alternate edges a=1, b=2, then this vertex will fall at the intersection of the lines. Mathematicians are not generally interested in the general case, and only discuss cases where a=0 or a=1.

The idea here is that the point (1,2) can be represented as a sum of two vectors (1,0) and (0,2). The vector (0,0) to (1,2) is the 'position vector' (1,2), and the image of this point and lines it creates is a 'position polytope'.

The idea here is that we could construct an octagon (1,1) from the squares (1,0) and (0,1), extends to a truncated cube (1,1,0) from a cube (1,0,0) and a cuboctahedron (0,1,0). If you look at a truncated cube, and imagine the edges between the octagons disappearing, the triangles would connect, so (1,1,0)-(1,0,0) = (0,1,0) the cuboctahedron. Alternately, now imagine the edges of the triangles getting smaller so (1,>,0), the edges formerly between the octagons now join up to make a cube.

We can use the coordinates given here as a proper coordinate scheme, if we take note that the vectors are at angles. The matrix used to find the heights of lace-layers is based on this coordinate system and a few matrix-vector tricks.

Lace towers and Lace cities

Below is the layers of a vertex-first {3,3,5}. The coordinates are given for one axis, giving one or more points in 3d. EPAC means even permutations, all change of sign. This means you can swap pairs of axies as one permutation, and do this an even number of times. The shape is abbreviated.

In the fourth column, we see the translation of a single shape into its CD. The edges are o=0, v=0.618, x=1, f=1.618 and 2=2. The letters are used consistantly through the scheme, using the table. You can see that the icosahedron in the 'f' row is size 'v=0.618', while the one in the v row is 1.618. But the coordinates are retained in the same order. That is f * 1,v,0 = f,1,0.

In the fifth column, we make a lace city, by expanding the figures in the fourth column along their pentagonal axis. When this is laid out correctly, you get a decagon with o5o, then inside this you draw a decagram with alternate v5o and o5v. Inside this you get another decagram marked o5x and x5o, and finally the centre, marked v5v, a decagon.

Code: Select all
     EPAC    shape
  ---------------------------------------------------------
  2  0,0,0   point      o5o3o                o5o
  f  1,v,0   icosa      o5o3v        o5o  v5o   o5v  o5o
  1  1,1,1   dodeca     v5o3o          o5v o5x x5o v5o
     v,f,0    ,,
  f  v,1,o   icosa      o5o3x       o5o  x5o     o5x   o5o
  0  2,0,0   ID         o5v3o        o5v  o5x v5v x5o v5o
     f,v,1
-f  v,1,0   icosa      o5o3x       o5o  x5o     o5x  o5o
-1  1,1,1   dodeca     v5o3o          o5v o5x x5o v5o
     v,f,0    ,,
-f  1,v,0   icosa      o5o3v        o5o  v5o   o5v  o5o
-2  0,0,0   point      o5o3o                o5o


Lace cities reveal a lot more of the internal structure of the polytope than a listing of layers. We have even used 3d lace cities, usually making a simple shape.

The notation with multiple letters between the numbers, is simply the values in column 4, read downwards, using the numbers as new-column markers.

So: oovo o ovoo 5 oooo v oooo 3 ovox o xovo & # tv = v3o3o5o

It would have nine letters bunched together, but i have seperated the middle row, they are at heights 2f1v 0 v1f2. If you write these letters in a vertical column, you get the appropriate column.

The & means we are adding an axis of additional symmetry. The t means that it's a tower, the 'v' means the bits of string (lacing) are 0.618 long.

The term 'lacing' comes from an antiprism, xo5ox&tx, means two pentagons are laced together with a string, so that the vertices are offset by a half-edge, and the string zigzag is a full edge. Kind of like lacing the top and bottom of a drum.

Incidence

An incidence means that something contains, or is part of something else. So if a vertex is part of a cube, then we could write "vertex ------> cube." The vertex is part of the cube, and the cube contains the vertex.

Because a vertex is also part of edges and faces, we would write these in too.

vertex--> edge ---> face ---> cube

The cube is itself an incidence (or Hass) diagram for the triangle. If you stand the cube on its vertex, the bottom vertex is the nulloid, the first layer is the three vertices, the second layer is three edges, and the top vertex is the triangle itself. Each hedron (2-face) of the incidence diagram is a rhombus, and shows the 'dyadic' nature of convex polytopes (the top and bottom surtopes are incident on exactly two middle ones).

Incidence diagrams can be written for any solid. Here is the cone.

Code: Select all
   vertex --------------> face -  cone
                    /           /
              edge -----> face



The vertex here is not incident on the edge, and there is no path from the vertex to the second face. It's not part of those things.

The incident matricies on individual elements gets quite large quite fast. What Klitzing does is to do incidence by type. Instead of listing each edge, all edges are grouped into a single example for each different kind. So in the truncated cube we have

1 vertex
2 kinds of edge (octagon-octagon, octagon-triangle)
2 kinds of face (triangle, octagon)
1 kind of body (tr. cube).

Each edge has two arrows, one pointing to the two faces it splits, so OO --> O (twice), and OT --> T and OT --> O.

Since the arrow-heads and tails are the same count, the OO --> O exists twice at each of the 12 OO edges, giving 24 tails. At the O end, there are 6 octagons, each with 4 OO -> O, or 24 arrows. Of the OT edges, there are 24 OT edges once by, to 6 O, four times, and 24 OT edges, onc by, to 8 T, three times.

The 24 vertices of this polytope are all OOT (two octagons, one triangle), so there is OOT -> OO., OOT to O_T, and OOT to _OT, so the 24 vertices are connected once to each OO edge, and twice to each OT edge.

The coordinate foursome, eg OOT T, describe the up and down incidence. We see OOT,OOT contains '24'. T,T contains '8'. OOT (a vertex) is incident on one triangle, so OOT, O contains '1'. T has three vertices, so T,OOT = 3. Now 24*1 = 8*3, means there are 24 arrows going from vertex to triangle. Likewise, there are 48 arrows going from OOT to OT, as 24*2 = 2*24.

The diagonal contains the number of individuals in the class, the rows below are the surtope count of each kind in an individual figure, so an octagon as x4x, has one kind of vertex, and two kinds of edge. The triangle as x3o has three vertices and two kinds of edges.

The same is true for the 'around' figure or 'vertex-figure'. This is an isoceles triangle OO:OT:OT. It has two kinds of vertex and two kinds of edge. The vertices and edges of the base are identical, which gives the '2', while the sloping sides gove '1'.

Code: Select all
         OOT  OO  OT    O   T
   OOT   24    1   2    2   1

   OO     2   12  --    2   0
   OT     2   --  24    1   1

   O      8    4   4    6   -
   T      3    0   3    -   8

The dream you dream alone is only a dream
the dream we dream together is reality.

\ ( \(\LaTeX\ \) \ ) [no spaces] at https://greasyfork.org/en/users/188714-wendy-krieger
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Re: Can anyone help with common notation?

Postby Klitzing » Mon Dec 31, 2018 9:40 am

 
There is even more to the incidence matrix.

Here is the same truncated cube, which Wendy just mentioned, as I would usually display it:
Code: Select all
o3x4x

. . . | 24 |  2  1 | 1 2
------+----+-------+----
. x . |  2 | 24  * | 1 1
. . x |  2 |  * 12 | 0 2
------+----+-------+----
o3x . |  3 |  3  0 | 8 *
. x4x |  8 |  4  4 | * 6

First you see the total counts of all different non-equivalent elements on the diagonal of the matrix, usually separated by lines in order to hint on the respective dimensionalities. (Non-equivalence here refers to the being used symmetry, for sure.)
But you see also the total counts of used subelements for each individual element in the subdiagonal parts of the corresponding row.
The superdiagonal parts of each row provide the corresponding counts for the vertex figures, edge figures (etc.).
(In the first added non-numeric column I usually just provide the corresponding subelements of the Dynkin diagram for additional info.)

All these 3 mentioned parts, the diagonal, the superdiagonals and the subdiagonals, individually bow to the (dimensionally extended) Euler rule (i.e. alternated sum, equating to 0 or 2).
Further, place any main-diagonally symmetric square on this matrix, and you'd get I_ii * I_ij = I_ji * I_jj.
Finally, choose any diagonal vertex or facet count and any sub- or superdiagonal staircase containing this very matrix element, then the corresponding product of those contained subdiagonal (superdiagonal) elements and that chosen diagonal element would provide the group order. Here eg. (24 : not selected) -down to- 2 -right to- (24 : not selected) -down to- 3 -right to- (8 : selected), yielding g = 2 * 3 * 8 = 48.

And, if you'd turn that very matrix within the paper (or screen) plane by 180 degrees, you would obtain the incidence matrix of the dual polytope for free. I.e. having here a polyhedron with 6 eightfold vertices plus 8 threefold vertices, and with 24 triangular faces of 2 edge sizes.

In fact, the incidence matrix even happens to be equivalent to the notion of abstract polytops.

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Re: Can anyone help with common notation?

Postby Klitzing » Mon Dec 31, 2018 10:20 am

Just as Wendy already mentioned, she exceeds the usual mathematicians view of zero and unit edges only for node types within the Dynkin symbol (then being displayed as unringed and ringed nodes respectively). In fact she often uses:
  • o : zero sized edge, which is to be discarded totally (thus corresponding to the usual unringed node symbol)
  • r : zero sized edge, which is to be maintained in the abstract polytopal sense
  • x : unit sized edge (thus corresponding to the usual ringed node symbol)
  • q : sqrt(2) sized edge, in fact the vertex figure of a square
  • f : (1+sqrt(5))/2 sized edge, in fact the vertex figure of a pentagon
  • v : (sqrt(5)-1)/2 sized edge, in fact the vertex figure of a pentagram
  • h : sqrt(3) sized edge, in fact the vertex figure of an hexagon
  • w : 1+sqrt(2) sized edge, in fact the side parallel width of an octagon
  • u : 2 units sized edge, in fact the vertex figure of an apeirogon
Using these length symbols for nodes in a Dynkin diagram, you would resize the corresponding edges accordingly. Eg. you could provide a v-scaled dodecahedron (refering to edge sizes) in mere brique subsymmetry as the convex hull of f2v2o + o2f2v + v2o2f + x2x2x. This then is what Wendy means when writing v5o3o = fovx-2-vfox-2-ovfx-&#zv ("-" : being introduced for visual aid only, "&" : and, "#" : being laced, "z" : in zero offset = no layer displacement, "v" : lacing edge size).

These symbols also could be used to provide lace city displays. In fact here the actual 2D positioning of symbols is just one of the subspace projections. The respective perp space then is given as the Dynkin diagram of the respective cross-section. Eg. the above v-sized dodecahedron (v5o3o) looks like
Code: Select all
         v         
                  
   x           x   
o                 o
                  
      f     f     
                  
o                 o
   x           x   
                  
         v         

or the (x-sized) truncated cube (x4x3o) of my previous post could be given in aproximate ASCII graphic as
Code: Select all
x w   w x
w       w
        
w       w
x w   w x

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