Negative nodes and Wythoff-constructions

[student91]

Changing nodes on CD-diagrams from negative to positive is an interesting process, and has shown to be very helpful at discovering new polytopes via Stott-expansion. Here I'll try to explain how node-changing works, why it works and how it can be used to calculate coordinates. (I guess Wendy finds this interesting)

How node-changing works:
Node-changing works on any diagram. Let's say we have a diagram x5u3o4q, then you can change any node from positive to negative according to the following rules, and you are guaranteed to get the same vertex-set.
1. You can only change a node from A to -A (A is allowed to be negative at the beginning)
2. when you change a node from A to -A, all other nodes must change accordingly. When you have a node B and the branch between A and B has value n (n can be a rational), B gets the value B+2Asin(90-180/n), or B+Astr(n) when str() takes the shortchord of a xno-polygon.
So when you change x to (-x), you get (-x)5(u+f)3o4q. When you change u to (-u), you get (x+uf)5(-u)3u4q. You can do iterative node-changing, so when you have (x+uf)5(-u)3u4q, you can change e.g. the u to -u, and you get (x+uf)5(-u+u)3(-u)4(q+uq) = (x+uf)5o3(-u)4(3q). Of course you can do this more times. in the given example, you can do this infinitely many times.

Why node-changing works:
You might know that the construction that makes a square from x4o starts with two mirrors with an angle of 180/4=45 degrees (or pi/4 radians). what you do next is place a vertex at a distance x of one mirror and a distance o of the other mirror. Then you keep reflecting in the mirrors until you have a square. In total this gives you four vertices: the original vertex, a vertex after reflection in the x4.-mirror, a vertex after reflection of the former vertex in the .4o-mirror, and a vertex after reflection of the former vertex in the x4.-mirror.

aardappel.png

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. So you have four vertices. what node-changing does is that you take a different vertex than the original vertex as seed-point, and then the construction gives the same vertex set. (That the vertex was given by the Wythoff-construction means it gives the same vertex-set, because the vertices more or less form a group). So when you take x4o, you can change this in (-x)4q. Changing q in (-q) gives you (-x+qq)4(-q)=(-x+u)4(-q)=x4(-q). A further changing of x in (-x) gives you (-x)4(-q+q)=(-x)4o.

sinaasappel.png

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. Changing node A in -A means you reflect the seed-point in the mirror of A.
So in short node-changing ables you to start the kaleidoscope-construction with any vertex of the polytope.

How node-changing can help finding coordinates:
The value of a node on a diagram gives the distance of the vertex to the corresponding mirror. This means the vertex lies on a (hyper)plane parallel to the mirror with corresponding distance to it. Every node gives a new hyperplane on which the vertex must lie, and thus the vertex must lie on the intersection of these hyperplanes. When you have all such hyperplanes, you have located a point. this means the decorated CD-diagram can be seen as a coordinate for a point in a special coordinate-system. The node-changing rules now give you tools to locate aal vertices of the polytope in the alternate coordinate system. I guess these node-changing rules aren't that hard to implement in a computer, and thus you can quite easilly find coordinates from the decorated CD-diagram. The "normal" coordinate system is given by the 222-symmetry. the only thing you need now is a transition from the alternate coordinate system to the 222-system. This can be done, and thus you can easilly obtain coordinates with the new node-changing devise.
so any Coxeter-group can be given a coordinate-system, and that can be used to find a polytopes' coordinates.

Taken into account that values on nodes haven't been published before, I think the new coordinate-system thing, together with the node-changing and the values-on nodes in general are sufficient for a small independent article in prefix of the article about pSe's. We might very well submit the two articles to the same paper to be published simultaneously as two independent but connected articles.

[wendy]

Negative nodes are provided for by placing the letter 'i' after the node, eg '4' = q, '4/3' = qi.

Most of the 'quit - x' are implementable by negative nodes.

[student91]

Negative nodes are provided for by placing the letter 'i' after the node, eg '4' = q, '4/3' = qi.

The mere use of negative nodes is not what I'm trying to explain. I understand completely that you can use any real number on the nodes of the CD-diagram. What I've found out is that changing nodes according to the rules I provided, you can find all vertices of the polytope, or conversely, start the Wythoff-construction from any vertex. Let me give an example: x3o5o. this is an icosahedron. the mere coxeter-graph .3.5. encodes three mirrors with respective angles of 90, 60 and 36 degrees. We can get such planes with the equations

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x=0
y=0
z=x/f+y/f^2

Now get back to the x3o5o. according to the rules provided in my previous post, we can change the x in (-x). The rules now tell us we also have to change the second node in o+x*str(3)=x. This gives us (-x)3x5o. With this, we can continue our node-changing, by changing the new x in a (-x). Now the rules tell us the (-x) must become -x+x*str(3)=-x+x=o, and the o must become o+x*str(5)=o+x*f=f, so we get o3(-x)5f. We can keep doing this, and now we get o3(-x+f*str(5))5(-f)=o3f5-f. All such outcomes are listed here:

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x3o5o
  ↓
-x3x5o
  ↓
o3-x5f
  ↓
o3f5-f
  ↓
f3-f5x => f3o5-x
   ↓        ↓
-f3o5x => -f3f5x
            ↓
  ---------
  ↓
o3-f5f
  ↓
o3x5-f
  ↓
x3-x5o
   ↓
-x3o5o

These are exactly twelve things. This might be a familiar number in combination with the icosahedron, and indeed these encode its vertices. The encoding is by giving the distance to three planes with respective angles of 90, 60 and 36 degrees. Using the formula's given above, we can decode these vertices to normal coordinates. For a polytope A3B5C, this gives you

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x=A
y=C
z=A/f+C/f^2+2fB

This gives you

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x3o5o => (1,0,f)
(-x)3x5o => (-1,0,f)
o3(-x)5f => (0,f,1)
o3f5(-f) => (0,-f,1)
f3(-f)5x => (f,1,0)
(-f)3o5x => (-f,1,0)
f3o5(-x) => (f,-1,0)
(-f)3f5x => (-f,-1,0)
o3(-f)5f => (0,f,-1)
o3x5(-f) => (0,-f,-1)
x3(-x)5o => (1,0,-f)
(-x)3o5o => (-1,0,-f)

Which means that the node-changing when done using the provided rules give exactly every vertex of the icosahedron. You could of course do this with a different polytope, such as a dodecahedron. I've written that down as well, but I'm not going to derive coordinates from it.

hoogmoed.png (9.97 KiB) Viewed 6836 times

Of course, this would just give the same coordinates as are on quickfurs site. Of course you could use other equations of planes with the desired angles, and those would give other coordinates, but that's just useless. My point is that the rules of node-changing are very interesting and could be used to find coordinates from the diagram by a computer. (you might even implement a program that gives equations for the desired planes, but that would give ugly coordinates for sure). Furthermore a thing with changed nodes might be interpreted as a encoding for an entire polytope rather than a vertex, and then the polytope would have the same vertex set as the starting polytope (so it's some kind of faceting). This has proven to be useful in some pSe's.

Most of the 'quit - x' are implementable by negative nodes.

What's a quit - x ?

[Klitzing]

Most of the 'quit - x' are implementable by negative nodes.

What's a quit - x ?

quit-... is a prefix found often in Hedrondude's OBSAs (official Bowers style accronyms). It derives from quasitruncated ...

When truncation was x-n/d-x-m/b-... then quasitruncation amounts just in x-n/(n-d)-x-m/b-...
You alternatively could just say: that is nothing but (-x)-n/d-x-m/b-...

For n/d=3 you would get x-3/2-x = (-x)-3-x, i.e. a doubly wound hexagon. Thus this Grünbaumian case usually is not considered any further.
But n/d=4 is much more interesting. Then x-4/3-x = (-x)-4-x becomes the octagram.
The abstractly equivalent thing (in fact the conjugate) to the truncated cube (tic = x4x3o) then would be the quasitruncated hexahedron (quith = x4/3x3o = (-x)4x3o).
Etc.

--- rk

[wendy]

You can walk the Cayley diagram to find the vertices.

The idea is that the dynkin matrix represents A-A', B-B', etc. You then do a script along the lines of

mirror_a ; parse arg a, b, c, d
return -a, b-a*ab, c-a*ac, d-d*ad

where ab, ac, ad are the elements in the dynkin matrix where row a crosses column b. ie ab = -2 cos (pi / branch)

A cayley walk is done by a series of mirrors (eg ABABA), where one applies the mirrors to the itermediate results.

Because you are using only the mirrors around the first cell, the steps in the walk jump across space, because you are in effect, stepping from 7 to -7, but the one-to-one correspondence with the omnitruncate (which is the cayley diagram) is preserved.

The complete set isfound by looking for all of the cosets etc.

Actually, quickfur's coordinates are for a rectangular system, not an oblique one. This is why he can use EPACS in his system, because the underlying system is the cubic group. The only place ye can do this sort of thing is in the equal arms of 3,3,A and 3,3,A,B, which do support EPACS over all nodees except the one between the two '3's . The idea in Quickfur is to reduce the icosahedral to epacs to cubic to all points.

The coordinates here are still in an oblique coordinate system, so ye have to use a stott matrix-dot to get the dot product, and a trimex to do the curl, but it still works.

[wendy]

When ye take a polytope x5o3o, and convert it into a coordinate, eg 1,0,0, ye are making it into a 'position polytope'. Ye can walk the vertex around by using columns of the dynkin matrix, but these are coordinates, not position polytopes.

For example the walk of a vertex around a hexagon is given here. The vertex 1,0,0 is repeatedly reflected in mirrors A, B, which add a vector -a, -b multiplied by the first, second coordinate. It requires six reflections to make this orbit, of which at two points, the vertex is already on the mirror.

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   1   0   0
  -1   1   0    A =  2  -1   0   * -1
   0  -1   f    B = -1   2  -f   * -1
   0  -1   f    A =  2  -1   0   *  0
  -1   1   0    B = -1   2  -f   *  1
   1   0   0    A =  2  -1   0   *  1
   1   0   0    B = -1   2  -f   *  0



Note however, that while x3o5o is an icosahedron, as given in the first coordinate, the edges of o3xi5f and xi3x5o would require the vertex edges to be connected to different mirrors than the one of the icosahedron. It has the same vertices but different edges.

As a solution to either of Quickfur or Roice3's pussles, it does not work well. What Quickfur is looking for is a cloud of points in a cubic coordinate system, and the idea of evaluating a,b.c five times in the EPACS coordinate system, is prehaps the optimum solution here.

Roice3 (Roice Nelson) does the fab piccies of the hyperbolic honeycombs. This process works in both euclidean and hyperbolic space, but i should actually have to sit down and play with a hyperbolic case to see what the numbers mean. It's clear in the euclidean clase.