Vertex-transitive

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Vertex-transitive

Postby wintersolstice » Tue Sep 01, 2009 11:24 am

I tried to understand what this means. I know it doesn't mean "the configuration of facets round a vertex is the same" All I've heard is, is that there's isomentry mapping one vertex onto another.

The only thing I understand about "isometry" is rotation and reflections etc. But I've tried to apply this to a "elongated square gyrobicupola" (which is not uniform) and I can't understand why it isn't. I know about the ring of squares in one way and triangles and squares in others, but uniform isn't about that type of symmetry (I don't think) it's about "vertices"

This and something else which I'll mention will be vital to my "polychoron search"

How do find a set of co-ordinates for a shapes vertices (not necessary uniform) given a "Hasse diagram" (or equivalent) and the lengths of edges? By moving and rotating etc there could be infinitly many sets of co-ordinates.
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Re: Vertex-transitive

Postby wendy » Wed Sep 02, 2009 8:23 am

I will try to explain some terms.

'vertex-transitive' means that the vertices are carried by symmetries. This means that all vertices are identical, upto reflection.

'isometry' is read as /iso/ equal in context + /metric/ = measure. It contrasts homometric (equal downwards), and equi (equal in measure only). An example of a homometric polyhedron can be formed by taking a rCO, and rotating one of the octagons relative to the other. The vertex figure is still (4,4,4,3) three squares and a triangle, but the figure is no longer uniform, since it contains a pair of faces that are rotated 45 degrees relative to each other, while the others are parallel to their opposite.

Isometry supposes that one can cut a hole in N+1 dimensions, and drop the figure so that to a given vertex of the hole, one can place the thing, either directly or upside down, so that vertex 1, 2, 3, ... all fit to the vertex A of the hole. This implies a kind of symmetry.

Figures like 'vertex-configuration' etc, means the arrangement of faces at a vertex. For example, the cube has three squares at each vertex. The 'vertex figure' at A might be constructed by joining the vertices connecting the vertices corrected to A, into a figure of N-1 dimensions. A cube gives here a square.

Hasse diagrams are not a good way of seeking coordinates. All triangles have the same hasse diagram, being the antitegum of a triangle, or a cube. Stand the cube on its vertex (so the long axis is vertical). The vertices of the cube represent the assorted surtopes, and the edges repersent the connections.

All vertices are directly incident on a surtope of -1 dimensions (nullitope, nulloid, wessian). These are the three vertices at the first layer (a triangle). The second layer is an other triangle, representing the edges of the triangle. These are connected to the vertices of the triangle, for which the vertices are connected to the represented edges. The top of the cube is now the content of the triangle, which is a point connected to the three points representing edges.

Turning the hasse antitegum upside down gives the hasse configuration of the dual.

Since all antitegums (and thus hasse configurations), have a main axis, and that every surtope of an antitegum is itself an antitegum, we get:

B = bottom of hasse configuration
T = top of hasse configuration
S = surtope node.

B=S = surtope itself
S-T = surtope configuration (surtopes that are bounded by the surtope S)
T-S = dual's surtope in that position.

The hasse diagram is a connection diagram, and does not give any metric associated with the polytope itself.
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Re: Vertex-transitive

Postby wintersolstice » Wed Sep 02, 2009 9:19 am

I forgot to mention. (in reference to the second question) You're not only given a Hasse diagram, you're also given that the faces are regular (which means the edges are the same length) and your given what the edge length is.

Anyway thanks I'll try and make sense of it :D
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