how to find spherical coordinates of adjacent vertices surrounding central vertex in A3/D3 lattice

Question

How could you define (using spherical coordinate system) all the adjacent vertices directly connected to a central vertex in a tetrahedral octahedral honeycomb? Alternatively it would be useful to get the answer in cartesian coordinates and I could convert it myself. If this is too much effort, please refer me to a site that has this answer.

Answer 1

Alright, what you are asking about is the http://en.wikipedia.org/wiki/Cuboctahedron around the origin which serves as the "vertex figure."

I had the locations exactly where you want them, sum of integer coordinates even. Now, the vertex figure has six square faces, these are sliced halfway through each octahedron with a vertex at the origin; what are these squares? More typing in a minute...

One of these six squares is $$ (1,0,1); \; (1,1,0); \; (1,0,-1); \; (1,-1,0). $$ $$ (-1,0,1); \; (-1,1,0); \; (-1,0,-1); \; (-1,-1,0). $$

$$ (0,1,1); \; (1,1,0); \; (0,1,-1); \; (-1,1,0). $$ $$ (0,-1,1); \; (1,-1,0); \; (0,-1,-1); \; (-1,-1,0). $$

$$ (0,1,1); \; (1,0,1); \; (0,-1,1); \; (-1,0,1). $$ $$ (0,1,-1); \; (1,0,-1); \; (0,-1,-1); \; (-1,0,-1). $$

The same collection of twelve points (each point in two of the squares) make up eight triangles. $$ (0,1,1); \; (1,0,1); \; (1,1,0). $$ And seven more.

I see how to describe the eight triangles in an orderly manner; for each of the following eight matrices, let the three rows be the corners of a triangle.

One $$ \left( \begin{array}{ccc} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array} \right), $$

Two $$ \left( \begin{array}{ccc} 0 & 1 & 1 \\ -1 & 0 & 1 \\ -1 & 1 & 0 \end{array} \right), $$

Three $$ \left( \begin{array}{ccc} 0 & -1 & 1 \\ 1 & 0 & 1 \\ 1 & -1 & 0 \end{array} \right), $$

Four $$ \left( \begin{array}{ccc} 0 & 1 & -1 \\ 1 & 0 & -1 \\ 1 & 1 & 0 \end{array} \right), $$

Five $$ \left( \begin{array}{ccc} 0 & -1 & -1 \\ 1 & 0 & -1 \\ 1 & -1 & 0 \end{array} \right), $$

Six $$ \left( \begin{array}{ccc} 0 & 1 & -1 \\ -1 & 0 & -1 \\ -1 & 1 & 0 \end{array} \right), $$

Seven $$ \left( \begin{array}{ccc} 0 & -1 & 1 \\ -1 & 0 & 1 \\ -1 & -1 & 0 \end{array} \right), $$

Eight $$ \left( \begin{array}{ccc} 0 & -1 & -1 \\ -1 & 0 & -1 \\ -1 & -1 & 0 \end{array} \right), $$

Answer 2

The easiest way to place the centers for a tiling by Voronoi cells, all rhombic dodecahedra, is all integer points $(x,y,z)$ such that $x+y+z$ is even. I think that is what you want, see http://en.wikipedia.org/wiki/Tetrahedral-octahedral_honeycomb#A3.2FD3_lattice

The center is the origin, and there are twelve immediate neighbors, one coordinate remains $0$ and the others are $1$ or $-1.$

Also see the "Basis" at http://www.math.rwth-aachen.de/~Gabriele.Nebe/LATTICES/D3.html