### M Theory Lesson 162

As a zonohedron, the 3 dimensional permutohedron is generated by the 12 edges of the octahedron, which become the edges shared by two hexagons. In this scheme, the required list of 6 generators (edges come in pairs) is

{1,1,0} {1,-1,0} {1,0,1} {1,0,-1} {0,1,1} {0,1,-1}

which corresponds to two $3 \times 3$ circulant sets. Now to obtain the cube from the permutohedron, observe that the 8 hexagons are vertices, the 6 squares are faces, and the 12 special edges ($\frac{1}{3} \times 36$) are the edges. The incidence relations for a cube may be read off the permutohedron:

* each hexagon touches 3 squares

* each edge joins 2 squares

* each edge lies on 2 hexagons

and so on. Note also that the 24 vertices represent 3 times the number of vertices of a cube, because each of 6 vertices of a hexagon is shared by 2 hexagons.

{1,1,0} {1,-1,0} {1,0,1} {1,0,-1} {0,1,1} {0,1,-1}

which corresponds to two $3 \times 3$ circulant sets. Now to obtain the cube from the permutohedron, observe that the 8 hexagons are vertices, the 6 squares are faces, and the 12 special edges ($\frac{1}{3} \times 36$) are the edges. The incidence relations for a cube may be read off the permutohedron:

* each hexagon touches 3 squares

* each edge joins 2 squares

* each edge lies on 2 hexagons

and so on. Note also that the 24 vertices represent 3 times the number of vertices of a cube, because each of 6 vertices of a hexagon is shared by 2 hexagons.

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