Recall that the three dimensional associahedron is obtained as the secondary polytope of a hexagon in the plane, defining vertices in terms of triangulations of the hexagon. In general, counting triangulations of planar point sets is a difficult combinatorial problem. Convex polygons with $d + 3$ sides lead to the associahedron in dimension $d$, whose vertices are enumerated by the Catalan numbers.

The tetractys diagram introduces internal points. How do we count the many triangulations of a tetractys? First observe that there are three natural ways to cut the triangle into square (or rhombus) blocks, as shown below in brown, blue and purple. Each square has two possible chords, leading to $24$ possible triangulations.

But we can mess up the pattern of square blocks to form further triangulations of the tetractys. This tends to create internal triangles around the central vertex. Counting incoming edges at a vertex, we recover the vertex weights of the qutrit paths.

8 years ago

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