The Octahemioctahedron

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This page has some information about an interesting uniform polyhedron, the octahemioctahedron.  This polyhedron is interesting because is serves as an elegant model for an elliptic curve and an important example in graph theory.  The caption is the Wythoff symbol for this polyhedron.

Uniform polyhedron.
The polyhedron has 4 regular haxagons and 8 equilateral triangles as faces.  Since 2 hexagons and 2 triangles meet in the same configuration at every vertex, this is an example of a uniform polyhedron.  Notice that the 24 edges and 12 vertices coincide with those of a cuboctahedron (2 | 3 4).  The Euler characteristic is
k = v - e + f = 12-24+12 = 0.
After checking that this surface is orientable, one sees that the octahemioctahedron is a polyhedral approximation of the 1-holed torus, the surface of a doughnut.

This model does not represent an imbedding of the torus, only an immersion.  In fact, there is a line of intersection for each of the 6 pairs of hexagons.  There is also point in the center of the model where four hexagons share a single point.  One way to imagine this model is as an assembly of 8 regular tetrahedra sharing particular edges.  However, this particular image does not faithfully reflect the properties of this polyhedron as being a model of a compact surface.

Elliptic curve.
This polyhedron shares some symmetry properties with the elliptic curve
X:  y^2 = x^3 + 1.
Although this is an equation in the complex affine plane, we regard it as defining a locus in the complex projective plane.  Thus, we regard a point (x,y) in X as being the same as the point [x:y:1] in the projective plane.  Technically, we regard X as the projective completion of the locus in C^2 defined the above equation.

The group law.
The torus is a direct product of two circles, and since the circle is a group under addition, so is the torus.  Since X is homeomorphic to a torus, one might suspect there is a natural way to "add" points of X, just as one adds angles.  The identity element corresponds to the point [0:1:0] at infinity.  Since this has no counterpart in the affine plane, we denote this point simply as "infinity".  (Fortunately this is the only point at infinity.)  Suppose (x1,y1) and (x2,y2) are points in X.  If x1 and x2 are not equal, let
m = (y1-y2)/(x1-x2),
but if x1 and x2 are both equal to x, then let
m = 3x^2/(y1+y2).
(If y1+y2=0, then the points are inverses of each other, and the sum is the point at infinity.)  Define (x1,y1)+(x2,y2)=(x3,y3) by
x3=m^2-(x1+x2),
and
y3=m(x1-x3)-y1.
One can then check that X is a group under this operation.  Moreover, one can show that this is an abelian operation.  We therefore refer to this operation as "addition".

Points of Finite Order.
The elliptic curve X has an interesting subgroup consisting of 12 points.  The points in this group are all algebraic.  Let w=[-1+sqrt(3)]/2 be a primitive 3rd root of unity.  The following table gives the elements of this group, listed according to their orders.
Order | Elements
------+--------------------------------
1   | point at infinity
2   | (-w^n,0), n=0,1,2
3   | (0,1), (0,-1)
6   | (2w^n,3), (2w^n,-3), n=0,1,2
The cyclic subgroup
{infinity,(2,3),(0,1),(-1,0),(0,-1),(2,-3)},
of order 6, comprises the group of all rational points on X.

The Covering.
Consider the map
F:  (x,y)   -->  x
infinity -->  infinity,
which maps points of the curve to the complex projective line.  The 4 points which have order 1 or 2 correspond to the centers of the hexagons, and the remaining 8 points correspond to the the centers of the triangles.  We can consider the 4 points of order 1 or 2 as the branch points of the map from the curve to the projective line.  In the model of the octahemioctahedron, these four points coincide at the center.

Shrikhande's graph.  This is an example of a strongly regular graph with parameters (16,6,2,2).  This notation means that there are 16 vertices, 6 edges at each vertex, 2 neighbors shared by every adjacent pair of vertices, and 2 neighbors shared by every non-adjacent pair of vertices.  (Vertices v and w are "neighbors" if the edge {v,w} lies in the graph.)

The octahemioctahedron can be used to demonstrate an imbedding of Shrikhande's graph on a torus.  In order to obtain this imbedding, one must adjoin 4 more vertices and 24 more edges.  Each of these new vertices is placed at the center of a hexagonal face.  The new edges are then obtained by connecting the new vertices to the vertices of the corresponding hexagon.