Projective polytope

A projective polytope of rank n + 1 is a tesselation of n-dimensional real projective space, which is equivalent to an n-hypersphere with antipodal points considered identical.

Hemi-polytopes

This section is about projective polytopes made from spherical polytopes. For Polytopes with facets that pass through their center, see Hemipolytopes.

A diagram illustrating a fundamental domain of the Hemicube. Identically labeled vertices and edges are identified.

Since the real projective plane is equivalent to the sphere with antipodal points identified, certain spherical polytopes naturally give rise to projective polytopes by identifying antipodal elements. This is a quotient of the original polytope.

For example using the cube we can build a hemicube, by identifying opposite elements. Since the cube has 6 faces the hemicube 3 faces, each corresponding to a pair of opposite faces in the original cube. Vertices and edges are identified as well leading to half as many of each in the hemicube.

This idea works well with a particular set of polytopes, namely those in which every element has an antipodal element. However the concept can be generalized by selecting a Petrie polygon over which the polytope is symmetric and performing the quotient by that symmetry. For example the cube has a skew hexagon as its Petrie polygon and the fundamental domain of the hemicube is also a hexagon.

When this is done on a regular polytope the result is regular as well. We give the hemipolytope the Schläfli symbol X_n where X is the Schläfli symbol of the original polytope and n is half the number of edges in the Petrie polygon. In the example of the hemicube the Schläfli symbol is $\{4,3\}_3$.

Regular hemipolyhedra

Sperical polyhedron	Projective quotient	Fundamental domain	Schläfli symbol
Cube	Hemicube		$\{4,3\}_3$
Octahedron	Hemioctahedron		$\{3,4\}_3$
Dodecahedron	Hemidodecahedron		$\{5,3\}_5$
Icosahedron	Hemiicosahedron		$\{3,5\}_5$
Great dodecahedron	Petrial hemiicosahedron		$\{5,5/2\}_3$

If we attempt to create a hemitetrahedron by associating opposite sides of a petrial square, the result's double cover is not the tetrahedron.

The tetrahedron has no proper hemipolytope. It does have a petrial polygon with an even number of sides, however to associate opposite sides of the petrial polygon to form a fundamental domain would not match any symmetry of the tetrahedron. And any fundamental domain of the petrial polygon would not produce a result that is doubly covered by the tetrahedron. In fact because of the size of the tetrahedron none of the results are valid abstract polytopes, having self edges or faces.

This is equivalent to the fact that the tetrahedron does not have antipodal faces.

Bibliography

• McMullen, Peter; Schulte, Egon (December 2002), Abstract Regular Polytopes (1st ed.), Cambridge University Press, ISBN 0-521-81496-0