## Saturday, May 1, 2010

### Polytopes: Part II

This is the second part of a four part post. For the first part, see here.

The space in which all possible planes reside is the third dimension. In three dimensional space, one can define six directions: up, down, left, right, forward and back. The structures in this space are known as polyhedra (singular: polyhedron). Polyhedra are constructed with polygons, with a certain number of polygons at each vertex. Regular polyhedra have a certain number of identical regular polygons at each vertex. By extending the system from previously, these are denoted {p,q}.

One can construct the polygon {p,q} by having q regular polygons {p} at each vertex. As with polygons, situations with p≤2 are trivial, and q≤2 also has this property. This is due to the fact that any polyhedron {p,1} equals {p}, and any {p,2} is two overlapping polygons {p}, again trivially equaling {p}.

For any {p,q}, the total angular measure around each vertex is q internal angles of the regular polygon {p}. Since the internal angle of a regular polygon with p≥3 is (1-2/p)180º, the total angular measure around each vertex is (1-2/p)q180º. If this value is less than 360º, than the polyhedron curves inward, and eventually intersects itself, forming a finite polyhedron. Therefore, for a finite polyhedron,

(1-2/p)q180<360
(1-2/p)q<2
1-2/p<2/q
1<2/p+2/q
1/p+1/q>1/2

The final formula sets a limit on the number of possible finite regular polyhedra (assuming p,q>2), and only five pairs of {p,q}, where p and q are whole numbers, meet these requirements. These are called the Platonic Solids:

{3,3}
1/3+1/3=2/3>1/2
The above pair {3,3} defines a polyhedron with three triangles at each vertex. The resulting figure has four triangular faces, six lines, four vertices, and is known as the tetrahedron. It is pictured below.

{3,4}
1/3+1/4=7/12>1/2
The pair {3,4} defines another polyhedron with four triangles at each vertex. The resulting figure has eight triangular faces twelve lines, six vertices, and is known as the octahedron. It is pictured below.

{4,3}
1/4+1/3=7/12>1/2
The pair {4,3} defines another polyhedron with three squares at each vertex. The resulting figure has six square faces, twelve lines, eight vertices, and is known as the cube, or hexahedron. It is pictured below.

{3,5}
1/3+1/5=8/15>1/2
The pair {3,5} defines yet another polyhedron with five triangles at each vertex. The resulting figure has twenty triangular faces, thirty lines, twelve vertices, and is known as the icosahedron. It is pictured below.

{5,3}
1/5+1/3=8/15>1/2
The pair {5,3} defines a final polyhedron with three pentagons at each vertex. The resulting figure has twelve pentagonal faces, thirty lines, twenty vertices, and is known as the dodecahedron. It is pictured below.

The above polyhedrons are the only convex regular polyhedrons that are finite. They are finite due to the fact that they are tilings of a sphere, and therefore exist on the elliptic plane in elliptic geometry. Elliptic space is simply space with positive curvature. For more information on the three geometries, see here. In addition, convex regular polyhedra in elliptic 2-space or Euclidean 3-space (depending on point of view because elliptic and hyperbolic planes in n dimensions can only be accurately represented in n+1 dimensions) also have the property that the number of vertices plus the number of faces minus the number of sides always equals two. Also, for the above polyhedra, if the number of vertices and the number of faces are switched, the result is another polyhedron, known as the original polyhedron's dual. The dual of the tetrahedron is the tetrahedron, since both numbers are the same, and the other four come in dual pairs, with the cube's dual being the octahedron and the icosahedron's dual being the dodecahedron. This is equivalent to switching the numbers in {p,q} to {q,p}. As a result, if a polyhedron is finite, its dual will be also.

Another case to consider is when 1/p+1/q=1/2. In this case, the curvature is exactly zero, and the polyhedron tiles the Euclidean plane, and the polyhedron is known as a tiling. Since the flat Euclidean Universe is infinite, so are polyhedra that tile it. There are only three regular tilings of the Euclidean plane:

{3,6}
(1/3+1/6=9/18=1/2)

{4,4}
(1/4+1/4=2/4=1/2)

{6,3}
(1/6+1/3=9/18=1/2)

The above are also known as the deltile, the quadrille, and the hextille, respectively.

The one remaining case, with 1/p+1/q<1/2,results in a tiling of negatively curved space, or hyperbolic space. It too, is infinite, and so are the polyhedra that tile it, but there are many projections of it into Euclidean geometry, and the one below squeezes the entire hyperbolic universe into one finite circle called the bounding circle.

In the above picture, there are seven triangles at each vertex. The triangles appear to be distorted and of different size, but from the viewpoint of hyperbolic space, all triangles are the same size, and the circle is infinite.

There also can be non-convex polyhedra, for which the faces of the polyhedra penetrate their interior. However, of these, there are only four finite star polyhedra, known as the Kepler-Poinsot polyhedra, that are regular. The four are shown below.

{5/2,5}
1/(5/2)+1/5=2/5+1/5=3/5>1/2
The pair {5/2,5} defines a non-convex polyhedron with pentagrams as faces and five at each vertex. The resulting figure has twelve faces, thirty sides, and twelve vertices. It is known as the small stellated dodecahedron and is pictured below.

{5,5/2}
1/5+1/(5/2)=1/5+2/5=3/5>1/2
The pair {5,5/2} defines a non-convex polyhedron with pentagons as faces and five at each vertex, although each cuts through the interior of the polyhedron. The resulting figure has twelve faces, thirty sides, and twelve vertices. It is known as the great dodecahedron and is pictured below.

{5/2,3}
1/(5/2)+1/3=2/5+1/3=11/15>1/2
The pair {5/2,3} defines a non-convex polyhedron with pentagrams as faces and three at each vertex. The resulting figure has twelve faces, thirty sides, and twenty vertices. It is known as the great stellated dodecahedron, and is pictured below.

{3,5/2}
1/3+1/(5/2)=1/3+2/5=11/15>1/2
The pair {3,5/2} defines a non-convex polyhedron with triangles as faces, with five at each vertex, but each penetrating the interior of the polyhedron. The resulting figure has twenty faces, thirty sides and twelve vertices. It is known as the great icosahedron, and is pictured below.

The four regular star polyhedra come in dual pairs with the small stellated dodecahedron and the great dodecahedron being one pair and the great stellated dodecahedron and the great icosahedron being the other. In addition, the vertices of each star polygon are in exactly the same orientation of one of the five Platonic Solids. The small stellated dodecahedron, the great dodecahedron, and the great icosahedron share the exact same vertex arrangement as the icosahedron, and the great stellated dodecahedron has the same arrangement as the dodecahedron. Also, the icosahedron and great dodecahedron have the same vertex and edge arrangements, and the small stellated dodecahedron and the great icosahedron form such a pair as well. However, in the first cases, the edges and faces are connected differently, and in the latter cases, just the faces. However, the star polyhedra are of less obvious construction than the Platonic solids, and it is often difficult to see the connections of the faces. For the sake of simplicity, a single face of each polyhedron is highlighted in yellow below to emphasize structure.

Unlike the five regular convex polyhedra, the four regular star polyhedra by no means cover all the possibilities of the equation 1/p+1/q>1/2. In fact, {5/2,4}, {4,5/2}, {3,7/2}, {7/2,3}, {3,7/3}, {7/3,3}, {5/2,5/2}... etc. haven't been dealt with at all! To solve this problem, one must consider the fact that all of the regular star polyhedra share the vertices of a Platonic Solid just as a pentagram shares the vertices of a pentagon. Since no heptagons exist in the five regular convex polyhedra, any pair involving {7/2}, {7/3} or any higher star polygon must be disregarded. Also, star polyhedra are based on the extensions of faces on the five Platonic Solids. However, for the tetrahedron, octahedron, and cube, extending their faces into planes will not allow additional intersections, while the icosahedron {3,5} and dodecahedron {5,3} do. This eliminates any polytopes with a 4, or any number greater than 5, because the icosahedron and dodecahedron do not have these numbers. This leaves only the four already shown, and {5/2,5/2}, but it can be shown that the midpoints of the edges of {5/2,5/2} are equivalent to the vertices of {5/2,4}, and this last possibility is discarded. One can always begin to construct these star polyhedra, but they will never close up in a finite time, and require infinite sides and faces, which is impossible in the elliptic plane.

A similar method states that there can be no tiling of the Euclidean plane with star polygons, despite there being possible pairs, such as {10,5/2}, because they are not based on the three existing Euclidean tilings, {3,6}, {4,4} and {6,3}. Since no continuous star polygon has 3, 4, or 6 sides, no star polyhedron can be based on these numbers. There do, however, exist an infinite number of hyperbolic star polygon tilings, such as {11/2,5}, and many others.

The above explores all possible regular polyhedra. However, there are other polyhedra with regular faces, but not necessarily made of just one regular polygon. Some are composed of regular triangles and squares, for example. These are called uniform polyhedra. To generate these figures, one can apply processes such as truncation and cantellation to regular polyhedra. Truncation, when applied to the polyhedron is the process of taking away a vertex and putting a face in its place, while cantellation is taking away a edge and replacing it with a face. Both processes as performed on the cube are shown below.

For the top picture, the slicing of the vertices creates triangles. When the vertices are cut to the point that the resulting faces intersect each other, but do not overlap, the polyhedron is rectified. The rectified cube is also known as the cubeoctahedron, which is last in the top row of pictures. If the cuboctahedron is truncated again, however, the resulting figure is the octahedron, which is the cube's dual. With cantellation, a slice is taken with respect to the edges. If the cube is half cantellated, as with the third picture in the second row, so that each polygon face is regular, the polyhedron is uniformly cantellated, and the resulting figure is known as the rhombicuboctahedron. If the polyhedron is cantellated to the point that the original faces disappear, the resulting figure is the original polyhedron's dual, in the case of the cube, the octahedron.

Using the symbols mentioned previously, fully truncated polyhedra are represented by t1{p,q}, where {p,q} is the regular polyhedron, also known as t0{p,q}. When truncated twice, the notation is t2{p,q}, but when a polyhedron is truncated twice, it becomes its dual, and therefore

t2{p,q}={q,p}

t0,2{p,q} is also the notation of the uniform cantellation of the polyhedron {p,q}.

Any regular polyhedron can undergo truncation and cantellation, including the star polyhedra. For example, the uniform truncation of a small stellated dodecahedron, as shown above, is simply a dodecahedron, because it is evident to see that if each point of the small stellated dodecahedron was cut off, there would only be pentagons left, and the figure would be a dodecahedron. If the small stellated dodecahedron is rectified, however, represented by t1{5/2,5}, the resulting figure is a dodecadodecahedron (a wonderful name) which is shown below.

In general, any polyhedra made up of two types of polygons, {p}, and {q}, and with the ordering around each vertex goes {p}, {q}, {p}, {q}, are called quasiregular polyhedra, which are expressed by {p;q} (p semicolon q). For example {3;5} is a polyhedron that has two triangles and two pentagons at each vertex, which is known as the icosidodecahedron.

In this figure, the pentagons and triangles alternate around each vertex.

In addition, a polyhedron {p,q} can always be found so that one of its truncations or cantellation results in any quasiregular polyhedron. Using the example above, the icosidodecahedron is the rectified dodecahedron. It is known to be finite because the angular measure around each vertex is 336º (108º+108º+60º+60º, which correspond to the angles of regular pentagons and triangles respectively), and this is less than 360º, making the polyhedron elliptic and finite, as a familiar formula from above states. Tilings of the Euclidean and hyperbolic planes can also be truncated and cantellated, and new tilings such as {3;6} and {3;7} arise from these processes. Both tilings, one the trihexagonal Euclidean tiling, and the triheptagonal hyperbolic tiling are shown below.

These tilings are only two of the many Euclidean and infinite hyperbolic quasiregular tilings.

There are infinite additional polyhedra that do not have regular polygonal faces, and vary from square-based pyramids to hexagonal prisms. However, these polyhedra are not relevant to this post, and are therefore omitted.

Sources: Wikipedia Titles: List of Regular Polytopes, Uniform Polytope, Quasiregular Polytope, etc.
Regular Polytopes by H.S.M. Coxeter

Notation Note: In the above post, the words side, edge, and line are generally interchangeable, all representing the one dimensional features of a polyhedron.

For the next part of this post, see Polytopes: Part III.