# Regular polyhedron

(Redirected from Kepler–Poinsot solids)

A regular polyhedron is a polyhedron whose flags are identical under its symmetry group. All regular polyhedra are uniform, isogonal, isotoxal, isohedral, and noble. In 3D Euclidean space, there are 9 finite regular polyhedra and 3 tilings that are not skew, and an additional 36 other regular polyhedra that are skew.

## Convex regular polyhedra

There are five strictly-convex regular polyhedra 3D Euclidean space. They are known collectively as the Platonic solids. These polyhedra are also the only finite regular polyhedra without self-intersection.

The Platonic solids
Solid Number of faces Schläfli symbol Image
Tetrahedron 4 {3,3}
Cube 6 {4,3}
Octahedron 8 {3,4}
Dodecahedron 12 {5,3}
Icosahedron 20 {3,5}

The existence and completeness of the Platonic solids has been known since at least the Ancient Greeks. Some sources attribute their discovery to Pythagoreas, however Theaetetus gives the first known description of all five.

## Regular star polyhedra

The Kepler-Poinsot polyhedra are a set of 4 non-convex, non-skew, finite polyhedra.

They can all be obtained as stellations of either the dodecahedron or the icosahedron.

The Kepler-Poinsot polyhedra
Solid Number of faces Schläfli symbol Image
Great dodecahedron 12 {5,5/2}
Small stellated dodecahedron 12 {5/2,5}
Great icosahedron 20 {3,5/2}
Great stellated dodecahedron 12 {5/2,3}

### Proof of completeness

The following proof is based on one given by Joseph Bertrand.[1]

The first result is that any non-degenerate regular polyhedron with finitely many vertices must share its vertices with one of the five Platonic solids:

Proof —

Let us consider a hypothetical regular polyhedron, 𝓟. We will show that its vertices correspond to one of the Platonic solids.

Because a regular polyhedron is necessarily vertex-transitive, its vertices must lie on a sphere and thus its convex hull, 𝓠, must share its vertices.

We will show that 𝓠 is transitive on its edges, vertices and faces. This demonstrates that 𝓠 is weakly-regular, and it is established that all weakly-regular convex polyhedra are also regular.

Since any symmetry of 𝓟 preserves its vertices and the convex hull is unique, any symmetry of P  must also be a symmetry of of 𝓠.

Thus 𝓠 must be vertex-transitive since its vertices are those of 𝓟.

To show the remainder we will superimpose 𝓠 on 𝓟 such that their vertices coincide. Let us consider an arbitrary vertex v  and two vertices, u  and w , both adjacent to v  in 𝓠. If we take the diameter of the circumsphere of 𝓟 and 𝓠 which passes through v  as an axis of rotation there is a rotation which takes u  to w . Furthermore, since 𝓟 is regular, this rotation is a symmetry of 𝓟 and thus 𝓠 as well. This means the edges vu  and vw  are identical under the symmetry of 𝓠. We can also see that all the faces in 𝓠 incident on v  must also be identical. Therefor 𝓠 is edge-transitive and face-transitive.

Since 𝓠 is transitive on each of its ranks it is weakly-regular, and all the weakly-regular convex polyhedra are regular.

This fact alone demonstrates there are only finitely many regular polyhedra with finite vertices, since there are only finitely many polyhedra that can be made from any finite vertex set. At this point it would be possible to enumerate all of these polyhedra and check that only 4 of them are regular, however there is a simpler way:

Proof —

For each Platonic solid we will find all regular polygons that can be drawn with its vertices and symmetries. We can do this by starting from an arbitrary vertex, then for every symmetry of the polyhedron we take the orbit traced out by that vertex. Some of these resulting polygons will be degenerate or skew, so we discard these and only look at those which are ordinary planar polygons. Since a regular polyhedron must have regular faces we know that any non-convex regular polyhedron must have faces made from these polygons. There are some cases where a regular polygon made from the vertices of a Platonic solid exist but can't be found using this method. In these cases there are necessarily symmetries of the regular polygon that are not symmetries of the polyhedron. However a regular polyhedron must have full symmetry on all of its faces and we know from the lemma that the symmetry of a finite regular polyhedron and its convex hull must be the same, so these cases can ignored, since they either generate a polyhedron which is not regular, or contains fewer vertices than the Platonic solid in question.

#### Tetrahedron

A regular polygon must have at least three vertices. Any three vertices of the tetrahedron form a face of the tetrahedron, thus for other polyhedra the faces must have at least 4 vertices, however there are only 4 vertices in total and they are not coplanar. Thus the tetrahedron is the only regular polyhedron that can be made with its vertices.

#### Cube

With the cube triangles can be made with its vertices. These either yield the stella octangula, if all such faces are taken, or a tetrahedron if only 4 are taken. The stella octangula is a compound and thus not a regular polyhedron, and the tetrahedron is already a platonic solid.

#### Octahedron

With the octahedron, in addition to its triangular faces, it has 3 square holes. These however do not share any edges with each other, only vertices, and thus no polyhedron at all can be made using only these squares.

#### Dodecahedron

The four ways to make regular planar polygons from the vertices and symmetries of the dodecahedron. Only the pentagrams on the far right form the faces of a regular polyhedron.

The dodecahedron has more available polygons than the previous polyhedra. In total there are four possible types regular polygons which can be made from the vertices and symmetries of the regular dodecahedron. Two types of triangle, a type of pentagon, and a type of pentagram.

• The first type of triangle can be made from any three vertices all mutually adjacent to a third. Since two vertices mutually adjacent to one vertex are not mutually adjacent to another vertex, these polygons do not share edges with each other, and thus don't form a polyhedron.
• The second type of triangle can be made by from three points each mutually distance 2 from a single vertex. These triangles, unlike the previous type of triangle, do close. However the polyhedron formed is a tetrahedron.
• Pentagons can also be made by selecting an arbitrary face and using the vertices opposite each edge of the face. Alternatively they are the alternation of the dodecahedron's Petrie polygons. No two of the pentagons share an edge and thus they do not form a regular polyhedron.
• Pentagrams can be made from the above pentagons. These pentagrams form the faces of the great stellated dodecahedron.

There is also a square which can be made from the vertices of the dodecahedron, however it can't be made using the orbit tracing method because, when the two are superimposed, the symmetries of the square are not symmetries of the dodecahedron. Turning the square a quarter turn would not map the dodecahedron to itself. Since a regular polyhedron must have full symmetry on all of its faces this means this case doesn't have to be considered. If we do consider it we find this face closes to a cube, which has fewer vertices than the dodecahedron. If we add all the possible squares we get the compound of 5 cubes, which is only weakly-regular.

#### Icosahedron

The three ways to make regular planar polygons from the vertices and symmetries of the icosahedron.

There are three ways to make regular planar polygons from the vertices of the icosahedron, all three of them correspond to the faces of a polyhedron.

• Regular triangles can be formed from the vertices of the icosahedron. These form the faces of the Great icosahedron.
• Regular pentagons can be formed from sets of 5 vertices all mutually adjacent to a single vertex. These pseudo-faces are holes of the icosahedron, and are the result of alternating its Petrie polygons. These form the faces of the great dodecahedron.
• The regular pentagons can be made into regular pentagrams. These form the faces of the small stellated dodecahedron.

### Regular dense polyhedra

Unlike with convex regular polyhedra, with non-convex regular polyhedra there are infinitely many Schläfli symbols with positive angular defect. As a result these are regular with their vertices lying of a sphere. However, only the Kepler-Poinsot solids are closed, all the other potential polyhedra never close back on themselves and thus have infinitely many elements. Thus these polyhedra have infinitely many vertices in finite space and thus are dense.

Even when infinite regular polyhedra are considered the regular dense polyhedra are not considered.

## Regular polyhedron compounds

There is a single finite planar polyhedron compound which is regular under the usual definition. The stella octangula, a compound of two tetrahedra, is flag transitive under its symmetry group. However the term "regular compound" is often used to refer to a broader class of polyhedron compounds, those which are vertex, edge and face transitive.

### Proof of completeness

This extended list including the stella octangula can be shown to be complete using the same method as in the proof there are 9 finite planar regular polyhedra. That method discards compound results, however since the initial lemma made no assumptions about connectivity we can be sure this search is exhaustive for compounds as well.

## Regular tilings

While classical definitions require polytopes to be finite, other definitions may relax this, allowing the regular tilings of the Euclidean plane to be considered as regular polyhedra. In Euclidean space the internal angles of the faces around a vertex must add up to 360 degrees. This puts a firm restriction on the number of possible non-dense regular tilings. Only four exist in Euclidean space:

Regular tilings of the Euclidean plane
Tiling Schläfli symbol Image
Triangular tiling {3,6}
Square tiling {4,4}
Hexagonal tiling {6,3}
Apeirogonal dihedron {∞,2}

The apeirogonal dihedron is often considered degenerate because its two faces share all their elements, and many authors do not count it for this reason.

### Dense tilings

Unlike with the Kepler-Poinsot polyhedra, where allowing self intersection created new polyhedra, allowing faces to overlap doesn't permit any additional tilings of the plane. However there are infinitely many dense tilings of the plane. Polyhedra of the form {p,q} where 1/p +1/q  = 1/2 tile the plane.[2] The cases not listed above are all dense. Alternatively dense tilings result when the angular defect is a multiple of 2π  less than 0. For example, {5,10/3} and {10/3,5} have an angular defect of -4π  and thus are dense regular apeirohedra.

### Spherical tilings

Due to the fact that all finite regular polyhedra have a circumsphere, they can also be embedded onto a sphere and preserve their regularity. Because of this, all tilings of the sphere have a corresponding polyhedron in Euclidean space, and vice versa. Note that all spherical tilings are 'flat': their internal angles add up to 360 degrees on the surface of the sphere.

### Hyperbolic tilings

In hyperbolic space, tilings can exist where in Euclidean space the internal angles would exceed 360 degrees. This leads to an infinite number of regular tilings of hyperbolic space. There is a tiling for every pair of convex regular polygons (including the apeirogon), with any pair that does not produce a tiling of spherical or Euclidean tiling producing a hyperbolic tiling.

#### Hyperbolic star tilings

There are only two infinite sets of star tilings of hyperbolic space: those of the form {p/2,p} and their duals {p,p/2}, where p  is odd and greater than 5 (e.g. the stellated heptagonal tiling {7/2,7}). The {p/2,p} tilings are stellations of the {p,3} tilings while the {p,p/2} dual tilings are second-order facettings of the {3,p} tilings and greatenings of the {p,3} tilings.

When p  is 5, the pattern produces the small stellated dodecahedron and great dodecahedron, which is regular but not hyperbolic, and when p  is 3, both patterns result in a tetrahedron. The other two Kepler–Poinsot polyhedra (the great stellated dodecahedron and great icosahedron) do not have analogues in the regular hyperbolic tilings.

Hyperbolic star tilings
Name Schläfli
symbol
CDD Image Face Vertex figure Density Symmetry Dual
Stellated heptagonal tiling {7/2,7} {7/2}
{7}
3 [7,3] Great heptagonal tiling
Great heptagonal tiling {7,7/2} {7}
{7/2}
3 [7,3] Stellated heptagonal tiling
Stellated enneagonal tiling {9/2,9} {9/2}
{9}
3 [9,3] Great enneagonal tiling
Great enneagonal tiling {9,9/2} {9}
{9/2}
3 [9,3] Stellated enneagonal tiling
Stellated hendecagonal tiling {11/2,11} {11/2}
{11}
3 [11,3] Great hendecagonal tiling
Great hendecagonal tiling {11,11/2} {11}
{11/2}
3 [11,3] Stellated hendecagonal tiling
Stellated p -gonal tiling {p/2,p} {p/2} {p} 3 [p,3] Great p -gonal tiling
Great p -gonal tiling {p,p/2} {p} {p/2} 3 [p,3] Stellated p -gonal tiling

### Tilings of other spaces

This notion of polytopes as tilings of a space can naturally extended to a great deal more spaces, such as the real projective plane (e.g. Hemiicosahedron) or the torus (e.g. square duocomb). Here regularity is defined in terms of the automorphism of the surface the polytope tiles, rather than the symmetries of ambient space.

In particular the regular polytopes on compact surfaces of small genus are an object of study. These have been largely classified.

Non-degenerate abstract regular polytopes by space[3]
Genus
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Orientable 5 4 12 12 18 17 14 12 46 36 17 13 31 14 37
Non-orientable 4 0 0 4 6 6 4 2 6 12 4 5 4 6 2

#### Projective space

There are four (non-degenerate) tilings of the real projective plane without overlaps:[4]

These are the hemi-polytopes of the Platonic solids other than the tetrahedron. They correspond to identifying elements with their image under central inversion. That is the element directly opposite them. The tetrahedron is not centrally symmetric, so the "hemitetrahedron" does not exist. Attempting to project the spherical tetrahedron onto the projective plane creates a double-cover.

The hemi-polytope of a Platonic soild {p,q}n  is {p,q}n /2.

This list can be seen as complete by showing that the hemi operation can be reversed to get a spherical polyhedron from a projective one. The orientation double cover of a projective polyhedron 𝓟 is spherical, as the orientation double cover of the projective plane is a sphere, and its hemi-polytope is 𝓟. It then follows if there was a non-degenerate tiling of the projective plane not on this list, there would need to be a sixth Platonic solid.

Additional regular tilings can be derived as the hemi-polytopes of some of the degenerate spherical tilings, the hosohedra and the dihedra. Polytopes of the form {n,2} and {2,n} are centrally symmetric, and thus have a hemi quotient, when n  is even. However these tilings are not only degenerate by do not have the diamond property. More permissive definitions of polyhedron such as maniplex permit these tilings, but they are not abstract polytopes.

#### Torus

There are infinitely many regular tilings of the torus. These come in 3 types, as quotients of the regular Euclidean tilings:

Other than the sphere, the torus is the only compact surface with infinitely regular tilings, and it is the only one with infinitely many non-degenerate tilings. This is a result of Hurwitz's automorphisms theorem which places a bound on the number of flags a regular polytope can have for surfaces of higher genus.

#### Bolza surface

The Bolza surface is the compact Riemann surface with the highest order automorphism group of orientable genus 2, with 96 automorphisms. All of the regular polyhedra tiling a genus 2 surface are all tilings of the Bolza surface with some suborder of its symmetries.

There are four abstract regular polyhedra on the Bolza surface[5]:

• A polytope of type {8,3}, maximal order
• A polytope of type {3,8}, maximal order
• A polytope of type {4,6}, half order
• A polytope of type {6,4}, half order

#### Orientable polyhedra of genus 3

The Klein quartic is the compact Riemann surface with the highest order automorphism group of orientable genus 3, with 336 automorphisms. This surface meets the Hurwitz bound on automorphisms, and is the lowest surface to do so. However unlike orientable polyhedra of genus 2, not all orientable surfaces of genus 3 are tilings of this maximal surface.

The Klein map and its dual both tile the Klein quartic with maximal order.

The Dyck map and its dual tile a differens surface of genus 2.

In total there are 12 abstract regular polyhedra of genus 3.

## Regular skew polyhedra

The definition of polyhedron can be further relaxed to include skew polyhedra. Skew polyhedra can have skew faces and vertex figures, however they are still required to be dyadic, non-dense, and non-degenerate. In total, there are 48 regular polyhedra in 3D Euclidean space, 36 of which are skew.

### Finite skew polyhedra

The Petrie dual, or Petrial, of a polytope can take any regular polyhedron and transform it into one sharing edges and vertices with the original, but with skew faces. Because of this, there is a Petrie dual to every previous regular polyhedron.

There are multiple extensions to Schläfli symbols which allow the Petrials to be represented. Where the Petrie dual operation can be represented with π, and {p,q}r is defined as a regular map, or equivalently a polyhedron with q  p -gons around a vertex, and an r-gonal Petrie polygon. The Petrie dual of a Petrial polyhedron gives the original polytope again.

### Blends

We can also create new regular skew polyhedra by taking the 6 2-dimensional polyhedra (the tilings of the Euclidean plane and their Petrials) and blending them with a 1-dimensional polygon, either a digon or an apeirogon (Note: Depending on the definition of blending, a line segment may be used as a blend component instead of a digon).

### Pure apeirohedra

In Euclid's proof a polyhedron with 6 square faces at each vertex is ruled out because its angular defect, ${\displaystyle 3\pi }$, would make it hyperbolic. However it is possible to arrange 6 squares at a vertex such that adjacent squares meet at the same angle.

This configuration alternates between "concave" and "convex" edges. In order for this to be the vertex of a regular polyhedron there has to be a symmetry taking the each convex edge to each concave edge. For a traditional polytope this is impossible, one side must be the "inside" and one side must be the "outside", and thus you can't bring a convex edge to a concave edge or vice versa. However for skew polyhedra there is no such requirement. If we take this vertex and repeat it indefinitely we indeed get a regular skew polyhedron, the mucube, ${\displaystyle \left\{4,6\mid 4\right\}}$.

In total, there are 3 possible configurations of planar polygons that work:

• Schläfli type {4,4} gives the mucube, ${\displaystyle \left\{4,6\mid 4\right\}}$
• Schläfli type {6,4} gives the muoctahedron, ${\displaystyle \left\{6,4\mid 4\right\}}$
• Schläfli type {6,6} gives the mutetrahedron, ${\displaystyle \left\{6,6\mid 3\right\}}$

These three regular polyhedra are called the Petrie-Coxeter polyhedra. The first two polyhedra were discovered by John Flinders Petrie and Coxeter found the final polyhedron shortly after.

Using finite skew polygons as faces gives three more apeirohedra:

• Schläfli type ${\displaystyle \left\{{\dfrac {6}{1,3}},6\right\}}$ gives the halved mucube, ${\displaystyle \left\{{\dfrac {6}{1,3}},6:{\dfrac {4}{1,2}}\right\}}$
• Schläfli type ${\displaystyle \left\{{\dfrac {4}{1,2}},6\right\}}$ gives the Petrial halved mucube, ${\displaystyle \left\{{\dfrac {4}{1,2}},6:{\dfrac {6}{1,3}}\right\}}$
• Schläfli type ${\displaystyle \left\{{\dfrac {6}{1,3}},4\right\}}$ gives the skewed Petrial muoctahedron, ${\displaystyle \left\{{\dfrac {6}{1,3}},4:{\dfrac {6}{1,3}}\right\}}$

#### Helical faces

In addition to the above there are 6 polyhedra with helical faces. Three of them are formed as Petrie duals of the pure apeirohedra with flat faces:

• ${\displaystyle \left\{{\dfrac {3}{0,1}},{\dfrac {6}{1,3}}\right\}_{4,4}}$ Petrial mucube
• ${\displaystyle \left\{{\dfrac {3}{0,1}},{\dfrac {4}{1,2}}\right\}_{6,4}}$ Petrial muoctahedron
• ${\displaystyle \left\{{\dfrac {3}{0,1}},{\dfrac {6}{1,3}}\right\}_{6,3}}$ Petrial mutetrahedron

There is one polyhedron obtained from skewing the muoctahedron that is self-Petrie:

• ${\displaystyle \left\{{\dfrac {3}{0,1}},4\right\}_{\cdot ,*3}}$ Skewed muoctahedron

And the remaining two are mutually Petrial:

Unlike the other regular polyhedra these last two cannot be expressed as a simple identification of points along holes and zigzags, so they are given ad-hoc Schläfli symbols.