S S symmetry

Article Regular and Irregular Chiral Polyhedra from Coxeter Diagrams via Quaternions

Nazife Ozdes Koca * and Mehmet Koca †

Department of Physics, College of Science, Sultan Qaboos University, P.O. Box 36, Al-Khoud 123, Muscat, Oman * Correspondence: [email protected]; Tel.: +968-2414-1445 † Retired professor: [email protected]

Received: 6 July 2017; Accepted: 2 August 2017; Published: 7 August 2017

Abstract: Vertices and symmetries of regular and irregular chiral polyhedra are represented by quaternions with the use of Coxeter graphs. A new technique is introduced to construct the chiral Archimedean solids, the snub cube and snub dodecahedron together with their dual Catalan solids, pentagonal icositetrahedron and pentagonal hexecontahedron. Starting with the proper subgroups of the Coxeter groups W(A1 ⊕ A1 ⊕ A1), W(A3), W(B3) and W(H3), we derive the orbits representing the respective solids, the regular and irregular forms of a tetrahedron, icosahedron, snub cube, and snub dodecahedron. Since the families of tetrahedra, icosahedra and their dual solids can be transformed to their mirror images by the proper rotational octahedral group, they are not considered as chiral solids. Regular structures are obtained from irregular solids depending on the choice of two parameters. We point out that the regular and irregular solids whose vertices are at the edge mid-points of the irregular icosahedron, irregular snub cube and irregular snub dodecahedron can be constructed.

Keywords: Coxeter diagrams; irregular chiral polyhedra; quaternions; snub cube; snub dodecahedron

1. Introduction In fundamental physics, chirality plays a very important role. A Weyl spinor describing a massless Dirac particle is either in a left-handed state or in a right-handed state. Such states cannot be transformed to each other by the proper Lorentz transformations. Chirality is a well-defined quantum number for massless particles. Coxeter groups and their orbits [1] derived from the Coxeter diagrams describe the molecular structures [2], viral symmetries [3,4], crystallographic and quasi crystallographic materials [5–7]. Chirality is a very interesting topic in molecular chemistry. Certain molecular structures are either left-oriented or right-oriented. In three-dimensional Euclidean space, chirality can be defined as follows: if a solid cannot be transformed to its mirror image by proper isometries (proper rotations, translations and their compositions), it is called a chiral object. For this reason, the chiral objects lack the plane and/or central inversion symmetries. In two earlier publications [8,9], we studied the symmetries of the Platonic–Archimedean solids and their dual solids, the Catalan solids, and constructed their vertices. Two Archimedean solids, the snub cube and snub dodecahedron as well as their duals are chiral polyhedral, whose symmetries are the proper rotational subgroups of the octahedral group and the icosahedral group, respectively. Non-regular, non-chiral polyhedra have been discussed earlier [10]. Chiral polytopes in general have been studied in the context of abstract combinatorial form [11–14]. The chiral Archimedean solids, snub cube, snub dodecahedron and their duals have been constructed by employing several other techniques [15,16], but it seems that the method in what follows has not been studied earlier in this context. We follow a systematic method for the construction of the chiral polyhedra. Let G be a rank-3 + Coxeter graph where W(G) represents the proper rotation subgroup of the Coxeter group W(G).

Symmetry 2017, 9, 148; doi:10.3390/sym9080148 www.mdpi.com/journal/symmetry Symmetry 2017, 9, 148 2 of 22

Symmetry 2017, 9, 148 2 of 22

For the snub cube and snub dodecahedron, the Coxeter graphs are the B3 and H3, respectively. , although they describe the achiral polyhedra such as the families of regular and irregular To describe the general technique, we first begin with simpler Coxeter diagrams A1 ⊕ A1 ⊕ A1 tetrahedron and icosahedron, respectively. We explicitly show that achiral polyhedra possess larger and A3, although they describe the achiral polyhedra such as the families of regular and irregular tetrahedronproper rotational and icosahedron, symmetries respectively.transforming We them explicitly to their show mirror that images. achiral We polyhedra organize possess the paper larger as properfollows. rotational In Section symmetries 2 we introduce transforming quaternions them toand their construct mirror the images. Coxeter We groups organize in theterms paper of 1 ⊕ 1 ⊕ 1 asquaternions follows. In[17]. Section We extend2 we introducethe group W quaternions (A A and A construct) to the octahedral the Coxeter group groups by the in symmetry terms of group (3) of the Coxeter diagram A1 ⊕ A1 ⊕ A1. In Section 3 we obtain the proper rotation quaternions [17]. We extend the group W (A1 ⊕ A1 ⊕ A1) to the octahedral group by the symmetry subgroup of the Coxeter group W (A1 ⊕ A1 ⊕ A1) and determine the vertices of an irregular group Sym(3) of the Coxeter diagram A1 ⊕ A1 ⊕ A1. In Section3 we obtain the proper rotation tetrahedron. In Section 4 we discuss a similar problem for the Coxeter-Dynkin diagram leading subgroup of the Coxeter group W (A1 ⊕ A1 ⊕ A1) and determine the vertices of an irregular tetrahedron. to an icosahedron and again prove that it can be transformed by the group W(B3)+ to its mirror image, In Section4 we discuss a similar problem for the Coxeter-Dynkin diagram A3 leading to an icosahedron which implies that neither the tetrahedron nor icosahedron are chiral+ solids. We focus on the irregular and again prove that it can be transformed by the group W(B3) to its mirror image, which implies icosahedra constructed either by the proper tetrahedral group or its extension pyritohedral group that neither the tetrahedron nor icosahedron are chiral solids. We focus on the irregular icosahedra and construct the related dual solids tetartoid and pyritohedron. We also construct the irregular constructed either by the proper tetrahedral group or its extension pyritohedral group and construct polyhedra taking the mid-points of edges of the irregular icosahedron as vertices. Section 5 deals with the related dual solids tetartoid and pyritohedron. We also construct the irregular polyhedra taking the the construction of irregular and regular snub cube and their dual solids from the proper rotational mid-points of edges of the irregular icosahedron as vertices. Section5 deals with the construction of octahedral symmetry W(B3)+ using the same technique employed in the preceding sections. The chiral irregular and regular snub cube and their dual solids from the proper rotational octahedral symmetry polyhedron+ taking the mid-points as vertices of the irregular snub cube is also discussed. In Section W(B3) using the same technique employed in the preceding sections. The chiral polyhedron taking 6 we repeat a similar technique for the constructions of irregular snub dodecahedra and their dual the mid-points as vertices of the irregular snub cube is also discussed. In Section6 we repeat a similar solids using the proper icosahedral group W(H3)+, which is isomorphic to the group of even technique for the constructions of irregular snub dodecahedra and their dual solids using the proper permutations of five letters+ (5). The chiral polyhedra whose vertices are the mid-points of the icosahedral group W(H3) , which is isomorphic to the group of even permutations of five letters Alt(5). edges of the irregular snub dodecahedron are constructed. Irregular polyhedra transform to regular The chiral polyhedra whose vertices are the mid-points of the edges of the irregular snub dodecahedron polyhedra when the parameter describing irregularity turns out to be the solution of certain cubic are constructed. Irregular polyhedra transform to regular polyhedra when the parameter describing equations. Section 7 involves the discussion of the technique for the construction of irregular chiral irregularity turns out to be the solution of certain cubic equations. Section7 involves the discussion of polyhedra. the technique for the construction of irregular chiral polyhedra.

2. Quaternionic Constructions of the Coxeter Groups Let = + , ( = 1, 2, 3) be a real unit quaternion with its conjugate defined by = − Let q = q0+ qiei,(i = 1, 2, 3) be a real unit quaternion with its conjugate defined by q = q0 − qiei, and, theand norm the normqq = qq== 1. The = 1 quaternionic. The quaternionic imaginary imaginary units satisfyunits satisfy the relations: the relations: =−δε + eei j ij ijk e k , (, , =1,2,3) (1) eiej = −δij + εijkek, (i, j, k = 1, 2, 3) (1) δ ε wherewhere δijijandand εijkijkare are the the Kronecker Kronecker and and Levi-Civita Levi-Civita symbols, symbols, and and summation summation overover thethe repeatedrepeated indicesindices isis understood.understood. The unit quaternions form a group isomorphic to thethe specialspecial unitaryunitary groupgroup SU((22). .Quaternions Quaternions generate generate the four-dimensional Euclidean space with the scalar product 1 1 ((p,, q )):≔= (pq++ qp))== (pq ++qp)) (2)(2) 2 2

The Coxeter diagram ⊕ ⊕ can be represented by its quaternionic roots as shown in The Coxeter√ diagram A1 ⊕ A1 ⊕ A1 can be represented by its quaternionic roots as shown in Figure1 1 where where 22 is is just just the the norm. norm.

Figure 1. The Coxeter diagram A1 ⊕ A1 ⊕ A1 with quaternionic simple roots. Figure 1. The Coxeter diagram ⊕ ⊕ with quaternionic simple roots.

The Cartan matrix and its inverse are given as follows:

    2 0 0 1 0 0 1 C =  0 2 0 , C−1 =  0 1 0  (3)   2   0 0 2 0 0 1

Symmetry 2017, 9, 148 3 of 22

The simple roots αi and the weight vectors ωi for a simply laced root system satisfy the scalar product [18,19]

 −1 (αi, αj = Cij, ωi, ωj = C , (αi, ωj) = δij, (i, j = 1, 2, 3). (4) ij

Note that one can express the roots in terms of the weights or vice versa:

 −1 αi = Cijωj, ωi = C αj (5) ij

If αi is an arbitrary quaternionic simple root and ri is the reflection generator with respect to the plane orthogonal to the simple root αi, then the reflection of an arbitrary quaternion Λ, an element in the weight space ω1, ω2, ω3 can be represented as [20]:  ∗ αi αi αi αi riΛ = − √ Λ √ := √ , − √ Λ. (6) 2 2 2 2

∗ It is then straight forward to show that riωj = ωj − δijαj. We will use the notations [p, q] Λ := pΛq and [p, q]Λ := pΛq for the rotoreflection (rotation and reflection combined) and the proper rotation, respectively, where p and q are arbitrary unit quaternions. If Λ is pure imaginary quaternion, ∗ then [p, q] Λ := pΛq = [p, −q]Λ. The Coxeter group W(A1 ⊕ A1 ⊕ A1) = < r1, r2, r3 > can be generated by three commutative group elements: ∗ ∗ ∗ r1 = [e1, −e1] , r2 = [e2, −e2] , r3 = [e3, −e3] . (7) 3 They generate the elementary abelian group W (A1 ⊕ A1 ⊕ A1) ≈ C2 × C2 × C2 := 2 of order 8. The scaled root system (±e1, ±e2, ±e3) represents the vertices of an octahedron and has more symmetries than the Coxeter group W (A1 ⊕ A1 ⊕ A1). The automorphism group of the root system can be obtained by extending the Coxeter group of order 8 by the Dynkin diagram symmetry Sym(3) of the Coxeter diagram in Figure1. This is obvious from the diagram where the generators of the symmetric group Sym(3) of order 6 can be chosen as:

 1 1  1 1 ∗ s = (1 + e1 + e2 + e3), (1 − e1 − e2 − e3) , d = [ √ (e1 − e2), − √ (e1 − e2)] , 2 2 2 2

s3 = d2 = 1, dsd = s−1. (8)

It is clear that the generators permute the quaternionic imaginary units as:

s : e1 → e2 → e3 → e1, d : e1 ↔ e2, e3 → e3 ,

−1 and satisfy the relations sris = ri+1, dr1d = r2, dr3d = r3 where the indices are considered modulo 3. It is clear that the Coxeter group < r1, r2, r3 > is invariant under the permutation group Sym(3) by conjugation so that the automorphism group of the root system (±e1, ±e2, ±e3) is an extension of 3 the group < r1, r2, r3 > by the group Sym(3) which is the octahedral group 2 : Sym(3) of order 48 (here: stands for the semi-direct product). It is clear from this notation that the Coxeter group 23 is an invariant subgroup of the octahedral group. We use a compact notation for the octahedral group in terms of quaternions as the union of subsets:

0 0 Oh = [T, ±T] ∪ [T , ±T ] (9) Symmetry 2017, 9, 148 4 of 22

={±1,±,± ,± , (±1 ± ± ± )}, Symmetry 2017, 9, 148 4 of 22 (10) ={ (±1 ± ), (± ± ), (±1 ± ), (± ± ), (±1 ± ), (± ± )}. √ √ √ √ √ √ where the sets of quaternions T and T0 are given by:

1 Here and ∪ Trepresent= {±1, ± thee1, ±binarye2, ±e 3tetrahedral, 2 (±1 ± e1 grou± e2p± ande3)} ,the binary octahedral group, (10) respectively.0 1 We have used1 a short-hand3
1 notation 1for the designation1 of the 1groups, e.g., [, ±] T = { √ (±1 ± e1), √ ±e2 ± e , √ (±1 ± e2), √ (±e3 ± e1), √ (±1 ± e3), √ (±e1 ± e2)}. means the set2 of all [, ±2]∈[,±] with2 ∈. The maximal2 subgroups2 of the octahedral2 group can be writtenHere T asand [21]:T ∪ T0 represent the binary tetrahedral group and the binary octahedral group, respectively. We have used a short-hand notation for the designation of the groups, e.g., [T, ±T] means Chiral octahedral group: =[,]∪[, ]≈() , the set of all [t, ±t] ∈ [T, ±T] with t ∈ T. The maximal subgroups of the octahedral group can be written as [21]: Tetrahedral group: =[,]∪[,−0], 0 + (11) Chiral octahedral group : O = [T, T] ∪ [T , T ] ≈ W(B3) , 0 0 Tetrahedral group : Td = [T, T] ∪ [T , −T ], (11) Pyritohedral group: =[,±]. Pyritohedral group : Th = [T, ± T]. The octahedral group, as we will see in what follows, can also be obtained as the () ≈ (The). octahedral group, as we will see in what follows, can also be obtained as the Aut(A3) ≈ W(B3). TheThe properproper rotationrotation subgroupsubgroup ofof thethe CoxeterCoxeter group W((A1 ⊕⊕A1 ⊕⊕A1)) is is thethe KleinKlein four-groupfour-group C2 ××C 2 represented represented by by the the elements: elements:

=[1,1], =[,−], =[,−], =[,−] (12) I = [1, 1], r1r2 = [e3, −e3], r2r3 = [e1, −e1], r3r1 = [e2, −e2] (12) We also note in passing that the Klein four-group in (12) is invariant by conjugation under the groupWe generated also note by in , passing with that=1 the. The Klein Klein four-group four-group in together (12) is invariant with the group by conjugation element under generate the 3 groupthe tetrahedral generated rotation by s, with grous p,= which1. The is Klein represented four-group by [, together ] in our with notation. the group element s generate the tetrahedral rotation group, which is represented by [T, T] in our notation. Next, we use the tetrahedral group ≈(). Its Coxeter-Dynkin diagram with its quaternionicNext, we roots use is the shown tetrahedral in Figure group 2. Td ≈ W(A3). Its Coxeter-Dynkin diagram A3 with its quaternionic roots is shown in Figure2.

Figure 2. The Coxeter diagram A3 with quaternionic simple roots. Figure 2. The Coxeter diagram with quaternionic simple roots.

TheThe Cartan Cartan matrix matrix of ofthe the Coxeter Coxeter diagram diagram andA3 itsand inverse its inverse matrix are matrix given are by giventhe respective by the respectivematrices: matrices:     2 −1 0 3 2 1 2−10 1 321   −1   C = =−1−1 2 2−1 −1,,C == 2422 4 2 (13)(13)   4   0 0−12−1 2 1231 2 3

TheThe generatorsgenerators ofof thethe CoxeterCoxeter groupgroupW ((A3))are are given given by: by: 1 1 ∗ ∗ =[ (1 +),− (1 +)] r1 √=2 [ √ (e1 + e2)√,2− √ (e1 + e2)] 2 2 1 1 ∗ =[ ( −),− ( −)] ∗ (14) 2 1 2 1 r2 √= [ √ (e3 − e2)√, − √ (e3 − e2)] (14) 2 2 1 1 ∗ ∗ =[ (1 −),− (1 −)] r3 √=2 [ √ (e2 − e1)√,2− √ (e2 − e1)] 2 2 =1 ( + +), =, = 1(− + +). ω = (e + e + e ), ω = e , ω = (−e + e + e ). 1 2 1 2 3 2 3 3 2 1 2 3 The generators in (14) generate the Coxeter group [22–24] which is isomorphic to the tetrahedral group Td of order 24. The automorphism group Aut(A3) ≈ W(A3) : C2 ≈ Oh where C2 is generated ∗ by the Dynkin diagram symmetry γ = [e1, −e1] , which exchanges the first and the third simple roots and leaves the second root intact in Figure2. Symmetry 2017, 9, 148 5 of 22 Symmetry 2017, 9, 148 5 of 22 The generators in (14) generate the Coxeter group [22–24] which is isomorphic to the tetrahedral groupThe generators of order 24.in (14) The generate automorphism the Coxeter group grou (p [22–24])≈ which(): is isomorphic≈ where to the is tetrahedral generated ∗ ( )≈( ): ≈ groupby the Dynkin of order diagram 24. The symmetry automorphism =[,− group] , which exchanges the first and where the third is simple generated roots Symmetry 2017, 9, 148 ∗ 5 of 22 byand the leaves Dynkin the diagram second root symmetry intact in=[ Figure,− 2. ] , which exchanges the first and the third simple roots and leavesThe Coxeter the second diagram root intact leading in Figure to the 2. octahedral group ()≈(4) × ≈ is shown ( )≈(4) × ≈ in FigureTheThe Coxeter 3. diagram B3 leading leading to to the the octahedraloctahedral groupgroup W(B3) ≈ Sym(4) × C2 ≈ Oh isis shownshown inin FigureFigure3 3..

Figure 3. The Coxeter diagram with quaternionic simple roots. Figure 3. The Coxeter diagram B3 with quaternionic simple roots. Figure 3. The Coxeter diagram with quaternionic simple roots.

The Cartan matrix of the Coxeter diagram and its inverse matrix are given by: TheThe CartanCartan matrixmatrix ofof thethe CoxeterCoxeter diagramdiagram B3 and and its its inverse inverse matrix matrix are are given given by: by: 2−10 222 = −12−10 2 −2, = 222244  (15) 2 −1 0 2 2 2 =−10−12 2 −2, = 1244123 (15) C =  −1 2 −2 , C−1 =  2 4 4  (15)  0−12 2 123  The generators below generate0 the−1 octahedral 2 group given1 2in 3(9) and the weight vectors are givenThe by: generators below generate the octahedral group given in (9) and the weight vectors are givenThe by: generators below generate the octahedral group given in (9) and the weight vectors are given by: 1 1 ∗ =[1 ( −),−1 ( −)] √2( ) √2( ) ∗ ∗ =[r = [√1 −(e −,−e ), − √1 −(e −]e )] 1√2 2 1 2√2 2 1 2 1 1 ∗ ∗ (16) =[r 1= [(√1 −(e −),−e )1, −(√1(−e −)e] )] (16)(16) 2 √2 2 2 3 √2 2 2 ∗3 =[ ( −),− ( −)] √2 √2 ∗ ∗r3 = [e3, −e3] =[,−] ∗ =[,−] =, = + , =1 ( + + ) ω = e ,ω = e + e , ω = (e + e + e ) 1 =1 , 2 =1 +2 , 3 = (1 +2 +3 ) 2 The group generated by the rotations and is isomorphic to the rotational octahedral The group generated by the rotations r1r2 and r2r3 is isomorphic to the rotational octahedral The group generated by the rotations and is isomorphic to the rotational octahedral group < , >≈ {,n ∪ ,h ].0 0 group << r 1r2,, r2r3>≈> ≈ {, T , ∪T ∪,T]., T ]}. groupThe Coxeter diagram leading to the icosahedral group is shown in Figure 4 with the The Coxeter diagram H3 leading to the icosahedral group is shown in Figure4 with the quaternionicThe Coxeter simple diagram roots: leading to the icosahedral group is shown in Figure 4 with the quaternionicquaternionic simplesimple roots:roots:

Figure 4. The Coxeter diagram of H3 with quaternionic simple roots. Figure 4. The Coxeter diagram of with quaternionic simple roots. √ √ Figure 4. The Coxeter diagram of with quaternionic simple roots. Here τ = 1+ 5 is the golden ratio and = 1− 5 . The Cartan matrix of the diagram H , its inverse 2√ 2√ 3 Here = is the golden ratio and = . The Cartan matrix of the diagram , its inverse and the weight vectors√ are given as follows: √ Here = is the golden ratio and = . The Cartan matrix of the diagram , its inverse and the weight vectors are given as follows:    2 3 3  and the weight vectors are given2 as− follows:τ 0 3τ 2τ τ 2−0 −1 1 3 2 =   =  3 2 2  C =−τ−2−02 2−1 −1,,C = 322τ 244τ 22τ  (17) 2 =0 −−1 2 2 −1, = τ3 2τ2 τ+ 2 (17) 0−12 2 42 +22 0−12 2 +2 √ =τ ( − ), =−√2 , = τ( − ) ω1 = √ √ (σe1 − τe3), ω2 = − 2τe3, ω3 = √√ (σe2 −e3) = 2 ( − ), =−√2 , = (2 − ) √ √ The quaternionic generators of the icosahedral group ≈ (5) × ≈() are given by: The quaternionic generators of the icosahedral group I ≈ Alt(5) × C ≈ W(H ) are given by: The quaternionic generators of the icosahedral group h ≈ (5) ×2 ≈()3 are given by: ∗ =[,−] , =∗ − ∗ =[,−r1] , [e1, e1] , (18) h i∗ ∗ 1 =[ ( + +1),− ( + +)] , (18) r2 = 2 (τe1+ e2 + σe3), − 2 (τe1+ e2 + σe3) , ∗ (18) =[ ( + +),− ( + +)] , ∗ r3 = [e2, −e2] ,

Symmetry 2017, 9, 148 6 of 22

∗ =[,−] , or shortly, () =,± where is the set of 120 quaternionic elements of the binary icosahedral group generated by the quaternions and ( + + ) [20]. The icosahedral rotation group is representedSymmetry 2017, by9, 148 the proper rotation subgroup () =, ≈ (5). All finite subgroups of the6 of 22 groups OO(3) and (4) in terms of quaternions can be found in the references [17,25].

A general vector in the dual space is represented by the vector Λ = + +. We will or shortly, W(H3) = I, ±I where I is the set of 120 quaternionic elements of the binary icosahedral use the notation Λ= + + ≔ (1), which are called the Dynkin indices in the Lie group generated by the quaternions e1 and 2 (τe1 + e2 + σe3) [20]. The icosahedral rotation group is algebraic representation theory [26]. We use the notation+  ()Λ: = () for the orbit of the represented by the proper rotation subgroup W(H3) = I, I ≈ Alt(5). All finite subgroups of the Coxeter group () generated from the vector Λ where the letter represents the Coxeter groups O(3) and O(4) in terms of quaternions can be found in the references [17,25]. diagram. A few examples could be useful to illuminate the situation by recalling the identifications A general vector in the dual space is represented by the vector Λ = a1ω1 + a2ω2 + a3ω3. We will [8]: use the notation Λ = a1ω1 + a2ω2 + a3ω3 : (a1a2a3), which are called the Dynkin indices in the Lie

algebraic representation(100)(tetrahedron), theory (111)[26]. We(truncated use the notation octahedron),W(G (100))Λ :=((octahedron),a1a2a3)G for the orbit of the Coxeter group W(G) generated from the vector Λ where the letter G represents the Coxeter diagram. (001) (cube), (100) (dodecahedron), (001) (icosahedron), and(010) , A few examples could be useful to illuminate the situation by recalling the identifications [8]: (icosidodecahedron). (100) (tetrahedron), (111) (truncated octahedron), (100) (octahedron), A3 A3 B3 3. The Orbit (001×) ((cube ),)( 100 as an) Irregular(dodecahedron Tetrahedron), (001) (icosahedron), and(010) , B3 H3 H3 H3

The proper rotation subgroup ×(icosidodecahedron of the Coxeter group). ( ⊕ ⊕) transforms a generic vector Λ as follows: 3. The Orbit C × C (a a a ) as an Irregular Tetrahedron 2 2 1 2 3 1 Λ= ( + +) The proper rotation subgroup2 C2 × C2 of the Coxeter group W(A1 ⊕ A1 ⊕ A1) transforms a generic vector Λ as follows: 1 Λ= (− − +), 2Λ = 1 (a e + a e + a e ) 2 1 1 2 2 3 3 (19) r r1Λ = 1 (−a e − a e + a e ), Λ=1 2 ( 2−1 1 −22), 3 3 2 (19) 1 r2r3Λ = 2 (a1e1 − a2e2 − a3e3), Λ= (− 1 + − ). r3r1Λ =2 (−a1e1 + a2e2 − a3e3).

TheseThese four four vectors vectors define define the the vertices vertices of ofan anirregular irregular tetrahedron tetrahedron with with four four identical identical scalene scalene p 2 2 p 2 2 p 2 2 trianglestriangles with with edge edge lengths lengths a+1 + a,2 , a2++ a,3 and, and a3++a. 1An. An irregular irregular tetrahedron tetrahedron with with a=1,1 = 1, a=2,2 = 2, a=33 = is3 isdepicted depicted inin Figure Figure 5.5 .

FigureFigure 5. An 5. An irregular irregular tetrahedron tetrahedron with with scalene scalene triangular triangular faces. faces.

Mid-points of the edges of an irregular tetrahedron are given by the vectors ±,±,± Mid-points of the edges of an irregular tetrahedron are given by the vectors ±a1e1, ±a2e2, ±a3e3 forming an irregular octahedron with eight identical scalene triangles ofp sides2 2p+2 , 2 forming an irregular octahedron with eight identical scalene triangles of sides a1 + a2 , a2 + a3 , p +2 , 2 + and the corresponding interior angles, say, , , . Four triangles with a3 + a1 and the corresponding interior angles, say, α, β, γ. Four triangles with identical face-angles identical face-angles surround the vertex . The remaining 4 triangles similarly meet at the α surround the vertex a1e1. The remaining 4 triangles similarly meet at the opposite vertex −a1e1. oppositeThis is vertex true for − every. This face-angle is true around for every every face-angle vertex. around every vertex. The mirror image of an irregular tetrahedron is obtained by applying any one of the reflection generators on these vectors, which lead to the vectors: Symmetry 2017, 9, 148 7 of 22

1 r1Λ = 2 (− a1e1 + a2e2 + a3e3), 1 r2Λ = 2 (a1e1 − a2e2 + a3e3), (20) 1 r3Λ = 2 (a1e1 + a2e2 − a3e3), 1 r1r2r3Λ = − 2 (a1e1 + a2e2 + a3e3). The vectors in (20) can also be obtained from those in (19) by quaternion conjugation. The eight vectors in (19) and (20) form a rectangular prism with edge lengths a1, a2 and a3 possessing the symmetry (C2 × C2) : C2 ≈ D2h of order 8. Assume now that we apply the one of the Dynkin diagram-symmetry operators d on vector Λ and assume that:

1 Λ = dΛ = (ae + ae + a e ). (21) 2 1 2 3 3 If we apply the rotation generators as in (19), we obtain an irregular tetrahedron with identical four isosceles triangles. The mirror copy of these vectors can be obtained similar to the procedure in (20) and when all are combined, we obtain a square prism with edge lengths a and a3. A more symmetric case is obtained by assuming:

1 Λ = sΛ = a(e + e + e ). (22) 2 1 2 3

In this case, the symmetry is the√ chiral tetrahedral group, and the vertices in (19) will represent a regular tetrahedron of edge length 2a which is also invariant under the larger tetrahedral symmetry 0 0 Td = [T, T] ⊕ [T , −T ], as expected. A mirror image of the regular tetrahedron is obtained either by ◦ reflections as described by (20) or by a rotation of 180 around the e1 − e2 axis, which can be obtained by the group element:   1 1 0 0 √ (e1 − e2), − √ (e1 − e2) ∈ [T , T ]. (23) 2 2 A tetrahedron is not a chiral solid since it can be converted to its mirror image by a rotation such as the one in (23). A regular tetrahedron with its mirror image constitutes a cube which has the full octahedral symmetry of order 48.

+ 4. The Regular and Irregular Icosahedron Derived from the Orbit W(A3) (a1a2a3) + The tetrahedral rotational subgroup W(A3) of the Coxeter group W(A3) is the tetrahedral group of order 12 isomorphic to even permutations of four letters, Alt(4), which can be generated 3 3 2 by the generators a = r1r2 and b = r2r3 satisfying the generation relations a = b = (ab) = 1. Let Λ = (a1a2a3) be a general vector in the weight space of A3. The following sets of vertices form two equilateral triangles. (Λ,r1r2Λ, r2r1Λ) and (Λ,r2r3Λ, r3r2Λ) (24) q q 2 2
2 2
with respective edge lengths 2 a1 + a1a2 + a2 and 2 a2 + a2a3 + a3 . Three more triangles can be obtained by joining the vertex r1r3Λ = r3r1Λ to the vertices Λ, r1r2Λ and r3r2Λ and r2r1Λ to r2r3Λ. The new edges are of the following lengths. q 2 2
|r1r2Λ − r1r3Λ| = 2 a2 + a2a3 + a3 , q 2 2
|r3r2Λ − r1r3Λ| = 2 a1 + a1a2 + a2 , (25) q 2 2
|Λ − r1r3Λ| = |r2r1Λ − r2r3Λ| = 2 a1 + a3

The vertices joined to vector Λ are illustrated in Figure6. Symmetry 2017, 9, 148 8 of 22

|Λ−Λ| = |Λ−Λ| = 2( +)

Symmetry 2017, 9, 148 8 of 22 The vertices joined to vector Λ are illustrated in Figure 6.

Λ r1r3

r r Λ 4 5 1 2 Λ Λ r3r2 3 1 2 Λ r2r1 r r Λ 2 3 Figure 6. The vertices connected to the general vertex Λ (note that all vertices are not in the same plane). Figure 6. The vertices connected to the general vertex Λ (note that all vertices are not in the same plane).

2 a1 a3 Factoring by an overall overall factor factor a2 6=≠0 0 and defining defining the the parameters parameters =x = a and y== a wewe obtain 2 2 three classes of triangles. After dropping the overall factor a2 we obtain, three classes of triangles. After dropping the overall factor we obtain, p equilateralequilateral triangle with edge length: length :A==2(12( +1 + +x + x)2, ), p equilateralequilateral triangle triangle with with edge edge length: length :=B =2(12( +1 + +y + y)2, ), p 2 2 3 scalene triangles with edge lengthslengths: : A,,=, B, C =2(2(x++)y ) as shown in Figure 6. The sum of the face-angles at the vertex Λ (at every vertex indeed) is 2π + as shown in Figure6. The sum of the face-angles at the vertex Λ (at every vertex indeed) is 3 +   (+ + + + ) = = 5whereπ , , and are the interior angles of the scalene triangles. The angular θ ϑ ϕ 3 where θ, ϑ, and ϕ are the interior angles of the scalene triangles. The angular deficiencydeficiency δ=2−= 2π − 5π == π isis the the same same as as in in the the regular regular icosahedron. Using Descartes’ formula, the 3 3 number of vertices cancan alsoalso bebe obtainedobtained asas inin [[27].27]. = =124π . (26) N = = 12. (26) 0 δ The vector Λ can be written in terms of quaternions as Λ=α + + where the new parametersThe vector are definedΛ can beby written in terms of quaternions as Λ = αe1 + βe2 + γe3 where the new parameters are defined by x−− y x++ y αα== , ,β = = ,,γ =+1.= β + 1. (27) (27) 22 2 ( )⁄ ≈,  Λ The orbit of the chiral tetrahedraltetrahedral groupgroupW (A3)/C2 ≈ T, T generatedgenerated from from the the vector vector Λ can be written as:as:

 ,  Λ = {± α ± ±, ± ± ±α,± ±α ±} T, T Λ = {± αe1 ± βe2 ± γe3, ± βe1 ± γe2 ± αe3, ± γe1 ± αe2 ± βe3} (28) (even number of (−) signs). (28) (even number of (−) signs). The quaternions in (28) with an odd number of (−) signs constitute the mirror image. A rotation element,The quaternionse.g., the one inin (28)(23), with tran ansforms odd numberthe quaternions of (−) signs in (28) constitute to theirthe mirror mirror images. image. This A rotation proves element,that the set e.g., in (28) the onedoes in not (23), represent transforms a chiral the polyhedron. quaternions inBefore (28) we to their discuss mirror the irregular images. Thisicosahedral proves thatstructures, the set inwe (28) point does out not that represent for >0,=0 a chiral polyhedron. or > 0, Before = 0, we the discuss set of the 12 irregularvectors describes icosahedral a structures,truncated tetrahedron. we point out For that ==0, for x > 0, y it= describes0 or y > an0, xoctahedron.= 0, the set More of 12 vectorsinteresting describes cases aarise truncated as we tetrahedron.will discuss below. For x = y = 0, it describes an octahedron. More interesting cases arise as we will discuss below. (1) ==, 1 + + =1++ = + 2 2 2 2 (1) Here,A = B all= theC, 1edges+ x + arex =equal1 + yleading+ y = tox the+ ysolution == −1⟹ −−1=0, ⟹= , or = . Substituting the first solution for = which results in α=0,=,= , the set of Here, all the edges are equal leading to the solution x = y = x2 − 1 ⇒ x2 − x − 1 = 0, ⇒ vectors in (28) can be written as: x = τ, or x = σ. Substituting the first solution for = τ which results in α = 0, β = τ, γ = τ2, the set of vectors in (28) can be written as:   T, T Λ = τ{± e1 ± τe2, ±e2 ± τe3, ±e3 ± τe1}. (29) Symmetry 2017, 9, 148 9 of 22

These vertices, which are also invariant under the pyritohedral symmetry, represent a regular icosahedron. Note that the vector Λ = τ(e2 + τe3) is invariant under the 5-fold rotation by 1 1 s = [ 2 (σ + e2 + τe3), 2 (σ − e2 − τe3)] while the other vectors in (29) are transformed to each other. This proves that the pyritohedral group T, ±T  can be extended to the icosahedral group I, ±I  by the generators so that the set of vertices in (29) is invariant under the icosahedral group of order 120. We emphasize that although (29) has a larger symmetry of the icosahedral group, it is obtained from its chiral tetrahedral subgroup. Its mirror image can be obtained by a rotation of 180◦ around the vector e1 − e2 implying that icosahedron is not a chiral solid. There is another trivial solution for A = B = C where x = −1, y = 0. The number of vertices in (28) reduces to 4 representing a regular tetrahedron.   1 p 2 (2) A = B 6= C, y = 2 −1 ± 1 + 4(x + x)

For various values of x and the corresponding y, one obtains an irregular icosahedron with (4 + 4) equilateral triangles and 12 isosceles triangles. An interesting case would be x = y = −τ with 12 vertices of {±τ e1 ± σe2, ±τe2 ± σe3, ±τe3 ± σe1} which represent an irregular icosahedron with (4 + 4) equilateral triangles with edges of length 2 and 12 isosceles triangles of sides 2, 2τ, 2τ (Robinson triangles). Any irregular icosahedron with x = y 6= 0 has a pyritohedral symmetry of order 24 as pointed out in (11). Another solution of the quadratic equation x = −τ, y = −σ leads to an irregular icosahedron√ of (4 + 4) equilateral triangles with edge lengths 2 and 12 isosceles triangles of edge lengths 2, 2, 6.

(3) A = C 6= B

This is another case with 12 isosceles triangles but here the 4 sets of equilateral triangles are not equal to the other set of 4 equilateral triangles. We have x = y2 − 1, x 6= τ or σ and x 6= 0, y 6= 0. For the values y = −τ, x = τ, the irregular icosahedron consists of 4 equilateral triangles with edge length 2, 4 equilateral triangles with edge length 2τ and 12 Robinson triangles with edge lengths 2, 2τ, 2τ as discussed in (2).

(4) A 6= C = B

Here, y = x2 − 1, x 6= τ or σ. For x = −τ, y = τ, the faces of the irregular icosahedron are identical to the one in case (3).

(5) A 6= C 6= B

2 2 Now we have x 6= y − 1 and y 6= x − 1. Taking√ x = 1, y = 2 the irregular icosahedron√ will consist of 4 equilateral triangles with√ edge√ of length√ 6, 4 equilateral triangles of edge length 14, and 12 scalene triangles with edges 6, 10 and 14.

4.1. Dual of an Irregular Icosahedron Now we discuss the construction of a dual of an irregular icosahedron. A dual of an irregular polyhedron can be obtained by determining the vectors orthogonal to its faces. Referring to Figure6, vectors orthogonal to the equilateral faces #1 and #3 can be taken as b1 := ω1 and b3 := ω3 as they are invariant under the rotations r2r3 and r1r2, respectively. The vectors orthogonal to the faces #2, #4 and #5 can be determined as: b2 = n1e1 + n2e2 + n3e3,

b4 = r1r2b2 = −n3e1 − n1e2 + n2, (30)

b5 = r3r2b2 = n3e1 + n1e2 + n2e3, where n1 = y − x, n2 = x + y + 2xy, n3 = x + y. Symmetry 2017, 9, 148 10 of 22

These vectors should be rescaled in order to determine the plane orthogonal to the vector Λ. Let Symmetry 2017, 9, 148 10 of 22 us redefine the vectors ≔, =, =, =, = . The scale factors can be determined as: These vectors should be rescaled2( in + order + 2)( to determine+ + 2) the plane orthogonal to the vector Λ. = Let us redefine the vectors d1 := λω1, d2 = 3++2b2, d3 = $ b3, d4 = b4, d5 = b5. The scale factors can be determined as: (31) 2(2 +(x + +y 2)(+ 2)( +x + +y 2)+ 2xy) =λ = . ( +3 3x + +y 2)+ 2 (31) The dual solid is an irregular dodeca2(xhedron+ y + 2with)(x +they +sets2xy of) vertices: $ = . (x + 3y + 2) , = (± ± ±),(even number of (−) signs) The dual solid is an2 irregular dodecahedron with the sets of vertices: , = (± =±λ ±±±), (odd± number of (−) signs) − T,T d1 2 ( e1 e2 e3), (even number of ( ) signs) (32) $   = (± ± ± ) ( (−) ) , ={±T, T d3 ±2 e±1 e2,±e3, ±odd number ± of,±signs± ±} (32)   (−) signs). T, T d2 = {±n1e1 ± n2e(even2 ± n3 numbere3, ±n2e 1of± n3e2 ± n1e3 , ±n3e1 ± n1e2 ± n2e3} (even number of (−) signs). We will not discuss in detail how an irregular dodecahedron varies with five different cases discussedWe will above. not discussA few examples in detail howwould an be irregular sufficient dodecahedron in the order variesof increasing with five symmetry different toward cases discussedregularity. above. The case A few (5) examplesabove corresponds would be sufficientto the invariance in the order under of increasingthe chiral symmetrytetrahedral toward group. regularity.Substituting The =1,=2 case (5) above in (32), corresponds we obtain to the an invariance irregular underdodecahedron the chiral tetrahedralcalled a tetartoid group. Substitutingcorrespondingx = to1, they = mineral2 in (32), cobaltite. we obtain The an vertices irregular dodecahedron,,, and called form a tetartoid an irregular corresponding pentagon towith the mineralthree different cobaltite. edge The lengths. vertices Thed1, d 2tetartoid, d3, d4 and withd5 formits dual an irregularirregular pentagon icosahedron with are three depicted different in edgeFigure lengths. 7. The tetartoid with its dual irregular icosahedron are depicted in Figure7.

(a) (b)

FigureFigure 7. 7.(a )(a A) faceA face transitive transitive irregular irregular dodecahedron dodecahedron (tetartoid) (tetartoid) under rotational under tetrahedralrotational tetrahedral symmetry; (bsymmetry;) the dual ( irregularb) the dual icosahedron. irregular icosahedron.

Since under the pyritohedral symmetry =, the vertices in (32) take a simpler form where: Since under the pyritohedral symmetry x = y, the vertices in (32) take a simpler form where: 4( + 1) == , =0, =2(+1), =2., (33) 4x(2(x + +1 1))2 λ = $ = , n = 0, n = 2x(x + 1), n = 2x. , (33) (2x + 1) 1 2 3 The edge lengths of the irregular pentagon take the values: The edge lengths of the irregular pentagon take the values: | −| = | −| = | −| = | −|

|d21 − d2| = |d2 − d3| = |d3 − d4| = |d1 − d5| (34) = (3 +6 +7 +4+1) (2 + 1) 2x p = (3x4 + 6x3 + 7x2 + 4x + 1) (34) (2x + 1) | −| =4||. |d4 − d5| = 4|x|. The irregular dodecahedron with the vertices of (32) with and from (33) is called the pyritohedron.The irregular They dodecahedron can also be rearranged with the an verticesd the set of can (32) be with givenλ inand its standard$ from (33)form: is called the pyritohedron. They can also be rearranged(± and± the± set), can be given in its standard form: (35) {±(1−ℎ ) ± (1+ℎ)},

Symmetry 2017, 9, 148 11 of 22

(±e1 ± e2 ± e3),  2
± 1 − h e1 ± (1 + h)e2 , (35) Symmetry 2017, 9, 148  2
11 of 22 ± 1 − h e2 ± (1 + h)e3 ,

 2
{±(1−ℎ± 1 −)h± (e1+ℎ3 ± ()1+}, h)e1 , x {±(1−ℎ ) ± (1+ℎ)}, where h = . where ℎ = . x +1 5 A specialA special case casex ===5y = 5 discussed discussedin in (2)(2) corresponding to to ℎ=h = in (35)in (35) is plotted is plotted in Figure in Figure 8. 8. 6

Figure 8. The pyritohedron. Figure 8. The pyritohedron. In the limit of either x = τ or x = σ, the dodecahedron is regular and the pentagon turns out In the limit of either = or =, the dodecahedron is regular and the pentagon turns out to 2 to bebe regular regular withwith all edges edges equal equal either either 44 τoror −4−4. σAfter. After rescaling rescaling by 4 by, 4theτ ,set the of set vertices of vertices of the of the dodecahedrondodecahedron with withx ==τ are: are: (1± ± ± ), {±1 ± ,± ± , ± ± }. (36) (± e ± e ± e), {± τe ± σe , ±τe ± σe , ±τe ± σe }. (36) 2 1 2 3 2 1 2 2 3 3 1 Each set above is invariant under the pyritohedral group, ± . The 20 vertices of (36) form a dodecahedronEach set above with is regular invariant faces. under As such the, they pyritohedral possess a larger group icosahedral[T, ±T]. Thesymmetry 20 vertices , ± of. (36) form a dodecahedronThe first 8 with vertices regular of (36) faces. represent As such, a cube they and possess the second a larger set of icosahedral 12 vertices, symmetryas we recall[ Ifrom, ±I ]. previousThe first discussions, 8 vertices represents of (36) representan irregular a icosahedron cube and thewith second (4+4) set equilateral of 12 vertices, triangles asand we 12 recall fromRobinson previous triangles. discussions, The dual represents of the irregular an irregularicosahedro icosahedronn in second set with of 12 vertices(4 + 4) inequilateral (36) is another triangles irregular dodecahedron whose vertices are the union of a cube and a regular icosahedron albeit with and 12 Robinson triangles. The dual of the irregular icosahedron in second set of 12 vertices in (36) different magnitudes of vectors. is another irregular dodecahedron whose vertices are the union of a cube and a regular icosahedron albeit4.2. with Regular different and Irregular magnitudes Icosidodecahedron of vectors.

4.2. RegularIt is andwell Irregular known that Icosidodecahedron mid-points of the edges of a regular icosahedron or dodecahedron form the Archimedean solid icosidodecahedron with 30 vertices 32 faces (12 pentagons + 20 triangles) and It is well known that mid-points of the edges of a regular icosahedron or dodecahedron form 60 edges which can be obtained from the Coxeter graph of as an orbit (010) [28]. An irregular the Archimedeanicosidodecahedron solid consists icosidodecahedron of irregular pentagonal with 30 verticesfaces and 32 scalene faces (12triangles pentagons in the +most 20 triangles) general and 60 edgescase and which will canbe derived be obtained from fromthe chiral the Coxetertetrahedral graph group of andH aswill an be orbit extended(010) by [pyritohedral28]. An irregular 3 H3 icosidodecahedrongroup representing consists a larger of irregularsymmetry. pentagonal One can de facesfine andfive scalenevectors trianglesas follows in representing the most general the case andvertices will be of derived the irregular from pentagon the chiral which tetrahedral is orthogonal group to andthe vertex will beΛ: extended by pyritohedral group representing aΛ+ largerΛ symmetry.Λ+ One canΛ define fiveη(Λ+ vectorsΛ) as followsη(Λ+ representingΛ) the vertices of the = , = , = , = , irregular pentagon2 which is orthogonal2 to the vertex Λ2: 2

η(Λ+Λ) κ(Λ + Λ) Λ + r r Λ = Λ + r r ,Λ = , 1 2 2 2 1 η2(Λ+r2r3Λ) η(Λ+r3r2Λ) c1 = , c2 = , c3 = 2 , c4 = 2 , 2 2 (37) +3 +2+2+4+2 η(Λ + r3r2Λ) κ(Λ + r1r3Λ) c =η= , c = , , 4 +32 +2+4+2+25 2 2 2 (37) x + 3y + 2xy + 2x + 4y + 2 η = +3 +2+2+4+2 κ= 2 2 . , y + 3x(++2)+ 2xy +4x + 2y + 2 x2 + 3y2 + 2xy + 2x + 4y + 2 In terms of and the verticesκ = read: 2 . 1 (x + y + 2) = [−(+1) + + (++1) ], (38) 2

Symmetry 2017, 9, 148 12 of 22

SymmetryIn 2017 terms, 9, 148 of x and y the vertices read: 12 of 22

1 1 = [−( + ) + + ( + + ) ] = c[−1 + (++1y 1 e1 )ye+2 (+1x )y ], 1 e3 , 2 2 1 c = [−ye + (x + y + 1)e + (y + 1)e ], 2 2 1 2 3 = [ + (++1) + (+1)], 2 η c = [xe + (x + y + 1)e + (x + 1)e ], (38) 3 2 1 2 3 = [( + 1) + + (++1) ], 2 η c = [(x + 1)e + xe + (x + y + 1)e ], 4 2 1 2 3 =κ(+1). c5 = κ(β + 1)e3. Here, with , with and define three orbits under the chiral tetrahedral symmetry =) of sizesHere, 12, 12 cand1 with 6, respectively.c2, c3 with c 4Impoandsingc5 define the pyritohedral three orbits group under invariance the chiral ( tetrahedral and symmetrymoreover, of lettingsizes = 12, 12 and and 6,dividing respectively. each vector Imposing by a the scale pyritohedral factor 2 group, one obtains invariance the (xusual= y) quaternionicand moreover, verticesletting ofx the= τ icosidodecahedronand dividing each vector[28] by a scale factor 2τ2, one obtains the usual quaternionic vertices of the icosidodecahedron1 [28] 1 1 (± ± ± ), (± ± ± ), (± ± ± ), 2 2 2 n 1 1 1 o (39) 2 (±e1 ± σe2 ± τe3), 2 (±σe1 ± τe2 ± e3), 2 (±τe1 ± e2 ± σe3) , ±,±,± (39) ±e , ±e , ±e They constitute a subset of the quaternionic 1binary2 icosahedral3 group . An icosidodecahedron consistsThey of regular constitute pentagons a subset and of theequilateral quaternionic triangles binary as icosahedralshown in Figure group 9a.I. AnA general icosidodecahedron irregular icosidodecahedronconsists of regular consists pentagons of 12 andpentagons equilateral of edges triangles , , as as shown shown inin FigureFigure9 9b.a. AIn generaladdition, irregular it has 4 equilateralicosidodecahedron triangles consists with edges of 12 pentagons, 4 equilateral of edges trianglesa, b, withc as shown edges in, Figure12 scalene9b. In triangles addition, with it has edges4 equilateral , , where triangles , , with can edges be expresseda, 4 equilateral in terms triangles of and with. edges b, 12 scalene triangles with edges a, b, c where a, b, c can be expressed in terms of x and y.

(a) (b)

FigureFigure 9. (a 9.) Regular(a) Regular icosadodecahedron(icosadodecahedron =( x = )y ;= (bτ) )irregular; (b) irregular icosidodecahedron icosidodecahedron ( = 1, (x == 2)1,. y = 2).

5. The Regular and Irregular Snub Cubes Derived from () (+ ) 5. The Regular and Irregular Snub Cubes Derived from W(B3) (a1a2a3) TheThe snub snub cube cube is a is chiral a chiral Archimedean Archimedean solid solid with with 24 24vertices, vertices, 60 60edges edges and and 38 38faces faces (8 squares, (8 squares, 6+24 6 + 24 equilateralequilateral triangles). triangles). Its verticesvertices andand its its dual dual can can be determinedbe determined by employingby employing the same the methodsame method described in Sections 3 and 4. The proper rotational subgroup of the Coxeter group () described in Sections3 and4. The proper rotational subgroup of the Coxeter group W(B3) is the is the rotational octahedral group (4)n ≈ {, ∪ [o,]}, of order 24, which permutes the rotational octahedral group Sym(4) ≈ T, T  ∪ [T0, T0] , of order 24, which permutes the diagonals diagonals of a cube [21]. The group is generated by two rotation generators = and = of a cube [21]. The group is generated by two rotation generators a = r r and b = r r satisfying the satisfying the generation relations = =() =1. When Λ is taken3 2 as a general2 1 vector, the generation relations a4 = b3 = (ab)2 = 1. When Λ is taken as a general vector, the following sets of following sets of vertices form an equilateral triangle and a square, respectively, vertices form an equilateral triangle and a square, respectively, (Λ, Λ, () Λ), (Λ, Λ, () Λ, Λ), (40) 2 2 (Λ, r2r1Λ, (r2r1) Λ), (Λ, r3r2Λ, (r3r2) Λ, r2r3Λ), (40) with respective edge lengths 2( + +) and 2( + + ). With the vertex Λ= q r 2 2
 2 1 2 withΛ, we respective obtain a edge figure lengths consisting2 a1 of+ a71 verticesa2 + a2 andas shown2 a 2in+ Figurea2a3 + 10.2 a3 . With the vertex r1r3Λ = r3r1Λ, we obtain a figure consisting of 7 vertices as shown in Figure 10.

Symmetry 2017, 9, 148 13 of 22 Symmetry 2017, 9, 148 13 of 22 Λ Λ r1r3 r3r2 5 4 2Λ r r Λ Λ 1 (r3r2) 1 2 3 2 r r Λ r r Λ 2 1 2 3 FigureFigure 10. 10. TheThe vertices vertices connected connected to to the the vertex vertex ΛΛ..

ThereThere are 3 different edge lengths among 11 edges which are given as:

q |Λ−Λ| = 2(2+ + 12), |r1r2Λ − r1r3Λ| = 2(a2 + a2a3 +2 a3), q | − | = (2 + + 2) (41) |r3r2ΛΛ−r1r3Λ| = 2(2 a1+a1a2+a2) , , (41) q 2 1 2 |Λ − r1r3Λ| = |r2r1Λ − r2r3Λ| = 2(a1 + 2 a3) 1 |Λ−Λ| = |Λ−Λ| = 2( + ) Factoring by a 6= 0 redefining x = a1 , y = a3 and α = x +2y + 1, β = y + 1, γ = y and 2 a2 a2 2 2 2 dropping a2, one obtains the following classes of faces of the irregular snub cube: Factoring by ≠ 0 redefining = , = and = + +1,= +1,= and p 6 equilateral triangles with edge length: A = 2(1 + x + x2), dropping , one obtains the followingp classes of faces of the irregular snub cube: 8 squares with edge length: B = 2(1 + y + y2), 6 equilateral triangles with edge length: =2(1r +  + ),  A B C = x2 + 1 y2 824 squares scalene with triangles edge withlength: edge = lengths:2(1++, , ), 2 2 . 24 scalene triangles with edge lengths: ,π , = 2( + ). Since the angular deficiency is δ = 6 , the number of vertices of an irregular snub cube is 24. The vertex Λ = a1ω1 + a2ω2 + a3ω3 can be written in terms of quaternions as Λ = αe1 + βe2 + γe3. Since the angular deficiency is = , the number of vertices of an irregular snub cube is 24. The orbit generated by the chiral octahedral group reads: The vertex Λ= + + can be written in terms of quaternions as Λ = α + + + γ. The orbit generatedW(B3) Λ by= {the±α chirale1 ± β octahedrale2 ± γe3, ±groupβe1 ± reads:γe2 ± αe3, ±γe1 ± αe2 ± βe3}, (42) + W((B3))ΛΛ={± α0 = {±βe1 ±± αe2 ±± γe3,, ±±αe1±± γe2±α± βe3,±, ±γe1±α± βe2±± αe3},} , 42 where Λ0 = r1(Λ is) theΛ′ mirror = {± image ±α of±Λ., The ±α vertices ± ± of (42),± represent ± (1)±α an}, octahedron for a1 = 1, a2 = a3 = 0; (2) truncated octahedron for x = 1, y = 0; (3) cube for a1 = a2 = 0, a3 = 1 and (4) where Λ′ = Λ is the mirror image of Λ. The vertices of (42) represent (1) an octahedron for = cuboctahedron for a1 = a3 = 0, a2 = 1. 1, = =0; (2) truncated octahedron for =1,=0; (3) cube for = =0, =1 and (4) The irregular chiral convex solid will have 24 vertices 38 faces (8 square, 6 equilateral triangles cuboctahedron for = =0, =1. and 24 scalene triangles) and 60 edges (24 of length A, 24 of length B, 12 of length C). In what follows The irregular chiral convex solid will have 24 vertices 38 faces (8 square, 6 equilateral triangles we classify them according to their edge lengths of triangles and squares. and 24 scalene triangles) and 60 edges (24 of length , 24 of length , 12 of length ). In what follows we classify them according to their edge lengths2 of triangles2 and2 squares.1 2 (1) The snub cube: A = B = C, 1 + x + x = 1 + y + y = x + 2 y (1) The snub cube: == ,1++ =1++ = + 2 When all edges are equal, one eliminates the variable satisfying y = x − 1 and obtains the cubic equation x3 − x2 − x − 1 = 0 which can also be written as x + x−3 = 2 The solution is When all edges are equal, one eliminates the variable satisfying = −1 and obtains the cubic Fn+1 the tribonacci constant x = lim ≈ 1.8393 where Fn is the n-th term in the Tribonacci series equation − − −1 =n 0→ ∞whichFn can also be written as + =2 The solution is the 0, 0, 1, 1, 2, 4, 7, 13, 24, 44, . . .. tribonacci constant = lim ≈ 1.8393 where is the n-th term in the Tribonacci series After dropping an overall→ factor 1 (x2+1), vertices of the snub cube and its mirror image are given by: 0, 0, 1, 1, 2, 4, 7, 13, 24, 44, …. 2 After dropping +an overall factor (−+1),1 vertices− of1 the snub cube−1 and its mirror image are W(B3) Λ = ±xe1 ± e2 ± x e3, ±e1 ± x e2 ± xe3, ±x e1 ± xe2 ± e3 (43) given by: +  −1 −1 −1 W(B3) Λ0 = ±e1 ± xe2 ± x e3, ±xe1 ± x e2 ± e3, ±x e1 ± e2 ± xe3 . () Λ={± ± ± ,± ± ±,± ± ±} (43)

Symmetry 2017, 9, 148 14 of 22 Symmetry 2017, 9, 148 14 of 22 () Λ′ = {± ± ± , ± ± ±,± ± ±}. Symmetry 2017, 9, 148 14 of 22 ( )Λ′ = {± ± ± , ± ± ± ,± ± ± }. For = 1.8393, the snub cube is shown in Figure 11. For = 1.8393, the snub cube is shown in Figure 11. For x = 1.8393, the snub cube is shown in Figure 11.

FigureFigure 11. 11.TheThe snub snub cube. cube. Figure 11. The snub cube. (2) =≠ , = −1 ± (1p + 2( +) (2) A = B 6= C, y = −1 ± (1 + 2(x2 + x) (2) =≠ , = −1 ± (1 + 2( +) For =1,=−2, the irregular snub cube consists of 6 equilateral triangles and 8 squares of ForFor6 x =1,=−2,= 1, y = −2τ , thethe irregular irregular snub snub cube cube consists consists6 of, of6 ,equilateral 62(2 equilateral + 3) triangles triangles and and 8 squares 8 squares of of edges √√ respectively and 24 isosceles triangles of edges √ √√ √ p as shown in Figure 12. edgesedges √66respectively respectively andand 24 isosceles triangles triangles of of edges edges √6,6√,6, 62(2, 2 +(2 3)τ +as3 shown) as shown in Figure in Figure 12. 12.

Figure 12. The irregular snub cube with squares, equilateral triangles, isosceles triangles and squares Figure 12. The irregular snub cube with squares, equilateral triangles, isosceles triangles and squares (for x = 1, y = −2τ ). (forFigure =1,=−2) 12. The irregular. snub cube with squares, equilateral triangles, isosceles triangles and squares (for =1,=−2). 1 2 3 2 (3)(3) =≠A = C 6=, B=, x = 2 y− 1− and1 and x3 − xx2 − xx − 1− ≠x 0 − 1 6= 0 (3) =≠, = − 1 and x3 − x2 − x − 1 ≠ 0 ForFor =4 and =7y = 4 and√ x = 7correspondingcorresponding to toα == 10,10, =√β 3,= 3, =γ 2, =the2, irregularthe irregular snub snubcube cubeconsists consists√ of √ of squaressquares√For of of edges=4 and =7 edges √1313, equilateral, equilateral corresponding√ triangles triangles ofto ofedges edges= 10, √57 = and57 3,and isosceles = isosceles2, the trianglesirregular triangles withsnub with edges cube edges √consists57, √57 ,of 57 andandsquares √1313;; allof all edgesedges edges √scaled scaled13, equilateral by by √2. 2. triangles of edges √57 and isosceles triangles with edges √57, √57

and √13; all edges scaled by √2. (4)(4) ≠=,A 6= B = Cy=, y =−1 andx2 − 1 andx3 − xx32 − xx −2 1− ≠ x0 − 1 6= 00 (4) ≠=, y= −1 and x3 − x2 − x − 1 ≠ 0 For =3,=8 the irregular snub cube after rescaling by √2 has√ equilateral triangles of sides For x = 3, y = 8 the irregular snub cube after rescaling by 2 has equilateral triangles of sides √√13, squaresFor =3,=8 of sides √ 41√ the, andirregular isosceles snub triangles cube after of edges rescaling √41 √,by√41 √2, √ has13 as equilateral√ faces. triangles of sides 13, squares of sides 41, and isosceles triangles of edges 41, 41, 13 as faces. √13, squares of sides √41, and isosceles triangles of edges √41, √41, √13 as faces. (5) ≠≠ A 6= C 6= B (5)(5) ≠≠ The variables should satisfy the inequalities ≠ −1 and ≠2(+1). 2 2 ForTheThe variables =variables 2 and =4 shouldshould, the satisfy irregular the the inequalities inequalitiessnub cube consists ≠y 6= x −1of− scaled 1and and equilateraly ≠26= 2((+1x + triangles1).) . of sides √7, √ squaresSymmetryForFor ofx 2017sides= =,2 29, and 148√and√13 y=4 and= 4 the,, thethe scalene irregularirregular triangles snub snub cubeof cube sides consists consists √7√, √of13 of√ scaled scaledand equilateral√10 equilateral√ as shown triangles triangles in Figure of sides of13.15 sides of√7 22, 7, squaressquares of of sidessides √1313 and and thethe scalenescalene triangles of of sides sides √77,, √1313 and and √1010 as as shown shown in in Figure Figure 13. 13 .

Figure 13. The irregular snub cube with equilateral triangles, scalene triangles and squares (for x = 2 and y = 4). Figure 13. The irregular snub cube with equilateral triangles, scalene triangles and squares (for = 2 and =4).

5.1. Dual of the Irregular Snub Cube The vectors orthogonal to the faces in Figure 10 can be determined as:

=,

=( + +),

1 = ( + + ), (44) 2

=( + +),

=( + −). Here the parameters are given by:

=( + 1) + , =, =(+1),

(45) = , = ()()()() Vertices of the dual of the irregular snub cube can be written as three sets of orbits under the chiral octahedral group {, ⊕ [,]},

(±,±,±),

1 (± ± ± ) (46) 2

{(± ± ±), (± ± ±), (± ± ±)}.

The mirror image can be obtained from (46) by exchanging ↔. The dual of the irregular snub cube consists of 24 irregular pentagons with three different edge lengths in general. However, for the special case = −1 and − −−1=0 which corresponds to the regular snub cube, the parameters are given by = , = , =(2 + 1), =, = . The lengths of the edges of the pentagon satisfy the relations | − | = | − | = , (47)

| −| = | −| = | −| = √2−, where ≈ 1.8393. The dual of the snub cube is shown in Figure 14.

Symmetry 2017, 9, 148 15 of 22

5.1. Dual of the Irregular Snub Cube The vectors orthogonal to the faces in Figure 10 can be determined as:

d1 = µe1,

d2 = ν(n1e1 + n2e2 + n3e3), 1 d3 = 2 (e1 + e2 + e3), (44)

d4 = ν(n3e1 + n1e2 + n2e3),

d5 = ν(n1e1 + n3e2 − n2e3).

Here the parameters are given by:

n1 = (x + 1) y + x, n2 = x, n3 = x(y + 1), (45) = 2x+3y+4 = 1 2x+3y+4 µ 4x+2y+4 , ν 2 (xy+x+y)(2x+y+2)+x(y+2)+yx(y+1)

Vertices of the dual of the irregular snub cube can be written as three sets of orbits under the n o chiral octahedral group T, T  ⊕ [T0, T0] ,

µ(±e1, ±e2, ±e3), 1 2 (±e1 ± e2 ± e3) (46)

ν{(±n1e1 ± n2e2 ± n3e3), (±n2e1 ± n3e2 ± n1e3), (±n3e1 ± n1e2 ± n2e3)}.

The mirror image can be obtained from (46) by exchanging e1 ↔ e2 . The dual of the irregular snub cube consists of 24 irregular pentagons with three different edge lengths in general. However, for the special case y = x2 − 1 and x3 − x2 − x − 1 = 0 which corresponds x x−3 3 to the regular snub cube, the parameters are given by µ = 2 , ν = 2 , n1 = x(2x + 1), n2 = x, n3 = x . The lengths of the edges of the pentagon satisfy the relations q x2−1 |d1 − d2| = |d1 − d5| = 4x , √ (47) |d2 − d3| = |d3 − d4| = |d4 − d5| = 2 − x, whereSymmetryx ≈ 20171.8393., 9, 148 The dual of the snub cube is shown in Figure 14. 16 of 22

FigureFigure 14. 14.Dual Dual of theof snubthe snub cube (pentagonalcube (pentagonal icositetrahedron) icositetrahedron) with itswith pentagonal its pentagonal face. If face. we substitute If we x =substitute1, y = =1,=00 in (46) we obtainin (46) thewe Catalanobtain the solid Catalan tetrakis solid hexahedron tetrakis hexahedron (dual of the (dual truncated of the octahedron). truncated octahedron). We shall not discuss all duals of the irregular snub cubes. They can be obtained by substituting x We shall not discuss all duals of the irregular snub cubes. They can be obtained by substituting = and yandin (45) in and(45) (46)and corresponding(46) corresponding to each to each case case above. above. We We will will illustrate illustrate only only the the case case for forx =2 and 2 y =and4 where =4 the where pentagon the pentagon has three has differentthree different edge lengths.edge leng Thisths. This is the is dualthe dual of the of the irregular irregular snub snub cube cube corresponding to the case of (5). It is depicted in Figure 15 with its pentagonal faces consisting

of three different edge lengths satisfying the relations | −| = | −| , | −| = | − |, | −|.

Figure 15. Dual of the irregular snub cube of Figure 13.

5.2. Chiral Polyhedra with Vertices at the Edge Mid-Points of the Irregular Snub Cube With a similar discussion to the case of irregular icosidodecahedron in Section 4, we may construct a chiral polyhedron assuming the mid-points of edges of an irregular snub cube as vertices. The vectors whose tips constitute the plane orthogonal to the vector Λ in Figure 10 which are constructed similar to (37) and (38) and are given by:

=[(++1) +(++2) + (+1)],

= [(++2) + (+1) + (++1)]

=′[(2++2) + + (+1)], (48) =′[(2++2) + (+1) −],

=′(++2)( +),

(+)+(+)+(+) (+)+(+) + ( + ) ′ = ,′= . 2 + + ( + )

The vertices and are in the same orbit of size 24 under the chiral octahedral group {, ∪ [ , ]}, and the orbit involving and is of size 24. The orbit of consists of cyclic permutations of the coefficients of the unit quaternions with all possible sign changes. The same

argument is valid for the orbit of . The orbit of comprises of 12 quaternions obtained as the cyclic permutations of the pair of quaternions with all sign changes. Actually, the orbit of by itself

Symmetry 2017, 9, 148 16 of 22

Figure 14. Dual of the snub cube (pentagonal icositetrahedron) with its pentagonal face. If we substitute =1,=0 in (46) we obtain the Catalan solid tetrakis hexahedron (dual of the truncated octahedron).

We shall not discuss all duals of the irregular snub cubes. They can be obtained by substituting Symmetry and2017 in, 9 ,(45) 148 and (46) corresponding to each case above. We will illustrate only the case for16 of 22 = 2 and =4 where the pentagon has three different edge lengths. This is the dual of the irregular snub cube corresponding to the case of (5). It is depicted in Figure 15 with its pentagonal faces consisting corresponding to the case of (5). It is depicted in Figure 15 with its pentagonal faces consisting of three of three different edge lengths satisfying the relations | −| = | −| , | −| = | − different edge lengths satisfying the relations |d − d | = |d − d |, |d − d | = |d − d |, |d − d |. |, | −|. 1 2 1 5 2 3 3 4 4 5

FigureFigure 15. 15.Dual Dual of of the the irregular irregular snub snub cube cube of of Figure Figure 13 .13.

5.2.5.2. Chiral Chiral Polyhedra Polyhedra with with Vertices Vertices at at the the Edge Edge Mid-Points Mid-Points of of the the Irregular Irregular Snub Snub Cube Cube WithWith a similara similar discussion discussion to the to case the of case irregular of irregular icosidodecahedron icosidodecahedron in Section in4, weSection may construct4, we may a chiralconstruct polyhedron a chiral assumingpolyhedron the assuming mid-points the of mid-points edges of an of irregular edges of snub an irregular cube as vertices.snub cube The as vectors vertices. whoseThe vectors tips constitute whose thetips plane constitute orthogonal the plane to the orthogonal vector Λ in Figureto the 10vector which Λ arein constructedFigure 10 which similar are toconstructed (37) and (38) similar and are to given (37) and by: (38) and are given by:

=[(++1) +(++2) + (+1)], c1 = [(x + y + 1)e1 + (x + y + 2)e2 + (y + 1)e3],

c2 = =[(x[(+++2y + 2)e1)+ +(y(++11)e2)+ (+x(+++1y + 1)e3)]]

c3 = µ0[(2x + y + 2)e1 + e2 + (y + 1)e3], =′[(2++2) + + (+1)], c = µ0[(2x + y + 2)e + (y + 1)e − e ], (48) 4 1 2 3 (48) =′[(2++2) + (+1) −], c5 = v0(x + y + 2)(e1 + e2), (α + β)α + (β +=′γ)(β++2+ (γ + α)()γ + ),( α + β)α + (β + γ)β + (γ + α)γ µ0 = , v0 = . 2α2 + β2 + γ2 (α + β)2 (+)+(+)+(+) (+)+(+) + ( + ) ′ = ,′= .   The vertices c1 and c2 are2 in the+ same+ orbit of size 24 under the chiral( + octahedral ) group { T, T ∪ [T0, T0]}, and the orbit involving c and c is of size 24. The orbit of c consists of cyclic permutations The vertices and are3 in the4 same orbit of size 24 under1 the chiral octahedral group of the coefficients of the unit quaternions with all possible sign changes. The same argument is valid {, ∪ [ , ]}, and the orbit involving and is of size 24. The orbit of consists of cyclic forpermutations the orbit of c 3of. Thethe orbitcoefficients of c5 comprises of the unit of 12quaternions quaternions with obtained all possible as the cyclicsign changes. permutations The same of the pair of quaternions with all sign changes. Actually, the orbit of c5 by itself represents the vertices argument is valid for the orbit of . The orbit of comprises of 12 quaternions obtained as the cyclic of a cuboctahedron. For all allowed x and y, the chiral polyhedra with 60 vertices can be displayed. permutations of the pair of quaternions with all sign changes. Actually, the orbit of by itself However, we will only display the polyhedron corresponding to the special case where the 60 vertices obtained from (48) represent the mid-points of edges of the regular case, namely the snub cube. This is obtained, as we discussed before, substituting y = x2 − 1 and using the real solution of the cubic equation x3 − x2 − x − 1 = 0. Then one obtains µ0 = v0 = 1 and the square lengths of sides of the irregular pentagon obtained from (48) are given by:

| − | √ c3 c4 3 |c1 − c2| = |c2 − c3| = |c4 − c5| = |c5 − c1| = √ = 2x ≈ 3.528 (49) 2

The vector orthogonal to this pentagon√ is represented by the vector√ Λ and around this pentagon there are 4 equilateral triangles of sides 2x3 and 1 square of sides 2 x3. With this pentagonal face, the chiral polyhedron is shown in Figure 16. It has 60 vertices, 62 faces (6 squares, (8 + 24) equilateral triangles and 24 irregular pentagons) and 120 edges (96 of sides 3.528 and 24 of length 4.989). Symmetry 2017, 9, 148 17 of 22

represents the vertices of a cuboctahedron. For all allowed and , the chiral polyhedra with 60 vertices can be displayed. However, we will only display the polyhedron corresponding to the special case where the 60 vertices obtained from (48) represent the mid-points of edges of the regular case, namely the snub cube. This is obtained, as we discussed before, substituting = −1 and using the real solution of the cubic equation − −−1=0. Then one obtains ′ = ′ = 1 and the square lengths of sides of the irregular pentagon obtained from (48) are given by: | − | | −| = | −| = | −| = | −| = = 2 ≈ 3.528 (49) √2 The vector orthogonal to this pentagon is represented by the vector Λ and around this pentagon there are 4 equilateral triangles of sides √2 and 1 square of sides 2√. With this pentagonal face, the chiral polyhedron is shown in Figure 16. It has 60 vertices, 62 faces (6 squares, (8 + 24) equilateral triangles and 24 irregular pentagons) and 120 edges (96 of Symmetrysides 3.528 and 24 of length 4.989)2017, 9, 148 . 17 of 22

Figure 16. Chiral polyhedron consisting of equilateral triangles, squares and irregular pentagons. Figure 16. Chiral polyhedron consisting of equilateral triangles, squares and irregular pentagons.

+ 6. The Regular and Irregular Snub Dodecahedron Derived from W(H ) (a a a ) 6. The Regular and Irregular Snub Dodecahedron Derived from (3) (123) The snub dodecahedron is a chiral Archimedean solid with 60 vertices, 92 faces (20 pentagons ++8080 triangles)triangles) andand 150150 edges. We will discuss how to obtain the snub dodecahedron from an irregular snub dodecahedron.dodecahedron. The The vertices vertices of anof irregular an irregula snubr dodecahedron,snub dodecahedron, its dual andits thedual chiral and polyhedronthe chiral obtainedpolyhedron from obtained the mid-points from the of edgesmid-points will be of constructed edges will by be employing constructed the sameby employing technique describedthe same intechnique Section 5described. The proper in Section rotational 5. The subgroup proper rotati of theonal icosahedral subgroup Coxeterof the icosahedral group is also Coxeter called group chiral is +   icosahedral group W(H3) = I, I ≈ Alt(5)which is a simple group of order 60. They can be generated also called chiral icosahedral group () =, ≈ (5) which is a simple group of order 60. 5 3 2 by the generators a = r1r2, b = r2r3 which satisfy the generation relations a = b = (ab) = 1. They can be generated by the generators = , = which satisfy the generation relations = Let Λ = a1ω1 + a2ω2 + a3ω3 be a general vector where ωi, (i = 1, 2, 3) can be obtained from (17). =() =1. Let Λ= + + be a general vector where , ( = 1, 2, 3) can be Theobtained following from sets(17). of The vertices following form sets a regular of vertices pentagon form and a regular an equilateral pentagon triangle, and an respectively:equilateral triangle, respectively: 2 3 4 2 (Λ, r1r2Λ, (r1r2) Λ, (r1r2) Λ, (r1r2) Λ), (Λ, r2r3Λ, (r2r3) Λ) (50) (Λ, Λ, ( )Λ, ( )Λ, ( )Λ), (Λ, Λ, ( )Λ) (50) q q with respective edge lengths 2a2 + τa a + a2
and 2a2 + a a + a2
. With the addition of the 2(1+1 2+2) 2(2+23 +3) with respective edge lengths and . With the additionq of the 2 2
vertex r1r3ΛΛ== r3rΛ1 Λ asas shown shown in inFigure Figure 17, 17 there, there are are three three edge edge lengths lengths including including 2(2+a1)+. Thea3 . Thediscussions discussions of Sections of Sections 4 and 45 willand 5be will repeated be repeated to obtain to the obtain regular the and regular irregular and snub irregular dodecahedra. snub dodecahedra.The dual and the The chiral dual polyhedron and the chiral based polyhedron on the mid-points based onof edges the mid-points follow the ofsame edges technique. follow the sameSymmetry technique. 2017, 9, 148 18 of 22 Λ Λ r2r1 r2r3 (r r )3Λ 1 2 2 1 r r Λ 3 Λ 3 2 5 2Λ 4 (r1 r2) r r Λ r r Λ 1 2 1 3

FigureFigure 17. 17. TheThe vertices vertices connected connected to to the the vertex vertexΛΛ..

Numbering of the faces has been done accordingaccording toto thethe H3 diagram of Figure4 4.. FactoringFactoring byby a3 a1 a2 a2 6= 0, redefining x = , y = and dropping √ one obtains an irregular snub dodecahedron ≠0, redefining = a,2 = aand2 dropping one2 obtains an irregular snub dodecahedron with √ p with 12 equilateral triangles with edge length: A = 2(1 + x + x2), 20 pentagons with edge length: 12 equilateralp triangles with edge length: =2(1 + + ), 20 pentagonsp with edge length: = B = ( + y + y2) A B C = (x2 + y2) 2(1 +2 1 + τ) , 60 ,scalene 60 scalene triangles triangles with with edge edge lengths: lengths: ,, ,, = 2(2 +) . .The The trianglestriangles numbered as 2, 4 and 5 are identical scalene triangles, but number 1 is an equilateral triangle. numbered as 2, 4 and 5 are identical scalene triangles, but number  1 is an equilateral triangle. The = − π + 3π + + + = π Theangular angular deficiency deficiency is =2−( is δ 2π+ +3 (+ + 5 θ ) )ϑ = whereϕ , and 15 where areθ, theϑ and interiorϕ are angles the interior angles of the scalene triangles. Descartes’ formula verifies the number of vertices N0 = 60. of the scalene triangles. Descartes’ formula verifies the number of vertices =60.

The orbit , Λ generated from the vector Λ= − − − ( + + 2) involves 60 vectors representing the vertices of an irregular snub dodecahedron. The mirror image , Λ′ can be obtained from the vector Λ = Λ= − − (++2). A few remarks are in order before we discuss the usual classification. For some special values of the parameters and , we obtain some of the Archimedean solids. For example, (i) =1 and = 0: Truncated icosahedron, (ii) =0 and =1: Truncated dodecahedron, (iii) = =0: Icosidodecahedron. Let us follow the sequence of Section 5 to discuss the regular and irregular snub dodecahedra. (1) The snub dodecahedron: = = , 1 + + =1++ = + Since all edges are equal, it leads to the relations =(1−) and =+1 resulting in the cubic equation − −−=0. The real solution is approximately = 1.943 . The snub dodecahedron is shown in Figure 18.

Figure 18. The snub dodecahedron.

(2) =≠ , = (− ± ( +4( +)) For =−1,=−, the irregular snub dodecahedron consists of 12 equilateral triangles and 20 pentagons of sides √2 and 60 isosceles triangles of sides √2, √2, 2( + 2)

(3) =≠, =+1 and − −−≠0

Symmetry 2017, 9, 148 18 of 22

Λ Λ r2r1 r2r3 (r r )3Λ 1 2 2 1 r r Λ 3 Λ 3 2 5 2Λ 4 (r1 r2) r r Λ r r Λ 1 2 1 3

Figure 17. The vertices connected to the vertex Λ .

Numbering of the faces has been done according to the diagram of Figure 4. Factoring by ≠0, redefining = , = and dropping one obtains an irregular snub dodecahedron with √ 12 equilateral triangles with edge length: =2(1 + + ), 20 pentagons with edge length: = 2(1 + + ) , 60 scalene triangles with edge lengths: , , = 2( +) . The triangles numberedSymmetry 2017 as, 92,, 148 4 and 5 are identical scalene triangles, but number 1 is an equilateral triangle.18 The of 22 angular deficiency is =2−( + + (+ + ) ) = where , and are the interior angles   of theThe scalene orbit triangles.I, I Λ Descartes’generated fo fromrmula the verifies vector theΛ =number− ye 1of− verticesxe2 − τ(τ=60y +. x + 2)e3 involves   60 vectorsThe orbit representing , Λ generated the vertices from of an the irregular vector snubΛ= − dodecahedron. − − (The mirror + + image 2) involvesI, I Λ0 can60 vectorsbe obtained representing from the the vector verticesΛ0 =of ran1Λ irregular= ye1 − snubxe2 − dodecahedron.τ(τy + x + 2) Thee3. A mirror few remarks image , are Λ′ in order can bebefore obtained we discuss from the usualvector classification. Λ = Λ= For − some special− (++2) values of the. parametersA few remarksx and arey, wein order obtain beforesome ofwe the discuss Archimedean the usual solids. classification. For example, For some special values of the parameters and , we obtain some of the Archimedean solids. For example, (i) x = 1 and y = 0 : Truncated icosahedron, (i) =1 and = 0: Truncated icosahedron, (ii) x = 0 and y = 1 : Truncated dodecahedron, (ii) =0 and =1: Truncated dodecahedron, (iii) x = y = 0 : Icosidodecahedron. (iii) = =0: Icosidodecahedron. LetLet us us follow follow the the sequence sequence of of Section Section 5 to disc discussuss the regular and irregularirregular snub dodecahedra.

(1)(1) TheThe snub snub dodecahedron: dodecahedron: A == B == ,C 1, + 1 + +x + x=1++2 = 1 + τy + y=2 = +x2 + y2 Since all edges are equal, it leads to the relations =(1−) and =+1 resulting in the Since all edges are equal, it leads to the relations y = σ1 − x2
and y2 = x + 1 resulting in cubic equation − −−=0. The real solution is approximately = 1.943 . The snub the cubic equation x3 − x2 − x − τ = 0. The real solution is approximately x = 1.943. The snub dodecahedron is shown in Figure 18. dodecahedron is shown in Figure 18.

FigureFigure 18. 18. TheThe snub snub dodecahedron. dodecahedron.

 p  (2)(2) =≠ A = B 6= , C=, y =(−1 − ±τ±( (+4(τ2 + 4(+)x2 +) x) 2

ForFor =−1,=−,x = −1, y = − τ the, the irregular irregular snub snub dodecahedron dodecahedron consis consiststs of 12 of equilateral 12 equilateral triangles triangles and and 20 √ √ √ p pentagons20 pentagons of sides of sides √2 and2 and 60 isosceles 60 isosceles triangles triangles of sides of sides √2, √2,2, 2,2( +2( 2)τ + 2)

(3)(3) =≠A = C 6=, B, y=+12 = x + and1 and x−3 − x−−≠02 − x − τ 6= 0

For y = 2 and x = 3, the corresponding irregular snub dodecahedron consists of pentagons of p √ edge length 2(2τ + 5), equilateral triangles of edge length 26 and isosceles triangles with edge √ √ p lengths 26, 26 and 2(2τ + 5)

(4) A 6= B = C, x2 = τy + 1 and x3 − x2 − x − τ 6= 0 p For y = 1, x = τ, the irregular snub dodecahedron has pentagons of sides 2(τ + 2), equilateral √ p p √ triangles of sides 2τ, and isosceles triangle of sides 2(τ + 2), 2(τ + 2) and 2τ.

(5) A 6= C 6= B

For x = 2 and y = 3, the irregular snub dodecahedron consists of equilateral triangles of sides √ p √ p √ 14, pentagons of sides 2(3τ + 10) and the scalene triangles of sides 14, 2(3τ + 10) and 26 as shown in Figure 19. Symmetry 2017, 9, 148 19 of 22

For = 2 and = 3, the corresponding irregular snub dodecahedron consists of pentagons of edge length 2(2 + 5), equilateral triangles of edge length √26 and isosceles triangles with edge lengths √26, √26 and 2(2 + 5)

(4) ≠=, =+1 and − −−≠0

For =1,=, the irregular snub dodecahedron has pentagons of sides 2( + 2), equilateral triangles of sides √2, and isosceles triangle of sides 2( + 2), 2( + 2) and √2. (5) ≠≠ For = 2 and =3, the irregular snub dodecahedron consists of equilateral triangles of sides √14, pentagons of sides 2(3 + 10) and the scalene triangles of sides √14, 2(3 + 10) and √26 asSymmetry shown2017 in, Figure9, 148 19. 19 of 22

Figure 19. IrregularIrregular snub dodecahedron with x=2,=3= 2, y = 3..

6.1. Dual of the Irregular Snub Dodecahedron We obtain 5 vectors as follows toto determinedetermine thethe planeplane orthogonalorthogonal toto thethe vertexvertex ΛΛ:: √ = √2 =( − ), d1 = 2ω1= τ(σe1 − τe3),

d2 =[= ν[((−y − σ)xe1 −−σy(+1x + 1)e2 ++ ((τxy + τx ++ y))e3] √ d3 = 2µω3 = µ(−e2 − τe3), = √2 =(− −), d4 = ν[σxe1 − σye2 + (2xy + τx + y)e3]

=[ − +(2++)] d5 = ν[−σxe1 + σye2 + (2xy + τx + y)e3],

=[−(3y + τx + +2τ) + (2 + (3 +y + )τx],+ 2τ) µ = , ν = . (51) τy + (σ + 2)x + 2 (σx − y + 2σ)(2xy + τx + y) (3 + + 2) (3 + + 2) = , = . (51) The orbits generated from + these ( + 2)five +vectors 2 represent( − + the 2 vertices)(2 + of the + dual) solid of the irregular snub dodecahedron. Faces of the dual consist of irregular pentagons, one of which is represented by The orbits generated from these five vectors represent the vertices of the dual solid of the the vertices of (51). Note that the orbits I, I d1 and I, I d3 represent the orbits of sizes 20 and 12, irregular snub dodecahedron. Faces of the dual consist of irregular pentagons, one of which is respectively. The other three vertices are in the same orbit; therefore, one single notation I, I d2 for representedthis orbit of sizeby the 60 suffices.vertices Therefore,of (51). Note the that dual the consists orbits of, 92 vertices, and , 60 irregular represent pentagons the orbits and 150 of Symmetrysizesedges. 20 The2017 and,regular 9 ,12, 148 respectively. case is obtained The other by substituting three verticyes= areσ(1 in− xthe2) andsamex =orbit;1.943 therefore,in (51). The one dual 20single of 22 of notationthe snub dodecahedron,, for this the orbit pentagonal of size 60 hexecontahedron, suffices. Therefore, is shown the dual Figure consists 20. of 92 vertices, 60 irregular pentagons and 150 edges. The regular case is obtained by substituting =(1−) and = 1.943 in (51). The dual of the snub dodecahedron, the pentagonal hexecontahedron, is shown Figure 20.

Figure 20. The pentagonal hexecontahedron (dual of snub dodecahedron). Figure 20. The pentagonal hexecontahedron (dual of snub dodecahedron).

We also illustrate the dual of the irregular snub dodecahedron obtained from 5) by substituting We also illustrate the dual of the irregular snub dodecahedron obtained from 5) by substituting the values x = 2 and y = 3 in (51). The 92 vertices with this substitution describe the chiral solid the values =2 and =3 in (51). The 92 vertices with this substitution describe the chiral solid depicted in Figure 21. depicted in Figure 21.

Figure 21. The dual of irregular snub dodecahedron with =2 and =3.

6.2. Chiral Polyhedra with Vertices at the Edge Mid-Points of the Irregular Snub Dodecahedron Similar to the previous discussions, we can construct a chiral polyhedron possessing the mid- points of the edges of the irregular snub cube as vertices. The vectors whose tips constitute the plane orthogonal to the vector Λ in Figure 17 are given by: 1 = [−(2 + ) + −(2 + 2 + + 2) ], 2

1 = [−(2 + + ) − (+1) −(2 + 2 + + 2) ], 2

= [−( + 1) − ( + 2 + 1) −(+2++2) ], 2 = [ − (2 + 1) −(2 + 2 + + 2) ], 2

=−(++2), 3 +2+ +4+(+2) ++2 = , +2+(+2) + (3 + 1)+4++2 (52) 3 +2+ +4+(+2) ++2 = ( + + 2) In the limit =(1−) , − −−=0 and ==1 ( = 1.943); for a snub dodecahedron, the vertices in (52) are simplified, and the edge lengths of the pentagon satisfy the relations:

| −| | − | = | − | = | − | = | − | = = ++1 (53)

Symmetry 2017, 9, 148 20 of 22

Figure 20. The pentagonal hexecontahedron (dual of snub dodecahedron).

We also illustrate the dual of the irregular snub dodecahedron obtained from 5) by substituting the values =2 and =3 in (51). The 92 vertices with this substitution describe the chiral solid Symmetry 2017, 9, 148 20 of 22 depicted in Figure 21.

FigureFigure 21. 21. TheThe dual dual of of irregular irregular snub snub dodecahedron dodecahedron with with =2x = 2 andand y=3= 3..

6.2. Chiral Polyhedra with Vertices at the Edge Mid-Points of the Irregular SnubSnub DodecahedronDodecahedron Similar to to the the previous previous discussions, discussions, we we can can construct construct a chiral a chiral polyhedron polyhedron possessing possessing the mid- the mid-pointspoints of the of edges the edgesof the ofirregular the irregular snub cube snub as cube vertices. as vertices. The vectors The whose vectors tips whose constitute tipsconstitute the plane theorthogonal plane orthogonal to the vector to the Λ invector FigureΛ in17 Figure are given 17 are by: given by: 1 1 = c[−=(2[− +( 2)y++τ)e −+ e(2− τ +(2 2τy ++ 2 +x + 2)τ+], 2)e ], 2 1 2 1 2 3 c = 1 [−(2y + τx + τ)e − (x + 1)e − τ(2τy + 2x + τ + 2)e ], 2 1 2 1 1 3 = α [−(2 + + ) − (+1) −(2 +2 2 + + 2)]
,  c3 = 22 −τ(τy + 1)e1 − (τy + 2x + 1)e2 − τ τ y + 2x + τ + 2 e3 , α c4 = [τe1 − (2x + 1)e2 − τ(2τy + 2x + τ + 2)e3], = [−(2 + 1) − ( + 2 + 1) −(+2++2) ], 2 c5 = −βτ(τy + x + 2)e3, = [ − (2 + 1) −(2 + 2 + + 2) ], 2 3y2 + 2τxy + x2 + 4τy + x(τ + 2) +τ + 2 α = , τ2y2 + 2τxy + x2(σ + 2) + (3τ + 1)y + 4x + τ + 2 =−(++2) , (52) 3y2 + 2τxy+ x2 + 4τy + x(τ + 2) + τ + 2 β = 3 +2+ +4+(+22 ) ++2 = (τy + x + 2) , +2+(+2) + (3 + 1)+4++2 2
3 2 (52) In the limit y = σ 31 − +2+x , x − x +4+− x − τ (=+20 and) ++2α = β = 1 (x = 1.943); for a snub = dodecahedron, the vertices in (52) are simplified,( + + 2) and the edge lengths of the pentagon satisfy the relations:In the limit =(1−) , − −−=0 and ==1 ( = 1.943); for a snub |c − c | p Symmetrydodecahedron, 2017, 9, 148 |thec − verticesc | = | cin− (52)c | are= | csimplified,− c | = | cand− cthe| = edge3 lengths4 = ofx 2the+ xpentagon+ 1 satisfy21 of(53)the 22 1 2 2 3 4 5 5 1 τ relations: This chiral polyhedron has 150 verticesvertices andand 152152 facesfaces (12(12 regularregular pentagonspentagons + + 60 irregular | − | pentagons and ((2020 + +60 60)) equilateral equilateral triangles triangles all all surroundingsurrounding 6060 irregularirregular pentagons),pentagons), a typical one | −| = | −| = | −| = | −| = = ++1 (53) of which is represented with the relations of vertices in (53). The chiral polyhedron with 300 edges is illustrated in Figure 2222..

Figure 22. ChiralChiral polyhedron polyhedron whose whose vertices vertices are are the mid-points of the edges of the snub dodecahedron.

7. Concluding Remarks In this paper we presented a systematic construction of regular and irregular chiral polyhedra, the snub cube, the snub dodecahedron and their duals using proper rotational subgroups of the octahedral group and the icosahedral group. Chiral polyhedra whose vertices are the mid-points of the chiral polyhedra were also constructed. We used the Coxeter diagrams B3 and H3 to obtain the quaternionic description of the relevant chiral groups. Employing the same technique for the diagrams A1 ⊕ A1 ⊕ A1 and A3 , the irregular tetrahedra and icosahedra have been included although they are not chiral as they can be transformed to their mirror images by the proper rotational subgroup of the octahedral group. The dual solids of the irregular icosahedron, the tetartoid and the pyritohedron are also constructed representing the chiral symmetries of chiral tetrahedral group and the pyritohedral group, respectively. This method can be extended to higher dimensional Coxeter groups to determine the irregular- regular chiral polytopes. For example, the snub 24-cell, a chiral polytope in the 4D Euclidean space can be determined using the D4 Coxeter diagram [29], and its irregular form can be obtained using the same technique used for the irregular icosahedron.

Author Contributions. The authors N.O.K. and M.K. equally contributed to the paper in conceiving, programming and writing.

Conflicts of Interest: The authors declare no conflict of interest.

References

1. Coxeter, H.S.M.; Moser, W.O.J. Generators and Relations for Discrete Groups; Springer: Berlin, Germany, 1965. 2. Cotton, F.A.; Wilkinson, G.; Murillo, C.A.; Bochmann, M. Advanced Inorganic Chemistry, 6th ed.; Wiley- Interscience: New York, NY, USA, 1999. 3. Caspar, D.L.D.; Klug, A. Cold spring harbor symp. Quant. Biol. 1962, 27, 1. 4. Twarock, R. Mathematical virology: A novel approach to the structure and assembly of viruses. Philos. Trans. R. Soc. 2006, 364, 3357–3373. 5. Jaric, M.V. (Ed.) Introduction to the Mathematics of Quasicrystals; Academic Press: New York, NY, USA, 1989. 6. Senechal, M. Quasicrystals and Geometry; Cambridge University Press: Cambridge, UK, 1995. 7. Suck, J.B.; Schreiber, M.; Haussler, P. (Eds.) Quasicrystals (An Introduction to Structure, Physical Properties, and Applications); Springer: Berlin/Heidelberg, Germany, 2002. 8. Koca, M.; Koc, R.; Al-Ajmi, M. Polyhedra obtained from Coxeter groups and quaternions. J. Math. Phys. 2007, 48, 113514. 9. Koca, M.; Koca, N.O.; Koc, R. Catalan solids derived from 3D-root systems. J. Math. Phys. 2010, 51, 043501. 10. Koca, M.; Al-Ajmi, M.; Al-Shidhani, S. Quasi regular polyhedra and their duals with Coxeter symmetries represented by quaternions II. Afr. Rev. Phys. 2011, 1006, 53.

Symmetry 2017, 9, 148 21 of 22

7. Concluding Remarks In this paper we presented a systematic construction of regular and irregular chiral polyhedra, the snub cube, the snub dodecahedron and their duals using proper rotational subgroups of the octahedral group and the icosahedral group. Chiral polyhedra whose vertices are the mid-points of the chiral polyhedra were also constructed. We used the Coxeter diagrams B3 and H3 to obtain the quaternionic description of the relevant chiral groups. Employing the same technique for the diagrams A1 ⊕ A1 ⊕ A1 and A3 , the irregular tetrahedra and icosahedra have been included although they are not chiral as they can be transformed to their mirror images by the proper rotational subgroup of the octahedral group. The dual solids of the irregular icosahedron, the tetartoid and the pyritohedron are also constructed representing the chiral symmetries of chiral tetrahedral group and the pyritohedral group, respectively. This method can be extended to higher dimensional Coxeter groups to determine the irregular-regular chiral polytopes. For example, the snub 24-cell, a chiral polytope in the 4D Euclidean space can be determined using the D4 Coxeter diagram [29], and its irregular form can be obtained using the same technique used for the irregular icosahedron.

Author Contributions: The authors N.O.K. and M.K. equally contributed to the paper in conceiving, programming and writing. Conflicts of Interest: The authors declare no conflict of interest.

References

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