Regular and Irregular Chiral Polyhedra from Coxeter Diagrams ... - MDPI

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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

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For the snub cube and snub dodecahedron, the Coxeter graphs are the B3 and H3 , respectively. althoughthethey describe the achiral polyhedra suchsimpler as the Coxeter families diagrams of regularAand irregular To ,describe general technique, we first begin with 1 ⊕ 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 proper rotational symmetries respectively. transformingWe them to theirshow mirror images. organizepossess the paper as tetrahedron and icosahedron, explicitly that achiralWe polyhedra larger follows. In Section 2 we introduce quaternions and construct the Coxeter groups in terms of proper rotational symmetries transforming them to their mirror images. We organize the paper quaternions [17]. We extend the group W (A1 ⊕ A1 ⊕ A1) to the octahedral group by the symmetry as follows. In Section 2 we introduce quaternions and construct the Coxeter groups in terms of (3) of the Coxeter diagram A1 ⊕ A1 ⊕ A1. In Section 3 we obtain the proper rotation group 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 Section 3 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 Section 4 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. Section 5 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 Section 6 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. Section 7 involves the discussion of polyhedra. the technique for the construction of irregular chiral polyhedra.

2. Quaternionic Quaternionic Constructions Constructions of of the the Coxeter Coxeter Groups Groups 2. Let q ==q0 ++q e , (, i (==1,1,2,2,33) bea areal realunit unit quaternion with conjugate defined Let quaternion with its its conjugate defined by qby = q0=− qi e−i , ) be i i , and the norm = = 1. The quaternionic imaginary units satisfy the relations: and the norm qq = qq = 1. The quaternionic imaginary units satisfy the relations:

e e j =−−δδij++ε εijke e, k(,i, (j,, k, ==1,1,2,2,33)) ei eij = ij ijk k

(1) (1)

ε ijkareare and ε ijk Kronecker and Levi-Civitasymbols, symbols,and andsummation summationover over the the repeated repeated where δδij ijand where thethe Kronecker and Levi-Civita indices is understood. The unit quaternions form a group isomorphic to the special unitary indices is understood. The unit quaternions form a group isomorphic to the special unitary group group SU(2). Quaternions generate the the four-dimensional four-dimensional Euclidean Euclidean space space with with the (2). Quaternions generate the scalar scalar product product 1 1 1 (2) + qp)) = = 1 ((pq + + qp)) = 2 ((pq + (2) ((p,,q)):≔ 2 2 2 can be represented by its quaternionic roots as shown in The Coxeter diagram ⊕ ⊕ The Coxeter diagram A1 ⊕ A1 ⊕ A1 can be represented by its quaternionic roots as shown in √ justthe thenorm. norm. Figure 11 where where 22isisjust Figure

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  C= 0 0

0 2 0

  0 1 1  0  , C −1 =  0 2 2 0

0 1 0

 0  0  1

(3)

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

 (αi , α j = Cij , ωi , ω j = C −1 , (αi , ω j ) = δij , (i, j = 1, 2, 3). ij

(4)

Note that one can express the roots in terms of the weights or vice versa:   αi = Cij ω j , ωi = C −1 α j ij

(5)

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]:   α α α α ∗ ri Λ = − √i Λ √i := √i , − √i Λ. 2 2 2 2

(6)

It is then straight forward to show that ri ω j = ω j − δij α j . 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: r 1 = [ e1 , − e1 ] ∗ , r 2 = [ e2 , − e2 ] ∗ , r 3 = [ e3 , − e3 ] ∗ . (7) They generate the elementary abelian group W (A1 ⊕ A1 ⊕ A1 ) ≈ C2 × C2 × C2 := 23 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 Figure 1. This is obvious from the diagram where the generators of the symmetric group Sym(3) of order 6 can be chosen as:  s=

 ∗ 1 1 1 1 (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 , and satisfy the relations sri s−1 = ri+1 , dr1 d = r2 , dr3 d = 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 the group < r1 , r2 , r3 > by the group Sym(3) which is the octahedral group 23 : 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: Oh = [ T, ± T ] ∪ [ T 0 , ± T 0 ]

(9)

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= {±1, ± , ± , ± , (±1 ±

±

±

)},

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(10)

= { (±1 ± ), (± ± ), (±1 ± ), (± √ √ √ √ where the sets of quaternions T and T 0 are given by:

±

),

√

(±1 ±

),

√

(±

±

)}.

± e2 ±and e3 )},the binary octahedral group, = {±1, ±the e1 , ± e2 , ±e3tetrahedral , 12 (±1 ± e1group Here and ∪ Trepresent binary (10)
respectively. We used short-hand 1 T 0 = { √1 (± ±ea2 ± e3 , √1 (±notation 1 ±have e1 ) , √ 1 ± e2 ), √1for e3 ±designation e1 ), √1 (±1 ±ofe3the e2 )}.[ , ± ] (±the ), √1groups, (±e1 ±e.g., 2 2 2 2 2 2 means the set of all [ , ± ] ∈ [ , ± ] with ∈ . The maximal subgroups of the octahedral group can be written [21]:T ∪ T 0 represent the binary tetrahedral group and the binary octahedral group, Here Tasand 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]: (11) Tetrahedral group: = [ , ] ∪ [ , − ], Chiral octahedral group : O = [ T, T ] ∪ [ T 0 , T 0 ] ≈ W ( B3 )+ , 0 0 (11) Tetrahedral group = : T[d ,= Pyritohedral group: ± [ T,]. T ] ∪ [ T , − T ], Pyritohedral group : Th = [ T, ± T ]. ( )≈ The octahedral group, as we will see in what follows, can also be obtained as the ). octahedral group, as we will see in what follows, can also be obtained as the Aut( A3 ) ≈ W (B3 ). ( The the Klein Klein four-group four-group ⊕ A1 ⊕ ⊕A1 )) isis the The proper proper rotation rotation subgroup subgroup of of the the Coxeter Coxeter group W ((A1 ⊕ The × C2 represented representedby bythe theelements: elements: C2 ×

= [1,1], = [ , − ], = [ , − ], = [ ,− ] (12) I = [1, 1], r1 r2 = [e3 , −e3 ], r2 r3 = [e1 , −e1 ], r3 r1 = [e2 , −e2 ] (12) We also note in passing that the Klein four-group in (12) is invariant by conjugation under the Wegenerated also noteby in passing (12) is invariant by conjugation the group , with that = 1.the TheKlein Kleinfour-group four-groupintogether with the group element under generate 3 = 1. The Klein four-group together with the group element s generate group generated by s, with s the tetrahedral rotation group, which is represented by [ , ] in our notation. the tetrahedral which group is represented T,Its T ] Coxeter-Dynkin in our notation. diagram ( [). Next, we rotation use thegroup, tetrahedral ≈ by with its Next, we use the tetrahedral group T ≈ W A . Its Coxeter-Dynkin diagram A with its ( ) 3 3 d quaternionic roots is shown in Figure 2. quaternionic roots is shown in Figure 2.

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

The Cartan matrix matrixofof Coxeter diagram and A3 its and its inverse matrix are by the The Cartan thethe Coxeter diagram inverse matrix are given bygiven the respective respective matrices: matrices:     2 −1 0 3 2 1 2 −1 0 −1 1 3 2 1   C =  =−1−1 2 2 −1−1 (13) ,,C = = 4 2 2 4 4 2 2  (13) 0 0 −1−1 2 2 1 1 2 2 33

aregiven givenby: by: Thegenerators generatorsof ofthe theCoxeter Coxetergroup groupW ( (A3 ))are The 1 1 ( 1 + )]∗ ∗ = [ ( 1 + ), − r1 √2 = [ √ (e1 + e2 )√2 , − √ (e1 + e2 )] 2 2 1 1 ( 1 − )]∗ ∗ = [ ( 1 − ), − , − √ (e3 − e2 )] r2 √2 = [ √ (e3 − e2 )√2 2 2 1 1 ( ), ( − =[ 1 − 1 − )]∗ ∗ r3 √2 = [ √ (e2 − e1 )√2 , − √ (e2 − e1 )] 2 2 = 1 ( + + ), = , = 1 (− + + ). ω1 = (e1 + e2 + e3 ), ω2 = e3 , ω3 = (−e1 + e2 + e3 ). 2 2

(14) (14)

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 Figure 2.

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The generators in (14) generate the Coxeter group [22–24] which is isomorphic to the tetrahedral The generators in (14) the Coxeter group ([22–24] ( ): is isomorphic group of order 24. Thegenerate automorphism group ) ≈ which ≈ whereto theistetrahedral generated ∗ ): ( group of order 24. The automorphism group ( ) ≈ ≈ where is generated by the Dynkin diagram symmetry = [ , − ] , which exchanges the first and the third simple roots Symmetry 2017, 9, 148 5 of 22 by the Dynkin symmetry = [ , −2. ]∗ , which exchanges the first and the third simple roots and leaves thediagram second root intact in Figure and leaves the second root intact in Figure 2. octahedral group ( ) ≈ The Coxeter diagram leading to the (4) × ≈ is shown The Coxeter diagram leading to the octahedralgroup group W ((B3 ))≈≈Sym((4) ≈ Oh is is shown shown in Figure 3. The Coxeter diagram B3 leading to the octahedral 4) ××C2 ≈ in Figure Figure 3. 3. in

Figure 3. 3. The The Coxeter Coxeter diagram diagram B with withquaternionic quaternionicsimple simpleroots. roots. Figure 3 Figure 3. The Coxeter diagram with quaternionic simple roots.

The Cartan matrix of the Coxeter diagram and its inverse matrix are given by: The The Cartan Cartan matrix matrix of of the the Coxeter Coxeter diagram diagram B3 and andits itsinverse inversematrix matrixare aregiven givenby: by: 2 −1 0 2 2 2     2 −1−1 0 , =2 −1 = 22 2 24 2 24 2 (15) 2 0 −2 (15) 2 −2 = −1 0 −1 2, −1 = 1 21 42 43  C =  −1 2 −2 , C =  2 4 4  (15) 0 −1 2 1 2 3 The generators below generate group2 given 0 the −1octahedral 2 1 2in 3(9) and the weight vectors are The given by:generators below generate the octahedral group given in (9) and the weight vectors are givenThe by:generators below generate the 1 given in (9) and the weight vectors are given by: 1 octahedral group = [ 1 ( − ), − 1 ( − )]∗ √2( 1 − )]∗ ∗ ), − =r[ √2 = ([ √1 −(e1 − e2 )√2 , − √ (e1 − e2 )] 1√2 2 2 1 1 (16) ∗ ∗ 1 ), ( − =r2[ 1= [(√1 − (16) e2 − e3 )1, − √ (− e2 −)]e∗3 )] ( (16) = [ √2( 2− ), − √2( 2 − )] √2 √2 ∗ = [ , − ]∗r3 = [e3 , −e3 ] = [ , − ]∗ = , = + , =1 ( + + ) ω1 = e1 , ω2 = e1 + e2 , ω3 = ( e1 + e2 + e3 ) = , = + , =2 ( + + ) The group generated by the rotations and is isomorphic to the rotational octahedral The group generated by the rotations r1 r2 and r2 r3 is isomorphic to the rotational octahedral n h The group generated by the rotations and is isomorphic to the rotational octahedral   , >≈ { , ∪ , ]. group < group < r1 r2,, r2 r3>≈ > ≈{ , T, ∪ T ∪, T]0 ,.T 0 ]}. group < leading to the icosahedral group is shown in Figure 4 with the The Coxeter diagram The diagram H leading to the icosahedral group is shown in Figure 4 with the The Coxeter Coxeter diagram quaternionic simple roots: 3 leading to the icosahedral group is shown in Figure 4 with the quaternionic quaternionic simple simple roots: roots:

Figure 4. The Coxeter diagram of H3 with quaternionic simple roots. Figure 4. The Coxeter diagram of with quaternionic simple roots. √ √ 1+ 5Figure 4. The Coxeter diagram 1−of 5 with quaternionic simple roots.

Here τ = 2 is the golden ratio and = 2 . The Cartan matrix of the diagram H3 , its inverse √ √ Here = vectors is the and = . The Cartan matrix of the diagram , its inverse and the weight are golden given asratio follows: √ √ Here = is the golden ratio and = . The Cartan matrix of the diagram , its inverse and the weight vectors aregiven as follows:   and the weight vectors are given 2 as−follows: τ 0 3τ 2 2τ 3 τ3 2 − 0 −1 1 3 3 2 2  2  C= − (17) 2 2 −2 −1−1 0, ,C = = 2 =τ− (17) 32 2τ 24 4τ 2 2τ  3 2 0 − 1 2 τ 2τ τ + 2 = 2 = −0 −1 (17) 2 −1 42 2+ 2 2 , 0 −1 2 2 +2 √ τ ( τ( = − ), = −√2 , = √ √ ω1 = √ (σe1 − τe3 ), ω2 = − 2τe3 , ω3 = √ (σe2−− e)3 ) =2 ( − ), = −√2 , = (2 − ) √ √ (5) × ≈ ( ) are given by: The quaternionic generators of the icosahedral group ≈ The quaternionic generators of the icosahedral group Ih ≈ Alt(5) × C2 ≈ W ( H3 ) are given by: (5) × ≈ ≈ ( ) are given by: The quaternionic generators of the icosahedral group = [ , − ]∗ , ∗ [e , −e ]∗ , = [ , − r1 ]= , 1 1 (18) i∗ h ∗ (18) )] + + + + , 1 =[ ( 1 ), − ( (18) r2 = 2 (τe1 + e2 + σe3 ), − 2 (τe1 + e2 + σe3 ) , ∗ ), − ( )] , =[ ( + + + + r 3 = [ e2 , − e2 ] ∗ ,

= [ , − ]∗ , or shortly, ( ) = , ± where is the set of 120 quaternionic elements of the binary icosahedral ) [20]. The icosahedral rotation group group generated by the quaternions and ( + + Symmetry 2017, by 9, 148 6 of 22 (5). All finite subgroups of the is represented the proper rotation subgroup ( ) = , ≈ groups O (3) and O (4) in terms of quaternions can be found in the references [17,25].  space is represented by the vector Λ = A general vector in the dual + + . We will or shortly, W ( H3 ) = I, ± I where I is the set of 120 quaternionic elements of the binary icosahedral ), which are called the Dynkin indices in the Lie + + ≔( 1 use the notation Λ = group generated by the quaternions e1 and 2 (τe1 + e2 + σe3 ) [20]. The icosahedral rotation group is  =( ) for the orbit of the algebraic representation theory [26]. We use the notation  ( )Λ: represented by the proper rotation subgroup W ( H3 )+ = I, I ≈ Alt(5). All finite subgroups of the ( ) generated from the vector Λ where the letter Coxeter group 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 = a1 ω1 + a2 ω2 + a3 ω3 . We will [8]: use the notation Λ = a1 ω1 + a2 ω2 + a3 ω3 : ( a1 a2 a3 ), which are called the Dynkin indices in the Lie (100) (tetrahedron), octahedron), algebraic representation theory (111) [26]. We(truncated use the notation W ( G(100) a1 a2 a3 )G for the orbit of the )Λ := ((octahedron), 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) A3 (tetrahedron), (111) A3 (truncated octahedron), (100) B3 (octahedron),

3. The Orbit

× ) B(3 (cube),)(100 as an Tetrahedron (001 ) H3Irregular (dodecahedron ), (001) H3 (icosahedron), and(010) H3 ,

). ( ⊕ ⊕ ) transforms a × (icosidodecahedron of the Coxeter group The proper rotation subgroup generic vector Λ as follows: 3. The Orbit C2 × C2 (a1 a2 a3 ) as an Irregular Tetrahedron 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 2 3 3 1 1 2 (19) 1 1 (− a1 e1 − − a2 e2 ),+ a3 e3 ), 1 r2 Λ Λ r= ( = 2− (19) 2 1 r 2 r 3 Λ = 2 ( a 1 e1 − a 2 e2 − a 3 e3 ) , Λr= (−= 1 (− +a1 e1 +−a2 e2 −).a3 e3 ). 3 r1 Λ 2 These four vectors define the vertices of of anan irregular tetrahedron with four identical scalene These four vectors define irregular p p the vertices p tetrahedron with four identical scalene 2 2 2 2 2 + , + , and irregular tetrahedron with triangles with edge lengths triangles with edge lengths a1 + a2 , a2 + a3 , and a3+2 + a. An irregular tetrahedron with 1 . An 1, 1, a=2 = 2, 2, a=3 = 3 is depicted inin Figure a=1 = 3 is depicted Figure5.5.

Figure 5. An irregular tetrahedron with scalene triangular faces. Figure 5. An irregular tetrahedron with scalene triangular 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 a 2 e2 , ± a 3 e3 p ± a1 e1 , ±p + forming anan irregular octahedronwith with eight identical scalene triangles 2 2 forming irregular octahedron eight identical scalene triangles of sidesof asides a2 2 + a, 3 2 , 1 + a2 , p ,2 + and the corresponding interior angles, say, , , . Four triangles with 2 a+ 3 + a1 and the corresponding interior angles, say, α, β, γ. Four triangles with identical face-angles identical face-angles surround vertex 4 triangles . The remaining triangles at − the α surround the vertex a1 e1 . Thethe remaining similarly4meet at thesimilarly oppositemeet vertex a 1 e1 . opposite is truearound for every face-angle This is vertex true for−every. This face-angle every 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:

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r1 Λ = 12 (− a1 e1 + a2 e2 + a3 e3 ), r2 Λ = 21 ( a1 e1 − a2 e2 + a3 e3 ), r3 Λ = 12 ( a1 e1 + a2 e2 − a3 e3 ),

(20)

r1 r2 r3 Λ = − 12 ( a1 e1 + a2 e2 + a3 e3 ). 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: Λ = dΛ =

1 ( ae + ae2 + a3 e3 ). 2 1

(21)

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: Λ = sΛ =

1 a ( e + e2 + e3 ) . 2 1

(22)

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 Td = [ T, T ] ⊕ [ T 0 , − T 0 ], 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 √ ( e1 − e2 ) , − √ ( e1 − e2 ) ∈ [ T 0 , T 0 ] . (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 )+ (a1 a2 a3 ) 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 by the generators a = r1 r2 and b = r2 r3 satisfying the generation relations a3 = b3 = ( ab)2 = 1. Let Λ = ( a1 a2 a3 ) be a general vector in the weight space of A3 . The following sets of vertices form two equilateral triangles. (Λ,r1 r2 Λ, r2 r1 Λ) and (Λ,r2 r3 Λ, r3 r2 Λ) (24) q q

with respective edge lengths 2 a21 + a1 a2 + a22 and 2 a22 + a2 a3 + a23 . Three more triangles can be obtained by joining the vertex r1 r3 Λ = r3 r1 Λ to the vertices Λ, r1 r2 Λ and r3 r2 Λ and r2 r1 Λ to r2 r3 Λ. The new edges are of the following lengths.

|r1 r2 Λ − r1 r3 Λ | =

q

|r3 r2 Λ − r1 r3 Λ | =

q


2 a22 + a2 a3 + a23 ,


2 a21 + a1 a2 + a22 , q
|Λ − r1 r3 Λ| = |r2 r1 Λ − r2 r3 Λ| = 2 a21 + a23 The vertices joined to vector Λ are illustrated in Figure 6.

(25)

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.

r1 r3 Λ

4

r1r2 Λ

3

Λ

2 r2 r1Λ

5

r3 r2 Λ

1 r2 r3 Λ

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).

Factoring by an overall overall factor factor a22 6= 0 and defining defining the the parameters parameters x = = ≠ 0 and three weobtain, obtain, three classes classes of of triangles. triangles. After After dropping dropping the the overall overall factor factor a2 we

a1 a2

and y = =

a3 a2

we we obtain

p equilateral triangle with edge length: length : A== 2(1 2(+ 1 ++ x + x),2 ), equilateral p equilateral triangle triangle with with edge edge length: length : B== 2(1 2(+ 1++ y + y),2 ), equilateral p 2 2 33 scalene , B, , C = = 2( 2( x+ + )y ) scalene triangles triangles with with edge edge lengths lengths:: A,

in Figure 6. 6. The sum of the the face-angles face-angles at at the the vertex vertex Λ + as Λ (at every vertex indeed) is 2π 3 +   shown in Figure 5π (θ+ , , and are the interior angles of the scalene triangles. The + ϑ + + ϕ) = = where 3 where θ, ϑ, and ϕ are the interior angles of the scalene triangles. The angular 2 − − 5π = π3 isisthe deficiency δ ==2π thesame same as as in in the the regular icosahedron. Using Descartes’ formula, the 3 = number of vertices can also be obtained as in [27]. can also be obtained as in [27]. = =4π12. (26) N0 = = 12. (26) δ + where the new The vector Λ can be written in terms of quaternions as Λ = α + The vector Λ can be written in terms of quaternions as Λ = αe + βe + γe where the new 2 3 1 parameters are defined by parameters are defined by + x −−y (27) = x + y ,, γ = α α== 2 , ,β = = β++1.1. (27) 2 22   ⁄ 2≈ ≈ T,, T generated Λ can can be be The orbit orbit of of the the chiral chiral tetrahedral tetrahedralgroup groupW ( (A3 ))/C The generatedfrom fromthe thevector vector Λ written as: written as:   Λ = {± α ± ± γe ,, ± ± ± α ,± ±α ± } T, T, Λ = {± αe1 ± βe2 ± 3 ± βe1 ± γe2 ± αe3 , ± γe1 ± αe2 ± βe3 } (28) (28) (even number of (−) signs). (even number of (−) signs). The quaternions in (28) with an odd number of (−) signs constitute the mirror image. A rotation The quaternions an odd number of (−) signs constitute mirror image.This A rotation element, e.g., the onein in(28) (23),with transforms the quaternions in (28) to theirthe mirror images. proves element, e.g., the one (23), transforms the polyhedron. quaternions in (28) we to their mirror images. This proves that the set in (28) doesinnot represent a chiral Before discuss the irregular icosahedral that the set in doesout notthat represent icosahedrala structures, we(28) point for a>chiral 0, =polyhedron. 0 or > 0,Before = 0,we thediscuss set ofthe 12 irregular vectors describes structures, we point out For that for 0 or y > an 0, xoctahedron. = 0, the setMore of 12 vectors describes truncated truncated tetrahedron. =x >= 0,0, yit=describes interesting cases aarise as we tetrahedron. For x = y = 0, it describes an octahedron. More interesting cases arise as we will will discuss below. discuss below. (1) = = , 1 + + =1+ + = + 2 A = Ball = the C, 1edges + x +are x2 = 1 + yleading + y2 =tox2the + ysolution Here, equal = = −1⟹ − − 1 = 0, ⟹ = , or = . Substituting the first solution for = which results in α 2= 0, = , 2 = , the set of Here, all the edges are equal leading to the solution x = y = x − 1 ⇒ x − 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 = τ 2 , the set of vectors in (28) can be written as:

(1)



 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 s = [ 21 (σ + e2 + τe3 ), 12 (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.   p (2) A = B 6= C, y = 21 −1 ± 1 + 4( x2 + 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

Now we have x 6= y2 − 1 and y 6= x2 − 1. Taking will √ x = 1, y = 2 the irregular icosahedron√ 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 Figure 6, 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 r2 r3 and r1 r2 , respectively. The vectors orthogonal to the faces #2, #4 and #5 can be determined as: b2 = n1 e1 + n2 e2 + n3 e3 , b4 = r1 r2 b2 = −n3 e1 − n1 e2 + n2 , b5 = r3 r2 b2 = n3 e1 + n1 e2 + n2 e3 , where n1 = y − x, n2 = x + y + 2xy, n3 = x + y.

(30)

Symmetry 2017, 9, 148

10 of 22

These vectors should be rescaled in order to determine the plane orthogonal to the vector Λ. Let 10 of 22 us redefine the vectors ≔ , = , = , = , = . The scale factors can be determined as:

Symmetry 2017, 9, 148

2( in + order + 2)(to determine + + 2 ) the plane orthogonal to the vector Λ. These vectors should be rescaled = Let us redefine the vectors d1 := λω1 , d2 = 3b2 , +d3 = + 2$ b3 , d4 = b4 , d5 = b5 . The scale factors can be (31) determined as: 2( 2+( x + +y2)( +x + +y2+ 2xy ) ) + 2 )( . =λ = ( +3x 3 + +y2)+ 2 (31) The dual solid is an irregular dodecahedron the sets of) vertices: 2( x + y + 2with y+ 2xy )( x + $ = . ( x + 3y + 2) = (± ± ± ), (even number of (−) signs) , The dual solid is an2irregular dodecahedron with the sets of vertices: = (±  ± λ ± ), (odd number of (−) signs) T, T d1 = (32) 2 (± e1 ± e2 ± e3 ), (even number of (−) signs) $   odd number of T d3 =± 2 (±e±1 ± e2 ,± , =T, {± ± e3 ), (± ± , ±(−) signs ± ) ± } (32)   (−) (even number of signs). T, T d2 = {±n1 e1 ± n2 e2 ± n3 e3 , ±n2 e1 ± n3 e2 ± n1 e3 , ±n3 e1 ± n1 e2 ± n2 e3 } (even number of (−) signs). We will not discuss in detail how an irregular dodecahedron varies with five different cases discussed above. A few examples would sufficient in the order varies of increasing symmetry We will not discuss in detail how an be irregular dodecahedron with five differenttoward cases regularity. The case (5) above corresponds to the invariance under the chiral tetrahedral group. discussed above. A few examples would be sufficient in the order of increasing symmetry toward SubstitutingThe case = 1, (5) = above 2 in (32), we obtain irregular under dodecahedron called a tetartoid regularity. corresponds to theaninvariance the chiral tetrahedral group. corresponding cobaltite. The , , , and called form an irregular pentagon Substituting x =to1,the y =mineral 2 in (32), we obtain anvertices irregular dodecahedron a tetartoid corresponding with three different edge with its dual irregularpentagon icosahedron depicted in to the mineral cobaltite. Thelengths. vertices The d1 , d2tetartoid , d3 , d4 and d5 form an irregular withare three different Figure 7. edge lengths. The tetartoid with its dual irregular icosahedron are depicted in Figure 7.

,

(a)

(b)

Figure7.7.(a)(a) A face transitive irregular dodecahedron (tetartoid) under tetrahedral rotational tetrahedral Figure A face transitive irregular dodecahedron (tetartoid) under rotational symmetry; symmetry; (b) the dual irregular icosahedron. (b) the dual 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) (33) = = , = 0, = 2 ( + 1), = 2 ., 4x(2 (x + )2 + 11) λ= $ = , n1 = 0, n2 = 2x ( x + 1), n3 = 2x. , (33) (2x + 1) 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 4 3 2 = (34) (3x + 6x + 7x + 4x + 1) (2x + 1) | − | = 4| |. | d4 − d5 | = 4| x |. The irregular dodecahedron with the vertices of (32) with and from (33) is called the pyritohedron. Theydodecahedron can also be rearranged the set given its standard form: The irregular with the and vertices ofcan (32)bewith λ in and $ from (33) 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 ),
± 1 − h 2 e1 ± ( 1 + h ) e2 , 
± 1 − h 2 e2 ± ( 1 + h ) e3 , 
2 e ± (1 + h ) e , ) h± (1 {±(1± − ℎ1 − 3 + ℎ) }, 1 

Symmetry 2017, 9, 148

where h =

x

{±(1 − ℎ )

.

11 of 22

(35)

± (1 + ℎ) },

where ℎ x=+ 1 .

5 A special case discussedin in(2) (2) corresponding corresponding totoℎh== in (35) is plotted in Figure 8. A special case x == y ==5 5discussed in (35) is plotted in Figure 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 the set of vertices of the to beberegular all edges edgesequal equaleither either After rescaling the ,set of vertices of the regular with with all 4 4τ oror −4−4σ. . After rescaling by 4by, 4τ dodecahedron with x == τ are: dodecahedron with are: 1 (±

± ± ), {±1 ± ,± ± ,± ± }. (±e1 ± e2 ± e3 ), {±τe1 ± σe2 , ±τe2 ± σe3 , ±τe3 ± σe1 }. 2 2

(36)

(36)

Each set above is invariant under the pyritohedral group , ± . The 20 vertices of (36) form a dodecahedron with faces. As such, possess a larger symmetry , ± of . (36) form Each set above isregular invariant under the they pyritohedral groupicosahedral [ T, ± T ]. The 20 vertices The first 8with vertices of (36) represent a cube and the second set of 12 vertices, symmetry as we recall[ I, from a dodecahedron regular faces. As such, they possess a larger icosahedral ± I ]. (4 previous discussions, represents an irregular icosahedron with + 4) equilateral triangles and 12 recall The first 8 vertices of (36) represent a cube and the second set of 12 vertices, as we Robinson triangles. The dual of the irregular icosahedron in second set of 12 vertices in (36) is another from previous discussions, represents an irregular icosahedron with (4 + 4) equilateral 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 albeit different magnitudes of vectors. 4.2.with Regular and Irregular Icosidodecahedron It is and wellIrregular known that mid-points of the edges of a regular icosahedron or dodecahedron form 4.2. Regular Icosidodecahedron

the Archimedean solid icosidodecahedron with 30 vertices 32 faces (12 pentagons + 20 triangles) and It edges is well known that mid-points of Coxeter the edges of aofregular dodecahedron 60 which can be obtained from the graph as anicosahedron orbit (010) or[28]. An irregular form the Archimedean solid icosidodecahedron with 30 vertices 32 faces (12 pentagons + 20 triangles) icosidodecahedron consists of irregular pentagonal faces and scalene triangles in the most general and case and willcan be derived from from the chiral tetrahedral group will extended pyritohedral 60 edges which be obtained the Coxeter graph of and H3 as anbeorbit An irregular (010) by H3 [28]. group representing a larger symmetry. One can define fivescalene vectors triangles as follows the case icosidodecahedron consists of irregular pentagonal faces and in representing the most general vertices of the irregular pentagon which is orthogonal to the vertex Λ: and will be derived from the chiral tetrahedral group and will be extended by pyritohedral group + + can η(Λvectors + η(Λ representing + Λsymmetry.ΛOne Λ define five Λ)as follows Λ) representing=aΛlarger the vertices of the , = , = , = , 2 which is orthogonal 2 2 2 irregular pentagon to the vertex Λ: η(Λ +

c1 =

In terms of

Λ)

κ(Λ +

Λ)

Λ + r1 r2 Λ = Λ + r2 r1,Λ = η( Λ +r r , Λ ) , c2 = 2 , c3 = 2 2 2 3 , c4 = 2 2 η(Λ++3r3 r2+Λ2) + 2 + κ(4Λ + + 2r r3 Λ) η= c4 = , c+5 4= + 2 + 2,1 , +3 + 2 2 2 x2 + 3y2 + + 2x 4y2+ 2 2 2xy +2 + 4+ + = 2+ 3 + κη = . 2, y + 3x(2 + 2xy + 4x + 2y + + + 2) and

x2 + 3y2 + 2xy + 2x + 4y + 2 the vertices read: κ=

1 = [−( + 1) 2

+

( x + y + 2)2 +( +

+ 1) ],

η( Λ +r3 r2 Λ ) , 2

(37)

(37)

. (38)

Symmetry 2017, 9, 148

SymmetryIn 2017, 9, 148 terms of

12 of 22

x and y the vertices read:

12 of 22

1c1 = 1 [−(y + 1)e1 + ye2 + ( x + y + 1)e3 ], = [− 2 + ( + + 1) + ( + 1) ], 2 1 c2 = [−ye1 + ( x + y + 1)e2 + (y + 1)e3 ], 2 ( + + 1) + ( + 1) ], = [ + 2 η (38) c3 = [ xe1 + ( x + y + 1)e2 + ( x + 1)e3 ], 2 = [( + 1) + + ( + + 1) ], η 2 c4 = [( x + 1)e1 + xe2 + ( x + y + 1)e3 ], 2 . = κ( + 1) c 5 = κ ( β + 1 ) e3 . with , with and define three orbits under the chiral tetrahedral symmetry Here, of sizesHere, 12, 12 cand 6, respectively. Imposing the pyritohedral group invariance ( = ) and moreover, 1 with c2 , c3 with c4 and c5 define three orbits under the chiral tetrahedral symmetry of letting = and dividing each vector by athe scale factor 2 group , one obtains the(xusual sizes 12, 12 and 6, respectively. Imposing pyritohedral invariance = y)quaternionic and moreover, 2 vertices [28] by a scale factor 2τ , one obtains the usual quaternionic vertices lettingofx the = τicosidodecahedron and dividing each vector of the icosidodecahedron [28] 1 1 1 (± ± ), (± ), ± ± ± ), (± ± ± o 2 2 2n (39) 1 1 1 e ± σe ± τe , σe ± τe ± e , τe ± e ± σe , (± ) (± ) (± ) 2 3 2 3 2 3 1 1 1 2 2± , ± , ± 2 (39) ± e1 , ± e2 , ± e3 They constitute a subset of the quaternionic binary icosahedral group . An icosidodecahedron consistsThey of regular pentagons equilateral triangles as icosahedral shown in Figure A general irregular constitute a subsetand of the quaternionic binary group9a. I. An icosidodecahedron icosidodecahedron consists of 12 pentagons of edges , , as shown in Figure 9b. In addition, it has consists of regular pentagons and equilateral triangles as shown in Figure 9a. A general irregular 4 equilateral trianglesconsists with edges , 4 equilateral triangles edges in,Figure 12 scalene icosidodecahedron of 12 pentagons of edges a, b, with c as shown 9b. Intriangles addition,with it has edges , , where , , can be expressed in terms of and . 4 equilateral triangles with edges a, 4 equilateral triangles 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)

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

5. The Regular and Irregular Snub Cubes Derived from ( (B) )(+ (a a a) ) 5. The Regular and Irregular Snub Cubes Derived from W 3 1 2 3 The snub cube is a chiral Archimedean solid with 24 vertices, 60 edges and 38 38 faces (8 squares, The snub cube is a chiral Archimedean solid with 24 vertices, 60 edges and faces (8 squares, 6 +6 + 2424equilateral verticesand anditsitsdual dual be determined by employing the method same equilateraltriangles). triangles). Its Its vertices cancan be determined by employing the same method described in Sections 3 and 4. The proper rotational subgroup of the Coxeter group described in Sections 3 and 4. The proper subgroup of the Coxeter group W ( B3 )( is )the n rotational o  is the rotational octahedral group (4) ≈ { , ∪ [ , ]} , of order 24, which permutes the rotational octahedral group Sym(4) ≈ T, T ∪ [ T 0 , T 0 ] , 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 = r3 r2 and b = r2 r1 satisfying the satisfying the generation4 relations = = ( ) = 1. When Λ is taken as a general vector, the generation relations a = 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) (Λ, r2 r1 Λ, (r2 r1 )2 Λ), (Λ, r3 r2 Λ, (r3 r2 )2 Λ, r2 r3 Λ), (40) with respective edge lengths 2( + + ) andr 2( + + ). With the vertex Λ= q  
1 2 with respective edge lengths 2 a21 of 2 a22in+Figure a2 a3 + 10. + a71vertices a2 + a22 and Λ, we obtain a figure consisting as shown 2 a3 . With the vertex r1 r3 Λ = r3 r1 Λ, we obtain a figure consisting of 7 vertices as shown in Figure 10.

Symmetry 2017, 9, 148 Symmetry 2017, 9, 148

13 of 22 13 of 22

r3r2Λ

r1r3Λ

r1r2 Λ

4 3 r2r1Λ

5

Λ 1

(r3r2)2Λ

2 r2r3Λ

Figure10. 10.The Thevertices verticesconnected connectedtotothe thevertex vertex Λ Λ. . Figure

There There are 3 different edge lengths among 11 edges which are given as: q Λ− − r r Λ| = 2(2( a2++ a a ++ 1 a2),), Λ = ||r1 r2 Λ | 3 2 3 1 2 2 3 q (41) Λ| = = 2(2( a21++ a1 a2++ a22),), ||r3 r2 Λ Λ− − r1 r3Λ| (41) q |Λ − r1 r3 Λ| = |r2 r1 Λ − r2 r3 Λ| = 2( a21 + 12 a23 ) 1 |Λ − Λ| = | a Λ − a Λ| = 2( + y ) y y 3 1 Factoring by a2 6= 0 redefining x = a2 , y = a2 and α = x +22 + 1, β = 2 + 1, γ = 2 and dropping a2 , one obtains the following classes of faces of the irregular snub cube: Factoring by ≠ 0 redefining = p , = and = + + 1, = + 1, = and 6 equilateral triangles with edge length: A = 2(1 + x + x2 ), dropping , one obtains the following classes of faces of the irregular snub cube: p 8 squares with edge length: B = 2(1 + y + y2 ), 6 equilateral triangles with edge length: = 2(1r +  + ),  1 2 2 24 scalene triangles with edge lengths: A, B, C = 8 squares with edge length: = 2(1 + + ), 2 x + 2 y .

24 scalene triangles with edge lengths: , number = 2( of +vertices ). of an irregular snub cube is 24. Since the angular deficiency is δ = π6 ,, the The vertex Λ = a1 ω1 + a2 ω2 + a3 ω3 can be written in terms of quaternions as Λ = αe1 + βe2 + γe3 . number of vertices of an irregular snub cube is 24. Sincegenerated the angular is = , the The orbit bydeficiency the chiral octahedral group reads: + + can be written in terms of quaternions as Λ = α + + The vertex Λ = γ . The orbit generated theαe chiral octahedral group reads: W ( B3 )+ Λ by = {± ± βe ± γe , ± βe ± γe ± αe , ± γe ± αe ± βe , } 2 3 2 3 2 3 1 1 1 (42) + W ( B(3 ) ) ΛΛ0 = = {± α ± αe2 ± ± γe3,, ±±αe1 ± ± γe2±±αβe3,,±±γe1±±αβe2±± αe3},}, {± βe1 ± 42 where Λ0 = r1 Λ( is )the of (42) (1) Λ′ mirror = {± image ± α of± Λ. ,The ±α vertices ± ± , ±represent ± ± αan},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 = = 0, = 1. cuboctahedron for 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 1 2 we them according (1) classify The snub cube: A = Bto = their C, 1edge + x +lengths x2 = 1of + triangles y + y2 = and x2 +squares. 2y = 1 +the+variable = satisfying + (1) The snub = equal, = , 1one + eliminates + When all cube: edges are y = x2 − 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 F the tribonacci constant x = lim nF+n 1 ≈ 1.8393 where Fn is the n-th term in the Tribonacci series equation − − − 1 = n0→∞ which can also be written as + = 2 The solution is the 0, 0, 1, 1, 2, 4, 7, 13, 24, 44, . . .. ≈ 1.8393 where is the n-th term in the Tribonacci series tribonacci constant = lim → factor 1 ( x2 +1), vertices of the snub cube and its mirror image are given by: After dropping an overall 2 0, 0, 1, 1, 2, 4, 7, 13, 24, 44, ….  −1 e , ± 1 e the −1 e ± snub image are After dropping anΛoverall W ( B3 )+ = ± xefactor e1 ± x −of xe2its ± mirror e3 3 vertices 2 ± xe 3 , ± xcube 1 ± e2 ±( x +1), 1 and (43)  given by: W ( B3 )+ Λ0 = ±e1 ± xe2 ± x −1 e3 , ± xe1 ± x −1 e2 ± e3 , ± x −1 e1 ± e2 ± xe3 . ( ) Λ = {± ± ± ,± ± ± ,± ± ± } (43)

Symmetry 2017, 9, 148

14 of 22

Symmetry 2017, 9, 148

14 of 22

(

) Λ′ = {± ± ± , ± ± ± ,± ( ) 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.

Symmetry 2017, 9, 148

±

± ±

}. ±

}.

14 of 22

Figure 11. 11. TheThe snub cube. Figure snub 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 = −2 −and 2τ, theisosceles irregular snubcube cube consists of 6√2(2 equilateral triangles 8 squares For , 24 the irregular snub consists 6 equilateral triangles 8 squares edgesFor triangles of edges shownand in and Figure 12. of of √6 respectively √6,of √ x ==1,1, y = √√6, p+ 3) as edges 6 respectively and isoscelestriangles trianglesofofedges edges√6,6, + 3shown in Figure (2τ ) as shown edges √6 respectively and 24 24 isosceles 3) as in Figure 12. 12. √6, 6,2(2 2+

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 ). (for = 1, = −2 ). Figure 12. The irregular snub cube with squares, equilateral triangles, isosceles triangles and squares (for

= 1,

= −2 ).

3 2 (3) A= = ≠ C 6=, B, =x = 21 y−21−and 1 and (3) x3 − xx2 − − xx− 1−≠x0 − 1 6= 0 (3) = ≠ , = − 1 and x3 − x2 − x − 1 ≠ 0 For y==4 and 4 and corresponding β 3,= 3,= γ2, the = 2,irregular the irregular consists For to to α == 10,10, =√ snub snub cube cube consists √ x==7 7corresponding √ of √ of squares of edges 13, equilateral triangles of edges 57 and isosceles triangles with edges 57,of 57 For = 4 and = 7 corresponding to = 10, = 3, = 2, the irregular snub cube consists triangles of edges √57 and isosceles triangles with edges squares of edges √57, √57 √13, equilateral √ √ 13; all edges scaled by 2. and and edges √13, equilateral scaled by √2. triangles of edges √57 and isosceles triangles with edges √57, √57 squares of edges √13; all and √13; all edges scaled by √2. (4) y=y =−x21 and x3 − xx32 −−xx−2 1−≠ x0 − 1 6= 00 (4) A≠ 6= = B =, C, − 1 and (4) ≠ = , y= − 1 and x3 − x2 − x − 1 ≠ 0 √ equilateral triangles of sides For = 3, = 8 the irregular snub cube after rescaling by √2 has For x = 3, y = 8 the irregular snub cube after rescaling by √2 has√equilateral triangles of sides √ √ √ of3,sides isosceles triangles of edges √41, faces. For = = 8√41, theand irregular snub cube after rescaling by √2 has equilateral triangles of sides √13, squares √41, √13 as 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) ≠ ≠ (5) = C ≠6= B (5) A 6≠ The variables should satisfy the inequalities ≠ − 1 and ≠ 2( + 1). The satisfy theinequalities inequalities y≠6= x2− 1and and equilateral y2 ≠6=2( 2( x++triangles 1). For = 2 and should = 4, the irregular snub cube consists of−1scaled of sides √7, √ Thevariables variables should satisfy the 1). For = y = 4, the irregular snub cube consists of scaled equilateral triangles of sides Symmetry 148 of 22 7, squares ofx2017, sides and the scalene triangles of sides and as shown in Figure 13.15 For = 229,and and = 4, the irregular snub cube consists of scaled equilateral triangles of sides √13 √13 √10 √7, √7, √ √ √ √ squares 13 and ofsides sides √7,7,√13 13and and √10 10 as asshown shownininFigure Figure13. 13. squaresof ofsides sides √13 andthe the scalene scalene triangles triangles of

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:

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 , d 2 = ν ( n 1 e1 + n 2 e2 + n 3 e3 ) , d3 = 12 (e1 + e2 + e3 ),

(44)

d 4 = ν ( n 3 e1 + n 1 e2 + n 2 e3 ) , d 5 = ν ( n 1 e1 + n 3 e2 − n 2 e3 ) . Here the parameters are given by: n1 = ( x + 1) y + x, n2 = x, n3 = x (y + 1), µ=

2x +3y+4 4x +2y+4 ,

ν=

(45)

2x +3y+4 1 2 ( xy+ x +y)(2x +y+2)+ x (y+2)+yx (y+1)

Vertices of the dualnof the irregular snub cube can be written as three sets of orbits under the o   0 0 chiral octahedral group T, T ⊕ [ T , T ] , µ(±e1 , ±e2 , ±e3 ), 1 2 (± e1

± e2 ± e3 )

(46)

ν{(±n1 e1 ± n2 e2 ± n3 e3 ), (±n2 e1 ± n3 e2 ± n1 e3 ), (±n3 e1 ± n1 e2 ± n2 e3 )}. 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 −3 to the regular snub cube, the parameters are given by µ = 2x , ν = x 2 , n1 = x (2x + 1), n2 = x, n3 = x3 . The lengths of the edges of the pentagon satisfy the relations

| d1 − d2 | = | d1 − d5 | =

q

x 2 −1 4x ,

| d2 − d3 | = | d3 − d4 | = | d4 − d5 | =

√

(47) 2 − x,

Symmetry 9, 148The dual of the snub cube is shown in Figure 14. where x ≈2017, 1.8393.

16 of 22

Figure Dual of snub the snub cube (pentagonal icositetrahedron) with its pentagonal If we Figure 14.14. Dual of the cube (pentagonal icositetrahedron) with its pentagonal face. If face. we substitute substitute = 1, 0 obtain in (46)the weCatalan obtain the Catalan solid tetrakis hexahedron of theoctahedron). truncated x = 1, y = 0 in (46)=we solid tetrakis hexahedron (dual of the(dual 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 yand in (45) to each casecase above. WeWe willwill illustrate only thethe case forfor x == 2 and in and (45) (46) and corresponding (46) corresponding to each above. illustrate only case 2 y= 4 where the pentagon has three different edge lengths. This is the dual of the irregular snub cube 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 of three different edge lengths satisfying the relations | − | = | − | , | − | = | − |, | − |.

substitute = 1, octahedron).

= 0 in (46) we obtain the Catalan solid tetrakis hexahedron (dual of the truncated

We shall not discuss all duals of the irregular snub cubes. They can be obtained by substituting and2017, in9,(45) =2 Symmetry 148 and (46) corresponding to each case above. We will illustrate only the case for 16 of 22 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 (5). It is depicted in the Figure 15 with |its pentagonal three |= | − of three different edgeof lengths satisfying relations , | − of − | = | faces − |consisting different edge lengths satisfying the relations d − d = d − d , d − d = d − d , d − d | | | | | | | | | 2 5 2 3 3 5 |. 1 1 4 4 |, | − |.

Figure 15.15. Dual of of thethe irregular snub cube of of Figure 13.13. Figure Dual irregular snub cube Figure

5.2. Chiral Polyhedra with Vertices at at thethe Edge Mid-Points of of thethe Irregular Snub Cube 5.2. Chiral Polyhedra with Vertices Edge Mid-Points Irregular Snub Cube With a similar discussion to thetocase irregular icosidodecahedron in Sectionin4, Section we may 4, construct With a similar discussion theofcase of irregular icosidodecahedron we may a chiral polyhedron theassuming mid-points edges of anofirregular cube as vertices. The construct a chiral assuming polyhedron theofmid-points edges ofsnub an irregular snub cube asvectors vertices. whose tips constitute the plane orthogonal to the vector Λ in Figure 10 which are constructed similar The vectors whose tips constitute the plane orthogonal to the vector Λ in Figure 10 which are to constructed (37) and (38)similar and aretogiven by: (38) and are given by: (37) and

+ ( + 1) ], c1 = = y ++ 1)1) e1 + + y + 2+)e2) [( x[(+ + ( x(+ + 2 + ( y + 1 ) e3 ] , c2 = = y ++ 2)2) e1 + + x (+ y++ 1+)e1) [( x[(+ + (y (+ 1+)e1) 2 + (+ 3] ] c3 = µ0[(2x + y + 2)e1 + e2 + (y + 1)e3 ], = ′[(2 + + 2) + + ( + 1) ], c4 = µ0[(2x + y + 2)e1 + (y + 1)e2 − e3 ], = ′[(2 + + 2) + ( + 1) − ], c5 = v0( x + y + 2)(e1 + e2 ),

(48) (48)

+ 2)( α)γ + ),(α + β)α + ( β + γ) β + (γ + α)γ (α + β)α + ( β + =γ)′(β ++(γ + . , v0 = 2α2 + β2 + γ2 ( α + β )2 ( + ) +( + ) +( + ) ( + ) +( + ) +( + ) , ′= .  ′=  ( + ) + same + orbit of size 24 under the chiral The vertices c1 and c2 are2in the octahedral group { T, T ∪ , and the orbit involving c3 and c4 issame of sizeorbit 24. The orbit24of under c1 consists of cyclic permutations [ T 0 , T 0 ]}The vertices and are in the of size the chiral octahedral group of {the, coefficients of the unit quaternions with all possible sign changes. The same argument ∪ [ , ]}, and the orbit involving and is of size 24. The orbit of consistsisofvalid cyclic forpermutations the orbit of c3of . The orbit of c comprises of 12 quaternions obtained as the cyclic permutations of 5 the coefficients of the unit quaternions with all possible sign changes. The same theargument pair of quaternions with all sign changes. Actually, the orbit of c by itself represents the vertices is valid for the orbit of . The orbit of comprises of 125 quaternions obtained as the cyclic ofpermutations a cuboctahedron. Forpair all allowed x and y,with the chiral polyhedra 60 vertices can be of the of quaternions all sign changes.with Actually, the orbit of displayed. 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: µ0 =

| c1 − c2 | = | c2 − c3 | = | c4 − c5 | = | c5 − c1 | =

| c3 − c4 | √ 3 √ = 2x ≈ 3.528 2

(49)

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).

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 Symmetry 2017, 9, 148 17 of 22 sides 3.528 and 24 of length 4.989).

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 Regular and and Irregular Irregular Snub Snub Dodecahedron Dodecahedron Derived 6. The Derived from from W((H3 )) ((a1 a2 a3 ))

The dodecahedron is The snub snub dodecahedron is aa chiral chiral Archimedean Archimedean solid solid with with 60 60 vertices, vertices, 92 92 faces faces (20 (20 pentagons pentagons + 80 triangles) and 150 edges. We will discuss how to obtain the snub dodecahedron from an irregular +80 triangles) and 150 edges. We will discuss how to obtain the snub dodecahedron from an irregular snub dodecahedron. The vertices of an irregular snub dodecahedron, its dual and the chiral polyhedron snub dodecahedron. The vertices of an irregular snub dodecahedron, its dual and the chiral obtained fromobtained the mid-points of edges will beof constructed bybe employing the same technique described polyhedron from the mid-points edges will constructed by employing the same in Section 5. The proper rotational subgroup of the icosahedral Coxeter group is also called chiral technique described in Section 5. The proper rotational subgroup of the icosahedral Coxeter group is   + ≈ Alt 5 icosahedral group W H = I, I which is a simple group of order 60. They can be generated ( ) ( ) 3 also called chiral icosahedral group ( ) = , ≈ (5) which is a simple group of order 60. 3 = ( ab)2 = 1. by thecan generators a =byr1the r2 , generators b = r2 r3 which generation a5 = brelations , = the which satisfyrelations the generation = They be generated = satisfy Let=Λ(= a)1 ω=1 1 +. a2Let ω2 +Λ a=3 ω3 be+a general ωi , (vector i = 1, 2,where 3) can be, (obtained from + vector bewhere a general = 1, 2, 3) can(17). be The following vertices form sets a regular pentagon an equilateral triangle, obtained fromsets (17).ofThe following of vertices formand a regular pentagon and anrespectively: equilateral triangle,

respectively:

(Λ, r1 r2 Λ, (r1 r2 )2 Λ, (r1 r2 )3 Λ, (r1 r2 )4 Λ), (Λ, r2 r3 Λ, (r2 r3 )2 Λ) (50) ) ) ) ) ( ( ( ( (Λ, Λ, Λ, Λ, Λ), (Λ, Λ, Λ) (50) q q

with respective edge lengths 2 a21 + τa1 a2 + a22 and 2 a22 + a2 a3 + a23 . With the addition of the with respective edge lengths 2( + + ) and 2( + + ). With the addition of the q
2 + a2 . vertex r1 r3 Λ = r r Λ as shown in Figure 17, there are three edge lengths including 2 a Λ = 3 Λ1 as shown in Figure 17, there are three edge lengths including 2( + 1 ). The 3 The discussions of Sections andbe5repeated will be to repeated to regular obtain and the irregular regular and snub discussions of Sections 4 and 54will obtain the snubirregular dodecahedra. dodecahedra. The dualpolyhedron and the chiral based on the mid-points edges follow the The dual and the chiral basedpolyhedron on the mid-points of edges follow theof same technique. Symmetry 2017, 9, 148 18 of 22 same technique.

r2 r1Λ 3

(r1r2 ) Λ

2 3

(r1 r2 )2 Λ

r2 r3 Λ

Λ

4 r1r2 Λ

1

r3 r2 Λ

5 r1 r3 Λ

Figure17. 17.The Thevertices verticesconnected connectedto tothe thevertex vertexΛ Λ. . Figure

according to to the the H3 diagram 4. Factoring by Numbering of the faces has been done according diagram of Figure 4. a1 a3 a2 √ a2 ≠ 6= 0, 0,redefining redefining x == a,2 , y= = aand and dropping oneone obtains an irregular snub dodecahedron dropping 2 2 obtains an irregular snub dodecahedron with √ p with 12 triangles length: = +2(1++ x),+20 x2 )pentagons , 20 pentagons length: 12 equilateral triangles with with edgeedge length: = A2(1 edgeedge length: = p equilateral p with with 2 ), 60 scalene triangles with edge lengths: A, B, C = 2 + y2 ). The triangles B2(1 = + 2 ( 1+ + τy) + y 2 x ( , 60 scalene triangles with edge lengths: , , = 2( + ) . The triangles numbered as 2, 2, 4 and 5 are 4 and 5 are identical triangles, an equilateral triangle.  scalene  number  numbered as identical scalene triangles, but but number 1 is 1anisequilateral triangle. The π 3π π The angular deficiency is δ = 2π − + + θ + ϑ + ϕ = where θ, ϑ and ϕ are the 3 5 15 ( ) ) = where , and are the interior angles angular deficiency is = 2 − ( + + + + interior angles of the scalene triangles. Descartes’ formula verifies the number of vertices N0 = 60. = 60. of the scalene triangles. Descartes’ formula verifies the number of vertices − − ( + + 2) involves 60 The orbit , Λ generated from the vector Λ = − vectors representing the vertices of an irregular snub dodecahedron. The mirror image , Λ′ can − − ( + + 2) . A few remarks are in order be obtained from the vector Λ = Λ = before we discuss the usual classification. For some special values of the parameters and , we obtain some of the Archimedean solids. For example,

Numbering of the faces has been done according to the ≠ 0, redefining

=

,

=

and dropping



√

diagram of Figure 4. Factoring by

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 Symmetry 2017, of 22 numbered as 9,2,148 4 and 5 are identical scalene triangles, but number 1 is an equilateral triangle.18The angular deficiency is = 2 − ( + + ( + + ) ) = where , and are the interior angles   = 60. of theThe scalene number I, I Λ Descartes’ orbittriangles. generated formula from theverifies vector the Λ = − ye1of−vertices xe2 − τ (τy + x + 2)e3 involves   from the vector snub Λ = − − − The ( mirror + +image 2) involves 60 The orbit , Λ generated 60 vectors representing the vertices of an irregular dodecahedron. I, I Λ0 can vectors representing vertices of ran dodecahedron. image are , in Λ′order can be obtained from thethe vector Λ0 = = ye1 −snub xe2 − τ (τy + x + 2)The e3 . Amirror few remarks 1 Λirregular special − ( values + +of 2)the. parameters A few remarks in order be obtained from the usual vectorclassification. Λ = Λ = For − before we discuss some x andare y, we obtain before wethe discuss the usual classification. For some special values of the parameters and , we some of Archimedean solids. For example, 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. Letus usfollow followthe thesequence sequenceof ofSection Section55to todiscuss discussthe theregular regularand andirregular irregular snub snub dodecahedra. dodecahedra. Let

+ + y=2 = + (1) ,C, 1 +1 ++x + x=2 1=+1 + τy (1) The Thesnub snubdodecahedron: dodecahedron: A== B== x 2 + 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 = σ 1 − x2 and y2 = x + 1 resulting in cubic equation − 3 − 2 − = 0 . The real solution is approximately = 1.943 . The snub the cubic equation x − x − 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.

Figure18. 18.The Thesnub snubdodecahedron. dodecahedron. Figure

(2) (2)

  p 1 A== B≠ 6= ,C, y== (− ±τ ±( (+τ 24(+ 4(+x2 + )) x ) 2 −

For thethe irregular snub dodecahedron consists of 12 equilateral triangles and 20 For x ==−1, −1, y==−√ −, τ, irregular snub dodecahedron √consists √ pof 12 equilateral triangles and pentagons of sides 60 isosceles triangles of sides 2( + √2 and √2, √2, 20 pentagons of sides 2 and 60 isosceles triangles of sides 2, 2, 2(2) τ + 2) 2 =+ 3 − x− 2 −− (3) ≠ 6=, B, y= 1 and 0 0 (3) A== C x+ 1 and x− x −≠ τ 6=

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.

√

triangles of sides √2 , and isosceles triangle of sides 2( + 2), (5)

≠

2( + 2) and √2 .

≠

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 Symmetry 2017, 9, 148 19 of 22 as shown in Figure 19.

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

6.1. Dual of the Irregular Snub Dodecahedron We obtain 5 vectors as follows follows to to determine determine the the plane plane orthogonal orthogonal to to the the vertex vertex ΛΛ:: √ = √2 =d1(= − ), τ (σe1 − τe3 ), 2ω1 = d2 = = ν[( − σ)) xe1 − + y))e3]] [(y − − σy((x + + 11))e2 + + ((τxy + + τx + √ d3 = 2µω3 = µ(−e2 − τe3 ), = √2 = (− − ), d4 = ν[σxe1 − σye2 + (2xy + τx + y)e3 ]

= d [ = ν[− − σxe + +σye (2 ++(2xy++ )τx ]+ y)e ], 5 2 3 1 = + (2 + (3y + +)τx],+ 2τ ) + τx + + 2τ ) (3y [− , ν= . (51) τy + (σ + 2) x + 2 (σx − y + 2σ)(2xy + τx + y) (3 + + 2 ) (3 + + 2 ) , =represent the vertices of the dual . (51) The orbits generated=from five + vectors ( − + 2 )(2 + + ) solid of the irregular +these ( + 2) 2 snub dodecahedron. Faces of the dual consist of irregular pentagons, one of which is represented by  vectors    The orbits generated from the vertices the of dual solid of the the vertices of (51). Note that thethese orbitsfive I, I d1 andrepresent I, I d3 represent the of orbits sizes 20 and 12,  which  irregular snub dodecahedron. Faces of the dual consist of irregular pentagons, one of is respectively. The other three vertices are in the same orbit; therefore, one single notation I, I d2 for represented by the vertices of (51). Note that the orbits , and , represent the orbits of this orbit of size 60 suffices. Therefore, the dual consists of 92 vertices, 60 irregular pentagons and 150 Symmetry 2017, 9,12, 148 respectively. The other three vertices are in the 20 of 22 sizes andregular same orbit; therefore, onedual single edges.20The case is obtained by substituting y = σ (1 − x2 ) and x= 1.943 in (51). The of for this orbit of size 60 suffices. Therefore, the dual consists of 92 vertices, 60 notation , the snub dodecahedron, the pentagonal hexecontahedron, is shown Figure 20. 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. µ=

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.

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.

Figure21. 21.The Thedual dualof ofirregular irregularsnub snubdodecahedron dodecahedronwith with x = = 22 and Figure and y ==3.3.

6.2. Chiral Polyhedra with Vertices at the Edge Mid-Points of the Irregular Snub Snub Dodecahedron Dodecahedron Similar to wewe cancan construct a chiral polyhedron possessing the midto the theprevious previousdiscussions, discussions, construct a chiral polyhedron possessing the points of theofedges of the of irregular snub cube vertices. The vectors tips constitute the plane mid-points the edges the irregular snubascube as vertices. Thewhose vectors whose tips constitute orthogonal to the vector Λ in Figure 17Figure are given by:given by: the plane orthogonal to the vector Λ in 17 are 1 ) ++τ )e1−+ e(2 = c1[−(2 + 2y 2 + ++ 2)τ + ], 2)e3 ], + 2x = 12 [−( (2τy 2 − τ+ 2 c2 = 12 [−(2y + τx + τ )e1 − ( x + 1)e1 − τ (2τy + 2x + τ + 2)e3 ], 1 = [−(2 + + ) − ( + 1) − (2 + 2 + + 2) ],
 c3 = α22 −τ (τy + 1)e1 − (τy + 2x + 1)e2 − τ τ 2 y + 2x + τ + 2 e3 ,

=

c4 = α [τe1 − (2x + 1)e2 − τ (2τy + 2x + τ + 2)e3 ], [− ( 2+ 1) − ( + 2 + 1) − ( + 2 + + 2) ], 2 c5 = − βτ (τy + x + 2)e3 ,

= [ − + 1) +−x2 (2 2 + 2τxy 3y(2 + 4τy++2 x (+τ ++2)2)+ τ],+ 2 2 , α= τ 2 y2 + 2τxy + x2 (σ + 2) + (3τ + 1)y + 4x + τ + 2 (52) = − ( +2 + 2) , 2 3y + 2τxy + x + 4τy + x (τ + 2) + τ + 2 β= 3 +2 + (+ τy4+ x++ 2( )2+ 2) + + 2 = , +
2 + ( + 2) + (3 + 1) + 4 + + 2 2 3 2 (52) In the limit y = σ 31 − + x 2 , x +− x +−4x −+τ (= +02)and + α+ = 2 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 | c3 − c4 | p 2 Symmetry 2017, 9, 148 21 of 22 c1 −vertices c2 | = |cin − c = c − c = c − c = = x + xpentagon +1 (53) | | | | | dodecahedron, |the (52) are simplified, and the edge lengths of the satisfy the 2 3 5 5 4 1 τ relations: This chiral polyhedron has 150 vertices vertices and and 152 152 faces faces (12 (12 regular regular pentagons pentagons + 60 + 60 irregular | 60 | −irregular + 60) equilateral triangles all surrounding 60 irregular pentagons), pentagons and ((20 20 + 60 equilateral triangles all surrounding pentagons), one ) | − |=| − |=| − |=| − |= = + + 1 a typical(53) of which is represented with the relations of vertices in (53). The chiral polyhedron with 300 edges is illustrated in Figure 22.

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

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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.

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