Catalan solids derived from three-dimensional-root systems and quaternions

Catalan solids derived from three-dimensional-rootsystems and quaternions

Mehmet Koca,1,a� Nazife Ozdes Koca,1,b� and Ramazan Koç2,c�

1Department of Physics, College of Science, Sultan Qaboos University, P.O. Box 36,Al-Khoud, 123 Muscat, Oman2Department of Physics, Faculty of Engineering, Gaziantep University, 27310 Gaziantep,Turkey

�Received 22 August 2009; accepted 11 February 2010; published online 2 April 2010�

Catalan solids are the duals of the Archimedean solids, the vertices of which can beobtained from the Coxeter–Dynkin diagrams A3, B3, and H3 whose simple roots canbe represented by quaternions. The respective Weyl groups W�A3�, W�B3�, andW�H3� acting on the highest weights generate the orbits corresponding to the solidspossessing these symmetries. Vertices of the Platonic and Archimedean solids resultfrom the orbits derived from fundamental weights. The Platonic solids are dual toeach other; however, the duals of the Archimedean solids are the Catalan solidswhose vertices can be written as the union of the orbits, up to some scale factors,obtained by applying the above Weyl groups on the fundamental highest weights�100�, �010�, and �011� for each diagram. The faces are represented by the orbitsderived from the weights �010�, �110�, �101�, �011�, and �111�, which correspond tothe vertices of the Archimedean solids. Representations of the Weyl groups W�A3�,W�B3�, and W�H3� by the quaternions simplify the calculations with no reference tothe computer calculations. © 2010 American Institute of Physics.�doi:10.1063/1.3356985�

I. INTRODUCTION

The discovery of the Platonic solids—tetrahedron, cube, octahedron, icosahedron, anddodecahedron—dates back to the people of Scotland who lived 1000 years earlier than the ancientGreeks. The models carved on the stones are now kept in the Ashmolean Museum at Oxford.1

It took nearly another century after Plato’s association of tetrahedron with fire, cube withearth, air with octahedron, and water with icosahedron that Archimedes discovered the semiregularconvex solids. However, several centuries passed before their rediscovery by the renaissancemathematicians. Finally, Kepler completed the work in 1620 by introducing prisms and antiprismsas well as four regular nonconvex polyhedra, now known as the Kepler–Poinsot polyhedra. Con-struction of the dual solids of the Archimedean solids was completed in 1865 by Catalan2 nearlytwo centuries after Kepler. Extension of the platonic solids to four dimensions was made in 1855by Schläfli3 and their generalizations to higher dimensions in 1900 by Gosset.4 Further importantcontributions were made by Wythoff5 among many others and, in particular, by the contemporarymathematicians Coxeter6 and Conway.7

The three-dimensional �3D� and four-dimensional �4D� convex polytopes single out as com-pared to the polytopes in higher dimensions. The number of Platonic solids is 5 in three dimen-sions and there exist six regular polytopes in four dimensions, contrary to the higher dimensionalcases where we have only three platonic solids.

a�Electronic mail: [email protected]�Electronic mail: [email protected]�Electronic mail: [email protected].

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Let G be a root system of rank n. The orbit O�a1a2¯an��W�G��a1a2¯an� defines a poly-tope �nonregular in general� possessing the W�G� symmetry, where �= �a1a2¯an� is the highestweight with ai �i=1,2 , . . . ,n�, non-negative integers,8 and W�G� is the Coxeter–Weyl group. Theorbits O�10¯0�=W�An��10¯0� and O�00¯1�=W�An��00¯1� each defines an �n+1�-cell �orn-simplex�. They can be transformed to each other because of the Dynkin-diagram symmetry ofthe root system of An. They are said to be self-dual. The orbits O�10¯0�=W�Bn��10¯0� andO�00¯1�=W�Bn��00¯1� represent hyperoctahedron and hypercube in n-dimensions, which aredual to each other. These are the only Platonic polytopes existing in arbitrary dimensions. In threedimensions, in addition to the self-dual tetrahedron, octahedron, and cube, there are two morepolyhedra, the icosahedron and dodecahedron described by the Coxeter diagram H3. In fourdimensions, in addition to the Platonic polytopes described by the symmetries of W�A4� andW�B4�, we have 120-cell and 600-cell described by the Coxeter group W�H4� and the 24-cellwhose vertices were described by the orbits O�1000�=W�F4��1000� and O�0001�=W�F4��0001�.The 24-cell is said to be self-dual because the F4 diagram is invariant under the Dynkin-diagramsymmetry.9,10

Having said these general statements for the Platonic polytopes in arbitrary dimensions, wereturn back to the semiregular polyhedra in three dimensions. In Ref. 11, we have studied, indetail, the Platonic and the Archimedean solids employing the above technique where the simpleroots are described by quaternions. Consequently, the Coxeter–Weyl groups were expressed usingthe binary tetrahedral, binary octahedral, and the binary icosahedral subgroups of quaternions. Inthis paper, we construct the vertices of the Catalan solids, duals of the Archimedean solids, usingthe same technique described in the paper.11 In Sec. II we set up the general frame by summarizingthe technique in Ref. 11 and construct the vertices of the first Catalan solid invariant under thetetrahedral group W�A3��Td. Section III is devoted to the discussion of the Catalan solids pos-sessing the octahedral symmetry W�B3��Oh. The Catalan solids described by the icosahedralsymmetry W�H3�� Ih will be studied in Sec. IV. Section V is based on the concluding remarksregarding the use of the Catalan solids in the description of viral structures and other applications.

II. QUATERNIONIC DESCRIPTIONS OF THE RANK-3 COXETER–DYNKINDIAGRAMS

Let q=q0+qiei �i=1,2 ,3� be a real unit quaternion with its conjugate defined by q̄=q0−qiei

and the norm qq̄= q̄q=1. Here the quaternionic imaginary units satisfy the relations

eiej = − �ij + �ijkek �i, j,k = 1,2,3� , �1�

where �ij and �ijk are the Kronecker and Levi-Civita symbols and summation over the repeatedindices is implicit. With the definition of the scalar product

�p,q� = 12 �p̄q + q̄p� = 1

2 �pq̄ + qp̄� , �2�

quaternions generate the 4D Euclidean space. The group of unit quaternions is isomorphic toSU�2�, which is the double cover of the proper rotation group SO�3�. Its finite subgroups are thecyclic groups of the order of n, the dicyclic groups of the order of 2n, binary tetrahedral group Tof the order of 24, the binary octahedral group O of the order of 48, and finally the binaryicosahedral group I of the order of 120.12 To set the scene, let us define the 4D orthogonaltransformations in terms of quaternions. Let p and q be two arbitrary unit quaternions and rrepresent any quaternion. Then the O�4� transformations can be defined by

�p,q�:r → r� = prq, �p,q��:r → r� = pr̄q . �3�

When the unit quaternions p and q take values from the binary icosahedral group I, the set ofelements

043501-2 Koca, Ozdes Koca, and Koç J. Math. Phys. 51, 043501 �2010�


W�H4� = ��p,q� � �p,q�� 4�

represents the Coxeter group of the order of 14 400, the symmetry of the 120-cell, and the600-cell.13 �The notation � stands for the disjoint union of sets.� When q= p̄ is substituted into �4�,we obtain the Coxeter group W�H3� representing the icosahedral symmetry Ih of the order of 120,

W�H3� = ��p, p̄� � �p, p̄�� A5 � C2, p � I , �5�

where the group of even permutations of five letters A5= ��p , p̄� , p� I� is the proper icosahedralgroup without inversion and the generator �1,1�� representing the inversion in three dimensionsgenerates the cyclic group C2.

We can represent the elements of the binary octahedral group O as follows:

O = T � T�. �6�

Here T represents the binary tetrahedral group given by

T = ��1, � e1, � e2, � e3, 12 ��1 � e1 � e2 � e3�� , �7�

which also represents the quaternionic vertices of the polytope 24-cell. The dual polytope isrepresented by the quaternions

T� = 12

��1 � e1�,12

��e2 � e3�,12

��1 � e2�,12

��e3 � e1�,12

��1 � e3�,12

��e1 � e2�� .

�8�

Let p ,q�O be arbitrary elements of the binary octahedral group, then the set of elements

Aut�F4� � W�F4�:C2 = ��p,q� � �p,q�� 9�

is the extension of the Coxeter–Weyl group W�F4� by the diagram symmetry.9 The Weyl groupW�F4� is the symmetry group of the 24-cell, the vertices of which can be represented either by setT or T�. The C2 group in Aut�F4� transforms T and T� to each other, where the C2 group representsthe Dynkin-diagram symmetry of the F4 diagram. If p ,q�T, then the group W�D4� :C3= ��p ,q�� p ,q�� of the order of 576 is a maximal subgroup in both groups W�F4� and W�H4� andrepresents the symmetry group of the snub 24-cell.14 The W�F4� group has several subgroupsacting in the 3D space. The Weyl group W�A3��S4 can be represented by the set of elements

W�A3� = ��p, p̄� � �t, t̄��, p � T, t � T�. �10�

It is a subgroup of both groups W�H4� and W�F4�. Another subgroup of the W�F4� group is theoctahedral group Oh�W�B3� described by the set of elements

W�B3� = ��p, p̄� � �t, t̄� � �p, p̄�� t, t̄�� . �11�

In Ref. 11 we have shown that the orbit O�a1a2a3�=W�A3��a1a2a3� can be written as

��e1 � �e2 � �e3, � �e1 � �e2 � �e3, � �e1 � �e2 � �e3,

��e1 � �e2 � �e3, � �e1 � �e2 � �e3, � �e1 � �e2 � �e3 �12�

�even number of �� sign�, where �= 12 �a1−a3�, �= 1

2 �a1+a3�, and �= 12 �a1+2a2+a3�.

The orbits O�100� and O�001� each represents a tetrahedron. The orbits O�010� and O�111�,respectively, represent octahedron and truncated octahedron which possess a larger octahedralsymmetry W�B3��Aut�A3��W�A3� :C2 and will be considered in Sec. III. Here we have only a

043501-3 Catalan solids and quaternions J. Math. Phys. 51, 043501 �2010�


truncated tetrahedron, an Archimedean solid, having tetrahedral symmetry, which can be repre-sented either by the orbit O�110� or the orbit O�011�. The vertices of the orbit O�110� are givenby the set of quaternions

12 ��e1 � e2 � 3e3�, 1

2 ��e1 � 3e2 � e3�, 12 ��3e1 � e2 � e3� �even number of �− � sign� .

�13�

It consists of 12 vertices, 18 edges, and 8 faces �4 equilateral triangles and 4 regular hexagons�.The dual of this solid is the Catalan solid called triakis tetrahedron consisting of 8 vertices, 18edges, and 12 faces. Before we proceed further, we note a very important property, namely, theorder of the Weyl–Coxeter group divided by the size of the orbit, �W�G�� / �O�a1a2¯an�� gives theorder of a subgroup of the group W�G� fixing one element of the orbit. In the above case, it is acyclic group C2 of the order of 2. In the Catalan solids one face is represented by the vertex of theArchimedean solid orthogonal to this face. Therefore, the symmetry fixing one vertex of theArchimedean solid is the symmetry of the face of the Catalan solid. Keeping this aspect of thesymmetry of the face of the Catalan solid in mind, we now describe the vertices and the faces ofthe triakis tetrahedron. A table of Catalan solids can be found in Ref. 15.

Let us take one of the vertices of the truncated tetrahedron, say, the highest weight

q = � = �110� = �100� + �010� = 12 �e1 + e2 + e3� + e3. �14�

The set of centers of the triangular faces of the truncated tetrahedron can be represented by ascaled copy of the set of quaternions of the orbit which represents the vertices of a tetrahedron

O�100� = � 12 �e1 + e2 + e3�, 1

2 �e1 − e2 − e3�, 12 �− e1 − e2 + e3�, 1

2 �− e1 + e2 − e3�� 15�

and the set of centers of the hexagonal faces are represented up to a scale factor by the set ofquaternions of the orbit representing the dual tetrahedron

O�001� = � 12 �− e1 − e2 − e3�, 1

2 �− e1 + e2 + e3�, 12 �+ e1 + e2 − e3�, 1

2 �+ e1 − e2 + e3�� . �16�

It is not difficult to see that the line joining two quaternions from �16�,

B = 12 �− e1 + e2 + e3�, C = 1

2 �e1 − e2 + e3� �17�

is orthogonal to the vertex of the truncated tetrahedron given in �14�.The quaternion A= 1

2 �e1+e2+e3� from �15� and the quaternions B and C form an isoscelestriangle; however, it is not orthogonal to the vertex of the truncated tetrahedron in �14�. In orderto obtain an isosceles triangle orthogonal to the vertex of �14�, one can change the scale of thevertex at A and redefine it as A= � /2��e1+e2+e3�. Now one can check that the isosceles triangleABC is orthogonal to the vertex in �14� provided = 3

5 =0.6. The relative magnitude of thesevertices is �A� / �B�= �A� / �C�=0.6. When the W�A3� group acts on A, it will generate the orbitO�100� and, similarly, when it acts on the B or C, it will generate the set of vectors in the orbitO�001�. This shows that the vertices of the triakis tetrahedron lie on two orbits of W�A3�. If werescale the quaternions on these two orbits as the unit quaternions, then the vertices of the triakistetrahedron lie on two spheres with radii of 0.6 and 1. To determine the dihedral angle betweentwo adjacent isosceles triangular faces, we rotate the system around the unit quaternion �1 /3��e1+e2+e3� by the element � 1

2 �1+e1+e2+e3� , 12 �1−e1−e2−e3�� of the W�A3� group, which leads

to the isosceles triangle ACD, where D= 12 �e1+e2−e3�. The vertex of the truncated tetrahedron

orthogonal to the triangular face ACD is the quaternion q�= 12 �e1+e2+e3�+e1. The obtuse angle

between the quaternions q and q� is the dihedral angle �=129°31�16� between two adjacent facesof the triakis tetrahedron. Since for the A vertex we have three isosceles triangular faces, we havealtogether 3�4=12 faces of the triakis tetrahedron and its eight vertices are in two orbits of theW�A3� group. One orbit cannot be transformed to the other orbit by the W�A3� group; therefore, thetriakis tetrahedron is not vertex transitive but it is face transitive because the faces are represented



by the vertices of the truncated tetrahedron. It has 18 edges; 12 of which are of equal length. Eachedge joins one element from O�100� to two elements of O�001� and six edges of equal lengthjoining elements of O�001�. The truncated tetrahedron and its dual triakis tetrahedron are depictedin Fig. 1.

III. THE CATALAN SOLIDS POSSESSING THE OCTAHEDRAL SYMMETRY W„B3…

The Archimedean solids with the octahedral symmetry can be represented by the orbits ob-tained by the highest weights �010� �cuboctahedron�, �110� �truncated octahedron�, �011� �trun-cated cube�, �101� �small rhombicuboctahedron�, and �111� �great rhombicuboctahedron�.11 ThePlatonic solids, octahedron and its dual cube, are represented by the orbits obtained from thehighest weights �100� and �001�, respectively. A general orbit derived from the highest weight�= �a1a2a3�=�e1+�e2+�e3 can be represented by the following 48 quaternions:

O�a1a2a3� = ��e1 � �e2 � �e3; � �e1 � �e2 � �e3; � �e1 � �e2 � �e3;

� �e1 � �e2 � �e3; � �e1 � �e2 � �e3; � �e1 � �e2 � �e3� . �18�

Here we have �= �a1+a2+a3 /2�, �= �a2+a3 /2�, and �=a3 /2. The orbit O�100�= ��e1 , �e2 , �e3� represents the vertices of an octahedron. The orbit O�001�= �1 /2��e1�e2�e3� represents the vertices of the cube and the orbit O�010�= ��e1�e2� , ��e2�e3� , ��e3�e1�� represents the vertices of a cuboctahedron. These orbits areessential for the determination of the vertices of all Catalan solids possessing the octahedralsymmetry. Next we discuss the construction of the vertices of the rhombic dodecahedron �dual ofcuboctahedron�.

A. Rhombic dodecahedron „dual of cuboctahedron…

The cuboctahedron is depicted in Fig. 2�a�, which shows that at one vertex two square facesand two triangular faces meet. The number of square faces is 6 and their centers are represented,

(a) (b)

FIG. 1. �Color online� Truncated tetrahedron �a� and its dual triakis tetrahedron �b�.

(a) (b)

FIG. 2. �Color online� The cuboctahedron �a� and its dual rhombic dodecahedron �b�.



up to a scale factor, by the orbit O�100�= ��e1 , �e2 , �e3�. Similarly, there are eight triangularfaces and their centers are represented by the orbit O�001�= �1 /2��e1�e2�e3� scaled by somefactor. The vertex represented by the highest weight q= �010�=e1+e2 is surrounded by two squarefaces, whose centers, up to a scale factor, are represented by the quaternions A=e1 and C=e2,which belong to the orbit O�100�. Similarly, the centers of the triangular faces meeting at the samepoint can be represented, up to a scale factor, by the vertices B=e1+e2+e3 and D=e1+e2−e3,which belong to the orbit O�001�. It is clear that the lines AC and BD are orthogonal to the vertexq. However, the vertex q is not orthogonal to the lines AB and AD. Keeping A=e1 and C=e2 asthey are but rescaling B=�e1+e2+e3� and D=�e1+e2−e3�, then one can determine the scalefactor by the requirement that AB and BD are orthogonal to the vertex q. This will determine thescale factor = 1

2 . One can easily check that the vertices ABCD form a rhombus orthogonal to thevertex q=e1+e2. This shows that the rhombic dodecahedron has 14 vertices lying on two orbitsgiven by the set of quaternions O�100�= ��e1 , �e2 , �e3� and the set �1 /2�O�001�= �1 /2��e1�e2�e3�. It is depicted in Fig. 2�b�. The system can be rotated by the element � 1

2 �1+e1

+e2+e3� , 12 �1−e1−e2−e3�� of the group W�B3� to determine the vertex q�=e2+e3, where A→C

=e2, C→E=e3. B remains fixed and D→F= 12 �−e1+e2+e3�. The new rhombus CBEF is orthogo-

nal to the vertex q�. The dihedral angle then is �=120°. The two vertex orbits lie on concentricspheres with radii of r1=1 and r2=3 /2�0.866. The rhombic dodecahedron has 14 vertices, 12faces �rhombus�, and 24 edges. The centers of edges are represented by the set of 24 quaternions

12 ��e1 � e2 � 3e3�, 1

2 ��e1 � 3e2 � e3�, 12 ��3e1 � e2 � e3� . �19�

The set of quaternions in �19� represents the vertices of the union of two truncated tetrahedra, onedefined by the orbit O�110� as given in �13� and the other by the orbit O�011�, which is obtainedby changing the sign of the elements of the orbit O�110� of �13�. Since the octahedral group isW�B3��Aut�A3�, the set of quaternions in �19� is a single orbit under the octahedral group.Therefore the rhombic dodecahedron is not only face transitive but also edge transitive. Since wehave �W�B3�� / �O�010��=4, the symmetry of the face is a group of the order of 4. The onlysymmetry in this case is the C2�C2 group generated by the elements �e3 ,−e3�� and ��1 /2��e1

−e2� ,−�1 /2��e1−e2�� which fixes the rhombus ABCD.

B. Tetrakis hexahedron „dual of truncated octahedron…

Vertices of the truncated octahedron can be determined from the highest weight �= �110�= �100�+ �010�=e1+ �e1+e2�=2e1+e2. Truncated octahedron is similar to the cuboctahedron; how-ever, triangular faces are replaced by the hexagonal faces. As in the previous case, the centers ofthe six square faces are represented by the elements of the orbit O�100�= ��e1 , �e2 , �e3�, andthe centers of the eight hexagonal faces are represented by the elements of the orbit O�001�= �1 /2��e1�e2�e3� up to some scale factors. Two hexagonal faces and one square face meetat a common vertex, as shown in Fig. 3�a�. If we take the common vertex to be the quaternionq=2e1+e2, then the surrounding vertices of the isosceles triangle can be taken to be A=e1, B

(a) (b)

FIG. 3. �Color online� The truncated octahedron �a� and its dual tetrakis hexahedron �b�.



=e1+e2−e3, and C=e1+e2+e3. It is obvious that the line BC is orthogonal to the vertex q. Wecheck that q will be orthogonal to the plane of the isosceles triangle ABC if = 3

2 . The action of theW�B3� group on these quaternions will generate the 14 vertices tetrakis hexahedron. This provesthat the set of eight vertices lie on a sphere with radius of r1=3�1.73 and the six vertices lie onthe sphere with radius of r2=1.5. The symmetry of the isosceles triangle ABC is the reflectiongroup C2 with respect to the x-y plane generated by �e3 ,−e3��.

The number of edges can be easily determined as 6�4+ ��8�3� /2�=36. A rotation by theelement �1 /2�1+e1� ,1 /2�1−e1�� around the unit quaternion e1 by � /2 will lead to a new vertexq�=2e1+e3 and the triangle ACD with D=e1−e2+e3. The dihedral angle between two adjacenttriangles is �=143°7�48�. The faces of the tetrakis hexahedron are represented by the 24 verticesof the truncated octahedron, which are given as

��2e1 � e2�, ��2e2 � e3�, ��2e3 � e1�, ��e1 � 2e2�, ��e2 � 2e3�, ��e3 � 2e1� .

�20�

The sets of vertices of the tetrakis hexahedron consist of two orbits of W�B3� and are given by

��32e1, �

32e2, �

32e3�, ��e1 � e2 � e3� . �21�

The tetrakis hexahedron with the above vertices is illustrated in Fig. 3�b�

C. Triakis octahedron „dual of truncated cube…

As shown in Fig. 4�a� the truncated cube consists of six octagonal and eight triangular faces.The centers of the octagonal faces are represented by the set of vertices of octahedron up to a scalefactor and the centers of the triangular faces are represented by the set of quaternions correspond-ing to the vertices of a cube subject to a change of scale. Let the vertex �= �011�=q3=e1+e2

+ �1 /2��e1+e2+e3� represent the quaternion orthogonal to the isosceles triangle ABC representedby the quaternions A=�e1+e2+e3�, B=e1, and C=e2. The line BC is orthogonal to q3 and isdetermined as =2−1 from the requirement that the triangle ABC be orthogonal to q3. Then theset of vertices of triakis octahedron are given by 14 quaternions

�2 − 1��e1 � e2 � e3�, ��e1, � e2, � e3� . �22�

This means that the eight vertices of triakis octahedron are on a sphere with radius of r1= �2−1�3�0.717 and the six vertices are on a sphere with radius of 1, as shown in Fig. 4�b�. Thesymmetry of the above triangle ABC is the cyclic group C2 generated by the element ��1 /2��e1−e2� ,−�1 /2��e1−e2�� which interchanges e1↔e2 and fixes e3. The vertices of the truncatedcube, which are orthogonal to the faces of the triakis octahedron are given by the set of 24quaternions

(a) (b)

FIG. 4. �Color online� The truncated cube �a� and its dual triakis octahedron �b�.



1 +12

��e1 � e2� �12

e3, 1 +12

��e2 � e3� �12

e1,

1 +12

��e3 � e1� �12

e2. �23�

The number of edges can be easily determined as 8�3+ ��6�4� /2�=36.Rotating the system around the unit quaternion �1 /3��e1+e2+e3�, the triangle ABC is rotated

to the triangle ACD, where D=e3 and the vertex q3 is transformed to q1= �1 /2�e1+ �1+ �1 /2��e2+e3�. The dihedral angle between the two triangular faces is �=147°21��0.4��.

D. Deltoidal icositetrahedron „dual of the small rhombicuboctahedron…

The small rhombicuboctahedron is displayed in Fig. 5�a� which consists of 24 vertices, 48edges, and 26 faces �8 equilateral triangles, 12+6 squares�. It is clear from the classification offaces that the dual solid deltoidal icositetrahedron will consist of vertices which will be repre-sented, up to scale factors, by three classes of orbits, namely, O�001�= �1 /2��e1�e2�e3�,O�010�= ��e1�e2� , ��e2�e3� , ��e3�e1��, and O�100�= ��e1 , �e2 , �e3�. The vertex corre-sponding to the highest weight of the orbit of the small rhombicuboctahedron can be given byq1=�= �101�=e1+ �1 /2��e1+e2+e3�. We have 24 vertices of the rhombicuboctahedron derivedfrom O�101�=W�B3��101�,

��e1 + ��e2 � e3�, � �e2 + ��e3 � e1�, � �e3 + ��e1 � e2� ,

� = 1 +12

�, � =12

. �24�

In small rhombicuboctahedron a vertex is surrounded by three squares and one equilateral triangle.It is easy to see that the centers of the squares and the triangle surrounding the vertex q1 givenabove can be, up to the scale factors, given by the quaternions A=e1, B= �1 /2��e1+e2�, C=�e1+e2+e3�, and D= �1 /2��e1+e3�. Note that the lines AB, AD, and BD are all orthogonal tothe quaternion q1. To have the vertex C in the same plane, we should have = �2+1� / �2+3��0.547. We obtain a kite with AB=AD and CB=CD. Rotating the system around the vertex C by120°, one obtains the vertices A�=e2, B�= �1 /2��e2+e3�, D�=B, and C�=C so that the BC is thecommon edge between the two kites. The new kite A�B�CB is orthogonal to the quaternion q2

=e2+ �1 /2��e1+e2+e3�; therefore, the dihedral angle between two faces is �=138°7�5�. The kiteABCD has the symmetry group C2 generated by the element ��1 /2��e1−e2� ,−�1 /2��e1−e2��.The 26 vertices of the deltoidal icositetrahedron are given by the sets of quaternions

(a) (b)

FIG. 5. �Color online� The small rhombicuboctahedron �a� and its dual deltoidal icositetrahedron �b�.



��e1, � e2, � e3�, 12

��e1 � e2�,12

��e2 � e3�,12

��e3 � e1��,2 + 12 + 3

��e1 � e2 � e3� ,

�25�

as shown in Fig. 5�b�. It is clear from the discussion that the 18=6+12 vertices lie on a sphere ofunit length and the remaining eight vertices lie on a sphere with radius of 0.547.

E. Disdyakis dodecahedron „dual of the great rhombicuboctahedron…

The great rhombicuboctahedron has 6 octagonal faces, 8 hexagonal faces, and 12 square faces,as shown in Fig. 6�a�. Since the great rhombicuboctahedron has 48 vertices the symmetry whichfixes one vertex is trivial, namely, just the unit element. This means the face of the disdyakisdodecahedron has no symmetry at all. The face will be a scalene triangle as we now discuss. Thehighest weight for the great rhombicuboctahedron is

q1 = � = �111� = e1 + �e1 + e2� +12

�e1 + e2 + e3� = �e1 + �e2 + �e3, �26�

where �=2+1 /2, �=1+1 /2, and �=1 /2. This vertex is surrounded by one octagon, onehexagon, and one square, centers of which are represented, respectively, by the quaternions A=e1, B=�e1+e2�, and C= �e1+e2+e3�. The parameters and are determined from the require-ment that the vertex q1 be orthogonal to the triangle ABC as = �22+1� / �32+2��0.613 and = �22+1� / �32+3��0.529. Therefore, the vertices of disdyakis dodecahedron lie on threedifferent concentric circles with radii of r1=1 �6 vertices�, r2=3 �0.916 �8 vertices�, and r3

=2�0.867 �12 vertices�. The 26 vertices can be determined by rescaling the orbits as O�100�,O�010�, and 2 O�001�. This is plotted in Fig. 6�b�. To determine the adjacent triangle and thequaternion orthogonal to it can be achieved by applying the group element �e3 ,−e3�� on thevertices of the triangle ABC to obtain the triangle ABC� where C�= �e1+e2−e3�. The same groupelement transforms q1 to q2=e1+ �e1+e2�+ �1 /2��e1+e2−e3�. Then the dihedral angle betweenthe faces is �=155°4�56�.

IV. THE CATALAN SOLIDS POSSESSING THE ICOSAHEDRAL SYMMETRY GROUPW„H3…

We have discussed the Archimedean solids in Ref. 11 possessing the icosahedral symmetryW�H3� defined in �5�. The orbits �O�100� and �O�001� represent two dual Platonic solids, namely,dodecahedron and icosahedron, respectively. Here, �= �1−5� /2 and the golden ratio �= �1+5� /2 satisfy the relations �+�=1, ��=−1, �2=�+1, and �2=�+1. An overall factor � comesfrom the projection of the six-dimensional Dynkin–Coxeter root system of D6 to three dimensions.Half of the 60 roots of D6 represent the vertices of the icosidodecahedron and the remaining 30 isits scaled copy by �. The Archimedean solids of the icosahedral symmetry are the orbits �O�010�,�O�110�, �O�011�, �O�101�, and �O�111�. They describe, respectively, the Archimedean solids,

(a) (b)

FIG. 6. �Color online� The great rhombicuboctahedron �a� and its dual disdyakis dodecahedron �b�.



icosidodecahedron, truncated dodecahedron, truncated icosahedron, small rhombicosidodecahe-dron, and great rhombicosidodecahedron. Now we construct their Catalan solids in turn.

A. Rhombic triacontahedron „dual of the icosidodecahedron…

The orbit describing the vertices of the icosidodecahedron is given by11

�O�010� = �W�H3��010�

= ��e1, � e2, � e3, 12 ��e1 � �e2� � e3�, 1

2 ��e2 � �e3� � e1�, 12 ��e3 � �e1� � e2�� .

�27�

Faces of the icosidodecahedron consist of 20 equilateral triangles and 12 regular pentagons, asshown in Fig. 7�a�. The rhombic triacontahedron has 32=20+12 vertices and 30 faces. The orderof the symmetry group of the faces, in other words, the order of the group fixing one vertex of theicosidodecahedron is �W�H3�� /30=4. As expected, the face is a rhombus having the symmetrygroup C2�C2. Let e1 represent the vertex, a common point of two pentagons and two triangles.One can easily determine the centers of the triangular faces as A=��e1+�e3� , C=��e1−�e3� andthe centers of the pentagonal faces as the quaternions B= ��e1+�e2� , D= ��e1−�e2�. These ver-tices define a golden rhombus, meaning that the ratio of the diagonals is �. The rhombus ABCD isorthogonal to the quaternion e1. The centers of the triangular faces belong to the orbit

�O�100� = � 12 ��e1 � �e3�, 1

2 ��e2 � �e1�, 12 ��e3 � �e2�, 1

2 ��e1 � e2 � e3�� . �28�

The centers of the pentagons belong to the orbit �O�001� given by

�O�001� = � 12 ��e1 � �e3�, 1

2 ��e2 � �e1�, 12 ��e3 � �e2�� . �29�

Therefore, the vertices of the rhombic triacontahedron consist of 32 vertices of �28� and �29� up tothe scale factors defined above. This indicates that the 20 vertices lie on a sphere with the radius�A�=�+2, and the 12 vertices lie on a sphere of radius �B�=3. The ratio between the two radiiis �A� / �B�=��+2� /3�1.098. If we rotate the system around the vertex A which can be done bythe group element �q , q̄� with q= 1

2 �1+�e1+�e2�, then e1→ 12 ��e1−e2−�e3� and the rhombus

ABCD is transformed to the adjacent one, which is orthogonal to the quaternion 12 ��e1−e2

−�e3�. Then one can determine the dihedral angle between two adjacent faces as �=144°. TheCatalan solid, rhombic triacontahedron is both face transitive and edge transitive. Indeed, therhombic dodecahedron and the rhombic triacontahedron are the only Catalan solids that are edgetransitive. The centers of the edges of the rhombus ABCD are given by the set of quaternions

(a) (b)

FIG. 7. �Color online� Icosidodecahedron �a� and its dual rhombic triacontahedron �b�.



A + B

2= �2�e1 + �e2 − e3�,

A + D

2= �2�e1 − �e2 − e3� ,

C + B

2= �2�e1 + �e2 + e3�,

C + D

2= �2�e1 − �e2 + e3� . �30�

These vertices belong to the orbit of size 60 edges obtained from the highest weight ��10��= 1

2 �−�e1+e2+2�e3�. The 60 quaternions representing the centers of the edges can be obtained byapplying W�H3� on any quaternion in �30� or on the highest weight. In short, the rhombic tria-contahedron has 30 faces, 32 vertices, and 60 edges.

B. Triakis icosahedron „dual of the truncated dodecahedron…

The truncated dodecahedron has 20 triangular and 12 decagonal faces with 60 vertices as wellas 90 edges, as shown in Fig. 8�a�. Therefore, the dual solid will have 32 vertices consisting of twoorbits and 60 faces. From the symmetry consideration, the order of the group fixing one vertex ofthe truncated dodecahedron is �W�H3�� /60=2. This indicates that the face of the dual solid has aface of cyclic symmetry C2 implying that the face is an isosceles triangle as we now discuss.

The highest weight of the truncated dodecahedral orbit is q1=�=��110�= 12 �−�e1+�e3�+e3,

which is the sum of the highest weights describing dodecahedron and the icosidodecahedron. Thecenters of two decagonal faces to this vertex can be taken as B=−�e2+e3 and C=�e2+e3, where�B�= �C� and the line BC is orthogonal to the vertex q1 represented by the highest weight. Thevertex A=�−�e1+�e3�, which represents the center of the triangular face up to a scale factor,determines the isosceles triangle ABC orthogonal to the vertex q1 provided = ��+2� / �2�+3�,which leads to the ratio of two radii of two concentric spheres �A� / �B�= ��+2� / �2�+3��3 / ��+2��0.855. This shows that in the outer sphere we have 12 icosahedral vertices, oneof which is connected to two vertices in the inner sphere which represent the dodecahedralvertices. They are given in �28� and �29�. When the system is rotated by 120° around the vertex A,one obtains an equilateral triangle BCD. Here, A represents the center of the triangle BCD up toa scale factor dividing it into three isosceles triangles. The normal vectors of these isoscelestriangles are represented by the quaternions

q1 = 12 �− �e1 + �� + 2�e3�, q2 = 1

2 ��e1 + �e2 + 2�e3�, q3 = 12 ��e1 − �e2 + 2�e3� . �31�

The dihedral angle can be determined as �=160°36�45� using the scalar product between any pairof vectors. The triakis icosahedron has been shown in Fig. 8�b� where three isosceles trianglesmeet at one vertex.

C. Pentakis dodecahedron „dual of the truncated icosahedron…

The truncated isosahedron with its 60 vertice and 32 faces �12 regular pentagons and 20regular hexagons� is a model of a C60 molecule. Its dual, the Catalan solid pentakis dodecahedron,

(a) (b)

FIG. 8. �Color online� Truncated dodecahedron �a� and its dual solid triakis icosahedron �b�.



has 32 vertices �12 from the icosahedral orbit and 20 from the dodecahedral orbit� and the 60 facesare isosceles triangles as expected from the fact that �W�H3�� /60=2. The isosceles triangle, deter-mined by the vertices A=�−�e2+e3�, B= �−�e1+�e3�, and C= �e1+e2+e3� is orthogonal to thevertex q= 1

2 �e1−2�e2+�2e3�, which determines the factor =3� / ��+4�. The ratio of radii of twospheres having 20 and 12 vertices, respectively, is �B� / �A�= ��+4� /3��+2��1.0265. This showsthat the dodecahedral vertex lie on the outer sphere and the icosahedral vertices lie on the innersphere. Five isosceles triangles meet at one vertex of the icosahedral vertices and six of them meetat one dodecahedral vertex. The dihedral angle between two adjacent faces can be determined as�=156°43�7� in a similar manner as explained in other cases. To count for the number of edgesis also easy. From each icosahedral vertex five equal edges originate, yielding 12�5=60, and foreach triangle, there exits one more extra edge shared by another triangle which is 60

2 =30. There-fore, the total number of edges is 90. The whole set of 32 vertices of the pentakis dodecahedroncan be written as

��e1 � �e3�,��e2 � �e1�,��e3 � �e2�,��e1 � e2 � e3�� ,

3�

� + 4��e1 � �e3�,��e2 � �e1�,��e3 � �e2�� . �32�

The truncated icosahedron as well as its dual pentakis dodecahedron are displayed in Fig. 9.

D. Deltoidal hexacontahedron „dual of small rhombicosidodecahedron…

The small rhombicosidodecahedron consists of 60 vertices, 62 faces �12 pentagons, 30squares, and 20 equilateral triangles�, and 120 edges, as shown in Fig. 10�a�. The Catalan solid,deltoidal hexacontahedron, as expected, consists of 62 vertices which will be in three sets oficosahedral, dodecahedral, and icosidodecahedral orbits. One vertex of the small rhombicosi-dodecahedron is shared by one pentagonal, one triangular, and two square faces. The symmetryfixing one vertex of the small rhombicosidodecahedron is a cyclic group C2. It indicates that the

(a) (b)

FIG. 9. �Color online� Truncated icosahedron �a� and its dual pentakis dodecahedron �b�.

(a) (b)

FIG. 10. �Color online� The small rhombicosidodecahedron �a� and its dual deltoidal hexacontahedron �b�.



face of the deltoidal hexacontahedron will be a kite of twofold symmetry. The highest weightleading to the orbit of vertices of the small rhombicosidodecahedron is q=�=��101�= 1

2 �−�e1

−�e2+�2e3�. The vertices of the kite which is orthogonal to this vertex is determined to be

A = �− �e2 + e3�, B = e3, C = �− �e1 + �e3�, D = 12 �e1 − �e2 + �e3� . �33�

The orthogonality of the vertex q to the kite ABCD determines =�2 /3 and =�2 / ��+3�. Verticesof the deltoidal hexacontahedron lie on three spheres with the radii of �A�=��+2 /3�1.0259,�B�= �D�=1, �C�=�23 / ��+3��0.982. On the outer sphere there exist 12 icosahedral vertices, inthe middle sphere there exist 30 icosidodecahedral vertices, and in the inner sphere there exist 20dodecahedral vertices. A rotation around the vertex A by 2� /5 would take q to q�= 1

2 �e1+�e2

+2e3�. Then the scalar product of these two vectors determine the dihedral angle as �=154°7�17�. The 60 faces of the deltoidal hexacontahedron are determined by the orbit �O�101�,which can be found in Ref. 11. It is shown in Fig. 10�b�.

E. Disdyakis triacontahedron „dual of the great rhombicosidodecahedron…

Vertices of the great rhombicosidodecahedron can be obtained by applying W�H3� on thehighest weight

� = ��111� = ��100� + ��010� + ��001� = 12 �− �e1 + �e3� + e3 + 1

2 �− �e2 + e3� . �34�

The whole set of 120 vertices can be found in Ref. 11. The great rhombicosidodecahedron, asshown in Fig. 11�a�, has 12 decagonal, 20 hexagonal, and 30 square faces. At one vertex, threedifferent faces meet. It is then obvious that the face of the dual polytope disdyakis triacontahedronis a scalene triangle without any nontrivial symmetry. One can argue that the three vertices A=�−�e2+e3�, B= �−�e1+�e3�, and C= 1

2 �e1−�e2+�e3� representing, respectively, the centers ofthe decagonal, hexagonal, and the square faces form a triangle. It is orthogonal to the vertex q= 1

2 ��e1−2e2+�3e3�, which is one of the vertices of the great rhombicosidodecahedron. The factorsare determined, as usual, from the orthogonality of the vertex q to the triangle ABC which resultsin = ��+3� /5 and = �2�+3� /3. This shows that the vertices of the disdyakis triacontahedron lieon three concentric spheres. The outermost sphere has a radius �A�= ��+3� /5��+2�1.0858; onits surface there lies the set of 12 vertices of icosahedron. The next sphere contains 20 vertices ofa dodecahedron with the radius equal to �B�= �2�+3� /3�1.0184. The inner sphere with radius�C�=1 contains 30 vertices of icosidodecahedron. When we rotate the system around the vertex Aby 2� /5, one obtains the vertex q�= 1

2 �2e1+5e2+ ��+2�e3�, which is orthogonal to the triangleAB�C�, where B� and C� are obtained from B and C, respectively, with the same rotation. Thedihedral angle between two adjacent faces is then �=164°53�16�. The disdyakis triacontahedronis shown in Fig. 11�b�.

(a) (b)

FIG. 11. �Color online� The great rhombicosidodecahedron �a� and its dual disdyakis triacontahedron �b�.



V. CONCLUSION

In this work, we displayed a systematic construction of the Catalan solids, the dual solids ofthe Archimedean solids, with the use of Coxeter–Weyl groups W�A3�, W�B3�, and W�H3�. Weemployed the highest weight method for the irreducible representations of Lie algebras to deter-mine the orbits. Catalan solids are face transitive, meaning that the faces are transformed to eachother by the Coxeter–Weyl group. The vectors orthogonal to the faces are the vertices of theArchimedean solids. The vertices of the Catalan solids are the unions of the orbits obtained fromthe fundamental weights. The vertices are on concentric spheres determined by the lengths of thefundamental weights up to some scale factors.

The Platonic solids and the Archimedean solids have been successfully applied to describe thecrystallography in physics, molecular symmetries in chemistry, and some virus structures in biol-ogy. In particular, the Coxeter group W�H3� representing the icosahedral symmetry with inversionin 3D Euclidean space is very useful in understanding the quasicrystallography in physics.16 Thepolyhedra possessing the icosahedral symmetry have been successfully used in chemistry17 andbiology18 for the symmetries of molecules and viral capsids which also requires the Catalansolids.19 Therefore, construction of the vertices of the polyhedra, through the Coxeter–Dynkindiagrams A3, B3, and H3 with quaternions, whether they are Platonic, Archimedean, or Catalansolids, will be very useful in the applications of the physical sciences displaying symmetries.

1 M. Atiyah and P. Sutcliffe, Milan J. Math. 71, 33 �2003�; e-print arXiv:math-ph/0303071v1.2 E. Catalan, J. Ec. Polytech. �Paris� 41, 1 �1865�.3 L. Schläfli, J. Math. 20, 359 �1855�.4 T. Gosset, Messenger of Mathematics 29, 43 �1900�.5 W. A. Wythoff, Proceedings of the Section of Sciences, 1907 �unpublished�, Vol. 9, p. 529.6 H. S. M. Coxeter, Regular Polytopes, 3rd ed. �Dover, New York, 1973�.7 J. H. Conway, H. Burgiel, and C. Goodman-Strass, The Symmetries of Things �A. K. Peters, Wellesley, 2008�.8 R. Slansky, Phys. Rep. 79, 1 �1981�.9 M. Koca, R. Koc, and M. Al-Barwani, J. Math. Phys. 47, 043507 �2006�.

10 M. Koca, M. Al-Ajmi, and R. Koc, African Phys. Rev. 2, 9 �2008�; 2, 10 �2008�.11 M. Koca, M. Al-Ajmi, and R. Koc, J. Math. Phys. 48, 113514 �2007�.12 P. du Val, Homographies, Quaternions and Rotations �Oxford University Press, Oxford, 1964�; H. S. M. Coxeter,

Regular Complex Polytopes �Cambridge University Press, Cambridge, 1974�.13 M. Koca, R. Koc, and M. Al-Ajmi, J. Phys. A: Math. Theor. 40, 7633 �2007�.14 M. Koca, M. Al-Ajmi, and N. O. Koca, e-print arXiv:0906.2109.15 See http://en.wikipedia.org/wiki/Catalan_solid.16 D. Shechtman, I. Blech, D. Gratias, and J. W. Cahn, Phys. Rev. Lett. 53, 1951 �1984�.17 H. W. Kroto, J. R. Heath, S. C. O’Brien, and R. E. Smalley, Nature �London� 318, 162 �1985�.18 T. S. Baker, N. H. Olson, and S. D. Fuller, Microbiol. Mol. Biol. Rev. 63, 862 �1999�.19 R. Zandi, D. Reguera, R. F. Bruinsma, W. M. Gelbart, and J. Rudnick, Proc. Natl. Acad. Sci. U.S.A. 101, 15556 �2004�;

R. Twarock, Philos. Trans. R. Soc. London 364A, 3357 �2006�.



http://dx.doi.org/10.1007/s00032-003-0014-1

http://dx.doi.org/10.1016/0370-1573(81)90092-2

http://dx.doi.org/10.1063/1.2190334

http://dx.doi.org/10.1063/1.2809467

http://dx.doi.org/10.1088/1751-8113/40/27/013

http://en.wikipedia.org/wiki/Catalan_solid

http://dx.doi.org/10.1103/PhysRevLett.53.1951

http://dx.doi.org/10.1038/318162a0

http://dx.doi.org/10.1073/pnas.0405844101

http://dx.doi.org/10.1098/rsta.2006.1900

Catalan solids derived from three-dimensional-root systems and quaternions

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