"Plato said God geometrizes continually." - Plutarch Convivialium disputationum liber 8,2 "We do not listen with the best regard to the verses of a man who is only a poet, nor to his problems if he is only an algebraist; but if a man is at once acquainted with the geometric foundation of things and with their festal splendor, his poetry is exact and his arithmetic musical." - Ralph Waldo Emerson Society and Solitude chapter 7, Work and Days SYMMETRY GROUPS OF THE PLATONIC SOLIDS This chapter requires a little more mathematical sophistication from the reader than the earlier chapters. However, the exercises are (I think) choosen to be doable. The "Platonic solids" are the 5 regular polyhedrons: polyhedron # faces # vertices # edges group p,q ------------------------------------------------------------ tetrahedron | 4 | 4 | 6 | T | 3,3 hexahedron | 6 | 8 | 12 | O | 4,3 octahedron | 8 | 6 | 12 | O | 3,4 dodecahedron | 12 | 20 | 30 | I | 5,3 icosahedron | 20 | 12 | 30 | I | 3,5 Here p, called the "face degree", denotes the number of edges bounding each face and q, called the "vertex degree", denotes the number of faces meeting each vertex. The three "Platonic groups" will be described below. Their names: T = symmetric group of the tetrahedron = "tetrahedral group", O = symmetric group of the octahedron (and the cube) = "octahedral group", I = symmetric group of the octahedron (and the dodecahedron) = "icosahedral group". These solids may be drawn in rectangular coordinates using polyhedron coordinates ------------------------------------------------- tetrahedron | (1,1,1), (1,-1,-1), (-1,-1,1), (-1,1,-1) hexahedron | (1,1,1), (1,1,-1), (1,-1,1), (-1,1,1), | (1,-1,-1), (-1,1,-1), (-1,-1,1), (-1,-1,-1) octahedron | (1,0,0), (0,0,1), (0,1,0), | (-1,0,0), (0,-1,0), (0,0,-1) dodecahedron | (see below) icosahedron | (1,0,phi), (1,0,-phi), (-1,0,phi), (-1,0,-phi), | (0,phi,1), (0,phi,-1), (0,-phi,1), (0,-phi,-1), | (phi,1,0), (phi,-1,0), (-phi,1,0), (-phi,-1,0) where phi denotes the golden ratio. If P1, P2, P3 are three vertices of an icosahedron which form a triangular face then (P1+P2+P3)/3 forms a vertex of the dual dodecahedron and every vertex of the dual dodecahedron arises in this way. Background on symmetries in 3-space This subsection presents, with some proofs, background on isometries in 3 dimensions necessary for understanding the symmetry groups of the Platonic solids. We fix once and for all the "right-hand-rule" orientation in 3-space. We call a distance-preserving transformation in 3-space which fixes the origin a "symmetry of 3-space". We say that such a symmetry is "orientation preserving" if it preserves the right-hand rule orientation. Example: Let s : R3 --> R3 denote the function which takes each vector v belonging to R3 and returns its reflection s(v) about the yz-plane. This is not orientation preserving since it reverses the direction of a counterclockwise moving circular path in the xy-plane. In terms of rectangular coordinates, s(x,y,z)=(-x,y,z). Let R3 = {(x,y,z) | x,y,z real numbers } denote "3-space". We also write this, when convenient, as column vectors [ x ] R3 = { [ y ] | x,y,z real numbers }. [ z ] The "distance function" on R3 is the function d(v1,v2) = sqrt((x1-x2)^2+(y1-y2)^2+(z1-z2)^2) where v1=(x1,y1,z1), v2=(x2,y2,z2). We call a function f : R3 --> R3 an "isometry" if it satisfies d(f(v1),f(v2))=d(v1,v2) for all v1 and v2 belonging to R3. We want to understand isometries a little better since they will preserve distances (and, in particular, preserve the shapes of solids) and therefore provide us with the kinds of symmetries of 3-space we want to consider. We can construct isometries using certain types of 3x3 matrices. First, a 3x3 matrix is a 3x3 table of real numbers [ a11 a12 a13 ] A = [ a21 a22 a23 ] [ a31 a32 a33 ] which acts on R3 by matrix multiplication: [ a11 a12 a13 ][ x ] Av = [ a21 a22 a23 ][ y ] [ a31 a32 a33 ][ z ] [ a11x + a12y + a13z ] = [ a21x + a22y + a23z ]. [ a31x + a32y + a33z ] For example, if A = I3, the 3x3 identity matrix, [ 1 0 0 ] I3 = [ 0 1 0 ], [ 0 0 1 ] then I3v = v for all v belonging to R3. In general, any such 3x3 matrix gives rise to a function A : R3 --> R3. Lemma: If A is a 3x3 matrix then the function A : R3 --> R3 is an isometry if and only if transpose(A)*A = I3, where I3 denotes the 3x3 identity matrix. proof: dot(v,w)=transpose(v)@w, where @ denotes the vector dot product. Since dot(Av,Bw)=transpose(v)@((transpose(A)*B)w), we have dot(Av,Aw) = dot(v,w), for all v, w in R3 if and only if transpose(v)@((transpose(A)*B)w)=transpose(v)@w, for all v, w in R3 if an only if transpose(A)*A=I3. QED You may have been wondering how one could construct an isometry. This lemma gives us lots of examples. Are there any isometries which do not come from such matrices? Yes: any translation gives rise to an isometry. Are there any examples of isometries which do not arise from a composition of a translation and an orthogonal matrix? Theorem: A function f : R3 --> R3 is an isometry fixing the origin if and only if f is left multiplication by an orthogonal matrix. This will not be proven here (see Artin [Ar], chapter 4, section 5, Proposition 5.16). As a consequence of this lemma, we see that if the matrix A gives rise to as isometry then det(A) is either equal to 1 or -1 (since det(A)^2= det(transpose(A)*A)=det(I3)=1). In particular, the determinant of such a matrix is non-zero, so the matrix is invertible. Lemma: The set of all 3x3 matrices A such that the function A : R3 --> R3 is an isometry forms a group under matrix multiplication. Exercise: Verify the group axioms needed to prove this lemma. Notation: This group will be denoted O3(R) and called the "orthogonal group" of R3. We denote by SO3(R) the following subset SO3(R) = { A in O3(R) | det(A) = 1 }. This is a subgroup of O3(R) (Exercise: Verify the group axioms for SO3(R)) which is called the "special orthogonal group" of R3. It is known that the number of cosets in O3(R)/SO3(R) is 2. In fact, it is known that (*) O3(R) = SO3(R) union s*SO3(R) (disjoint union) where s is the reflection in the above example (this follows from [Ar], chapter 4, section 5). Lemma: The isometry A in O3(R) is orientation preserving if and only if det(A) = 1. We will not prove this lemma here. Symmetries of the tetrahedron Fix a tetrahedron centered at the origin, with one vertex along the z-axis. Each edge has an "opposite" edge on the tetrahedron (which is actually perpendicular to it if you look at it straight on). Each vertex has an "opposite" face. There are orientation preserving symmetries (called "rotations") of the tetrahedron and orientation reversing symmetries of the tetrahedron. The orientation preserving symmetries of the tetrahedron will be denotes ST. They are obtained as follows: * the 4 axes of symmetry through the centers of the faces yield 2 elements each (120 degree clockwise rotation when viewed from outside and a 240 degree rotation), for a total of 8 elements, * the 3 pairs of edges (formed by an edge and its opposite) yield one element each (a 180 degree rotation), for a total of 3 elements. These, plus the identity, give 12 elements in ST. Symmetries of the cube We fix a cube centered about the origin in 3-space. The set of centers of the faces of a cube forms a set of vertices of an octahedron drawn inside the cube. This octahedron is called the "dual" polyhedron. These two polyhedra have the same symmetry group, which we denote by O. There are orientation preserving symmetries (called "rotations") of the cube and orientation reversing symmetries of the cube. The orientation preserving symmetries of the cube will be denotes SO. They are obtained as follows: * the 3 axes of symmetry through the centers of the faces yield 3 elements each (90 degree clockwise rotation when viewed from outside, a 180 degree rotation, and a 270 degree rotation), for a total of 9 elements, * the 4 axes through the opposing vertices yield 2 elements each (all of order 3), for a total of 8 elements, * the 6 axes through the opposing mid-edge points yield 1 element each (of order 2), for a total of 6 elements. These elements, plus the identity, yield 24 elements. Lemma: There are 24 orientation preserving elements in O, i.e., |SO|=24. The above sketch is one way to see why this is true. Here's another proof: Let V be the set of vertices of the cube. The group SO acts on the set V. Fix a v belonging to V and let H=stab_SO(v). One can check that |H|=3 (since the only symmetry which fixes v is a rotation g about the line therough v and its opposite vertex. Since g is order 3, H= is order 3 as well). We have |V|=8, so by a lemma in the previous chapter on orbits and stabilizers, we have |SO/H| = |V|. By Lagrange's theorem, |SO|=|SO/H||H|=8x3=24. QED Now we know SO, what is O? Note that s, the reflection in the Example in the previous section, belongs to O. Using (*) of the previous section, we have (*) O = SO union s*SO (disjoint union). We know that |s*SO|=|SO|=24, so Lemma: The order of the octahedral group is |O|=48. Symmetries of the dodecahedron The set of centers of the faces of a dodecahedron forms a set of vertices of an icosahedron drawn inside. This icosahedron is called the "dual" polyhedron. We fix a dodecahedron in 3-space so that the vertices of the dual icosahedron are as listed in section 1 above. Let SI denote the group of orientation preserving symmetries of the dodecahedron. Note SI is a finite subgroup of SO3(R). Let I denote the group of all symmetries of the dodecahedron. Note I is a finite subgroup of O3(R) and that SI is a subgroup of I. Let F denote the set of faces of the dodecahedron, so |F|=12. SI acts on F. Lemma: SI acts on F transitively. We won't prove this. If you look at a dodecahedron it follows "by inspection". The reason why this is useful is that it tells us that if x is any face then any other face can be obtained from x by applying some element of SI. In other words, the orbit of x is all of F: SI*x = F. If x is any face then the only orientation preserving symmetries which don't send x to a different face is a rotation by an integer multiple of 72 degrees about the line passing through the center of x and the center of its opposite face. There are, for each face x, exactly 5 distinct rotations of this type. Therefore, |stab_SI(x)| = 5. By a lemma in the chapter on orbits, we have SI/stab_SI(x) = SI*x, so |SI| = |stab_SI(x)||SI*x|=5x12=60. The elements of SI include: * rotation by 2*Pi*k/5, k=0,1,2,3,4, about the line passing through the center of a face and its opposite, * rotation by 2*Pi*k/3, k=0,1,2, about the line passing through a vertex and its opposite, * rotation by Pi about the center of an edge. Subgroups include: * stabilizer of a vertex. These are all cyclic of order 3, and they are all conjugate. There are 10 distinct such subgroups since a vertex and its opposite share the same stabilizer. * stabilizer of a face. These are all cyclic of order 5, and they are all conjugate. There are 6 distinct such subgroups since a face and its opposite share the same stabilizer. * stabilizer of an edge. These are all cyclic of order 2, and they are all conjugate. There are 15 distinct such subgroups. Exercise: Verify all these. 