The symmetries of the cube and the octahedron

 In the previous post, I stated that the cube and the octahedron are duals of one another. A cube has six faces and eight vertices, while an octahedron has eight faces and six vertices. Since six is less than eight, my natural laziness says it should be represented as a subset of S6 instead of S8, which means for a cube, we are permuting the faces, while for an octahedron, we are permuting the vertices. Same diff, the group will still have 24 elements.

 

Six-sided dice have a convention for the placement of the numbers represented, opposite sides have a sum of 7, meaning 1 is opposite 6, 2 is opposite 5 and 3 is opposite 4. 

 

The 12 even permutations first.

 

The identity is always even.

(1)

 

Pick two opposite faces to be fixed, and turn the cube 180° on the axis that runs through the centers of the fixed faces. There are three pairs of opposite faces, so three elements here.

(25)(34), (16)(34), (25)(16)

 

 

Pick two opposite vertices to be the axis of rotation, and turn the cube 120° and 240°. There are four pairs of opposite vertices, so this accounts for eight elements.

 

Spin 1-2-3 and 4-5-6

(123)(654), (132)(645)

 

Spin 1-3-5 and 2-4-6

(135)(642), (153)(624)

 

Spin 1-5-4 and 3-2-6

(154)(623), (145)(632)

 

Spin 1-4-2 and 5-3-6

(142)(635), (142)(653) 

 

Composing two even permutations creates an even permutation since even+even = even. This means these 12 permutations are a subgroup because of closure under the group operation and the existence of the identity in the set. Because 12 = 24/2, This must be a normal subgroup of the the larger group.

The 12 odd permutations.

Pick two opposite faces to be fixed, and turn the cube 90° or 270° on the axis that runs through the centers of the two fixed faces. There are three pairs of opposite faces, so this accounts for six permutations.

(2354), (2453), (1364), (1463), (2156), (2651)

 

Swap two opposite faces, and turn the cube 90° or 270° on the axis that runs through the centers of the swapped faces. There are three pairs of opposite faces, so this accounts for six permutations.

(16)(24)(35)

(16)(23)(45)

(25)(13)(46)

(25)(14)(36)

(34)(12)(56)

(34)(15)(26)

 

If we add diagonals to all six faces with the proviso that every vertex is either the endpoint of three diagonals or the endpoint of none, we get a regular tetrahedron embedded in our cube. Even permutations will permute the pink tetrahedron, while odd permutations will send the pink tetrahedron's vertices to the vertices that have zero diagonals in the original position.

 

This means the symmetries of the tetrahedron is a normal subgroup of the symmetries of the cube, and by duality, a normal subgroup of the symmetries of the octahedron.

 

Symmetries within symmetries. The beauty of group theory.

 

We learned about the group representation trick that tells us if we have k conjugacy classes, we can find k perfect squares that add up to the order of the group, with the proviso that at least one of the squares is 1², which is matched with the identity, always standing alone as a singleton conjugacy class. We have at least five conjugacy classes, based on the five different permutation shapes listed above.

Question 1.

Can you find five perfect squares such that 1²+a²+b²+c²+d²=24?

Question 2.

If yes to Question 1, does this mean there are exactly five conjugacy classes?

 

Question 3.

Find the conjugacy classes of the symmetries of the cube, and find the k squares that sum to 24.

 

Answers on Friday, when I will post more information about the structure of this group.

 

 

 

 

The symmetry groups of the Platonic solids

 I love the Platonic solids. Just writing this reminds me of a joke on Pee Wee's Playhouse, where Pee-Wee said "I love fruit salad!" and all his friends yell "If you love it so much, why don't you marry it?"

And he replies "All right, I will!"

A Platonic solid is a three dimensional shape whose faces are all equal sized regular polygons. There are exactly five, the 4-sided tetrahedron, the 6-sided cube, the 8-sided octahedron, the 12-sided dodecahedron and the 20-sided icosahedron.

The cube is clearly the best known of the five shapes. I like to say the cube is the Justin Timberlake of the Platonic solids. You might not remember any other member of NSYNC, but you probably know Timberlake's name. The cube is so commonly known, it doesn't have a fancy Greek based name like hexahedron. The cube is a special case of a rectangular solid, and rectangular solids are everywhere. Most boxes and buildings are rectangular solids, and if you see a 90° angle somewhere, it's a good bet that some human being put it there.

Let's count the faces, edges and vertices of these shapes.

Tetrahedron: 4 faces, 6 edges, 4 vertices

Cube: 6 faces, 12 edges, 8 vertices

Octahedron: 8 faces, 12 edges, 6 vertices 

Dodecahedron: 12 faces, 30 edges, 20 vertices

Icosahedron: 20 faces, 30 edges, 12 vertices

 

Notice that vertices + faces = edges + 2. This is called Euler's formula and it is true for any three dimensional shape with polygons as faces that does not have a "hole" in it. To deal with holes, there is the more complex Euler-Poincaré formula, which I won't discuss here.

 

If we switch the face and vertex numbers, we will see the octahedron is related to the cube, the dodecahedron is related to the icosahedron and the tetrahedron is related to itself. This relationship is called duality. One way to think of this is to put a point in the middle of every face of a Platonic solid and connect those points with edges. If you do this to any of the shapes, you will get its dual. For our purposes, this means that the symmetries of the cube is the same group and the symmetries of the octahedron, and likewise the symmetries of the icosahedron is the same as the symmetries of the dodecahedron.

The tetrahedron stands alone, it is its own dual. I have already mentioned in passing that the group of symmetries of the tetrahedron is isomorphic to A4, the alternating group. Let's look at this in greater detail, using the permutation notation to identify the group elements.

The identity

(1)

 

Physically, this means leaving the object alone.

The double transpositions

(12)(34), (13)(24), (14)(23)

If we swap the position of any two vertices, the other two vertices must swap as well.

The 3-cycles

(123), (132), (124), (142), (134), (143), (234), (243)

In these physical movements, one vertex remains fixed and we rotate the opposite face either clockwise or counterclockwise.

If we look at A4 as a subgroup of S4, there would be three conjugacy classes, the lists I enumerated, but conjugacy in a subgroup of the permutation group is not this easy. Because we no longer have all the values of x, the form xax⁻ⁱ will no longer make every 3-cycle conjugate to every other 3-cycle. We still have the property that a 3-cycle can't be conjugate to a double transposition, but these large conjugacy classes may find themselves partitioned into more than one class.

 

In A4, here are the conjugacy classes.

 

(1)

The identity always stands alone.

(12)(34), (13)(24), (14)(23)

Every double transposition can be shown to be conjugate to every other double transposition by using a 3-cycle and its inverse as the x and x⁻ⁱ values in the conjugacy form xax⁻ⁱ.

 

(123), (134), (142), (243)

The conjugacy class that numbers 8 in S4 is split in half, and no 3-cycle is conjugate to its inverse.

 

(132), (143), (124), (234)

A small preview of group representation theory: the order of the finite group G must equal to the sum of k squares, where k is the number of conjugacy classes of G and at least one of the squares is 1², corresponding to the conjugacy class of the identity alone. In this case, k = 4, and the four squares that add up to 12 are 1² + 1² + 1² + 3².

 

Commentary

 

Since I mentioned the name of Henri Poincaré, it's only good manners to include a link to his biography. If people argued about the top ten mathematicians the way they argue about the top ten athletes in any sport, Henri Poincaré would get a lot of votes. My personal top eight are:

Archimedes

Newton

Euler

Gauss

Riemann

Von Neumann

Hilbert 

Poincaré

 

There are multiple people I might consider to fill this list to ten, but I could not make a list that didn't include these eight.

Later this week we will look at the other symmetry groups of the Platonic solids.