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.

Enhanced Hasse Diagrams and Subnormal Subgroups.

 So far, we have talked about subgroups being either normal or not normal, but there is a refinement of not normal groups called subnormal. A subgroup H is subnormal if it is not normal in G, but in a string of nested subgroups ending in G, each one normal in the supergroup directly above it.

We just started talking about the quaternion group, an eight element non-abelian group where all the subgroups are normal. The other eight element non-abelian group we have discussed is D4, the symmetries of the square. The order of a subgroup has to divide the order of the group, so the only possible sizes of subgroups of a group of order 8 are 4, 2 and 1.  In D4, the 8 element subgroup is the group itself, and the only subgroup of order 1 is the identity. While normality can be difficult to prove in some cases, one of the easiest cases is that if a group has order 2n, and subgroup of order n is normal. This means all the subgroups of order 4 must be normal, but we have to check on each of the subgroups of order 2. It turns out some are not normal, but if we look at them as 2 element subgroups of a group of order 4, they must be normal because 2 is half of 4.

 

Here is the Enhanced Hasse Diagram for D4. There are three subgroups of order 4, all of them normal, so they are represented by ovals. Of the five subgroups of order 2, only the one generated by R180° is normal, while the other four are subnormal. Subnormal groups are represented by rectangles with rounded corners.

 

Commentary

 

I am learning about subnormality on the fly as I put it up on the blog. D4 is a subgroup of S4, but it is not normal or subnormal, since subgroups of a symmetric group must be a union of conjugacy classes.  Written in cycle notation, D4 looks like this.

 

Rotations: (1), (1234), (13)(24), (1423)

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

 

Because the symmetric group has 24 elements and the dihedral group has 8, there is no subgroup "between" them, because there is no number between 8 and 24 that is divisible by 8 and also divides 24. D4 is not normal in S4 because it doesn't include all the 4-cycles and doesn't include all the 2-cycles. The chain of subgroups that defines subnormality is supposed to go all the way up to the group itself, and this is not the case here.

 

I expect there is a name for such a situation since it can be found so easily using well known small finite groups. I will search for it over the next few weeks, and I hope to report back.

 

In any case, learning new stuff is fun, but I fully expect I am reinventing the wheel here. It wouldn't be the first time.

 

The invariants of conjugacy classes

 There is a direct link between groups represented by permutations and groups represented by permutation matrices. For example, if we have the permutation (12)(45) and we assume it belongs to S5, it can also be represented by the 5x5 matrix

0 1 0 0 0

1 0 0 0 0

0 0 1 0 0

0 0 0 0 1

0 0 0 1 0

In Sn, the conjugacy classes are all permutations with the same cycle structure, so the definition is necessary and sufficient. If we deal with subgroups of Sn, two conjugates must have the same cycle structure, but having the same cycle structure is not sufficient to state that the elements are conjugate. For example, the permutation (1234) generates a four element abelian group.

 

(1234)    

(1234)² = (13)(24)   

(1234)³ = (1432)

(1234)⁴ = (1)   

 

(1234) and (1432) have the same cycle structure, but this group is abelian and every element is in a conjugacy class by itself. In S4, (24)(1234)(24) = (1432), but (24) isn't available here.

 

A group of matrices also has invariants under conjugacy, the determinant, which was discussed in the last post and the trace, which is the sum of the elements along the main diagonal. Let's look at a four element group of 2x2 matrices that represent rotations of 0°, 90°, 180° and 270° in the xy-plane. This is an abelian group isomorphic to the earlier permutation group, so every conjugacy class is a singleton.

0° matrix

1  0 

0  1

determinant = 1, trace = 2

90° matrix

0 -1

1  0

determinant = 1, trace = 0

180° matrix

-1  0 

 0 -1

determinant = 1, trace = -2

270° matrix

0   1

-1  0

determinant = 1, trace = 0

Note that the 90° matrix and the 270° matrix have the same determinant and trace, which means there is a 2x2 non-singular matrix M such that M(90° matrix)M⁻¹ = 270° matrix. One such matrix is

 

-1  0

 0  1

 

which is not one of the elements of our defined group.

 

Commentary

 

As I have stated earlier, group theory is a generalization of the concept of symmetry. It is taught as an abstract field and the applications seem remote, with the possible exception of Rubik's Cube solutions. In fact, having symmetry in a physical problem makes it easier to solve. According to my professor Stu Smith, all solved differential equations rely on symmetry except for one. The differential equation that solves the solitary wave, also known as a soliton, does not have symmetry.

Most waves have peaks and valleys, and when the hit other waves, they can add to each other or cancel each other out, but a soliton only is almost all peak with a tiny valley. The bigger a soliton is, the faster it moves. In the ocean, we call the biggest solitons tsunamis.

 

A basic tenet of physics is big + fast = fuck you up. This is why a tsunami can wreak havoc when it hits land thousands of miles away from the source, because it will nearly the same speed and size it had when it was formed.

 

Determinants and finite groups of real-valued nxn matrices

 Every square matrix has a determinant, a number that corresponds to the area, volume or hypervolume of the shape formed by the n-vectors defined by the rows of the matrix. The columns can also be used, and though the shape would be different, the determinant is unchanged.

 

Example: for any 2x2 matrix

 

a b

c d

 

the determinant equals ad - bc.

 

This is the area of a parallelogram defined by the points (0,0), (a,b), (c,d) and (a+c,b+d), or by the points (0,0), (a,c), (b,d) and (a+b,c+d).

 

We get a positive area if the vector (c d) is on the left hand side of the vector (a b) in the two-dimensional plane. Conversely, a negative area means the vector (c d) is on the right hand side of the vector (a b) in the two-dimensional place. A determinant of zero means there is a scalar k such that c = ak and d = bk.

 

There is a problem with determinants. The number of terms increases factorially, which is a growth rate even faster than exponential growth. 3x3 matrices have 6 terms in the determinant, 4x4 have 24, 5x5 have 120, etc. I will show the 3x3 formula, but I will avoid the general formula for larger cases.

Example: for any 3x3 matrix

a b c

d e f

g h i

 

the determinant equals aei + bfg + cdh - afh - bdi - ceg.

 

Why these terms and why add some while subtracting others?

Every term in an nxn determinant is the product of n entries, with exactly 1 term in every row and every column. The term aei corresponds to the main diagonal matrix

1 0 0

0 1 0

0 0 1

This is also the identity matrix. Every other term corresponds to the positions of the 1 entries in one of the permutation matrices. If the permutation consists of an even number of transpositions from the identity, the corresponding term is added to the total, and the terms that correspond to odd permutations are subtracted from the total.

Thinking in the language of group theory, the group even and odd under addition is mapped onto the group 1 and -1 under multiplication.

 

A matrix M is singular if det(M) = 0.

 

There is a standard method for finding the inverse of a matrix and one of the steps is to divide by the determinant. We can't divide by zero, which means singular matrices do not have inverses. 

Stated without proof: det(MN) = det(M)det(N).

If the singular nxn matrices are removed from the set of all nxn matrices, we have a group, and determinant is a homomorphism from the group of non-singular nxn matrices under multiplication to the group (ℝ-{0}, ✕).

If a matrix M has real entries, det(M) will also be real, since the operations used to compute the determinant are multiplication, addition and subtraction. This brings us to a simply proved statement.

Lemma: If G is a finite group of nxn matrices under multiplication, the determinants of the matrices must be either 1 or -1.

 

Proof by contradiction: Assume one of our matrices M has a determinant x, not equal to 1 or -1. This would mean det(M²)=x² and det(M³)=x³, and so on, ad infinitum. It could not be a finite group, so instead we must assume det(M) must be either 1 or -1 for the group to be finite.

 

Stated without proof: If B is a matrix created by switching two rows of the matrix A, det(B)=-det(A).

 

Since every nxn permutation matrix can be created by multiple transpositions of rows of the identity matrix In, and the determinant of In = 1, all permutation matrices have a determinant of either 1 or -1. Having a lot of zeros in a matrix can make the computation of the determinant much simpler, since any product that includes a zero must be zero. Permutation matrices have so many zeros that there is only term that is non-zero and that term is the product 1ⁿ=1, which will be multipled by 1 or -1, depending on the parity of the number of row switches away from the idenity.

Theorem: If two matrices M and N are conjugates, det(M)=det(N).

Proof: Conjugacy means N = XMX⁻¹. If det(X)=x, then
det(X⁻¹)=1/x.

 

det(N)= det(XMX⁻¹) = x*det(M)*1/x= det(M).

 

The determinant of a matrix is invariant under conjugacy. 

 

Stu Smith, who taught me the majority of my postgraduate math classes, said that one should always pay attention to invariance, because it usually implied something of a physical nature. 

 

Like Tracewell, Stu Smith is someone I tried to emulate when I taught, but I admit I fell short on both counts. I was never as enthusiastic as Ted Tracewell, and I was was never as erudite as Stu Smith.

Commentary

In the middle of the 19th Century, a great number of mathematicians were working on ways to simplify the calculation of large matrix determinants, including Leopold Kronecker, who was introduced in the last post, and Charles Dodgson, better known to the general public as Lewis Carroll. 

 

I learned a story in a math class, I forget what professor, that Queen Victoria loved Alice in Wonderland, published in 1865, after the death of her husband Prince Albert in 1861, an event that left her nearly inconsolable. She wrote to Lewis Carroll and asked him to send her the next book he published. Following her wishes, she sent him a book on determinants.

 

The story is true until the punch line. Dodgson knew the Queen wanted his next Lewis Carroll book Alice Through the Looking Glass, and he sent it to her as soon as it was published in 1871. He did have a quirky sense of humor, but he would never disrespect his monarch in such a fashion.