Group Representation

A game called “Group Representation”, the purpose of which is to show the value of using matrices to represent symmetry operations.

Assume we decide to build a group and we pick two operations. A set and an operation isn’t a group unless several rules are followed and one of those rules is that applying the operation to any two elements of the set will result in something that is also an element of the set. Now, here’s where the “representation” part comes in–if we represent every element of the set by a matrix then it is also true that all those matrices can be put into a set and matrix multiplication on any two of those matrices should result in a matrix that is in the set. With all this in mind, let the game begin with the set below.

$G = (C_2, \sigma_{yz})$

We now need to decide if this is really a group or if we might have missed one or more operations that are needed in the group (if you have some experience you’ll immediately notice that we forgot the identity operation.)

Assume a transformation that we give the name “Matricize”. It will take a symmetry operation as input and give a matrix as the output.

$Matricize(C_2) = \begin{bmatrix} cos(\pi) & - sin(\pi) & 0 \\ sin(\pi) & cos(\pi) & 0 \\ 0 & 0 & +1 \end{bmatrix} = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & +1 \end{bmatrix}$

$Matricize(\sigma_{yz}) = \begin{bmatrix} -1 & 0 & 0 \\ 0 & +1 & 0 \\ 0 & 0 & +1 \end{bmatrix}$

Our test procedure is to take two operations, get the matrices for those two operations and then multiply the matrices together and ask “is what we get something we have?” If it isn’t then we know we missed something.

$C_2 * \sigma_{yz} = ?$

$Matricize(C_2) Matricize(\sigma{yz} = Matricize(?)$

$\begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & +1 \end{bmatrix} \begin{bmatrix} -1 & 0 & 0 \\ 0 & +1 & 0 \\ 0 & 0 & +1 \end{bmatrix} = \begin{bmatrix} +1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & +1 \end{bmatrix}$

First thing–this is a new matrix so we know that G as we defined it above was incomplete. So, what did we get?

Having x stay the same and z staying the same, with y changing, is a reflection through the xz plane. We now know that $\sigma_{xz}$ must be a part of the set G if we want G to become a group, so we update G:

$G = (C_2, \sigma_{yz}, \sigma_{xz})$

The set of corresponding matrices is also updated:

$G = ( \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & +1 \end{bmatrix}, \begin{bmatrix} -1 & 0 & 0 \\ 0 & +1 & 0 \\ 0 & 0 & +1 \end{bmatrix}, \begin{bmatrix} +1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & +1 \end{bmatrix}$ )

We’re going to take advantage of a rule that you may have not seen before. Matrix Multiplication is not commutative, but the matrix multiplication of two diagonal matrices is commutative. This helps us because it means we don’t have to test $\sigma_{yz} * C_2$ .

Let’s now test $C_2$ and $\sigma_{xz}$ :

$Matricize(C_2) Matricize(\sigma_{xz}) = Matricize(?)$

$\begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & +1 \end{bmatrix} \begin{bmatrix} +1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & +1 \end{bmatrix}=\begin{bmatrix} -1 & 0 & 0 \\ 0 & +1 & 0 \\ 0 & 0 & +1 \end{bmatrix}$

We recognize the answer as being the matrix for $\sigma_{yz}$ and we also see it in the set showing the matrices. This time we don’t have to add something to G. Let’s try another combination:

$Matricize(\sigma_{yz}) Matricize(\sigma_{xz}) = Matricize(?)$

$\begin{bmatrix} +1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & +1 \end{bmatrix} \begin{bmatrix} +1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & +1 \end{bmatrix}=\begin{bmatrix} +1 & 0 & 0 \\ 0 & +1 & 0 \\ 0 & 0 & +1 \end{bmatrix}$

We recognize the answer as the Identity Matrix (usually given the symbol E). Let’s add it to G:

$G = (E, C_2, \sigma_{yz}, \sigma_{xz})$

$G = ( \begin{bmatrix} +1 & 0 & 0 \\ 0 & +1 & 0 \\ 0 & 0 & +1 \end{bmatrix}, \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & +1 \end{bmatrix}, \begin{bmatrix} -1 & 0 & 0 \\ 0 & +1 & 0 \\ 0 & 0 & +1 \end{bmatrix}, \begin{bmatrix} +1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & +1 \end{bmatrix}$ )

We don’t have to worry about our new element leading to yet another new element. Since the work prior to that gave us a matrix we already had, we can believe we are done. The only way this isn’t true is if the above work contains a mistake.

$G = (E, C_2, \sigma_{yz}, \sigma_{xz})$

Work coming soon:

We will build a multiplication table with the matrices for the four elements above, and while filling it out, confirm that each matrix we get from matrix multiplication corresponds to one of the four symmetry elements.

Keep in mind that we want to twist this story to say that we’re going to all this trouble because it is easier to do the work with the matrices [[I’m typing them all in, I can tell you that it is not! : – )]]

Well, try this, we could have a computer do the work to make the multiplication table with the matrices, but our computer can’t do it with all those symbols for symmetry operations. Also, try this, we only have four elements, but if had a system with sixteen elements, we’d be begging for it to be possible to have the computer build the multiplication table.

Once the full computer-generated multiplication table is completed (in 0.027 seconds), we could have the computer translate each matrix back to the symmetry operation and give us a Cayley Table (multiplication of the symmetry elements).

$.$

Appendix A — Beyond Matrices

Someone might point out that if we say “matrices”, we are limiting ourselves, and that the concept can be used with things that aren’t matrices, so that we should replace “matrices” with a term that is more general, a term that would include all those wonderful, marvelous things which are not matrices and they could be used.

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