## Group theory – Week 04

During this week we study the Dirac group and its representations. Recall that the order of the Dirac group is

with being the number of spacetime dimensions. We will consider separate cases when is either even or odd. The number of classes depends on :

- for even, and
- for odd, .

This can be understood by doing a few examples. The general result is this: the identity and the negative of the identity form individual classes, and then we have one class for +/- each member of the group. The subtlety lies on the “last” element, : for even it forms a class together with it’s negative, but for odd dimensions they form separate classes.

Note that the center of the group only has 2 elements,

The number of one-dimensional representations can be found by finding the ratio of the order of the group to the order of the center :

With this result we can find the dimension of the spinor representation by using some of the theorems from last week. Recall that the order of the group is given by the sum of the squares of the dimension of each irreducible representation. For even spacetime dimensions we have

This is the number of components of spinors in spacetime dimensions. For odd dimension we have

This equation has a solution when

The square brackets denote the integer part. So we have two inequivalent representations. An explanation for this comes from the fact that in odd dimensions the last Dirac matrix is proportional to the product of the others. We fix the constant of proportionality such that . There are always two choices.

An example: the Dirac matrices in 3 dimensions look like

and

We will write this result as a **theorem**:

- In
*odd*dimensions one gets an inequivalent irreducible representation from the another irreducible representation by changing the sign of one of the Dirac matrices, usually .

We now discuss *faithful irreducible representations*. First we mention the **claim**:

- If and only if a group is simple then all, non-trivial irreducible representations are faithful.

Here is the proof. Let be an invariant subgroup of , so that we can have the group . Then we can find representation +1 on and -1 on the rest. Remember that any homomorphism has a kernel when there is an invariant subgroup. If this happens then the representation is unfaithful, which means that if the map from the group to the unit is not one-to-one, then some elements will be many-to-one.

Some other important remarks:

- Every group has at least one reducible faithful representation.
- Not every group has a faithful irreducible representation.

Next we come to the charge conjugation matrices. Consider the Dirac matrices in even-dimensional Euclidean space. They satisfy the Clifford algebra,

and

Out of nowhere we mention that Stony Brook University will not graduate someone who does not know that a *normal matrix* obeys and it is always diagonalizable.

The charge conjugation matrices are defined by

For even dimensions one can write

with the *chiral* *matrix*.

The square of the chiral matrix is equal to the unit matrix. For odd dimensions we do not have a chiral matrix (it will be proportional to the identity matrix) and there will only be only one of the charge conjugation matrices. Some properties of :

- When both exists, they will be either both symmetric or both antisymmetric,
- the definition of the charge conjugation matrices can be written as

With the charge conjugation matrices in our minds we mention Schur’s lemmas:

- Given an irreducible representation with matrices , any matrix with for all group elements can be written as proportional to the identity matrix.
- If and are two irreducible representations (with matrices m x m and n x n respectively) and there exist a rectangular matrix that intertwine the two representations

TO BE CONCLUDED

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