# Index Concordia

## Group theory – Week 03

Posted in Group theory by Index Guy on September 21, 2007

We started the week by defining what a semi-direct product is:

• A group $G$ is the semi-direct product of two subgroups if one of those subgroups is normal and only shares the identity with the other.

An example is the Poincaré group, which is a semi-direct product of translations (the invariant subgroup) and Lorentz transformations (not an invariant subgroup of Poincaré).

It turns out that any group can be written as the semi-direct product of a semi-simple group and a solvable group. Some definitions:

• A simple group is one that only has trivial normal subgroups.
• A semi-simple group is one without an abelian invariant subgroup

For example, the gauge group of the Standard Model is not semi-simple since it contains $U(1)$, which is abelian.

Next we turned to the center of a group:

• The center $Z$ of a group is the set of all elements $z$ such that $zg = gz$.

The center is a group. Another concept is the commutator subgroup $C(G)$ of a group $G$,

$\left\{c = aba^{-1}b^{-1}| a,b \in G\right\}.$

Without proof we

• $C(G)$ is an invariant subgroup of $G$,
• $G/C(G)$ is an abelian group,
• $C(G)$ is the smallest invariant subgroup such that $G/N$ is abelian.

Next we turned to the concept of (conjugacy) classes.

• For a fixed $x \in G$ a class is the set of all elements of the form $gxg^{-1}$.

This way we can split a group into classes. The product of two classes is again a class,

$C_{i}C_{j} = c_{ij}^{k}C_{k}$

This product is symmetric, and hence the coefficients are also symmetric in the down indices. Some lemmas,

• $x \in G$ and $xCx^{-1} = C$ for any fixed $x$ then $C$ is a class,
• If $gCg^{-1} = C$ for all elements $g \in G$ then the set $C$ consists of a set of classes.

Finally, we state another theorem:

• An invariant subgroup consists of entire classes.

After all these, we moved forward to representations of a group, mainly matrix representations:

• A matrix representation is a set of $n \times n$ matrices with the same group multiplication table as some abstract group $G$.

We denote the elements of this set as $m(g_{i}) = D_{i}$. By definition,

$m(g_{i})m(g_{j}) = m(g_{i}g_{j}).$

It then follows that the representation of the identity is the identity matrix, and the representation of the inverse of an element is the inverse of the matrix representing that element. Some more definitions:

• A representation is faithful if $m(g_{i}) \neq m(g_{j})$ if $g_{i} \neq g_{j}$,
• A representation is reducible if all the elements map a subspace into itself,
• A representation is completely reducible if all elements can be reduced to diagonal form.

For finite groups, reducibility implies complete reducibility. A more important statement concerning finite groups and representations is

• Any representation of a finite group $G$ is equivalent to a unitary representation.

This is so important that I might come back later and write the proof that was mentioned in class. For now let us mention some examples of representations, mainly those of the symmetry group $S_{3}$. The elements of this group are

$e , (12) , (13) , (23) , (123) , (132).$

We can visualize this group as the set of rotations of a tetrad (a set of three 3-dimensional vectors with the same origin and perpendicular to each other). From mechanics (or elsewhere) we know that rotations of a vector can be written as $3 \times 3$ matrices. But there exists a vector that is left invariant under the set of all this rotations, namely the diagonal vector. Perpendicular to this vector we have a plane that contains a triangle that intersect the tetrad at each of its vertices. Then we can represent our group as the set of $2 \times 2$ matrices that rotate this triangle. So we have just gone from 3 dimensions to 2. It turns out that the determinant would be a 1-dimensional representation of our group too.

Now we list three important theorems regarding counting representations:

1. The order of 1-dimensional irreducible representations is equal to the order of $G / C(G)$,
2. The number of inequivalent irreducible representations is equal to the number of classes,
3. The order of $G$ is the sum of the squares of the dimensions of each class.

Using these theorems we learn that abelian groups have only 1-dimensional representations. Now we apply the theorems to our example of the group $S_{3}$:

• The order of the group is $g = 6$,
• We have three classes,
• Then $6 = x^2 + y^2 + z ^2$. The solution of this equation is two classes of dimension 1 and one class of dimension 2. This does not include the reducible representations.
• The other theorems tells me that the commutator group has 3 elements.

Next week we apply all this theory to the Dirac group.