Yang Mills theory and SU(N) groups [duplicate]

Trying to get a better understanding of the relation between a SU(N) Yang Mill theory and its number of "color" space.

Most of the description I've found so far are either way to complex/specific. Yiannis answer on this post is almost what I'm looking for, except I was hoping someone could provide addition sources and readings corresponding precisely to what he is describing.

• This is essentially a duplicate of physics.stackexchange.com/q/102941/2451 Jul 25 '14 at 23:59
• This is quite blatantly a duplicate of [hysics.stackexchange.com/q/102941/2451](physics.stackexchange.com/q/102941/2451)
– zzz
Jul 26 '14 at 3:19
• @Qmechanic Are you sure? Because here he seems to specifically asking for resources? Maybe that makes it off-topic, but I wouldn't consider it as a duplicate, especially since he already linked this question in his post. Jul 26 '14 at 6:26
• @Bernhard: Asking for resource recommendations is implicitly implied in any question. Pure resource recommendation questions are restricted. Jul 26 '14 at 8:15

The SU(n) representation refers to the space of the unitary matrices with determinant 1. By using the exponential form we can write an arbitrary SU(n) representation as $$U = e^{it_{n}a_{n}},$$ where $t_{n}$ are the generators of the SU(n) group, $a_{n}$ are continuous parameters. Let's get the relations for them.

$$\tag{1} U^{\dagger} = 1 \Rightarrow t_{n}^{\dagger} = t_{n}.$$

$$\tag{2} det U = 1 \Rightarrow Tr (t_{n}) = 0.$$

By $(1)$ and $(2)$ we can reduce the number of independent parameters (or the generators) by which we can represent the arbitrary SU(n) matrix. The starting number is $2n^{2}$ (matrix rank $n\Rightarrow n^{2}$ parameter, the complexity of matrix - multiplication of $n^{2}$ by two). $(1)$ reduce this number to $$2n^{2} - n - (n^{2} - n) = n^{2}$$ (here we have the first "-" summand as the result of the bounds for diagonal component and the second summand as the result for nondiagonal components).

$(2)$ reduce the number to $n^{2} - 1$.

Let's have the example: the SU(3) group. Let's use approximate representation: $U \approx E + it_{n}a_{n} = E + A$ (it is more convenient for finding the generators).

$(1)$ gives the zero imaginary part for each diagonal component and the connection $$A_{ij} = a_{ij} + ib_{ij}, \quad A_{ji} = a_{ij} - ib_{ij}.$$ The $(2)$ fixes $\sum_{i} A_{ii} = 0$.

So one of possible parametrization of the $U$-matrix is

$$U = E + i\begin{pmatrix} a_{3} + a_{8} & a_{1} - ia_{2} & a_{4} - ia_{5} \\ a_{1} + ia_{2} & a_{8} - a_{3} & a_{6} - ia_{7} \\ a_{4} + ia_{5} & a_{6} + ia_{7} & -2a_{8} \end{pmatrix}.$$

By the expanding the second matrix to a sum $t_{n}a_{n}$ you will get Gell-Mann matrices.