vanhees71 said:
For SU(2), the matrix M in the previous post is 2 \times 2, and the identity is just the following relation for Pauli matrices:
\frac{1}{2} \sigma^{a}_{ij}\sigma^{a}_{kl} = \delta_{il} \delta_{jk} - \frac{1}{2} \delta_{ij} \delta_{kl} , \ \ \forall (i,j,k,l) \in \{1,2\} .
From this, as you can check, we obtain the following correct relations
<br />
\begin{align*}<br />
\left( \sigma^{a}\sigma^{a} \right)_{il} &= \delta_{jk} \sigma^{a}_{ij}\sigma^{a}_{kl} \\<br />
&= 2(2\delta_{il} - \frac{1}{2}\delta_{il}) \\<br />
&= 3 \delta_{il}, \\<br />
\mbox{Tr}(\sigma^{a}\sigma^{a}) &= 6 .<br />
\end{align*}<br />
Of course, the adjoint representation is for spin 1 and thus 3-dimensional,
Yes, the adjoint representation of SU(2) is 3-dimensional. This means that the generators matrices in the adjoint representation are 3 \times 3, and there are 3 of them. So, your objects (T^{a})_{ij} = - i \epsilon_{aij} with a \in \{1,2,3 \} and (i,j) \in \{1,2 \} are not correct because: (1) \epsilon_{aij} is a 2 \times 2 matrix, and (2) T^{1} = T^{2} = 0 and T^{3} = \sigma^{2}. So, (T^{a})_{ij} = - i \epsilon_{aij} is not a representation, let alone adjoint representation.
So, how is the discrepancy between the two formulas resolved?
You created the discrepancy by wanting the identity to apply to the single 2\times2 matrix -i \epsilon_{aij} which is not even a representation.
As far as I see from your derivation, it's only valid in the fundamental representation(s) of SU(N).
Yes, and I was stressing this fact all along, by specifying the range of different indices. However, this does not mean that the identity is good only for the fundamental representation (quarks). In fact, it is very useful in calculating the colour factors in gluon loop.
The sum over colour indices in gluon loop can be represented by either the trace factor t_{2}(A):
\mbox{Tr}(A^{b}A^{c}) = t_{2}(A) \delta^{bc} ,
or by the eigenvalue of the quadratic Casimir operator in the adjoint representation C_{2}(A)
(A^{a}A^{a})_{bc} = C_{2}(A) \delta_{bc} ,
where A^{a} is the generator in the adjoint representation, as required by the gluon field
(A^{a})_{bc} = - i f_{abc} = - \frac{1}{4} \mbox{Tr}\left( \tau_{c} [ \tau_{a} , \tau_{b} ] \right) .
Since the structure constant f_{abc} of SU(n) is totally antisymmetric, one can easily show that
\mbox{Tr}(A^{a}A^{b}) = \left( A^{c} A^{c} \right)_{ba} = f_{acd}f_{bcd} .
So, in order to calculate the value of the factor t_{2}(A) = C_{2}(A), we need to compute
<br />
\begin{align*}<br />
f_{acd}f_{bcd} &= - \frac{1}{16} \mbox{Tr}\left( \tau_{d}[ \tau_{b} , \tau_{c} ] \right) \mbox{Tr}\left( \tau_{d}[ \tau_{a} , \tau_{c} ] \right) \\<br />
&= - \frac{1}{16} \tau^{d}_{ij} \ [ \tau^{b} , \tau^{c} ]_{ji} \ \tau^{d}_{kl} \ [ \tau_{a} , \tau_{c} ]_{lk} .<br />
\end{align*}<br />
Now, using our identity
\tau^{d}_{ij} \ \tau^{d}_{kl} = 2 \left( \delta_{il}\delta_{jk} - \frac{1}{n} \delta_{ij}\delta_{kl} \right) ,
we obtain
<br />
f_{acd}f_{bcd} = \frac{-1}{8} \mbox{Tr} \left( [ \tau^{b}, \tau^{c}] [ \tau^{a}, \tau^{c}] \right) - \frac{1}{n} \mbox{Tr}\left( [ \tau^{b} , \tau^{c}] \right) \mbox{Tr}\left( [ \tau^{a} , \tau^{c}] \right) .<br />
The second term is zero because the trace of [ A , B ] is zero for any two matrices A and B. The first term, when expanded, consists of 4 terms each of which has a factor of the form \tau^{e}_{ij}\tau^{e}_{kl}. So, we can apply our identity to each term. I leave it for you as exercise to show that
<br />
\begin{align*}<br />
\mbox{1st term} &= \mbox{Tr}(\tau^{b}\tau^{c}\tau^{a}\tau^{c}) = - \frac{4}{n} \delta^{ab} \\<br />
\mbox{2nd term} &= \mbox{Tr}(\tau^{b}\tau^{a}\tau^{c}\tau^{c}) = 4( n - \frac{1}{n}) \delta^{ab} \\<br />
\mbox{3rd term} &= 4(n - \frac{1}{n}) \delta^{ab} \\<br />
\mbox{4th term} &= - \frac{4}{n} \delta^{ab} .<br />
\end{align*}<br />
This gives us
f_{acd}f_{bcd} = \frac{-2}{8} \Big \{ - \frac{4}{n} - 4 \left( n - \frac{1}{n} \right) \Big \} \delta^{ab} = n \delta^{ab} .
Which means that t_{2}(A) = C_{2}(A) = n.
