so suposing $(t^m)_{m \in M}$ are the generators of a Lie group, and we have the following relation in the Lie algebra:
$$[t^a,t^b]= i f^{abc}t^c$$
Where the $f^{abc}$ are the structure constants of the algebra.
And in a irreducible representation ($r$): $ \,\,\,\,\,\,t_r^at_r^a = C_2(r) I$
where $C_2(r)$ is the the quadratic Casimir operator of the representation $r$.
In a book of QFT the following equalities are written but I can't seem to understand why they are correct:
$$t^bt^at^b= t^bt^bt^a+t^b[t^a,t^b] \,\,\,\,\,\,\,\, \text{this one really is obvious}$$ $$\;\;\;\;\;\;=C_2(r)t^a + i t^bf^{abc}t^c\,\,\, \text{this one is also obvious}$$ $$\hspace{3.1cm}=C_2(r)t^a + \frac{1}{2}if^{abc}if^{bcd}t^d \,\,\, \text{??? what just happened here?}$$
In the text it says that "In the third line we have used the antisymmetry of $f^{abc}$ to rewrite the matrix product as a commutator;" and I can't see how that would be possible.
One cannot write $t^bt^c$ as a simple sum of commutators, if we were to try:
$t^bt^c-t^ct^b+t^ct^b-t^bt^c = [t^b,t^c] + [t^c,t^b]=0$
$t^bt^c-t^ct^b+t^ct^b+t^bt^c = [t^b,t^c] + \{t^b,t^c\} = 2t^bt^c $
So only if $\{t^b,t^c\}=0$ we would get that result, but I can't see why would that be.
