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I'm trying to prove equation 25.20 in Schwartz: $$T^a T^b=\frac{1}{2N}\delta ^{ab}+\frac{1}{2}d^{abc}T^c + \frac{1}{2}if^{abc}T^c,\tag{25.20}$$ where $T^a$ are the fundamental representation generators, but I'm having some trouble. I understand that we can decompose the product into its commutator and anti-commutator, the former of which gives me the last term in the identity. The anti-commutator can be split into a trace and a hermitian, traceless part. The generators themselves form a basis for the hermitian traceless bit, so we can write this as $\frac{1}{2}d^{abc}T^c.$

My confusion is about the exact form of these coeffecitients $d$. Schwartz gives them as $$d^{abc}=2 tr(T^a\{T^b,T^c\}),$$ which he claims is the unique totally symmetric group invariant (up to a constant) for $SU(N)$. I can see that this is indeed totally symmetric. I don't understand why we need such an object, as opposed to one just symmetric in its first two indices? What is fixing the constant, and how do we know that it is unique?

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  • $\begingroup$ Maybe related? See also links therein. $\endgroup$ – MannyC Apr 22 at 14:37
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You can find the coefficient by taking the above equation (which can be understood as an Ansatz) and add to it the symmetrized form. That gets rid of the $f_{abc}$ piece. Now you multiply both sides with $T_c$ and take the trace to find $$ \text{Tr}[T_c\{T_a, T_b\} = \frac{1}{N} \delta^{ab} \text{Tr}[T_c] + d_{abd} \text{Tr}[T_c T_d], $$ which implies your second equation.

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