Product of traceless hermitian $N \times N$ matrices

In Srednicki's textbook Quantum Field Theory, section 80 shows that, for a complete set of $$N^{2}$$ tracelss hermitian $$N \times N$$ matrices normalized according to $$Tr(T^{a}T^{b}) = \delta^{ab}$$, $$$$(T^{a})_{i}^{j}(T^{a})_{k}^{l} = \delta_{i}^{l} \delta_{k}^{j} - \frac{1}{N} \delta_{i}^{j} \delta_{k}^{l}. \tag{80.17}$$$$ However, under the solution to Problem 97.3 (on page 165 of the solution book), it states

For $$SU(N)$$, we have $$(T^{a})_{\alpha'}^{\alpha}(T^{a})_{\beta'}^{\beta} = \frac{1}{2}\left( \delta_{\alpha'}^{\beta} \delta_{\beta'}^{\alpha} - \frac{1}{N} \delta_{\alpha'}^{\alpha} \delta_{\beta'}^{\beta}\right).$$

Why is there an additional $$\frac{1}{2}$$ on the right-hand side for $$SU(N)$$?

• In the book, in the formulation of the problem 97.3, I cannot see the presumed wrong formula. I have seen it on problems' solutions book by the author. Of course, you can ask directly to him for clarifications. – Jon Nov 20 '19 at 8:40
• @Jon - Do you mean that the formula for $SU$($N$) is wrong? I didn't mean it's wrong. I guess the $\frac{1}{2}$ appears because the $SU$($N$) matrices are symmetric or antisymmetric. Am I right? But I can't figure out the detailed steps that yield $\frac{1}{2}$. – Shen Nov 20 '19 at 10:04
• @Shen I have not seen the solution to the problem, but it seems that this is a simple matter of normalisation. You can also normalise the generators as Tr(T^a T^b) = (1/2) \delta^{a b}, see for example scipp.ucsc.edu/~haber/ph218/sunid17.pdf – Darth_Bane Nov 20 '19 at 10:59
• @A_user_with_NoName - I have revised my text and added a link to the solution book. – Shen Nov 20 '19 at 14:06

Set $$i=l$$ and $$j=k$$ in 80.17 and sum over the repeated indices. The LHS becomes $$\sum_a {\rm tr}\{T^a T^a\}$$ the RHS becomes $$N^2-1$$. There are $$N^2-1$$ traceless hermitian matrices, so if $${\rm tr}\{T^aT^b\}=\delta_{ab}$$ the LHS is $$N^2-1$$ also. So everything is consistent.
• If the normalization is $Tr(T^{a}T^{b}) = \frac{1}{2}\delta^{ab}$, then the formula for $SU(N)$ is valid. – Shen Nov 22 '19 at 4:04
This is indeed a matter of normalization, as A_user_with_NoName correctly commented. If the normalization is $$Tr(T^{a}T^{b}) = \frac{1}{2}\delta^{ab}$$, then the formula for $$SU(N)$$ is valid. The proof is given in the link provided by A_user_with_NoName.