My question is from the last paragraph in p.466.$P_L=\frac{1}{2}-\frac{1}{2}\gamma_5$.Why is $P_L\rightarrow \frac{1}{2}$part coming up with an extra factor $\frac{1}{2}Tr([T_R^a,T_R^b]T_R^c)=\frac{i}{2}T(R)f^{abc}$,while $P_L\rightarrow-\frac{1}{2}\gamma_5$ part coming with an extra factor of $\frac{1}{2}Tr(\{T_R^a,T_R^b\}T_R^c )=A(R)d^{abc}$?
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1$\begingroup$ You will have a higher chance to get a reply if you add all the relevant information in the body of the question. $\endgroup$– tbtCommented Aug 31, 2019 at 15:14
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$\begingroup$ Doesn't it simply follow from the fact that taking the $\gamma^5$ through the $\gamma^\mu$'s generates a sign and changes the commutator into an anti-commutator? $\endgroup$– OбжорoвCommented Sep 1, 2019 at 12:18
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$\begingroup$ I still don't understand. There are two diagrams and they have the same structure of $\gamma^\mu$ and $\gamma_5$. $\endgroup$– heheheheheheCommented Sep 1, 2019 at 14:43
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$\begingroup$ From a cursory reading it seems that it follows from evaluating (75.17). Without the gauge factor the only nonzero part comes from the $\gamma^5$ in $P_L$. Including the gauge factor both the $1$ and the $\gamma^5$ contribute and I expect the $\gamma^5$ generates and extra sign between the two diagrams in Fig 75.2. $\endgroup$– OбжорoвCommented Sep 2, 2019 at 8:01
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