# Chrial gauge theories and anomalies(Srednicki ch75-76)

• LAHLH
In summary, Srednicki says the triangle diagrams analyzed now come with the extra factor Tr(T^a T^b) , why not Tr(T^a T^b T^c) ? Because now one of the three external lines corresponds to the current of the global symmetry, and this does not couple to the nonabelian charge represented by a T^a matrix.

#### LAHLH

Hi,

Don't know if anyone can help me but have a bit of confusion with Srednicki ch75 p466 just above (75.55). I understand why in non-Abelian gauge theory we get extra factors $Tr(T^aT^bT^c)$ and so on, but I don't understand why the $P_{L}\to1/2$ diagrams then end up with the extra factor $1/2Tr([T^a,T^b],T^c)$, does anyone know?

Also then in ch77, Srednicki says the triangle diagrams analyzed now come with the extra factor $Tr(T^a T^b)$, why not $Tr(T^a T^b T^c)$?, after all they are the same diagrams he talks about of p466 (except for some changes he notes on p470 that don't seem to make a difference to this argument)

Thanks, would be really grateful if anyone is familiar with this...

LAHLH said:
I don't understand why the $P_{L}\to1/2$ diagrams then end up with the extra factor $1/2Tr([T^a,T^b],T^c)$
In QED, the two diagrams just cancel. In nonabelian theory, one diagram gets an extra factor of $Tr(T^a T^b T^c)$, and the other gets an extra factor of $Tr(T^b T^a T^c)$. These two factors do not cancel, and their difference gives the expression you wrote (up to factors of 2, which I'm not trying to get right here).

LAHLH said:
Also then in ch77, Srednicki says the triangle diagrams analyzed now come with the extra factor $Tr(T^a T^b)$, why not $Tr(T^a T^b T^c)$?
Because now one of the three external lines corresponds to the current of the global symmetry, and this does not couple to the nonabelian charge represented by a T^a matrix.

In nonabelian theory, one diagram gets an extra factor of Tr(TaTbTc), and the other gets an extra factor of Tr(TbTaTc). These two factors do not cancel, and their difference gives the expression you wrote (up to factors of 2, which I'm not trying to get right here).

Calling the first term in (75.16) : $D^{\mu\nu\rho}(p,q,r)$ for brevity ,and also taking the $P_L \to 1/2$ part of it, then in nonabelian gauge theory I believe the PL 1/2 term of (75.16) generalizes to:

$$iV^{\mu\nu\rho}(p,q,r)=1/2\text{Tr}\left(T^aT^bT^c\right)D^{\mu\nu\rho}(p,q,r)+1/2\text{Tr}\left(T^bT^aT^c\right)D^{\nu \mu \rho}(q,p,r)+\mathcal{O}(g^5)$$

Now if $D^{\mu\nu\rho}(p,q,r)=-D^{\nu \mu \rho}(q,p,r)$ you could take this D out as a common factor, and be left with $1/2 r\left(T^aT^bT^c\right)-1/2r\left(T^bT^aT^c\right)=1/2Tr([T^a,T^b],T^c)$. But is this relation between the D's (integrals over the loops) true? It's not completely obvious to me if so...

Yes. Without the group matrices, and without a $\gamma_5$, the two diagrams are just the ones of QED, and they cancel, as discussed in the paragraph above eq.(75.16).

Avodyne said:
Yes. Without the group matrices, and without a $\gamma_5$, the two diagrams are just the ones of QED, and they cancel, as discussed in the paragraph above eq.(75.16).

Yes, I can see your logic must be correct, as if these D's didn't behave this way the cancellation wouldn't happen in QED, as you say. Seems hard to show directly just from the integrals themselves that they cancel though...

thanks again for the help

Cancel wasn't the right word, my point is in order to do the manipulations shown by me in post #3, the relation between the $D^{\mu\nu\rho}(p,q,r)=-D^{\nu\mu\rho}(q,p,r)$ must hold. I found it hard to prove this relation just from the integrals alone, yet it must be true since as Avodyne noted in QED we know the expression is zero, and the expression is exactly the same except for lacking the group factors as coefficients, thus the relation $D^{\mu\nu\rho}(p,q,r)=-D^{\nu\mu\rho}(q,p,r)$ simply must be true, even if I can't seem to show it directly. This means you can pull out the D as a common factor as in post #3 and all the manipulations go through to arrive as the Srednicki result etc etc..
In QED, you can use charge-conjugation symmetry to show that there can't be a 3-photon amplitude. So inserting $C^{-1}C$ between every pair of gamma matrices in one expression might help you to turn it into minus the other one. ( I have not tried this myself.)