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I'm trying to show that the value of the R-charge $r$ for which the R-symmetry is non-anomalous is given by $r=\frac{F-N}{F}$.

To do this we must calculate the triangle diagrams for the quarks $\psi_i$ and the gauginos $\lambda^a$. For the quarks, everything seems right. What I get is that

$$ (r-1)\sum_{i,j,k}^F \langle 0 |T\{ (\bar{\psi}^i\psi^i) (\bar{\psi}^jT^a\psi^j) (\bar{\psi}^kT^b\psi^k )\} |0\rangle = F(r-1)\delta_{ab}$$

where $T_a$ is in the fundamental of the gauge group $SU(N)$. However, when it comes to the triangle diagram for the gauginos I get that

$$\sum_{a,b,c}^{N^2-1} \langle 0 |T\{ (\bar{\lambda}^a\lambda^a) (\bar{\lambda}^bt^A\lambda^b) (\bar{\lambda}^ct^B\lambda^c )\} |0\rangle = (N^2-1)(N)\delta_{AB}$$

where $t^A$ are in the adjoint rep.

Here is the problem

The $N^2-1$ should not be there. It came from $tr(I_{N^2-1})$, just like the $F$ came from $tr(I_F)$. This came from the fact that I assumed that the contraction of

$$ \overbrace{\lambda^a \bar{\lambda}^b} = \delta_{ab}. $$

So this must have been incorrect, but why? And what happens to the sums over $a,b,c$ then?

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  • $\begingroup$ There is no $(N^2-1)$. $N^2-1$ is just the dimension of the adjoint. You also do not get the dimension of the fundamental, $N$, in the quark contribution. There are many issues with your indices. The generators $t^A$ also carry indices, which you suppress. If you do things properly, you arrive at traces tr$t^At^B$ in the respective representations, which of course gives just the Dynkin indices, $1/2$ for the fundamental and $N$ for the adjoint. Recalling that the the quarks come in conjugate pairs, i.e. you have $F$ generations of $Q+\bar Q$, then gives the correct result. $\endgroup$ – user178876 Aug 11 '18 at 5:04
  • $\begingroup$ Fair enough.. I did suppress the indices of the generators and the things they contract with.. So you're saying that I forgot to sum over the color indices in the quark contribution? Also, what do you mean that there is no $N^2-1$ doesn't the sum over the $a,b,c$ run from 0 to $N^2-1$? I guess I'm wondering why I explicitly write out the flavor sum but not the sum over the $a,b,c$? $\endgroup$ – InertialObserver Aug 11 '18 at 5:09
  • $\begingroup$ Yes. The indices are very different. The propagators are diagonal in color space, you will get contractions of the form $T^A_{ab}T^B_{bc}\delta_{ca}$. $\endgroup$ – user178876 Aug 11 '18 at 5:12
  • $\begingroup$ Isn't that where the $N\delta_{AB}$ comes from? $\endgroup$ – InertialObserver Aug 11 '18 at 5:13
  • $\begingroup$ Yes, precisely. But no $N^2-1$. $\endgroup$ – user178876 Aug 11 '18 at 5:14

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