Anomaly of the $\text{U}(1)$-$\text{SU}(2)$-$\text{SU}(3)$ triangle diagram

In Srednicki's textbook "Quantum Field Theory", Problem 89.3 asks us to show that the Standard Model is anomaly free. I am puzzled by the triangle-vertex diagram whose external lines are a combination of $\text{U}(1)$-$\text{SU}(2)$-$\text{SU}(3)$ gauge fields. As stated in section 75 of this book, in chiral gauge theories, triangle-vertex diagrams have an extra factor of $$\frac{1}{2}Tr(\{T^{a}_{R}, T^{b}_{R}\}T^{c}_{R}) = A(R)d^{abc} \tag{75.55}$$ where $d^{abc}$ is a completely symmetric tensor that is independent of the representation, $A(R)$ is the anomaly coefficient of R, and $T^{a}_{R}, T^{b}_{R}, T^{c}_{R}$ are generators of the gauge fields. For the case of $\text{U}(1)$-$\text{SU}(2)$-$\text{SU}(3)$, the three generators have different dimensions, i.e., $T^{a}_{R}, T^{b}_{R}$ and $T^{c}_{R}$ are $1\times 1, 2\times 2$ and $3\times 3$ matrices respectively. How can they multiply each other in eq. (75.55)? In the solutions manual, it is said that the combination $\text{U}(1)$-$\text{SU}(2)$-$\text{SU}(3)$ involves the trace of a single $\text{SU}(2)$ or $\text{SU}(3)$ generator, and this vanishes. I don't understand how it comes about?

• I do not understand why you seem to think that the $T_R^a,T_R^b,T_R^c$ belong to the three different groups, respectively. When we write $T_R^a$, we just mean any generator of the group you're talking about in the given representation $R$, and they all have the same size as matrices, namely the dimension of the vector space $R$ acts on, regardless of which of the three factors they belong to. Jun 16 '17 at 8:24
• @ACuriousMind - For the case of U(1)-SU(3)-SU(3), the left-hand side of eq. (75.55) becomes $\frac{1}{2}Tr(\{T^{a}_{R}, T^{b}_{R}\}Y) = \frac{1}{4}\delta^{ab} \sum Y = 0$. This led me to think that $T^{a}_{R}, T^{b}_{R}, T^{c}_{R}$ belong to three different groups. But this is not right according to what you said. Then, in the case of U(1)-SU(2)-SU(3), how can we have the trace of a single SU(2) or SU(3)? How to apply eq. (75.55) to this case? Can you write it out? I am puzzled.
– Shen
Jun 16 '17 at 13:32

You are only asking a question about usage: The problem's answer is trivial, once you appreciate the language involved.

For a generation of left-handed quarks, consider the 6-d representation. So your 6-vector would have the up quark in its upper 3 components (representing the 3 colors thereof), and the down in the lower 3. The 12 SM generators then correspond to a dozen 6×6 matrices acting on such vectors.

The SU(2) generators amount to $\tau^i\otimes 1\!\!1_3$, that is, 2×2 Pauli matrices with 3×3 identity ones in each of their entries. The SU(3) ones likewise amount to $1\!\!1_2\otimes \lambda^a$, that is, Gell-Mann matrices acting on the u triplet block and the very same ones on the d block. The hypercharge commutes with everything and is 1/3 or 1/6 (depending on conventions: the average charge of the doublet or twice that) times $1\!\!1_2 \otimes 1\!\!1_3$.

Except for the hypercharge, the trace of each of the remaining 11 matrices vanishes—do you see that? Moreover, the 1-2-3 product you are asked to consider is thus proportional to $\tau^i\otimes \lambda^a$, also traceless.

The trace of the Kronecker product is the product of the traces of the tensor factors, cf the last equation here. (If this were not painfully obvious to you, consider the diagonal one $\tau^3\otimes \lambda^3$.) Now appreciate how this is also true for all reps, including the right-handed ones where the SU(2) generators vanish. (It does not matter their hypercharge is more complicated: the trace of 0 for SU(2) is zero.)

It is also true for the 2-2-3 anomalies of your problem, as the color part trace will always vanish—and the anticommutator of two Pauli matrices is the identity, in case you needed to know that, but you shouldn't: the duplexed trace of the color space, the right factor of the tensor product, always vanishes.

In practice, you introduces the relevant indices. For example, the gluon vertex features the mere number $(T_R^a)_{\alpha\beta}$ where $\alpha$ and $\beta$ are the colour indices of the quarks in the loop. Each quark propagator has a $\delta_{ab}$. So you end up with a sum over all colours that results into the trace of $T_R^a$, which is 0. The photon vertex in this viewpoint is just a number. The $Z$ vertex is more arcane but the same idea applies: introduce indices. But to prove the result asked in the exercise, you don't even need to explicit the $Z$ vertex since the above inspection of the gluon vertex gives you the result.