$SU(5)$ representation and higher anti-symmetric traces

Question

In Zee QFT book v2 p.411 eq.16-17, he shows the SU(5) gauge theory anomaly cancellation by the following:

The 1st line in fundamental of SU(5) $$ tr(T^3)=3(+2)^3+2(-3)^3=30, $$ is easy to follow, which sums over 3 U(1) charge 2 fermions and 2 U(1) charge -3 fermions, with the cubic polynomial for the anomaly.

The 2nd line in anti-symmetric 10 of SU(5), my understanding is the following $$ tr(T^3)=3(2+2)^3+6(2-3)^3+(-3-3)^3=-30, $$ , which sums over 3 U(1) charge 2+2=4 fermions, 6 U(1) charge 2-3=-1 fermions, and 1 U(1) charge -3-3=-6 fermion with the cubic polynomial for the anomaly.

I kind of give an answer for the above. But I wonder whether there is another easier or succinct way to interpret this $$ tr(T^3)|_{5^*} $$ and $$ tr(T^3)|_{10} $$ and also the $T$ generator for them in the precise representation theory manner?

This post imported from StackExchange Physics at 2020-11-09 09:42 (UTC), posted by SE-user annie marie heart

Answer 1

Yes, there is a representation-theoretic way to do this in one step. Zee is taking a slightly different approach, showing the cancellation of the $U(1)^3$ anomaly where $U(1)$ is the hypercharge subgroup of the SM gauge group embedded in $SU(5)$. But actually it's easier to do the whole calculation in one go, looking at the whole gauge group $SU(5)$.

We want to avoid an $SU(5)^3$ gauge anomaly, which means we need to consider triangle diagrams with three $SU(5)$ currents, i.e. evaluate the correlator $\langle j^\mu_a j^\nu_b j^\rho_c \rangle$. For fermions in the loop transforming in a representation $R$, the result is proportional to $$\text{tr}(T^a_R T^b_R T^c_R) = \frac{i}{2} T_R f^{abc} + \frac14 d^{abc}_R, \quad d_R^{abc} = 2 \text{tr}(T^a_R \{T^b_R, T^b_C\})$$ where I replaced $T^a_R T^b_R$ with the average of their commutator and anticommutator, and $T_R$ is the Dynkin index. One can show the first term does not contribute to the anomaly. As for the second term, there is a unique totally symmetric rank $3$ tensor associated with the algebra up to scaling, so $$d^{abc}_R = A(R) d^{abc}_F$$ where $F$ stands for the fundamental representation, and $A(R)$ is called the anomaly coefficient.

The total anomaly is proportional to the sum of all the anomaly coefficients. By definition, $$A(5) = A(F) = 1.$$ From this fact we can construct other anomaly coefficients by the results $$A(R) = - A(\bar{R}), \quad A(R_1 \oplus R_2) = A(R_1) + A(R_2), \quad A(R_1 \otimes R_2) = A(R_1) d(R_2) + d(R_1) A(R_2)$$ where $d(R)$ is the dimension of the representation $R$. By the first rule, $$A(\bar{5}) = -1.$$ As for $A(10)$, you need to invoke an additional rule, which is that if something is taking more than two minutes you can look it up, because some mathematician has already computed it for you. (Now you can look them up too, since you know the name!) I looked up $A(10) = 1$, so $$A(\bar{5}) + A(10) = -1 + 1 = 0$$ as desired. For more details, see Srednicki, Schwartz, or a representation theory book like Ramond.

This post imported from StackExchange Physics at 2020-11-09 09:42 (UTC), posted by SE-user knzhou

Comments

For completeness, you may want to include the general formula for the anomaly coefficient (cf. doi.org/10.1103/PhysRevD.14.1159), to wit, $$A(\lambda)=\mathrm{dim}(\lambda)\sum_{i,j,k=1}^{N-1}a_{ijk}\lambda_i\lambda_j\lambda_k,$$ where $\lambda$ are the Dynkin labels of the representation, $a_{ijk}$ is a totally-symmetric symbol with $\lambda_{i\le j\le k}\equiv \frac{2(N-3)!}{(N+2)!}i(N-2j)(N-k)$, and $$\mathrm{dim}(\lambda)=\prod_{j=1}^{N-1}\frac{1}{j!}\prod_{k=j}^{N-1}\sum_{i=k-j+1}^k\lambda_i.$$ For the anti-symmetric, $\mathrm{dim}(\lambda)=\frac12N(N-1)$ and $A(\lambda)=N-4$.

This post imported from StackExchange Physics at 2020-11-09 09:42 (UTC), posted by SE-user AccidentalFourierTransform

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^ This here is exactly the “some mathematician” I was referring to!

This post imported from StackExchange Physics at 2020-11-09 09:42 (UTC), posted by SE-user knzhou