Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

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

+ 5 like - 0 dislike
695 views

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

enter image description here

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
asked Aug 1, 2018 in Mathematics by annie marie heart (1,205 points) [ no revision ]

1 Answer

+ 5 like - 0 dislike

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
answered Aug 1, 2018 by knzhou (115 points) [ no revision ]
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
^ 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

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$y$\varnothing$icsOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...