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
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  The 6-j symbol and intersecting Wilson loops, redux

+ 4 like - 0 dislike
745 views

This is a quite specific question continuing the problems I have with computing the expectation value of intersecting Wilson loops I laid out here. Using the tools from the answer there, I quite quickly arrive at the following expression for the local factor associated to a vertex, where two Wilson loops with reps $\alpha_1$ and $\alpha_2$ meet, and where the four surrounding regions have reps $\beta_1$ to $\beta_4$:

$$ G(\alpha_1,\alpha_2,\beta_{1,2,3,4})_{\mu\nu}^{\sigma\rho} := \epsilon_\mu^{ijk}(\alpha_1,\beta_1,\beta_4)\epsilon_\nu^{lmn}(\alpha_2,\beta_1,\beta_2){\epsilon^*}^\sigma_{ijk}(\alpha_1,\beta_2,\beta_3){\epsilon^*}^\rho_{lmn}(\alpha_2,\beta_3,\beta_4)$$

The Greek indices are the indices incurred from decomposing tensor products as $a^i \otimes b^j \otimes b^k = \epsilon_\mu^{ijk}e^\mu$, leading to integral results like

$$\int \alpha_i(V_b)^i_{i'} \beta_c(V_b)^j_{j'} \beta_{c'}(V_b)^k_{k'} \mathrm{d} V_b = {\epsilon^*}^\mu_{i'j'k'}\epsilon^{ijk}_\mu$$ (see previous answer). Since the $\epsilon^*$ that has the $\mu$ this is summed with lives on the opposite end of the (part of) the Wilson line, the Greek indices must necessarily remain open at the vertices. I am totally fine with this being the result of the computation, but I am still puzzled why the relation to the 6j symbol is so casually tossed about.

Let me first remark that the above equation is already suspiciously similar to the very first equation in the definition of $6j$ symbols, but the free indices are irritating me. If $\epsilon_\mu^{ijk}(\alpha_l,\beta_m,\beta_n)$ is the $3jm$ symbol (with $i,j,k$ playing the role of the $m$ and the reps corresponding to the $j$), what is the additional index $\mu$ doing here? If it is not the $3jm$ symbol (which I am currently thinking), then why would the $G$ defined above be the $6j$ symbol (and why has it free indices)? (If these are neither $3jm$ nor $6j$ symbols, then why do Witten, Ramgoolam, Moore, etc. insist they are?)

Note that the $6j$ symbol cannot arise after summing the Greek indices, since the $G$s the second index belongs to are, in general, at other vertices, and so have not exactly the same 6 reps as arguments.

Furthermore, the $3jm$ symbols are, if I understand them correctly, essentially the Clebsch-Gordan coefficients for expanding a tensor product of two irreducible reps in a third, and the $\epsilon$ above expand the tensor product of three irreducible reps in all possible fourths (which are then summed over in form of the Greek indices).

Something does not add up here, and I heavily suspect it is only in my understanding of the symbols, so I would really appreciate someone clearing up my confusion.

This post imported from StackExchange Physics at 2014-06-25 21:01 (UCT), posted by SE-user ACuriousMind
asked Jun 22, 2014 in Theoretical Physics by ACuriousMind (910 points) [ no revision ]

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$\hbar$ysicsO$\varnothing$erflow
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
...