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The 6-j symbol and intersecting Wilson loops, redux

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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 (820 points) [ no revision ]

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