0
$\begingroup$

I somewhat advanced since 6j symbols trouble when irrep multiplicity >1 (because I found the decades old papers where the Racah algebra is dealt with, e.g. "Coupling Coefficients and Tensor Operators for Chains of Groups", P. H. Butler) but now it's even worse: I just don't call my finding "antinomy" because I reserve the possibility I'm too flamin' stupid :-)
So, I need just one irrep with multiplicity >=2 (take for example the "fusion mini" $A\bigotimes{A}=1\bigoplus{2A}\bigoplus{B},A\bigotimes{B}=A,B\bigotimes{B}=1$ and just use $A$ for "coloring") and asking the innocent question: Is >1-2< equal to >2-1< ?
Excuses for the ASCII art. These are two connected cubic nodes where both nodes are labeled with different multiplicities (I don't know what's the standard way to label with multiplicity, but "coloring" the node seems logical).
A: They are not equal. But "mutant logic" says they are equal anyway, expressing them as linear combinations of 4-tangles that ARE mutant-symmetric (like >1-1<) makes them obviously equal (weasel out: the set of mutant-symmetric 4-tangles is not a complete vector space basis) and doing a 6j recoupling to the rotated projectors makes them obviously equal too (a multiplicity flip in a 6j symbol built from just one irrep has no effect).
B: They are equal. But then the vector basis gets the wrong dimension (in my example, e.g. 6 projectors, but then two of them are the same) and even worse, "multiply" the equation from the right and reduce the loop by Schur: >1-2<>1- = >2-1<>1- . Complete nonsense results.

If someone would have the complete 6j set for a fusion ring with multiplicity, I might find out where the bug sits and how to encode the 6j formulas (Biedenharn-Elliott, Racah) correctly in case of multiplicity (and non-self-duality, which also annoys :-). Or can you tell me directly what I am doing wrong?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

The fusion ring you consider is known as (even part of) $E_6-$subfactor. The corresponding 6j symbols (also called associativity matrices) are given in the paper by Hagge and Hong, see Appendix.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Downloaded :-) I hope I can find the hole in my logic now myself, but still would appreciate a direct answer, although I checked the "answered" button. (Note added in reading: Argl, how do associtativity matrices relate to 6j symbols? In any case, Ax|xxx is 6-dimensional so this should mean my case A is true. Shrug.) $\endgroup$
    – Hauke Reddmann
    Commented Dec 2, 2014 at 12:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .