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With my uncanny guessing abilities :-) I finally derived the size of the vector space where a n-valent node of a graph, edge-colored with the irreps $R_k, 1\le{k}\le{n}$ "lives". (I.e. in how many linear independent diagrams you decompose it, see e.g. "Birdtracks".) It's (heavy notation abuse) $\sharp{(1)}\in\Pi^\bigotimes_n{R_k}$ or, since this likely causes a syntax crash, in words: Just tensor multiply all the $R_k$ and count how many copies of the trivial irrep $1$ you find in the product. (Can you verify my finding?)

Especially, for n=3 this means (excuse my ASCII) a triangle =|>- reduces either to 6j* >- or zero if >- is inadmissible (the triple product doesn't contain $1$). But haaaalt! Assume all irreps here are $(1,1)$ from $A_2$. $(1,1)\bigotimes(1,1)$ contains TWO copies of $(1,1)$ and thus the triple product two $1$, i.e. there exist two linear independent graphs with three open ends.

Which means =|>- should be something like 6j* >- + 6j* >- ...but there are no two >-, you can only build this one acyclic graph! On the other hand: 6j symbols being undefined due to multiplicity >1 - I would have heard of that. Where is the hole in my logic?

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    $\begingroup$ There is nothing wrong with your logic. This is something you have to live with. $\endgroup$
    – Bruce Westbury
    Commented Sep 10, 2012 at 12:57
  • $\begingroup$ I've always found 6j symbols when there's multiplicities very confusing. $\endgroup$
    – Noah Snyder
    Commented Sep 10, 2012 at 15:05
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    $\begingroup$ 6j symbols should really be called 6j+4k symbols, where the 4 k's index bases of the trivalent vertex spaces. It's a weird, special fact about SU(2) that you can ignore the k's. In general this is almost never the case. $\endgroup$
    – Kevin Walker
    Commented Sep 10, 2012 at 16:42
  • $\begingroup$ @Kevin (or anybody else) - do you know a good reference to start? (From my spectroscopic work, I'm accustomed to additional quantum numbers, like "seniority" or an additional counter index for equal $^{2S+1}L_J$ terms, but luckily, I'm exclusively working in SU(2) :-) $\endgroup$
    – Hauke Reddmann
    Commented Sep 11, 2012 at 8:49
  • $\begingroup$ Not to bump a dead thread, but in case it is still needed and for anyone else who may run across this, On Arithmetic Modular Categories gives a through description and derivation for moving between 10j-Symbols and fusion categories. $\endgroup$
    – Matthew Titsworth
    Commented May 19, 2015 at 22:49

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