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Consider a Clebsch-Gordan expansion $R_i\bigotimes{R_j}=\bigoplus_p{R_p}$. Assume the irrep $R_k$ does NOT appear in the sum on the right side. Does it now follow that the "triangle" ${R_i,R_j,R_k}$ is "inaccessible" and consequently the 6j symbol $ \begin{Bmatrix} R_i & R_j & R_k\\\ R_l & R_m & R_n \end{Bmatrix} $ vanishes for any entries ${R_l,R_m,R_n}$ ?

(And what about a converse? After all, 6j symbols have accidental zeroes. But can ALL 6j symbols with some fixed upper row accidentally vanish even if the upper row forms an accessible triangle?)

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  • $\begingroup$ For the converse, if $Hom(R_k, R_i \otimes R_j)$ is nonzero, then take $l=0$, $m=k$ and $n=j$ and the map in Bugs Bunny's answer will be nontrivial (in fact, an isomorphism). $\endgroup$
    – David E Speyer
    Jul 23, 2012 at 14:52

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Doc, not sure what "inaccessible" but the second question is a tautology. Your $6j$ symbol is a linear map $\hom(R_k,R_i\otimes R_j)\otimes\hom(R_m,R_k\otimes R_l)\rightarrow \hom(R_n,R_j\otimes R_l)\otimes \hom(R_m,R_i\otimes R_n)$. It is zero because the left vector space is zero.

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  • $\begingroup$ "Inaccessible" as in "violating the triangle inequality" (in the SU2 case), resp. the analogon thereof. (Something with Weyl alcoves or suchlike, I'm just a dummy :-) (Background: I was computing a closed form for a load of 6j symbols with small irreps and the group belonging to the E7 family, and observed {SAA|VVV}=0, where S,A,V are the symmetric, adjoint and defining irreps. In the SU2 case it's obvious: 311 forms no triangle. For the other groups it's magic :-) $\endgroup$
    – Hauke Reddmann
    Jul 23, 2012 at 16:28

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