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Is there any reference that discuss the $F$ symbols for finite groups?. Maybe they are known by other name in this context. I'm looking for a discussion (and actually a table if it exist), of the matrices that relate two different tensor decompositions of representations. In the case of the representation of $SU(2)$ Lie algebra, these objects are also known as the 6j symbols.

Does it make sense to talk about such object for finite groups?

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    $\begingroup$ I doubt it. $6j$ is about decomposing $A\otimes B\otimes C$ in two different ways, and the crucial thing about $SU(2)$ is that inside a tensor product $A\otimes B$ of irreps, each $D$ occurs at most once. So you get, essentially, two different canonical bases of $Hom(D,A\otimes B\otimes C)$ and $6j$ compares them. This multiplicity-freeness won't hold for almost any nonabelian finite groups. $\endgroup$ Jan 24, 2017 at 12:03
  • $\begingroup$ Allen, thanks for your input. I am a little puzzled now. In principle, I can take the Clebsch-Gordan coefficients of a tensor decomposition and use them to compute the $6j$ symbols. This amounts to choose a basis of $A\otimes B$. Tensoring the result with another representation in $C$, and computing the Clebsch-Gordan coefficients, will give me a basis of $(A\otimes B)\otimes C$. Doing the same in the opposite order gives a basis in $A\otimes(B\otimes C)$. Comparing the two should give the $6j$. A related thread in which you also participated is mathoverflow.net/q/15800/103992 $\endgroup$ Jan 25, 2017 at 11:46
  • $\begingroup$ Yep, same issue: how to pick these bases? In particular, if you and some author pick them differently, then their table won't be good for you. For $SU(2)$ everybody knows how to do this. For other connected Lie groups one can in principle use Lusztg's extremely-difficult-to-compute canonical basis. For general finite groups there's nothing analogous. $\endgroup$ Jan 26, 2017 at 6:50
  • $\begingroup$ There is a draft put online in 2015 concerning 3j and 6j symbols for finite groups with $dimHom(A,BC)=1$. What doesn't make sense to me is also the arbitrariness of the choice of basis.. I'd love to discuss further if you wish. $\endgroup$
    – Student
    Sep 1, 2020 at 20:32

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