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Which $6j$-symbols for quantised enveloping algebras are known explicitly?

The quantum $6j$-symbols for $sl(2)$ are well-known. The references are Masbaum and Vogel and Frenkel and Khovanov.

What is known for other simple Lie algebras?

In case this seems somewhat vague here is a precise question. The data for a $6j$-symbol starts with a tetrahedral graph with edges labelled by highest weights. Then usually there is additional information needed at the vertices which I want to avoid. Take the example of $sl(n)$ and use partitions instead of highest weights. Label two opposite edges by a partition of the form $1^k$ (corresponding to an exterior power of the vector representation) and label the other four edges by partitions. Then associated to this data is a scalar.

Then I would expect this function to be determined by linear recurrence relations (i.e. D-finite or holonomic). Is this correct? and if so can you give recurrence relations?

Ideally we would also regard $n$ as an indeterminate.

Background In general $6j$-symbols arise for any semisimple abelian category which is also monoidal. They are the components of the natural transformation $(\otimes)(\otimes \times 1)\cong (\otimes)(1\times \otimes)$.

In more down to earth terms. If you know the Grothendieck ring of a semisimple abelian monoidal category and you attempt to construct this then the information you are missing is the $6j$-symbols. You can construct the abelian category and you can construct the tensor product functor but you don't have the associator.

For example it is well-known that the character table of a finite group does not determine the group. It does determine the category of representations as a semisimple abelian category. The $6j$-symbols are needed to make this a monoidal category.

For further discussion see:
Character table does not determine group Vs Tannaka duality

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I think that the result for sl(2) is due to Kirillov and Reshetikhin, and that these other references are later, although they may give new arguments. –  Greg Kuperberg Feb 19 '10 at 14:43
    
I agree. Kirillov and Reshetikhin stated the result. Masbaum and Vogel proved it by showing both sides satisfied the same linear recurrence relations. Frenkel and Khovanov derive the result. The Frenkel and Khovanov method is the only one which looks as though it could be applied in other examples. –  Bruce Westbury Feb 19 '10 at 15:22
    
In SL(2) the decomposition of 2-fold tensor products is unique, prompting one to compare the decomposition of $(A \otimes B) \otimes C$ with $A \otimes (B\otimes C)$. But for other groups the decomposition isn't unique, without some canonical basis input perhaps. So it's never been clear to me how to define 6j symbols, much less compute them. –  Allen Knutson Feb 20 '10 at 15:05
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Allen, The decomposition of an exterior power with any representation is unique. The question was phrased to avoid this problem. I defined the 6j-symbols as the components of a functor. In general this would mean specifying bases in vector spaces and there is no consensus on how to do this (but there are several ways it can be done). You put a vector at each vertex of the tetrahedron graph to define a scalar. –  Bruce Westbury Feb 22 '10 at 9:06
    
Allen: Yes, you can define a numerical 6j-symbol given any system of canonical bases of $\text{Inv}(A \otimes B \otimes C)$, where "canonical" is taken in the informal sense. More formally, you can (a) use Lusztig's dual canonical bases; or (b) simply ask for any self-consistent set of formulas. A consistent set of formulas would define the bases with respect to which they are correct. –  Greg Kuperberg Feb 23 '10 at 21:21
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1 Answer

up vote 13 down vote accepted

We can also use partitions ($k$) (symmetric powers) instead of ($1^k$) on one or two of the edges. This still gives just scalars, but includes the full story for sl(2).

This problem seems to be equivalent to the problem of computing the exchange operator in the tensor product of two (quantum) symmetric or exterior powers of the vector representation of the quantum sl(n), e.g. $S^kV\otimes S^mV$, in the standard basis of the symmetric (resp., exterior) powers (about exchange operators see my paper with Varchenko arXiv:math/9801135 and my ICM talk arXiv:math/0207008). This can be computed if you know the fusion operator for these representations, which can be computed efficiently using the ABRR (Arnaudon-Buffenoir-Ragoucy-Roche) equation, see e.g. the appendix to arXiv:math/9801135. I am not sure if the answer is completely worked out anywhere, but there are at least some answers. For instance, see the paper arXiv:q-alg/9704005 where something is done even in the elliptic setting (which relates to elliptic 6j symbols of Frenkel-Turaev). What they do is compute the matrix elements for $m=1$, but the general $m$ can be obtained using the fusion procedure. This should be a really nice computation with a nice answer of the type you are expecting. In particular, in a special case you'll get coefficients of Macdonald's difference operators attached to symmetric powers. In the exterior powers case (or a product of a symmetric and an exterior power) the answer will be simpler, since $k$ cannot get larger than $n$ (in this case you should perhaps get a pure product, and it should be completely covered in the above paper).

EDIT: Remark. Let $V,W$ be representations of the quantum group with 1-dimensional zero weight spaces. Then a natural basis in ${\rm Hom}(L_\lambda, (V\otimes L_\mu)\otimes W)$ is the compositions of intertwiners in one order, while a natural basis of ${\rm Hom}(L_\lambda, V\otimes (L_\mu\otimes W))$ is the composition of intertwiners in the other order. Thus, the 6j-symbol matrix (which is by definition the transition matrix between these two bases) is the exchange operator for intertwiners.

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It has taken me a while to respond as I had not come across fusion operators or exchange operators. I have found a formula which I almost understand which says that the 6j-symbols can be calculated from 3j-symbols using the fusion operator. I have not seen a relation between the 6j-symbols and the exchange operator. –  Bruce Westbury Mar 4 '10 at 21:17
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I edited my answer to explain the relation between 6j symbols and exchange operators. –  Pavel Etingof Mar 4 '10 at 22:13
    
I need to clear my desk so I can work through this. Meanwhile; if you use Lusztig's tensor product then the fusion operators are identity maps. This means the 6j-symbols are equal to the 3j-symbols (with respect to this tensor product). Then I think the exchange operator is the R_0-matrix. I am not sure how this helps calculate 6j-symbols. –  Bruce Westbury Mar 5 '10 at 8:04
    
I am not sure what you mean by "Lusztig's tensor product". By fusion operator I mean the one defined in my paper with Varchenko "Exchange dynamical quantum groups", and I don't see how it can be equal to the identity map. Also the fusion and exchange operators depend on the dynamical parameter $\lambda$ which is one of the six j-labels. –  Pavel Etingof Mar 5 '10 at 14:30
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