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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).

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.

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).

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).

This problem seems to be equivalent to the problem of computing the exchange operator in the tensor product of two (quantum) symmetric powers $S^kV\otimes S^mV$ of the vector representation of the quantum sl(n), in the standard basis of the symmetric 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.

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).