# Triangle condition for quantum 6j symbols?

For SU2 and even SU2(q) the triangle condition is, well, the triangle condition (conveniently, all irreps are described by (half)integer J completely). Additionally, all three J of a triple must add to integer.

But what is the analogue to that for some arbitrary (quantum) group? As usual, I don't have a clue but an assumption :-) - random example in group A3: Let the reps be 1R1+2R2+1R3,2R1+R2+1R3 and 5R1+2R2+2R3, then you check every irrep separately: 125 - triangle fail, 212 - parity fail, 112 - pass, overall: fail.

Of course, I could be totally wrong...What is the correct condition? (Maybe it's not even a simple one: outer multiplicity and that.)

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## 1 Answer

As you probably guessed the answer is "its complicated." I guess you mean what is the rule for when the tensor product of two irreps $V_\lambda \otimes V_\gamma$ contains an irrep $V_\delta$. The answer involves identifying the irrep with its highest weight vector $\lambda$ and viewing it in the Weyl chamber. Humphreys "Intro to Lie Algebra and ..." is the a good place to learn this stuff. Roughly the analog of the triangle inequality says that if you add all images of $\lambda$ under the Weyl group to $\gamma$, $\delta$ will fall in the convex hull. the analog of the integrality condition is that $\lambda+\gamma - \delta$ must be in the root lattice: in general the weight lattice mod the root lattice is a small abelian group that Humphreys will tell you about. The full story for the tensor product decomposition is given by the beautiful Racah formula, which can be visualized in rank 2 thus: mark the dimensions of the weights of $V_\lambda$ in tracing paper in the weight lattice. Slide it to put the 0 over $\gamma$. Fold the tracing paper along the walls of the extended Weyl chamber until it lies entirely in that chamber. add the numbers over $\delta$, with minus signs on the ones written backwards (i.e. an odd number of folds). That is the multiplicity of $V_\delta$.

Similar story holds for quantum groups at roots of unity (you have to throw away pesky nonirreducible reps), and there is a perfectly analogous quantum Racah formula with the Weyl alcove taking the role of the chamber. To toot my own horn, see http://arxiv.org/abs/math/0308281 on archive for a pretty full treatment.

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Too bad it's not as simple as SU2. BTW, I read about Weyl chambers before, but I always thought they contained instruments of math torture designed for amateurs like me :-) – Hauke Reddmann Oct 5 '11 at 9:31