From: Dan Christensen jdc [snail] Apr 25, 2009 Question about evaluating 3n-j symbols I've been look at your Group Theory book and at your paper with Elvang and Kennedy, and I have a question about one of the examples. I have attached the relevant page of the book below to give the context. I understand that P_Y P_X P_Y = (3/4) \alpha_X P_Y and P_Y P_Z P_Y = (3/4) \alpha_Z P_Y and therefore that P_Y P_X P_Y P_Z P_Y = (3/4)^2 \alpha_X \alpha_Z P_Y but I don't understand how it follows that P_Y P_X P_Z P_Y = (3/4)^2 \alpha_X \alpha_Z P_Y . In general it is *not* true that m_{\sigma_1 \sigma_2} = m_{\sigma_1} m_{\sigma_2} (e.g. this fails for the three non-trivial sigma displayed earlier on that page) so in general you can't treat P_X and P_Z independently. But is something like this true if the sigma's involve disjoint sets of strands? Or is there some other reason why an extra P_Y can be inserted into the formula above? I've checked that this isn't always true, even in this special case. For example, take Y to be the tableaux from page 102 of your book, with 1,2,4 in the first row and 3,5,6 in the second. For sigma_1 = (2,3) and sigma_2 = (1,5) we have m_{sigma_1} = 0, m_{sigma_2} = 0 but m_{sigma_1 sigma_2} = 1. So in general one can't do the computation separately for each group of strands and then multiply the answers. I haven't found a counterexample where the permutations sigma_1 and sigma_2 don't "overlap", i.e. one involves only numbers <= some m while the other involves only numbers > m. Is there a result along these lines that justifies the calculation in your book? I'm implementing software for these sorts of computations, so if there are tricks like this that make the computations easier, that would be very helpful. In general, there can be so many terms in the projectors sandwiched between the two copies of P_Y, and anything that can avoid treating each term separately would be wonderful. Dan