From: Dan Christensen  jdc [snail] uwo.ca        Apr 25, 2009
      http://jdc.math.uwo.ca

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