1 Mar 2008
From: KENNEDY Anthony
To: Jutho.Haegeman
Subject: so(n) 3-j and 6-j coefficients
Hi Jutho,
I would also like to know an efficient way of computing so(n) 3-j and
6-j coefficients: in fact I would even like a simple construction for
the projection operators. I have some hope that this may be possible as
there is a simple formula for the dimension of the so(n) irreps due to
King.
I have also written a program to compute the projectors and hence 3-j
and 6-j coefficients by brute force, namely by removing all traces. I
have a slightly more efficient method which uses the Clebsch-Gordan
series for reducing an su(n) irrep into so(n) irreps (using the result
in Littlewood's book), but it is still slow because I need to use a
brute-force expansion to compute the normalization factors.
The problem of computing the 3-j and 6-j coefficients for sp(n) is also
open, and is of course equivalent to the so(n) problem using the
negative dimensional duality.
On the other hand I can compute the 3-j and 6-j coefficients for su(n)
much more efficiently: as noted our paper in JMP these are always
proportional to the dimension of Y, the highest-weight (most boxes!)
irrep occurring, and the coefficient is just the trace of the matrix
representing the projectors for each irrep tied together with
permutations (which can be read off from the Littlewood-Richardson
rule), where the matrices are in the irrep of the symmetric group
corresponding to the Young diagram for Y. This work is in the process of
being written up, so I hope to produce a more comprehensible explanation
of it soon.
I believe that the solution to this problem is not known, at least none
of the experts I have asked (such as Ron King) know of one.
If you have any ideas about how to do things better please let me know!
-Tony-