From: Tony Kennedy 30 Jan 2008
I haven't had a chance to read M. Bergolet's letter very carefully, but my
immediate reaction is that projectors correspond to Young DIAGRAMS, not
Young tableaux.
In other words if P is the projector corresponding to the diagram
corresponding to partition (3,2) and s is any permutation in S[5] then P.s
can be written as a linear combination of the five birdtracks P.t which
correspond to standard tableaux (using Garnir relations), but P.t for t not
equal to the identity is NOT in general a projector.
Indeed, Young already noted that P.s.P = m[s] P where m[s] = -1,0,1 for any
permutation s, and m can be zero even for s corresponding to a standard
tableau (e.g., consider the permutation (1,2,3,4,5) -> (1,2,5,3,4)), in
which case (P.s).(P.s) = (P.s.P).s = 0. On the other hand, as M. Bergolet
notes, m[s] is not always zero for non-trivial s, so P.s.P need not vanish.
This is not a problem, as "our" theorem just says that the projectors P and
P' corresponding to DIFFERENT Young diagrams are orthogonal.
If one looks a little deeper the "non-orthongonality" of the P.t for
standard tableaux corresponds to the fact the irreps. of S[n] we
construct using Garnir relations as above is not unitary (although it must
be equivalent to a unitary irrep., as this is true for all finite groups).
It also means that the matrix representing P itself in the representation
constructed on the subspace projected by P is not zero everywhere except in
the top left corner, but can have a non-trivial top row (or left column,
depending on your conventions). This is the case in the notes I sent around
a while ago where I derived the formula for the 3j and 6j coefficients in
Gl(N) (or U(n) if you prefer) as the product of the Gl(N) trace of P and the
S[n] trace of the projectors.
-Tony-