“…Pick ξ (1,2,3) and ξ (1,3,2) in such a way that (1, 2)(ξ (1,2,3) ) = ξ (1,3,2) ; ξ (1,2,3)j f and ξ (1,3,2)j f in such a way that (1, 2)(ξ (1,2,3)j f ) = ξ (1,3,2)j f and so on. Then the basis of B(f, G) is ξ id and x 3 1 x 3 2 x 3 3 x 3 4 x 3 5 ξ id in bidegrees (0, 0) and (3,3), respectively;…”