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Example text
224 merely multiplies it b y a constant. 4), (6) which is identical with X1 of Eq. (III-1). Proceeding analogously, one finds that the component belonging to F(5) b y this procedure is just X2 of (111-1). If the ligands are allowed to be chiral, we have ~ = ~n. In this case, a single scalar parameter is not sufficient to characterize a ligand, since that could not distinguish between the tigand and its mirror image. In addition to our scalar parameter ~, therefore, we need a pseudoscalar one, ~.
Thus, according to the criterion of Theorem 3, Subsection B, Y and Y" belong to the same IR. '~ The proof again proceeds in steps, which are quite analogous to those for Theorem 1. These are as follows: i') Eq. (40) holds for P+, P-, P; Q+, Q_, -Q. We also have 3+T+ = T+ ; (51) T_T_ = e(r-)T-. (52) Eqs. ). (54) 31 c. A. Mead All of these assertions follow immediately from the definitions. ii') A permutation p-q does not move any ligand from a site in one part of ] to a site in the other; nor does it move two ligands initially in the same column of t<+> or t<-> into the same row. All permutations having this property are expressible in the form ~ .