Splitting Orbits

In the case when is a finite action, we can apply the sign map to , the permutation group induced by on . Its kernel

is either itself or a subgroup of index 2, as is easy to see. Denoting its inverse image by

we obtain a useful interpretation of the alternating sum of fixed point numbers:

. Lemma   For any finite action such that , the number of orbits of on which split over (i.e. which decompose into more than one - and hence into two - -orbits) is equal to

Proof: As , and hence , we have

Each orbit of on is either a -orbit or it splits into two orbits of , since . Hence is the number of orbits which split over . Finally the stated identity is obtained by an application of the homomorphism theorem.

. Corollary   In the case when , the number of -orbits on which do not split over is equal to

Note what this means. If acts on a finite set in such a way that , then we can group the orbits of on into a set of orbits which are also -orbits. In figure we denote these orbits by the symbol . The other -orbits split into two -orbits, we indicate one of them by , the other one by , and call the pair an enantiomeric pair   of -orbits. Hence gives us the number of enantiomeric pairs of orbits, while yields the number of selfenantiomeric   orbits of on . The elements belonging to selfenantiomeric orbits are called achiral   objects, while the others are called chiral . These notions of enantiomerism   and chirality   are taken from chemistry, where is usually the symmetry group of the molecule while is its subgroup consisting of the proper rotations. We call a chiral action if and only if .

  
Figure: Enantiomeric pairs and selfenantiomeric orbits

Using this notation we can now rephrase and in the following way:

. Corollary   If is a finite chiral action, then the number of selfenantiomeric orbits of on is equal to

while the number of enantiomeric pairs of orbits is

The sign of a cyclic permutation is easy to obtain from the equation and the homomorphism property of the sign, described in :

But in fact we need not check the lengths of the cyclic factors of since an easy calculation shows (exercise ) that, in terms of the number of cyclic factors of , we have

Exercises

E .   Check ().