First, note that for the -element subsets and of are symmetry patches, since the translation maps one into the other. Hence, for , there are more than symmetry 3-patches. This implies the following estimate for the translational symmetry rate:
This rough estimate suffices to prove the first statement.
Claim: Each subset of not containing one tone from as well as one from is a symmetry patch with respect to . Every subset of not containing a point of is transformed by into another subset of . A similar argument using applies to each subset disjoint from . Thus, only subsets of containing at least one point from as well as one from can be symmetry patches for . In the case of 2-symmetry patches this implies that there are at most 3-element subsets of which are no symmetry patches. For we thus have the following bounds
For this quotient tends to 0. Hence we obtain .
Remark 2Perhaps the dullest and at the same time highly translationallysymmetric melodywith-symmetry rateis thenatural candidate to realize the least upper bound for the-symmetry rates of all-tone melodies. In addition, the proofs of both propositions show that thesame results also hold in the case of higher symmetry rates.
The SR discussed above only took into account the number of elements in , ignoring any additional structure. The new numerical characteristic to be defined and discussed below is based on the orbit structure in this set.
Definition 6Letbe a melody withtones,a paradigmatic group,anda positiv integer. Two symmetry patchesandfrom,, belong to the same-orbit if there exists a transformationintakingintounder its action in.