User:Inthar/Endoparticular extensions: Difference between revisions

From Xenharmonic Reference
Line 8: Line 8:


== Conjecture: An endoparticular extension of a one-comma temperament is unique ==
== Conjecture: An endoparticular extension of a one-comma temperament is unique ==
Possible strategy: Since the definition of endoparticular requires consecutive square-particulars, I do have that constraint fortunately and I can, say, lower-bound the number of consecutive S-commas that must appear in a factorization into consecutive S-commas
Possible strategy: Since the definition of endoparticular requires consecutive square-particulars, maybe I can bound the number of consecutive S-commas that must appear in a factorization into consecutive S-commas

Revision as of 03:01, 7 February 2026

This is a problematic page or section. It lacks sufficient justification, content, or organization, and is subject to future overhaul or deletion.

Suppose the base temp tempers out a comma that is a product of powers of consecutive S-commas with nonzero exponents. An extension is endoparticular if it additionally tempers out the individual S-commas.

Conjecture: There is at most one S-expression for a comma in a given extended subgroup. As a corollary, an endoparticular extension of a temperament tempering out one given comma to a given extended subgroup is unique if it exists.

Stronger conjecture: An endoparticular extension for a given temperament of any rank and a given extended subgroup is unique if it exists (regardless of the choice of comma basis).

Conjecture: An endoparticular extension of a one-comma temperament is unique

Possible strategy: Since the definition of endoparticular requires consecutive square-particulars, maybe I can bound the number of consecutive S-commas that must appear in a factorization into consecutive S-commas