User:Inthar/Epimorphic temperament

An epimorphic temperament of an epimorphic scale S on a JI group A is a temperament supported by its epimorphic val (linear map v: A → ℤ such that v(S[i]) = i) on G. Some exotemperaments (including vals for small edos) can be used as epimorphic temperaments for small CS scales:

• The 2.3.5 temperament dicot supports nicetone (3L2M2s), blackdye (5L2M3s) and superzarlino (a 17-form) scale structures.
• The 2.3.7 temperament semaphore supports archylino (2L3M2s), diasem (5L2M2s), and other scales in the Tas series.

Facts

Definition: constant structure (CS)

Given a periodic scale S, let [math]\displaystyle{ C_k }[/math] be the set of k-steps of S. Then S is constant structure (CS) if for any [math]\displaystyle{ i, j \in \mathbb{Z} }[/math] we have [math]\displaystyle{ C_i \cap C_j = \varnothing. }[/math]

Epimorphic scales are CS

Proof

Let [math]\displaystyle{ x \in C_j. }[/math] Then there exists [math]\displaystyle{ i > 0 }[/math] such that [math]\displaystyle{ S[i+j]/S[i] = x. }[/math] Suppose by way of contradiction there exist [math]\displaystyle{ k \neq j }[/math] and [math]\displaystyle{ i > 0 }[/math] such that [math]\displaystyle{ S[i+k]/S[i] = x. }[/math] Then [math]\displaystyle{ v(x) = v(S[i+j]/S[i]) = v(S[i+j]) - v(S[i]) = i + j - i = j, }[/math] but also [math]\displaystyle{ v(x) = v(S[i^\prime+k]/S[i^\prime]) = v(S[i^\prime+k]) - v(S[i^\prime]) = k, }[/math] a contradiction. [math]\displaystyle{ \square }[/math]

If the steps of a CS scale are linearly independent, then the scale is epimorphic

Theorem: Suppose S is a 2/1-equivalent increasing constant structure JI scale of length n. Let [math]\displaystyle{ C_1 }[/math] be the set of 1-steps of S, and suppose that [math]\displaystyle{ C_1 }[/math] is a basis for the JI group A generated by it. Then there exists an epimorphic val [math]\displaystyle{ v: A \to \mathbb{Z} }[/math] which is a val of n-edo (and a similar statement holds for other equaves).

The condition of [math]\displaystyle{ C_1 }[/math] being a basis rather than merely a generating set cannot be omitted, since otherwise the scale {5/4, 32/25, 2/1} is a counterexample.

Proof

Define [math]\displaystyle{ v:A \to \mathbb{Z} }[/math] by defining [math]\displaystyle{ v(\mathbf{s}) = 1 }[/math] for any step [math]\displaystyle{ \mathbf{s} \in C_1 }[/math] and extending uniquely by linearity. Then for any [math]\displaystyle{ i \in \mathbb{Z} }[/math] we have

[math]\displaystyle{ v(S[i]) = v(S[i]/S[i-1]\cdots S[1]) = v(S[i]/S[i-1]) + \cdots + v(S[1]) = i. }[/math]

That [math]\displaystyle{ v(2) = n }[/math] is also automatic. [math]\displaystyle{ \square }[/math]