Epimorphic scale

A JI scale S is epimorphic if on the JI subgroup A generated by the intervals of S, there exists a linear map, called an epimorphism, v: A → ℤ such that v(S[i]) = i for all i ∈ ℤ.

An epimorphic temperament of an epimorphic scale S on a JI group A is a temperament supported by its epimorphism on A. Some exotemperaments (including vals for small edos) can be used as epimorphic temperaments for small epimorphic 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.

Example

Consider the Ptolemaic diatonic scale, {9/8, 5/4, 4/3, 3/2, 5/3, 15/8, 2/1}, which is nicetone with L = 9/8, M = 10/9, and s = 16/15. This scale is epimorphic because we can apply ⟨7 11 16], the 7edo patent val, to map the intervals into the number of scale steps:

[math]\displaystyle{ \left(\begin{array} {rrr} 7 & 11 & 16 \end{array} \right) \left(\begin{array}{rrrrrrr} -3 & -2 & 2 & -1 & 0 & -3 & 1 \\ 2 & 0 & -1 & 1 & -1 & 1 & 0 \\ 0 & 1 & 0 & 0 & 1 & 1 & 0 \end{array}\right) = \left(\begin{array}{rrrrrrr} 1 & 2 & 3 & 4 & 5 & 6 & 7 \end{array}\right) }[/math]

where the columns of the 3×7 matrix are the scale intervals written in monzo form. Hence, 7edo (equipped with its patent val) is an epimorphic temperament of the Ptolemaic diatonic scale.

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}, i \neq j, }[/math] we have [math]\displaystyle{ C_i \cap C_j = \varnothing. }[/math]

Epimorphic scales are CS

Proof

Let v: A → ℤ be the epimorphism for s. Let [math]\displaystyle{ x \in C_j. }[/math] Then there exists [math]\displaystyle{ i \gt 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 \gt 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 the [math]\displaystyle{ C_1 }[/math] is a basis of the JI subgroup A generated by it. Then there exists an epimorphism [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 the scale {5/4, 32/25, 2/1} is CS but not epimorphic. The converse of this conditional also fails, as {9/8, 5/4, 3/2, 25/16, 2/1} is epimorphic under 5edo's patent val.

Proof

Define the linear map [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] whence v is an epimorphism. That [math]\displaystyle{ v(2) = n }[/math] is also automatic. [math]\displaystyle{ \square }[/math]