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#redirect [[Periodic scale#Epimorphism]]
A JI scale is '''epimorphic''' if on the JI subgroup ''A'' generated by the scale's intervals, there exists a linear map, called an '''epimorphism''' or '''epimorphic val''', ''v'': ''A'' → ℤ such that ''v''(''S''[''i'']) = ''i''.
 
An '''epimorphic temperament''' of an [[epimorphic]] scale ''S'' on a JI group ''A'' is a temperament supported by its epimorphic val on ''A''. Some [[exotemperament]]s (including vals for small edos) can be used as epimorphic temperaments for small epimorphic scales 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>C_k</math> be the set of ''k''-steps of ''S''. Then ''S'' is ''constant structure'' (CS) if for any <math>i, j \in \mathbb{Z}</math> we have <math>C_i \cap C_j = \varnothing.</math>
 
=== Epimorphic scales are CS ===
{{proof|contents=
Let ''v'' be the val witnessing the epimorphicity of ''s''. Let <math>x \in C_j.</math> Then there exists <math>i > 0</math> such that <math>S[i+j]/S[i] = x.</math> Suppose by way of contradiction there exist <math>k \neq j</math> and <math>i > 0</math> such that <math>S[i+k]/S[i] = x.</math>
 
Then <math>v(x) = v(S[i+j]/S[i]) = v(S[i+j]) - v(S[i]) = i + j - i = j,</math> but also <math>v(x) = v(S[i^\prime+k]/S[i^\prime]) = v(S[i^\prime+k]) - v(S[i^\prime]) = k,</math> a contradiction.
}}
 
=== 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>C_1</math> be the set of 1-steps of ''S'', and suppose that <math>C_1</math> is a ''basis'' for the JI group ''A'' generated by it. Then there exists an epimorphic val <math> 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>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.
 
{{proof|contents=
Define <math>v:A \to \mathbb{Z}</math> by defining <math>v(\mathbf{s}) = 1</math> for any step <math>\mathbf{s} \in C_1</math> and extending uniquely by linearity. Then for any <math>i \in \mathbb{Z}</math> we have <math>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>v(2) = n</math> is also automatic.
}}
 
[[Category:Scale]]

Revision as of 02:39, 31 January 2024

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

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

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 v be the val witnessing the epimorphicity of 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 [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 the scale {5/4, 32/25, 2/1} is CS but not epimorphic.

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]