User:Inthar/Epimorphic temperament: Difference between revisions

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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''.
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 CS scales:
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 2.3.5 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]].
* The 2.3.7 temperament [[semaphore]] supports [[archylino]] (2L3M2s), [[diasem]] (5L2M2s), and other scales in the [[Tas series]].
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=== Epimorphic scales are CS ===
=== Epimorphic scales are CS ===
{{proof|contents=
{{proof|contents=
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>
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.
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.