Ternary parallelogram scales are MOS substitution: Difference between revisions
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This article proves the following theorem: | This article proves the following theorem: | ||
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=== Pitch-class group === | === Pitch-class group === | ||
The ''pitch-class group'' of a scale word ''w'' in letters {{nowrap|'''x'''<sub>1</sub>, ..., '''x'''<sub>''r''</sub>}} with [[step signature]] {{nowrap|'''e''' ∈ ℤ<sup>''r''</sup>{{angbr|'''x'''<sub>1</sub>, ..., '''x'''<sub>''r''</sub>}}}} is the abelian group {{nowrap|C(''w'') :{{=}} ℤ<sup>''r''</sup>{{angbr|'''x'''<sub>1</sub>, ..., '''x'''<sub>''r''</sub>}}/{{angbr|'''e'''}}.}} The pitch-class group is associated with a canonical map π that sends every step vector to its pitch class. | The ''pitch-class group'' of a scale word ''w'' in letters {{nowrap|'''x'''<sub>1</sub>, ..., '''x'''<sub>''r''</sub>}} with [[step signature]] {{nowrap|'''e''' ∈ ℤ<sup>''r''</sup>{{angbr|'''x'''<sub>1</sub>, ..., '''x'''<sub>''r''</sub>}}}} is the abelian group {{nowrap|C(''w'') :{{=}} ℤ<sup>''r''</sup>{{angbr|'''x'''<sub>1</sub>, ..., '''x'''<sub>''r''</sub>}}/{{angbr|'''e'''}}.}} The pitch-class group is associated with a canonical map π that sends every step vector to its pitch class. | ||
Below we take it as known that the gcd of the coordinates ''v''<sub>''i''</sub> of '''v''' ∈ ℤ<sup>''r''</sup> is 1 iff the quotient group ℤ<sup>''r''</sup>/{{angbr|'''v'''}} is torsion-free; this can be proven using Bézout's identity. | |||
=== Parallelogram scale === | === Parallelogram scale === | ||
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== Proof == | == Proof == | ||
=== Step 1: Get a surjective homomorphism <math>\varphi:\mathbb{Z}^2 \to \mathbb{Z}/mn\mathbb{Z}</math> === | === Step 1: Get a surjective homomorphism <math>\varphi:\mathbb{Z}^2 \to \mathbb{Z}/mn\mathbb{Z}</math> === | ||
The π-image of any ''k''-step interval (abelianized slice) {{nowrap|ab(''w''[''i'' : ''i'' + ''k''])}} is identical to that of {{nowrap|ab(''w''[''i'' : ''i'' + ''k'' + ''mn'']).}} Hence there is a well-defined map from the pitch classes of intervals of ''w'' to {{nowrap|ℤ/''mn''ℤ.}} | The π-image of any ''k''-step interval (abelianized slice) {{nowrap|ab(''w''[''i'' : ''i'' + ''k''])}} is identical to that of {{nowrap|ab(''w''[''i'' : ''i'' + ''k'' + ''mn'']).}} Hence there is a well-defined map from the pitch classes of intervals of ''w'' to {{nowrap|ℤ/''mn''ℤ.}} Because ''w'' is a parallelogram scale, traversing ''w'' step by step gives a traversal of {{nowrap|[0 : ''m''] × [0 : ''n'']}} where we label each grid point with the index of the current note in ''w''. We also recall that the pitch-class vector '''v''' has a representative that is a ''k''<sub>'''v'''</sub>-step interval in ''w'', {{nowrap|0 < ''k''<sub>'''v'''</sub> < ''mn'',}} and similarly for '''w'''. | ||
We thus wish to constrain ways of labeling {{nowrap|[0 : ''m''] × [0 : ''n'']}} with {{nowrap|ℤ/''mn''ℤ}} elements such that | We thus wish to constrain ways of labeling {{nowrap|[0 : ''m''] × [0 : ''n'']}} with {{nowrap|ℤ/''mn''ℤ}} elements such that | ||
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This pattern of movements is in fact the same as the one produced by taking the circular word {{nowrap|''v'' {{=}} "1 1 1 ... 1 (1 - ''n'')"}} ((''n'' - 1)-many 1's) and stacking sums <math>\sum_{i=0}^{c-1} v[i_0+i]</math> of ''c''-step subwords. As the resulting word has only one bad position per period, the filling word can easily be seen to be MOS by stacking ''kc''-step subwords of ''v'' for {{nowrap|2 ≤ ''k'' ≤ length - 1.}} {{Qed}} | This pattern of movements is in fact the same as the one produced by taking the circular word {{nowrap|''v'' {{=}} "1 1 1 ... 1 (1 - ''n'')"}} ((''n'' - 1)-many 1's) and stacking sums <math>\sum_{i=0}^{c-1} v[i_0+i]</math> of ''c''-step subwords. As the resulting word has only one bad position per period, the filling word can easily be seen to be MOS by stacking ''kc''-step subwords of ''v'' for {{nowrap|2 ≤ ''k'' ≤ length - 1.}} {{Qed}} | ||
== Open problems == | |||
# Conjecture: Ternary scales that are parallelogram substrings with full row length ''m'', full column length ''n'', and cardinality {{nowrap|''mn'' - 1}} are MOS substitution. | |||
# Conjecture: Ternary scales that are parallelogram substrings with full row length ''m'', full column length ''n'', and cardinality {{nowrap|''mn'' - 2}} are MOS substitution. | |||
# Prove a converse to this theorem. | |||
#* <s>If the template MOS of a MOS substitution scale has m > 1 periods and the step signature's gcd = 1 => m × n parallelogram scale.</s> This is false as stated, since no 5L(3m7s) scale is a parallelogram scale. | |||
#* Conjecture: If the template MOS of a MOS substitution scale has m > 1 periods, the step signature's gcd = 1, and the filling MOS is a multiMOS, then the scale is a parallelogram scale. (But the filling MOS need not be a multiMOS for a parallelogram scale: LLmLLsLLmLLsLLs is of type 10L(2m3s)) | |||
#* Conjecture: If a MOS substitution scale's template MOS is a multiMOS, and its step signature's gcd = 1, then its generator structure is a (possibly) *shifted* full parallelogram, i.e. either a full parallelogram or the first and last rows' lengths add up to the full row length. | |||
[[Category:Pages with proofs]] | [[Category:Pages with proofs]] | ||
[[Category:Combinatorics on words]] | [[Category:Combinatorics on words]] | ||
[[Category:Pages with open problems]] | |||