Ternary parallelogram scales are MOS substitution: Difference between revisions

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This article proves the following theorem:

''[[Ternary]] parallelogram scale ~~words~~ are [[MOS substitution]] scale words, where the period count of the template MOS is the number of rows of the parallelogram parallel to the unique step size parallel to a side of the parallelogram.''

''[[Ternary]] parallelogram scale [[word]]s are [[MOS substitution]] scale words, where the period count of the template MOS is the number of rows of the parallelogram parallel to the unique step size parallel to a side of the parallelogram.''

== Definitions ==

=== Pitch-class group ===

The ''pitch-class group'' of a scale word ''w'' in letters {{nowrap|'''x'''1, ..., '''x'''''r''}} with [[step signature]] {{nowrap|'''e''' ∈ ℤ''r''{{angbr|'''x'''1, ..., '''x'''''r''}}}} is the abelian group {{nowrap|C(''w'') :{{=}} ℤ''r''{{angbr|'''x'''1, ..., '''x'''''r''}}/{{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''''i'' of '''v''' ∈ ℤ''r'' is 1 iff the quotient group ℤ''r''/{{angbr|'''v'''}} is torsion-free; this can be proven using Bézout's identity.

=== Parallelogram scale ===

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=== MOS substitution scale ===

~~See~~ [[MOS ~~substitution~~]].

A ternary scale word ''w''('''x'''1, '''x'''2, '''x'''3) is a ''MOS substitution'' scale word if there exists a permutation <math>\pi \in S_3</math> such that the following holds:

* identifying '''x'''π(1) and '''x'''π(2) results in a [[MOS]] scale word (called the ''template MOS'') and

* deleting all instances of '''x'''π(3) (called the ''slot letter'') results in a MOS scale word (called the ''filling MOS'').

== Proof ==

=== 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.}} ~~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'''''v'''-step interval in ''w'', {{nowrap|0 < ''k'''''v''' < ''mn'',}} and similarly for '''w'''.

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'''''v'''-step interval in ''w'', {{nowrap|0 < ''k'''''v''' < ''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

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Under the above assumption, since {{nowrap|φ(''t'''''v''') {{=}} 1}} we have that {{nowrap|φ('''v''') {{=}} ''k'''''v'''}} is a cyclic generator of the group {{nowrap|ℤ/''mn''ℤ,}} such that {{nowrap|''tk'''''v''' {{=}} 1 mod ''mn''.}} Multiplication by ''t'' is a group automorphism ψ of {{nowrap|ℤ/''mn''ℤ}} such that ψ(''k'''''v''') = 1, so the first row {{nowrap|[0 : ''m''] × {0}}} is mapped as {{nowrap|ψφ((''i'', 0)) {{=}} ''i''}}; so the image of the first row under ψφ is {{nowrap|{0, ..., ''m'' - 1}.}}

Claim: ''k'''''w''' has order ''n'' in {{nowrap|ℤ/''mn''ℤ.}}

Claim: ψ(''k'''''w''') (thus ''k'''''w''' as well) has order ''n'' in {{nowrap|ℤ/''mn''ℤ.}}

Proof: The order cannot be less than ''n'', lest we have {{nowrap|φ((0, ''uk'''''w''')) {{=}} 0}} for some {{nowrap|0 < ''u'' < ''n'',}} contradicting injectivity of φ within a fundamental domain (following from Step 2). If the order is ''N'' > ''n'', we have two cases. If ''n'' = 2, then ~~clearly~~ the image of the (0, 1) translation of [0 : ''m''] must be [''m'' : 2''m''], forcing ''k'''''w''' = ''m'' which is of order 2. If ''n'' > 3, ~~then~~ by ~~the lemma,~~ {{nowrap|[0 : ''m''], [0 : ''m''] + ''a'', ..., [0 : ''m''] + (''n'' - 1)''a''}}, where ''a'' = ψ(''k'''''w'''), are ~~not~~ disjoint, ~~contradicting injectivity~~.

Proof: The order cannot be less than ''n'', lest we have {{nowrap|φ((0, ''uk'''''w''')) {{=}} 0}} for some {{nowrap|0 < ''u'' < ''n'',}} contradicting injectivity of φ within a fundamental domain (following from Step 2). If the order is ''N'' > ''n'', we have two cases.

* If ''n'' = 2, then by disjointness the image of the (0, 1) translation of [0 : ''m''] must be [''m'' : 2''m''], forcing ψ(''k'''''w''') = ''m'' which is of order 2.

* If ''n'' ≥ 3, assume by way of contradiction that {{nowrap|[0 : ''m''], [0 : ''m''] + ''a'', ..., [0 : ''m''] + (''n'' - 1)''a''}}, where ''a'' = ψ(''k'''''w'''), are disjoint. Then {{nowrap|[0 : ''m''], [0 : ''m''] + ''a'', ..., [0 : ''m''] + (''n'' - 2)''a''}} are disjoint. We ask: where do we place {{nowrap|[0 : ''m''] + (''n'' - 1)''a''?}}

** The number of remaining slots on all of {{nowrap|ℤ/''mn''ℤ}} is ''m''.

** There is no contiguous region of ''m'' unoccupied slots. If there were, then the (''n'' - 1) windows that have already been placed must have no gap between them. This implies that ''a'' generates {{angbr|''m''}} and has order ''n''. Hence there is no way to place {{nowrap|[0 : ''m''] + (''n'' - 1)''a''}} without it overlapping with one of the other windows.

The claim establishes that

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Without loss of generality assume that {{nowrap|'''u''''''y''' {{=}} (''b'', ''c''), ''c'' > 0,}} and {{nowrap|'''u''''''z''' {{=}} (''b'', ''c'' - ''n'').}} As the '''v'''-coordinates of both vectors are equal, we only need to look at the '''w'''-coordinate. Since the '''w'''-coordinate of a point must stay within {{nowrap|[0 : ''n''],}} at any point it must follow the rule: "If the current '''w'''-coordinate + c ≥ ''n'', then move by ''c'' - ''n'' units (using the letter '''z'''). Otherwise, move by ''c'' units (using the letter '''y''')."

This pattern of movements is in fact the same as the one produced by taking the circular word {{nowrap|"1 1 1 ... 1 (1 - ''n'')"}} ((''n'' - 1)-many 1's) and stacking (sums of) ''c''-step subwords. As ~~there is~~ only one bad position per period, the filling word can easily be seen to be MOS by stacking ''kc''-step subwords of ~~the latter word~~ 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:Math]]

[[Category:Pages with proofs]]

[[Category:Combinatorics on words]]

[[Category:Pages with open problems]]

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+{{expert}}
 This article proves the following theorem:
-''[[Ternary]] parallelogram scale words are [[MOS substitution]] scale words, where the period count of the template MOS is the number of rows of the parallelogram parallel to the unique step size parallel to a side of the parallelogram.''
+''[[Ternary]] parallelogram scale [[word]]s are [[MOS substitution]] scale words, where the period count of the template MOS is the number of rows of the parallelogram parallel to the unique step size parallel to a side of the parallelogram.''
 == Definitions ==
 === 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.
+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 ===
@@ Line 17: / Line 20: @@
 === MOS substitution scale ===
-See [[MOS substitution]].
+A ternary scale word ''w''('''x'''<sub>1</sub>, '''x'''<sub>2</sub>, '''x'''<sub>3</sub>) is a ''MOS substitution'' scale word if there exists a permutation <math>\pi \in S_3</math> such that the following holds:
+* identifying '''x'''<sub>π(1)</sub> and '''x'''<sub>π(2)</sub> results in a [[MOS]] scale word (called the ''template MOS'') and
+* deleting all instances of '''x'''<sub>π(3)</sub> (called the ''slot letter'') results in a MOS scale word (called the ''filling MOS'').
 == Proof ==
 === 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.}} 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'''.
+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
@@ Line 53: / Line 59: @@
 Under the above assumption, since {{nowrap|φ(''t'''''v''') {{=}} 1}} we have that {{nowrap|φ('''v''') {{=}} ''k''<sub>'''v'''</sub>}} is a cyclic generator of the group {{nowrap|ℤ/''mn''ℤ,}} such that {{nowrap|''tk''<sub>'''v'''</sub> {{=}} 1 mod ''mn''.}} Multiplication by ''t'' is a group automorphism ψ of {{nowrap|ℤ/''mn''ℤ}} such that ψ(''k''<sub>'''v'''</sub>) = 1, so the first row {{nowrap|[0 : ''m''] × {0}}} is mapped as {{nowrap|ψφ((''i'', 0)) {{=}} ''i''}}; so the image of the first row under ψφ is {{nowrap|{0, ..., ''m'' - 1}.}}
-Claim: ''k''<sub>'''w'''</sub> has order ''n'' in {{nowrap|ℤ/''mn''ℤ.}}
+Claim: ψ(''k''<sub>'''w'''</sub>) (thus ''k''<sub>'''w'''</sub> as well) has order ''n'' in {{nowrap|ℤ/''mn''ℤ.}}
-Proof: The order cannot be less than ''n'', lest we have {{nowrap|φ((0, ''uk''<sub>'''w'''</sub>)) {{=}} 0}} for some {{nowrap|0 < ''u'' < ''n'',}} contradicting injectivity of φ within a fundamental domain (following from Step 2). If the order is ''N'' > ''n'', we have two cases. If ''n'' = 2, then clearly the image of the (0, 1) translation of [0 : ''m''] must be [''m'' : 2''m''], forcing ''k''<sub>'''w'''</sub> = ''m'' which is of order 2. If ''n'' > 3, then by the lemma, {{nowrap|[0 : ''m''], [0 : ''m''] + ''a'', ..., [0 : ''m''] + (''n'' - 1)''a''}}, where ''a'' = ψ(''k''<sub>'''w'''</sub>), are not disjoint, contradicting injectivity.
+Proof: The order cannot be less than ''n'', lest we have {{nowrap|φ((0, ''uk''<sub>'''w'''</sub>)) {{=}} 0}} for some {{nowrap|0 < ''u'' < ''n'',}} contradicting injectivity of φ within a fundamental domain (following from Step 2). If the order is ''N'' > ''n'', we have two cases.
+* If ''n'' = 2, then by disjointness the image of the (0, 1) translation of [0 : ''m''] must be [''m'' : 2''m''], forcing ψ(''k''<sub>'''w'''</sub>) = ''m'' which is of order 2.
+* If ''n'' &ge; 3, assume by way of contradiction that {{nowrap|[0 : ''m''], [0 : ''m''] + ''a'', ..., [0 : ''m''] + (''n'' - 1)''a''}}, where ''a'' = ψ(''k''<sub>'''w'''</sub>), are disjoint. Then {{nowrap|[0 : ''m''], [0 : ''m''] + ''a'', ..., [0 : ''m''] + (''n'' - 2)''a''}} are disjoint. We ask: where do we place {{nowrap|[0 : ''m''] + (''n'' - 1)''a''?}}
+** The number of remaining slots on all of {{nowrap|ℤ/''mn''ℤ}} is ''m''.
+** There is no contiguous region of ''m'' unoccupied slots. If there were, then the (''n'' - 1) windows that have already been placed must have no gap between them. This implies that ''a'' generates {{angbr|''m''}} and has order ''n''. Hence there is no way to place {{nowrap|[0 : ''m''] + (''n'' - 1)''a''}} without it overlapping with one of the other windows.
 The claim establishes that
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 Without loss of generality assume that {{nowrap|'''u'''<sub>'''y'''</sub> {{=}} (''b'', ''c''), ''c'' > 0,}} and  {{nowrap|'''u'''<sub>'''z'''</sub> {{=}} (''b'', ''c'' - ''n'').}} As the '''v'''-coordinates of both vectors are equal, we only need to look at the '''w'''-coordinate. Since the '''w'''-coordinate of a point must stay within {{nowrap|[0 : ''n''],}} at any point it must follow the rule: "If the current '''w'''-coordinate + c &ge; ''n'', then move by ''c'' - ''n'' units (using the letter '''z'''). Otherwise, move by ''c'' units (using the letter '''y''')."
-This pattern of movements is in fact the same as the one produced by taking the circular word {{nowrap|"1 1 1 ... 1 (1 - ''n'')"}} ((''n'' - 1)-many 1's) and stacking (sums of) ''c''-step subwords. As there is only one bad position per period, the filling word can easily be seen to be MOS by stacking ''kc''-step subwords of the latter word for {{nowrap|2 &le; ''k'' &le; 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 &le; ''k'' &le; 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:Math]]
 [[Category:Pages with proofs]]
 [[Category:Combinatorics on words]]
+[[Category:Pages with open problems]]