Ternary scale theorems: Difference between revisions

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==== Statement (2) ====
==== Statement (2) ====
In case 2, let (2, 1) &minus; (1, 1) = g<sub>1</sub>, (1, 2) &minus; (2, 1) = g<sub>2</sub> be the two alternants. Let g<sub>3</sub> be the leftover generator after stacking alternating g<sub>1</sub> and g<sub>2</sub>. Then the generator circle is (g<sub>1</sub> g<sub>2</sub>)<sup>floor(''n''/2)</sup> g<sub>3</sub>. If a step is formed by stacking ''k'' generators, ''k'' must be odd, and the combinations of alternants corresponding to a step come in exactly 3 sizes:
In case 2, let ''n'' ≥ 3 and let (2, 1) &minus; (1, 1) = g<sub>1</sub>, (1, 2) &minus; (2, 1) = g<sub>2</sub> be the two alternants. Let g<sub>3</sub> be the leftover generator after stacking alternating g<sub>1</sub> and g<sub>2</sub>. Then the generator circle is (g<sub>1</sub> g<sub>2</sub>)<sup>floor(''n''/2)</sup> g<sub>3</sub>. If a step is formed by stacking ''k'' generators, ''k'' must be odd, and the combinations of alternants corresponding to a step come in exactly 3 sizes:
# ceil(''k''/2)g<sub>1</sub> + floor(''k''/2)g<sub>2</sub>
# ceil(''k''/2)g<sub>1</sub> + floor(''k''/2)g<sub>2</sub>
# floor(''k''/2)g<sub>1</sub> + ceil(''k''/2)g<sub>2</sub>
# floor(''k''/2)g<sub>1</sub> + ceil(''k''/2)g<sub>2</sub>
# floor(''k''/2)g<sub>1</sub> + floor(''k''/2) g<sub>2</sub> + g<sub>3</sub>
# floor(''k''/2)g<sub>1</sub> + floor(''k''/2) g<sub>2</sub> + g<sub>3</sub>
(since the scale size is odd, we can always ensure this by taking octave complements of all the generators). By counting the length-k subwords of the (linear) word (g<sub>1</sub> g<sub>2</sub>)<sup>floor(''n''/2)</sup>, we see that the first two sizes must both occur (''n'' &minus; ''k'') times. This proves (2).
(since the scale size is odd, we can always ensure this by taking octave complements of all the generators). By counting the length-k subwords of the (linear) word (g<sub>1</sub> g<sub>2</sub>)<sup>floor(''n''/2)</sup>, we see that the first two sizes must both occur (''n'' &minus; ''k'') times. This proves (2).
(The above holds for any odd ''n'' ≥ 3.)


==== Statement (3) ====
==== Statement (3) ====
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  x x x ... x  
  x x x ... x  
  x x x ... x x
  x x x ... x x
and use the vectors (-1, 2) and (ceil(n/2), 1) as the Fokker block chromas. A rank-3 Fokker block has the property that tempering out by each of the chromas gives two mosses. These correspond to two of the temperings X = Y, Y = Z and X = Z. The third tempering follows by symmetry (by taking the other chirality).<!--
and use the vectors (-1, 2) and (ceil(n/2), 1) as the Fokker block chromas. A rank-3 Fokker block has the property that tempering out by each of the chromas gives two mosses. These correspond to two of the temperings X = Y, Y = Z and X = Z. The third tempering follows by symmetry (by taking the other chirality). {{qed}}<!--


==== Statement (6) ====
==== Statement (6) ====
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Assume that the generator is a ''k''-step and ''k'' is even. (If ''k'' is not even, invert the generator.) On some note ''p'' we have a chain of (''n'' + 1)/2 notes and on ''p′'' = ''p'' + offset we'll have (''n'' &minus; 1)/2) notes.
Assume that the generator is a ''k''-step and ''k'' is even. (If ''k'' is not even, invert the generator.) On some note ''p'' we have a chain of (''n'' + 1)/2 notes and on ''p′'' = ''p'' + offset we'll have (''n'' &minus; 1)/2) notes.


If gcd(''k'', ''n'') > 1. since ''n'' is odd, gcd(''k'', ''n'') is an odd number at least 3, and by well-formedness with respect to the generator, the generators must form more than 2 parallel chains. Thus, gcd(''k'', ''n'') = 1. By modular arithmetic, if 0 &le; ''r'' < ''n'', ''rk'' ≡ ''k''/2 mod ''n'' iff ''r'' ≡ (''n'' + 1)/2 mod ''n''. (Note that both 2 and ''k'' are coprime with ''n'', hence multiplicatively invertible mod ''n''.) This proves that the offset, which must be reached after (''n'' + 1)/2 ''k''-steps, is a ''k''/2-step, as desired. (As [''k''] is a generator of '''Z'''/''n'', stacking (''n'' &minus; 1)-many ''k''-steps must visit every note exactly once. Thus if the offset wasn't reached in (''n'' + 1)/2 steps, the two generator chains either wouldn't be disjoint or wouldn't have the assumed lengths.)
If gcd(''k'', ''n'') > 1. since ''n'' is odd, gcd(''k'', ''n'') is an odd number at least 3, and by well-formedness with respect to the generator, the generators must form more than 2 parallel chains. Thus, gcd(''k'', ''n'') = 1. By modular arithmetic, if 0 &le; ''r'' < ''n'', ''rk'' ≡ ''k''/2 mod ''n'' iff ''r'' ≡ (''n'' + 1)/2 mod ''n''. (Note that both 2 and ''k'' are coprime with ''n'', hence multiplicatively invertible mod ''n''.) This proves that the offset, which must be reached after (''n'' + 1)/2 ''k''-steps, is a ''k''/2-step, as desired. (As [''k''] is a generator of '''Z'''/''n'', stacking (''n'' &minus; 1)-many ''k''-steps must visit every note exactly once. Thus if the offset wasn't reached in (''n'' + 1)/2 steps, the two generator chains either wouldn't be disjoint or wouldn't have the assumed lengths.) {{qed}}


== Proposition 3 (Properties of even generator-offset ternary scales) ==
== Proposition 3 (Properties of even generator-offset ternary scales) ==
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# If ''M'' = ''M''(''y'', ''z'') is the primitive MOS necklace above, then ''s'' = ''M''(XY, XZ) for some assignment of variable names X, Y, and Z to the three letters of ''s''.
# If ''M'' = ''M''(''y'', ''z'') is the primitive MOS necklace above, then ''s'' = ''M''(XY, XZ) for some assignment of variable names X, Y, and Z to the three letters of ''s''.
=== Proof ===
=== Proof ===
(1) and (2) were proved in the proof of Proposition 1 (the part that we appeal to, from "all multiples of the generator g must be even-steps ..." to "These are all distinct by Z-linear independence", does not rely on ''S'' having the SGA property). (3) and (4) are easy to check using (1).
(1) and (2) were proved in the proof of Proposition 1 (the part that we appeal to, from "all multiples of the generator g must be even-steps ..." to "These are all distinct by Z-linear independence", does not rely on ''S'' having the SGA property). (3) and (4) are easy to check using (1). {{qed}}


== Theorem 4 (Classification of pairwise well-formed scales) ==
== Theorem 4 (Classification of pairwise well-formed scales) ==
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* '''Case 3.9''': (''x'', ''y'') = (floor(''n''/μ), 2*floor(''n''/μ) + 2)
* '''Case 3.9''': (''x'', ''y'') = (floor(''n''/μ), 2*floor(''n''/μ) + 2)
* '''Case 3.10''': (''x'', ''y'') = (floor(''n''/μ), 2*floor(''n''/μ) + 3)
* '''Case 3.10''': (''x'', ''y'') = (floor(''n''/μ), 2*floor(''n''/μ) + 3)
 
{{qed}}
== Theorem 5 (Classification of MV3 scales) ==
== Theorem 5 (Classification of MV3 scales) ==
# A primitive MV3 is either (1) equivalent to XYZYX, (2) equivalent to XYXZXYX, (3) constructed from ''a''X ''b''Z with ''a'' even and gcd(''a'', ''b'') = 1 by replacing every other X with Y, (4) constructed from 2''a''X 2''b''Z with ''a'' odd and gcd(''a'', ''b'') = 1 by replacing every other X with Y, or (5) a "twisted" word constructed as follows:
# A primitive MV3 is either (1) equivalent to XYZYX, (2) equivalent to XYXZXYX, (3) constructed from ''a''X ''b''Z with ''a'' even and gcd(''a'', ''b'') = 1 by replacing every other X with Y, (4) constructed from 2''a''X 2''b''Z with ''a'' odd and gcd(''a'', ''b'') = 1 by replacing every other X with Y, or (5) a "twisted" word constructed as follows:
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# Primitive MV3 scales of type (3) are odd generator-offset scales.
# Primitive MV3 scales of type (3) are odd generator-offset scales.
=== Proof ===
=== Proof ===
Proven by Bulgakova, Buzhinsky and Goncharov (2023), "[https://www.sciencedirect.com/science/article/pii/S0304397522006417 On balanced and abelian properties of circular words over a ternary alphabet]" (and Theorem 4).
Proven by Bulgakova, Buzhinsky and Goncharov (2023), "[https://www.sciencedirect.com/science/article/pii/S0304397522006417 On balanced and abelian properties of circular words over a ternary alphabet]" (and Theorem 4). {{qed}}


[[Category:Math]]
[[Category:Math]]
[[Category:Ternary scale]]
[[Category:Ternary scale]]
[[Category:Scale]]
[[Category:Scale]]