Ternary scale theorems: Difference between revisions

* We leave the distinction between linear words (words in the ordinary sense) and circular words up to context. We usually also elide the distinction between subwords and the dyad sizes that subtend them.

# If ''n'' is odd, ''s'' is pairwise-MOS. That is, the following operations each result in a [[MOS]]: setting {{nowrap|'''L''' {{=}} '''M'''}}, setting {{nowrap|'''L''' {{=}} '''s'''}}, and setting {{nowrap|'''M''' {{=}} '''s'''}}.

# If ''n'' is odd, {{nowrap|''s'' {{=}} ''a'''''X''' ''b'''''Y''' ''b'''''Z'''}} is obtained from some mode of the (primitive) MOS ''a'''''X''' 2''b'''''W''' by replacing all the '''W'''s successively with alternating '''Y'''s and '''Z'''s (or alternating '''Z'''s and '''Y'''s for the other chirality, fixing the mode of ''a'''''X''' 2''b'''''W'''). The two alternants differ by replacing one '''Y''' with a '''Z'''.

 

In particular, odd generator-offset scales always satisfy these properties (see Proposition 2 below).

 

[Note: This is not true with AGS replaced with generator-offset; [[blackdye]] is a counterexample that is MV4.]

Since ''s'' is generator-offset it is well-formed with respect to the aggregate generator {{nowrap|'''g''' {{=}} ('''g'''₂ + '''g'''₁)}}. Since '''g'''₁ and '''g'''₂ subtend the same number of steps by the AGS assumption, each is an odd-step. All multiples of the aggregate generator '''g''' must be even-steps, and those dyads that are "offset" by '''g'''₁ must be odd-steps. Letting ''M'' be the subset consisting of all even-numbered notes (which are generated by '''g''') and considering ''M'' as a scale by dividing degree indices in ''M'' by two, ''M'' is well-formed with respect to '''g''', thus ''M'' (and its offset) must be a MOS subset. Hence {{nowrap|('''g'''₃ + '''g'''₁)}}, the imperfect generator of the MOS generated by '''g''', subtends the same number of steps as '''g'''. Thus '''g'''₂ and '''g'''₃ subtend the same number of steps, a fact we need in order to be able to substitute one instance of '''g'''₂ with '''g'''₃ in the next part.

Let ''r'' be odd and ''r'' ≥ 3. Consider the following abstract sizes for the dyad class of ''k''-steps reached by stacking ''r'' generators:

In case 2, let {{nowrap|''n'' ≥ 3}} and let {{nowrap|(2, 1) − (1, 1) {{=}} '''g'''₁|(1, 2) − (2, 1) {{=}} '''g'''₂}} be the two alternants. Let '''g'''₃ be the closing generator after stacking alternating '''g'''₁ and '''g'''₂. Then the generator circle is {{nowrap|('''g'''₁ '''g'''₂)^{{{floor|''n''/2}}}}} '''g'''₃. If a step is formed by stacking ''k'' generators, we may assume that ''k'' is odd, and the combinations of alternants corresponding to a step come in exactly 3 sizes:

We only need to see that if len(''s'') is odd and ''s'' is AGS, ''s'' is abstractly SV3. But the argument in case 2 above works when you substitute any odd-step dyad classes in ''s'' instead of a 1-step (abstract SV3 wasn't used). To get even-step dyad classes, we can take octave complements. Hence any dyad class in such a scale comes in (abstractly) exactly 3 sizes.

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 MOSes. These correspond to two of the temperings {{nowrap|'''X''' {{=}} '''Y'''|'''Y''' {{=}} '''Z'''}}, and {{nowrap|'''X''' {{=}} '''Z'''}}. The third tempering follows by symmetry (by taking the other chirality).

 

Consider the two generators in the GS of ''s'', which are detemperings of the generator {{nowrap|''i'''''X''' + ''j'''''W'''}} of ''T''('''X''', '''W'''), where {{nowrap|gcd(''j'', 2''k'') {{=}} 1}}. Assume, possibly after inverting the generator, that the imperfect generator of ''T'' has {{nowrap|''j'' + 1}} '''W'''s and the perfect generator has ''j'' '''W'''s.

'''Claim 1''': Deleting '''X'''s from the generator subwords of ''s'' gives every ''j''-step subword in the scale ''E''_X(''s'')('''Y''', '''Z'''), the scale word obtained by deleting all '''X''''s from ''s''.  

Proof: Consider the subword for the closing generator of ''s'' on some index ''p'', which is {{nowrap|''I'' {{=}} ''s''[''p'' : ''p'' + ''i'' + ''j'']}}, and suppose the result of deleting all '''X''''s from ''I'' has a ''j''-step subword ''w''. Shifting ''I'' one step to the left and one step to the right, {{nowrap|''s''[''p'' − 1: ''p'' − 1 + ''i'' + ''j'']}} and {{nowrap|''s''[''p'' + 1 : ''p'' + 1 + ''i'' + ''j'']}} are both detemperings of perfect generators of ''T'', and have one fewer non-'''X''' step than ''I'' by our assumption. Thus the word ''I'' must both begin and end in a non-'''X''' letter. Removing all the '''X''''s from ''I'' results in a word that is {{nowrap|''j'' + 1}} letters long and is the ''j''-step word ''w'' with just one extra letter appended. Thus one of the two perfect generators above, namely the one that removes the extra letter, must contain this ''j''-step. The rest of the ''j''-step subwords of ''s'' can all be obtained by deleting '''X'''s from detempered perfect generators; take {{nowrap|''q'' ≠ ''p''}} to be the index of the first letter of one such ''j''-step subword (as contained in ''s'') and use {{nowrap|''s''[''q'' : ''q'' + ''i'' + ''j'']}}.

Proof: Write '''u''' and '''v''' for the two sizes of ''j''-steps. Since {{nowrap|gcd(''j'', 2''b'') {{=}} 1}}, there exists ''m'' such that stacking ''m''-many ''j''-steps yields scale steps of ''U'', and ''m'' is odd because {{nowrap|gcd(''m'', 2''b'') {{=}} 1}}. Hence the scale steps of ''U'' are {{nowrap|('''uv''')^{{{sfrac|''m'' − 1|2}}}'''u''' (mod '''e''')}} and {{nowrap|('''vu''')^{{{sfrac|''m'' − 1|2}}}'''v''' (mod '''e''')}}, and the step sizes alternate because '''u''' and '''v''' do.

== Theorem 2 (Odd generator-offset scales are AGS) ==

 

 

 

 

 

 

 

== Theorem 4 (Classification of pairwise well-formed scales) ==

We can write sizes of dyads in ''s'' as vectors {{nowrap|(''p'', ''q'', ''r'')}} using the basis {{nowrap|('''a''', '''b''', '''c''')}}.  

These give exactly three distinct sizes for every dyad class. Hence ''s'' is SV3.

 

In this case ''s'' has two chains of '''Q''', one with {{floor|''n''/2}} notes and one offset by {{nowrap|'''Q'''^{(''f'' − 1)}R}} with {{ceil|''n''/2}} notes. Every instance of Q must be a ''k''-step, since by ℤ-linear independence {{nowrap|'''Q''' {{=}} α'''a''' + β'''b''' + γ'''c'''}} is the only way to write '''Q''' in the basis {{nowrap|('''a''', '''b''', '''c''')}}; so ''s'' is well-formed with respect to '''Q'''. Thus ''s'' also satisfies the generator-offset property with generator '''Q'''.

Since Λ₃ has more β than β′, Λ₂ is {{floor|''n''/2}}β {{ceil|''n''/2}}β′, and Λ₃ is {{ceil|''n''/2}}γ {{floor|''n''/2}}γ′. There is a unique mode of {{ceil|''n''/2}}γ {{floor|''n''/2}}γ′ that both begins and ends with γ, namely γγ′γγ′…γγ′γ. Thus Λ₂ is ββ′ββ′…ββ′β′. It is now easy to see that if the number of ''k''-steps stacked is odd, then there are two sizes that do not contain '''T''' and one size that contains '''T'''; if the number of ''k''-steps stacked is even, then there is one size that does not contain '''T''' and two sizes that contain T. Hence ''s'' is SV3.

In this case we have {{nowrap|Σ {{=}} '''QRQR'''…'''QRT'''}}, and ''s'' is well-formed with respect to the generator {{nowrap|'''Q''' + '''R'''}}, thus ''s'' satisfies the generator-offset property. By Proposition 1, ''s'' is SV3.

Λ₂ has a chunk of β (after the first β′) of size ''x'' where {{nowrap|''x'' {{=}} {{floor|''n''/μ}}}} {{nowrap|≥ {{floor|''n''/{{floor|''n''/2}}}}}} = 2 or {{nowrap|''x'' {{=}} {{ceil|''n''/μ}}}} {{nowrap|{{=}} {{floor|''n''/μ}} + 1}}. Hence Λ₃ has a chunk of γ of size ''x''. Λ₃ also has a chunk that contains {{nowrap|Λ₃[''n'' − : 1]}} as a subword. This chunk must be of size ''y'', where  

== Theorem 5 (PMOS scales are balanced) ==

== Theorem 6 (Generator-offset structure of even-regular scales) ==

# the generator has the same dyad class as some generator of the MOS 2''a'''''W''' 2''c'''''Z''';

# the two generator chains are offset by a len(''s'')/2-step dyad;

== Theorem 7 (Classification of MV3 scales) ==

=== Theorem 7.1 (Classification of ternary balanced scales) ===

# A primitive [[balanced]] ternary scale ''s'' is pairwise-MOS, and satisfies one of the following:

For 7.1.1: We showed previously that the Fraenkel, odd-regular, and even-regular circular words are balanced. Thus it remains to show that (a) ternary balanced words are pairwise-MOS (b) if ''a'' > ''b'' > ''c'', then ''s'' is equivalent to the Fraenkel word (c) assuming ''a'' != ''b'' = ''c'' any ''s'' that is not odd-regular or even-regular is not balanced.

For 7.1.2: Suppose ''s'' is balanced and has at least three sizes for ''k''-steps, {{nowrap|''a''_''i'''''X''' + ''b''_''i'''''Y''' + ''c''_''i'''''Z''' {{=}} (''a''_''i'', ''b''_''i'', ''c''_''i'')}} for {{nowrap|''i'' ∈ {{(}}1, 2, 3{{)}}}}. We may assume {{nowrap|(''a''₂, ''b''₂, ''c''₂) {{=}} (''a''₁, ''b''₁ + 1, ''c''₁ − 1)}}. Then either {{nowrap|(''a''₃, ''b''₃, ''c''₃) {{=}} (''a''₁ + 1, ''b''₁, ''c''₁ − 1)}} or {{nowrap|(''a''₃, ''b''₃, ''c''₃) {{=}} (''a''₁ − 1, ''b''₁ + 1, ''c''₁)}}. In both cases, by balancedness applied to subwords of length ''k'', the three vectors represent the only possible dyad sizes.

For 7.1.3: The ternary Fraenkel word may be verified as SV3 by inspection, and we have already shown in Theorem 1 that odd-regular balanced scales are SV3. To show that even-regular balanced scales are ''not'' SV3, observe that {{nowrap|(''a'' + ''c'')}}-steps come in only 2 sizes in such a scale ''s'': {{nowrap|{{floor|''a''/2}}'''X''' + {{ceil|''a''/2}}'''Y''' + ''c'''''Z'''}} and {{nowrap|{{ceil|''a''/2}}'''X''' + {{floor|''a''/2}}'''Y''' + ''c'''''Z'''}}, since the underlying MOS 2''a'''''X'''2''c'''''Y''' only has the {{nowrap|(''a'' + ''c'')}}-step {{nowrap|''a'''''X''' + ''c'''''Z'''}}. The construction replaces the '''X'''s in these subwords with alternating '''X'''s and '''Y'''s; either of '''X''' or '''Y''' may occur first, corresponding to the two possible sizes, since ''a'' is odd and thus the {{nowrap|(''a'' + ''c'')}}-step subword {{nowrap|''s''[''k'' − 1 : ''k'' + ''a'' + ''c'' − 1]}} becomes the subword {{nowrap|''s''[''k'' + ''a'' + ''c'' − 1 : ''k'' + 2''a'' + 2''c'' − 1]}} via interchanging '''X''' and '''Y'''.

Claim 7.1.4 can be verified by noting that such scales are PWF and using Theorem 4. {{Qed}}

=== Theorem 7.2 (Classification of MV3 scales) ===

== Theorem 8 (Even-regular scales as (contra)interleavings) ==

* In the doubly even case, start from the mode of ''s'' whose template MOS is the brightest mode. Let ''s''₂ be offset at a generator of the even-regular scale, which by Theorem 6 we choose to have the same dyad class as a bright generator of the MOS ''a'''''x''' 2''k'''''X'''. This is what induces the equality of ''s''₁ and ''s''₂ (in particular, the two scales have the same period, thus they are both primitive): Let ''s''_''t'' be the period of the brightest mode of the template MOS, and let ''g'' be its bright generator class. Then the slice {{nowrap|''s''_''t''[-''g'' +1 : 1]}} is the imperfect generator of the MOS. Now when we "darken" the mode by one generator, which is the difference between the template MOSes of ''s''₁ and ''s''₂, we turn that slice into the bright generator, hence swapping ''s''_''t''[-''g''] and ''s''_''t''[-''g'' + 1]. Note that ''g'' must be odd since it generates a 2-period MOS. So (under 0-indexing) the first letter's index is even and the second letter's index is odd, which is what we want since the letters are within a stacked 2-step. While the generator might have to be higher by an (''n''/2)-step, that doesn't affect the parity since ''n''/2 is even.

We prove that ''s''₁ and ''s''₂ are MOS substitution scales with a filling MOS of period 2. The number the 2-step (1) occurs must be the same in both ''s''₁ and ''s''₂. The word of stacked 2-steps of the template MOS (which is of the form {{nowrap|''w''('''x''', '''X''', '''X''')''w''('''x''', '''X''', '''X''')}}), which is itself a MOS word, consists of letters (1) '''x''' + '''X''' and (2) 2'''X''' if more '''X''''s than '''x''''s, 2'''x''' if more '''x''''s than '''X''''s. The word of stacked 2-steps from our chosen offset is also this same MOS word. Thus it remains to handle the cases (1) and (2) above. IWhenever the letter '''x''' + '''X''' is encountered, the number of the last letters that are equated to '''X''' that are consumed is 1, which is odd. Whenever the other letter is encountered, that number is even (0 or 2). Hence (since ''n'' > 4) the letter 2'''X''' resp. 2'''x''' serves as the non-slot letter, and the letters ('''x''' + '''X''') serve as the slot letters where a 2-period filling MOS word (a repetition of {{nowrap|('''x'''+'''y''')('''x'''+'''z''')}}) is substituted.