Ternary scale theorems: Difference between revisions

Consider the two generators in the GS of ''s'', which are lifts 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.

* C.3. We need only stack {{nowrap|2''b'' ≤ ''n'' − 1}} generators (to get {{nowrap|2''b''-many}} ''j''-step windows downstairs) to witness the alternation. Under the ordering induced by this stacking, the 1st ''j''-step subword of ''U'' and the {{nowrap|2''b''-th}} ''j''-step window differ due to parity. Since {{nowrap|gcd(''j'', 2''b'') {{=}} 1}}, this visits every note of ''U''.

These give exactly three distinct sizes for every interval 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.

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

For 5.1.1: 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.

* 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 interval 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.