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.

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:

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.

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.

# 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;

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.

* 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.