Generator-offset property: Difference between revisions

Assuming SGA, we have two chains of generator ''g''₀ (going right). The two cases are:

In case 1, let ''g''₁ = (2, 1) − (1, 1) and ''g''₂ = (1, 2) − (2, 1). We have the chain ''g''₁ ''g''₂ ''g''₁ ''g''₂ ... ''g''₁ ''g''₃.  

# from ''g''₁ ... ''g''₁, we get ''a''₁ = (''r'' − 1)/2*''g''₀ + ''g''₁ = (''r'' + 1)/2 ''g''₁ + (''r'' − 1)/2 ''g''₂

# from ''g''₂ ... ''g''₂, we get ''a''₂ = (''r'' − 1)/2*''g''₀ + ''g''₂ = (''r'' − 1)/2 ''g''₁ + (''r'' + 1)/2 ''g''₂

# from ''g''₂ (...even # of gens...) ''g''₁ ''g''₃ ''g''₁ (...even # of gens...) ''g''₂, we get ''a''₃ = (''r'' − 1)/2 ''g''₁ + (''r'' − 1)/2 ''g''₂ + ''g''₃  

# from ''g''₁ (...odd # of gens...) ''g''₁ ''g''₃ ''g''₁ (...odd # of gens...) ''g''₁, we get ''a''₄ = (''r'' + 1)/2 ''g''₁ + (''r'' − 3)/2 ''g''₂ + ''g''₃.  

Choose a tuning where ''g''₁ and ''g''₂ are both very close to but not exactly 1/2*''g''₀, resulting in a scale very close to the mos generated by 1/2 ''g''₀. (i.e. ''g''₁ and ''g''₂ differ from 1/2*''g''₀ by ε, a quantity much smaller than the chroma of the ''n''/2-note mos generated by ''g''₀, which is |''g''₃ − ''g''₂|). Thus we have 4 distinct sizes for ''k''-steps:

# ''a''₁, ''a''₂ and ''a''₃ are clearly distinct.

# ''a''₄ − ''a''₃ = ''g''₁ − ''g''₂ ≠ 0, since the scale is a non-degenerate GO scale.  

# ''a''₄ − ''a''₁ = ''g''₃ − ''g''₂ = (''g''₃ + ''g''₁) − (''g''₂ + ''g''₁) ≠ 0. This is exactly the chroma of the mos generated by ''g''₀.

# ''a''₄ − ''a''₂ = ''g''₁ − 2 ''g''₂ + ''g''₃ = (''g''₃ − ''g''₂) + (''g''₁ − ''g''₂) = (chroma ± ε) ≠ 0 by choice of tuning.

By applying this argument to 1-steps, we see that there must be 4 step sizes in some tuning, a contradiction. Thus ''g''₁ and ''g''₂ must themselves be step sizes. Thus we see that an even-length, unconditionally SV3, GO scale must be of the form ''xy...xyxz''. But this pattern is not abstractly SV3 if ''n'' ≥ 6, since 3-steps come in 4 sizes: ''xyx'', ''yxy'', ''yxz'' and ''xzx''. Thus ''n'' = 4 and the scale is ''xyxz''. This proves (3).

In case 2, let (2, 1) − (1, 1) = ''g''₁, (1, 2) − (2, 1) = ''g''₂ be the two alternants. Let ''g''₃ be the leftover generator after stacking alternating ''g''₁ and ''g''₂. Then the generator circle looks like ''g''₁ ''g''₂ ''g''₁ ''g''₂ ... ''g''₁ ''g''₂ ''g''₃. Then the combinations of alternants corresponding to a step come in exactly 3 sizes:

# ''kg''₁ + (''k'' − 1)''g''₂

# (''k'' − 1)''g''₁ + ''kg''₂

# (''k'' − 1)''g''₁ + (''k'' − 1) ''g''₂ + ''g''₃,

For (4), assume S is aX bY bZ, a odd. If b = 1, there’s nothing to prove. So assume b > 1. Suppose for the sake of contradiction that Y’s and Z’s don't alternate perfectly. Assume that the perfect generator of aX 2bW is iX + jW with j ≥ 2. (If j = 1, we can invert the generator to make j ≥ 2, since b > 1.)

In ''S'', (i+j)-steps (representing the generator) are always one of the following:

Since a + 2b ≥ 5, there are at least 4 perfect generators, so there must be at least one of each of (a), (b), and (c), giving a contradiction to SV3. [Whenever the root of the (i+j)-step are moved within S, the numbers of Y's and Z's change one at a time and reach a maximum at some choice of the root and a minimum with another choice, guaranteeing that intermediate values are reached. We can use this "intermediate value theorem" argument because (d) occurs at only one note.]

Any generator of aX 2bW must have an odd number of W steps (Otherwise, intervals with an odd number of W steps can't be generated.). This with (4) immediately gives (5).

For (7), consider the mos aX 2bW as chunks of X separated by W (tempering Y and Z together into W). Eliminating every other W turns it into a mos, because the sum of sizes of consecutive chunks of X (1st chunk with 2nd chunk, 3rd with 4th, ...) must form a mos. This is because the chunk sizes of X form a mos, and taking every kth note of an n-note mos where k divides n yields a mos. Since the result of setting X = 0 is the mos bY bZ, ''S'' is elimination-mos.