Generator-offset property: Difference between revisions

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# from g1 (...odd # of gens...) g1 g3 g1 (...odd # of gens...) g1, we get a4 = (''r'' + 1)/2 g1 + (''r'' − 3)/2 g2 + g3 ≡ (''r'' − ''n''/2 − 1/2)g1 + (''r'' − ''n''/2 − 3/2)g2 mod e.

~~Choose a tuning where g1 and g2~~ are ~~both very close to but not exactly 1/2*g0, resulting in a scale very close to the mos generated~~ by ~~1/2*g0 (i.e. g1 and g2 differ from 1/2*g0 by ε, a quantity much smaller than the chroma of the '~~'n''~~/2-note mos generated by g0, which is |g3 − g2|). Thus we have 4 distinct sizes for~~ ''k''-~~steps:~~

These are all distinct by '''Z'''-linear independence. By applying this argument to 1-steps, we see that there must be 4 step sizes in some tuning, a contradiction. Thus g1 and g2 must themselves be step sizes. Thus we see that an even-length, unconditionally SV3, GO scale must be of the form (xy)''r''xz. (Note that (xy)''r''xz is not SV3, since it has only two kinds of 2-steps, xy and xz.) This proves (1).

~~# a1, a2 and a3 are clearly distinct.~~

~~# a4 − a3 = g1 − g2 ≠ 0, since the scale is a non-degenerate GO scale.~~

# a4 − a1 = g3 − g2 = (g3 + g1) − (g2 + g1) ≠ 0. This is exactly the chroma of the mos generated by g0.

# a4 − a2 = g1 − 2g2 + g3 = (g3 − g2) + (g1 − g2) = (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 g1 and g2 must themselves be step sizes. Thus we see that an even-length, unconditionally SV3, GO scale must be of the form (xy)''r''xz. (Note that (xy)''r''xz is not SV3, since it has only two kinds of 2-steps, xy and xz.) This proves (1).

In case 2, let (2, 1) − (1, 1) = g1, (1, 2) − (2, 1) = g2 be the two alternants. Let g3 be the leftover generator after stacking alternating g1 and g2. Then the generator circle looks like g1 g2 g1 g2 ... g1 g2 g3. Then the combinations of alternants corresponding to a step come in exactly 3 sizes:

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 # from g<sub>1</sub> (...odd # of gens...) g<sub>1</sub> g<sub>3</sub> g<sub>1</sub> (...odd # of gens...) g<sub>1</sub>, we get a<sub>4</sub> = (''r'' + 1)/2 g<sub>1</sub> + (''r'' &minus; 3)/2 g<sub>2</sub> + g<sub>3</sub> ≡ (''r'' &minus; ''n''/2 &minus; 1/2)g<sub>1</sub> + (''r'' &minus; ''n''/2 &minus; 3/2)g<sub>2</sub> mod e.
-Choose a tuning where g<sub>1</sub> and g<sub>2</sub> are both very close to but not exactly 1/2*g<sub>0</sub>, resulting in a scale very close to the mos generated by 1/2*g<sub>0</sub> (i.e. g<sub>1</sub> and g<sub>2</sub> differ from 1/2*g<sub>0</sub> by ε, a quantity much smaller than the chroma of the ''n''/2-note mos generated by g<sub>0</sub>, which is |g<sub>3</sub> &minus; g<sub>2</sub>|). Thus we have 4 distinct sizes for ''k''-steps:
+These are all distinct by '''Z'''-linear independence. By applying this argument to 1-steps, we see that there must be 4 step sizes in some tuning, a contradiction. Thus g<sub>1</sub> and g<sub>2</sub> must themselves be step sizes. Thus we see that an even-length, unconditionally SV3, GO scale must be of the form (xy)<sup>''r''</sup>xz. (Note that (xy)<sup>''r''</sup>xz is not SV3, since it has only two kinds of 2-steps, xy and xz.) This proves (1).
-# a<sub>1</sub>, a<sub>2</sub> and a<sub>3</sub> are clearly distinct.
-# a<sub>4</sub> &minus; a<sub>3</sub> = g<sub>1</sub> &minus; g<sub>2</sub> ≠ 0, since the scale is a non-degenerate GO scale.
-# a<sub>4</sub> &minus; a<sub>1</sub> = g<sub>3</sub> &minus; g<sub>2</sub> = (g<sub>3</sub> + g<sub>1</sub>) &minus; (g<sub>2</sub> + g<sub>1</sub>) ≠ 0. This is exactly the chroma of the mos generated by g<sub>0</sub>.
-# a<sub>4</sub> &minus; a<sub>2</sub> = g<sub>1</sub> &minus; 2g<sub>2</sub> + g<sub>3</sub> = (g<sub>3</sub> &minus; g<sub>2</sub>) + (g<sub>1</sub> &minus; g<sub>2</sub>) = (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<sub>1</sub> and g<sub>2</sub> must themselves be step sizes. Thus we see that an even-length, unconditionally SV3, GO scale must be of the form (xy)<sup>''r''</sup>xz. (Note that (xy)<sup>''r''</sup>xz is not SV3, since it has only two kinds of 2-steps, xy and xz.) This proves (1).
 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 looks like g<sub>1</sub> g<sub>2</sub> g<sub>1</sub> g<sub>2</sub> ... g<sub>1</sub> g<sub>2</sub> g<sub>3</sub>. Then the combinations of alternants corresponding to a step come in exactly 3 sizes: