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.

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).

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, abstractly 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. Assuming that a step is an odd number of generators, the combinations of alternants corresponding to a step come in exactly 3 sizes:

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For (3), we now only need to see that if ''S'' has an odd number of notes and is SGA, ''S'' is abstractly SV3. But the argument in case 2 above works when you substitute any interval class in ''S'' instead of a 1-step (abstract SV3 wasn't used), hence any interval class comes in (abstractly) exactly 3 sizes.

For (4), assume ''S'' is ''a''X ''b''Y ''b''Z, 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 ''a''X 2''b''W is ''i''X + ''j''W with ''j'' ≥ 2. (If ''j'' = 1, we can invert the generator to make ''j'' ≥ 2, since ''b'' > 1.)

For (4), assume ''S'' is ''a''X ''b''Y ''b''Z, 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, i.e. YX''t''Y (for some ''t'' ≥ 0) occurs in ''S''. Assume that the perfect generator of ''a''X 2''b''W is ''i''X + ''j''W 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:

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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.
-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).
+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, abstractly 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>. Assuming that a step is an odd number of generators, the combinations of alternants corresponding to a step come in exactly 3 sizes:
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 For (3), we now only need to see that if ''S'' has an odd number of notes and is SGA, ''S'' is abstractly SV3. But the argument in case 2 above works when you substitute any interval class in ''S'' instead of a 1-step (abstract SV3 wasn't used), hence any interval class comes in (abstractly) exactly 3 sizes.
-For (4), assume ''S'' is ''a''X ''b''Y ''b''Z, 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 ''a''X 2''b''W is ''i''X + ''j''W with ''j'' ≥ 2. (If ''j'' = 1, we can invert the generator to make ''j'' ≥ 2, since ''b'' > 1.)
+For (4), assume ''S'' is ''a''X ''b''Y ''b''Z, 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, i.e. YX<sup>''t''</sup>Y (for some ''t'' &ge; 0) occurs in ''S''. Assume that the perfect generator of ''a''X 2''b''W is ''i''X + ''j''W 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: