Generator-offset property: Difference between revisions

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In case 1, let g1 = (2, 1) − (1, 1), g2 = (1, 2) − (2, 1), and g3 = (1, 1) − (''n''/2, 2) = (−''n''/2*g1 − g1 − ''n''/2*g2) mod e. We assume that g1, g2 and e are '''Z'''-linearly independent. We have the chain g1 g2 g1 g2 ... g1 g3 which visits every note in ''S''.

Since ''S'' is GO it is well-formed with respect to g = (g2 + g1). Since g1 and g2 subtend the same number of steps, all multiples of the generator g must be even-steps, and those intervals that are "offset" by g1 must be odd-steps. Letting ''M'' be the subset 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 (g3 + g1), the imperfect generator of the mos generated by g, subtends the same number of steps as g. Thus g2 and g3 subtend the same number of steps, a fact we need in order to be able to substitute one instance or g2 with g3 in the next part.

Since ''S'' is GO it is well-formed with respect to g = (g2 + g1). Since g1 and g2 subtend the same number of steps, all multiples of the generator g must be even-steps, and those intervals that are "offset" by g1 must be odd-steps. Letting ''M'' be the subset 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 (g3 + g1), the imperfect generator of the mos generated by g, subtends the same number of steps as g. Thus g2 and g3 subtend the same number of steps, a fact we need in order to be able to substitute one instance of g2 with g3 in the next part.

Let ''r'' be odd and ''r'' ≥ 3. Consider the following abstract sizes for the interval class (''k''-steps) reached by stacking ''r'' generators:

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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).

===== Statement (2) =====

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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 In case 1, let g<sub>1</sub> = (2, 1) &minus; (1, 1), g<sub>2</sub> = (1, 2) &minus; (2, 1), and g<sub>3</sub> = (1, 1) &minus; (''n''/2, 2) = (&minus;''n''/2*g<sub>1</sub> &minus; g<sub>1</sub> &minus; ''n''/2*g<sub>2</sub>) mod e. We assume that g<sub>1</sub>, g<sub>2</sub> and e are '''Z'''-linearly independent. We have the chain g<sub>1</sub> g<sub>2</sub> g<sub>1</sub> g<sub>2</sub> ... g<sub>1</sub> g<sub>3</sub> which visits every note in ''S''.
-Since ''S'' is GO it is well-formed with respect to g = (g<sub>2</sub> + g<sub>1</sub>). Since g<sub>1</sub> and g<sub>2</sub> subtend the same number of steps, all multiples of the generator g must be even-steps, and those intervals that are "offset" by g<sub>1</sub> must be odd-steps. Letting ''M'' be the subset 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 (g<sub>3</sub> + g<sub>1</sub>), the imperfect generator of the mos generated by g, subtends the same number of steps as g. Thus g<sub>2</sub> and g<sub>3</sub> subtend the same number of steps, a fact we need in order to be able to substitute one instance or g<sub>2</sub> with g<sub>3</sub> in the next part.
+Since ''S'' is GO it is well-formed with respect to g = (g<sub>2</sub> + g<sub>1</sub>). Since g<sub>1</sub> and g<sub>2</sub> subtend the same number of steps, all multiples of the generator g must be even-steps, and those intervals that are "offset" by g<sub>1</sub> must be odd-steps. Letting ''M'' be the subset 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 (g<sub>3</sub> + g<sub>1</sub>), the imperfect generator of the mos generated by g, subtends the same number of steps as g. Thus g<sub>2</sub> and g<sub>3</sub> subtend the same number of steps, a fact we need in order to be able to substitute one instance of g<sub>2</sub> with g<sub>3</sub> in the next part.
 Let ''r'' be odd and ''r'' ≥ 3. Consider the following abstract sizes for the interval class (''k''-steps) reached by stacking ''r'' generators:
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 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).
 ===== Statement (2) =====
 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: