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 an even ~~number of~~ steps, and those intervals that are "offset" by g1 must be an odd number of steps. Letting ''M'' be the subset of all even-numbered notes (which are generated by g) and considering ''M'' as a scale by dividing interval 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.

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 an odd number of steps. Letting ''M'' be the subset of all even-numbered notes (which are generated by g) and considering ''M'' as a scale by dividing interval 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.

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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# It is ''not'' chiral

==== Proof ====

(1) and (2) were proved in the proof of Proposition 1 (the part that we appeal to, from "all multiples of the generator g must be an even ~~number of~~ steps ..." to "These are all distinct by Z-linear independence", does not rely on SGA property). (3) is easy to check using (1).

(1) and (2) were proved in the proof of Proposition 1 (the part that we appeal to, from "all multiples of the generator g must be even-steps ..." to "These are all distinct by Z-linear independence", does not rely on SGA property). (3) is easy to check using (1).

=== Theorem 4 (Classification of PWF scales) ===

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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 an even number of steps, and those intervals that are "offset" by g<sub>1</sub> must be an odd number of steps. Letting ''M'' be the subset of all even-numbered notes (which are generated by g) and considering ''M'' as a scale by dividing interval 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.
+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 an odd number of steps. Letting ''M'' be the subset of all even-numbered notes (which are generated by g) and considering ''M'' as a scale by dividing interval 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.
 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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 # It is ''not'' chiral
 ==== Proof ====
-(1) and (2) were proved in the proof of Proposition 1 (the part that we appeal to, from "all multiples of the generator g must be an even number of steps ..." to "These are all distinct by Z-linear independence", does not rely on SGA property). (3) is easy to check using (1).
+(1) and (2) were proved in the proof of Proposition 1 (the part that we appeal to, from "all multiples of the generator g must be even-steps ..." to "These are all distinct by Z-linear independence", does not rely on SGA property). (3) is easy to check using (1).
 === Theorem 4 (Classification of PWF scales) ===