Generator-offset property: Difference between revisions
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Label the notes (1, ''j'') and (2, ''j''), 1 ≤ ''j'' ≤ (number of notes in the chain), for notes in the upper and lower chain respectively. | Label the notes (1, ''j'') and (2, ''j''), 1 ≤ ''j'' ≤ (number of notes in the chain), for notes in the upper and lower chain respectively. | ||
===== Statement (1) ===== | |||
In case 1, let g<sub>1</sub> = (2, 1) − (1, 1), g<sub>2</sub> = (1, 2) − (2, 1), and g<sub>3</sub> = (1, 1) − (''n''/2, 2) = (−''n''/2*g<sub>1</sub> − g<sub>1</sub> − ''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''. | In case 1, let g<sub>1</sub> = (2, 1) − (1, 1), g<sub>2</sub> = (1, 2) − (2, 1), and g<sub>3</sub> = (1, 1) − (''n''/2, 2) = (−''n''/2*g<sub>1</sub> − g<sub>1</sub> − ''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''. | ||
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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). | 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) − (1, 1) = g<sub>1</sub>, (1, 2) − (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: | In case 2, let (2, 1) − (1, 1) = g<sub>1</sub>, (1, 2) − (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: | ||
# ''k''g<sub>1</sub> + (''k'' − 1)g<sub>2</sub> | # ''k''g<sub>1</sub> + (''k'' − 1)g<sub>2</sub> | ||
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(The above holds for any odd ''n'' ≥ 3.) | (The above holds for any odd ''n'' ≥ 3.) | ||
===== Statement (3) ===== | |||
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. | |||
===== Statement 4 ===== | |||
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. | |||
Consider the two alternants, detemperings of the generator iX + jW of the ''a''X 2''b''W mos ''T''(X, W) = ''S''(X, W, W) where gcd(''j'', 2''k'') = 1. | |||
'''Claim 1''': Deleting X's from the alternants of ''S'' gives every ''j''-step in the scale ''E''<sub>X</sub>(''S'')(Y, Z), the scale word obtained by deleting all X's from X. | |||
Assume that the imperfect generator of ''T'' has ''j'' + 1 W's and the perfect generator has ''j'' W's. Suppose that one ''j''-step word ''R'' on note ''p'' of ''E''<sub>X</sub>(''S'') is "contained in" the corresponding "imperfect alternant" of ''S'', the unique (''i'' + ''j'')-step, which is ''I'' = ''S''[''p'' : ''p'' + ''i'' + ''j'']. By this we mean that ''E''<sub>X</sub>(''I'') has ''R'' as a substring. Then ''S''[''p'' - 1: ''p'' - 1 + ''i'' + ''j''] and ''S''[''p'' + 1 : ''p'' + 1 + ''i'' + ''j''] are both perfect, and have one fewer step that is Y or Z. Thus the word ''I'' must both begin and end in a letter that is either Y or Z. Removing all the X's from ''I'' results in a word that is ''j'' + 1 letters long and is the ''j''-step we started with, with just one extra letter. Thus one of the two perfect generators above, namely the one that removes the extra letter, must contain this ''j''-step. | |||
Proof: Every (''i'' + ''j'')-step in S except one involves *a* j-step in ''E''<sub>X</sub>(''S'')(Y, Z). | |||
'''Claim 2''': If a scale ''U'' has ''k'' X's and ''k'' Y's, gcd(''j'', 2''k'') = 1, and consecutively stacked ''j''-steps occur in 2 alternating sizes, the ''k'' Y's and ''k'' Z's alternate. | |||
Proof: write η and ζ for the two sizes of ''j''-steps. Since gcd(''j'', 2''k'') = 1 there exists ''m'' such that stacking ''m''-many ''j''-steps yields scale steps of ''U'', and ''m'' is odd. Hence the scale steps of ''U'' are ηζ…ζη mod e and ζη…ηζ mod e, and the step sizes clearly alternate. | |||
These two claims prove that ''E''<sub>X</sub>(S) = (YZ)<sup>''k''</sup>, which also proves that the two alternants' sizes differ by replacing one Y for a Z. | |||
===== Statement (5) ===== | |||
For (5), odd-numbered SGA scales are [[Fokker block]]s (in the 2-dimensional lattice generated by the generator and the offset). To see this, consider the following lattice depiction of such a scale: | For (5), odd-numbered SGA scales are [[Fokker block]]s (in the 2-dimensional lattice generated by the generator and the offset). To see this, consider the following lattice depiction of such a scale: | ||
x x x ... x | x x x ... x | ||
x x x ... x x | x x x ... x x | ||
and use the vectors (-1, 2) and (ceil(n/2), 1) as the Fokker block chromas. A Fokker block has the property that tempering out by each of the chromas gives two mosses. These correspond to two of the temperings X = Y, Y = Z and X = Z. The third tempering follows by symmetry (by taking the other chirality). | and use the vectors (-1, 2) and (ceil(n/2), 1) as the Fokker block chromas. A Fokker block has the property that tempering out by each of the chromas gives two mosses. These correspond to two of the temperings X = Y, Y = Z and X = Z. The third tempering follows by symmetry (by taking the other chirality). | ||
===== Statement (6) ===== | |||
For (6), consider the mos ''a''X 2''b''W 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 ''k''th note of an ''n''-note mos where ''k'' divides ''n'' yields a mos. Since the result of setting X = 0 is the mos ''b''Y ''b''Z, ''S'' is elimination-mos. | For (6), consider the mos ''a''X 2''b''W 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 ''k''th note of an ''n''-note mos where ''k'' divides ''n'' yields a mos. Since the result of setting X = 0 is the mos ''b''Y ''b''Z, ''S'' is elimination-mos. | ||