Rank-3 scale theorems: Difference between revisions

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# from g2 (even) g1 g3 g1 (even) g2, get a3 = (n/2-1) g1 + (n/2-1) g2 + g3

# from g1 (odd) g1 g3 g1 (odd) g1, get a4 = n/2 g1 + (n/2-2) g2 + g3.

Choose a tuning where ~~g0 is different enough from g3 + g1 (the imperfect gen of the mos generated by g0), and where~~ g1 and g2 are both very close to but not exactly 1/2*g0. We have 4 distinct sizes for n/2-steps, a contradiction to unconditional-MV3:

Choose a tuning where: g1 and g2 are both very close to but not exactly 1/2*g0 (i.e. they differ from 1/2*g0 by ε, a quantity much smaller than the chroma of the m-note mos generated by g0, which is |g3 - g2|). We have 4 distinct sizes for n/2-steps, a contradiction to unconditional-MV3:

(1) a1, a2 and a3 are clearly distinct.

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(2) a4 - a3 = g1 - g2 != 0, since the scale is a non-trivial AG.

(3) a4 - a1 = g3 - g2 = (g3 + g1) - (g2 + g1) != 0. ~~By choice of tuning this~~ is ~~very close to~~ the chroma of the mos generated by g0.

(3) a4 - a1 = g3 - g2 = (g3 + g1) - (g2 + g1) != 0. This is exactly the chroma of the mos generated by g0.

(4) a4 - a2 = g1 - 2 g2 + g3 != 0, by choice of tuning.

(4) a4 - a2 = g1 - 2 g2 + g3 = chroma ± ε > 0; by choice of tuning, this is very close to the chroma of the mos generated by g0, with at most ε error.

In case 2, let (2,1)-(1,1) = g1, (1,2)-(2,1) = g2 be the two alternating generators. 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 generators corresponding to a step are:

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 # from g2 (even) g1 g3 g1 (even) g2, get a3 = (n/2-1) g1 + (n/2-1) g2 + g3
 # from g1 (odd) g1 g3 g1 (odd) g1, get a4 = n/2 g1 + (n/2-2) g2 + g3.
-Choose a tuning where g0 is different enough from g3 + g1 (the imperfect gen of the mos generated by g0), and where g1 and g2 are both very close to but not exactly 1/2*g0. We have 4 distinct sizes for n/2-steps, a contradiction to unconditional-MV3:
+Choose a tuning where: g1 and g2 are both very close to but not exactly 1/2*g0 (i.e. they differ from 1/2*g0 by ε, a quantity much smaller than the chroma of the m-note mos generated by g0, which is |g3 - g2|). We have 4 distinct sizes for n/2-steps, a contradiction to unconditional-MV3:
 (1) a1, a2 and a3 are clearly distinct.
@@ Line 125: / Line 125: @@
 (2) a4 - a3 = g1 - g2 != 0, since the scale is a non-trivial AG.
-(3) a4 - a1 = g3 - g2 = (g3 + g1) - (g2 + g1) != 0. By choice of tuning this is very close to the chroma of the mos generated by g0.
+(3) a4 - a1 = g3 - g2 = (g3 + g1) - (g2 + g1) != 0. This is exactly the chroma of the mos generated by g0.
-(4) a4 - a2 = g1 - 2 g2 + g3 != 0, by choice of tuning.
+(4) a4 - a2 = g1 - 2 g2 + g3 = chroma ± ε > 0; by choice of tuning, this is very close to the chroma of the mos generated by g0, with at most ε error.
 In case 2, let (2,1)-(1,1) = g1, (1,2)-(2,1) = g2 be the two alternating generators. 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 generators corresponding to a step are: