Rank-3 scale theorems: Difference between revisions

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In case 1 (even scale size), we have the chain g1 g2 g1 g2... g1 g3. Consider the sizes of the n/2-step (which is an odd number of generator steps):

# g1 ... g1, (n/2-1)*g0 + g1 = n/2 g1 + (n/2-1) g2

# from g1 ... g1, get a1 = (n/2-1)*g0 + g1 = n/2 g1 + (n/2-1) g2

# g2 ... g2, (n/2-1)*g0 + g2 = (n/2-1) g1 + n/2 g2

# from g2 ... g2, get a2 =(n/2-1)*g0 + g2 = (n/2-1) g1 + n/2 g2

# g2 (even) g1 g3 g1 (even) g2 ~~(inverse of #1)~~ = (n/2-1) g1 + (n/2-1) g2 + g3

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

# g1 (odd) g1 g3 g1 (odd) g1 ~~(inverse of #2)~~ = n/2 g1 + (n/2-2) 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 g1 = 1/2*g0 + ε, g2 = 1/2*g0 - ε. We have 4 distinct sizes for n/2-steps, a contradiction to MV3:

(1) #1, #2 and #3 are clearly distinct.

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

(2) #4 - #3 = g1 - g2 != 0, since the scale is a non-trivial AG.

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

(3) #4 - #1 = 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. By choice of tuning this is very close to the chroma of the mos generated by g0.

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

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

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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 In case 1 (even scale size), we have the chain g1 g2 g1 g2... g1 g3. Consider the sizes of the n/2-step (which is an odd number of generator steps):
-# g1 ... g1, (n/2-1)*g0 + g1 = n/2 g1 + (n/2-1) g2
+# from g1 ... g1, get a1 = (n/2-1)*g0 + g1 = n/2 g1 + (n/2-1) g2
-# g2 ... g2, (n/2-1)*g0 + g2 = (n/2-1) g1 + n/2 g2
+# from g2 ... g2, get a2 =(n/2-1)*g0 + g2 = (n/2-1) g1 + n/2 g2
-# g2 (even) g1 g3 g1 (even) g2 (inverse of #1) = (n/2-1) g1 + (n/2-1) g2 + g3
+# from g2 (even) g1 g3 g1 (even) g2, get a3 = (n/2-1) g1 + (n/2-1) g2 + g3
-# g1 (odd) g1 g3 g1 (odd) g1 (inverse of #2) = n/2 g1 + (n/2-2) 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 g1 = 1/2*g0 + ε, g2 = 1/2*g0 - ε. We have 4 distinct sizes for n/2-steps, a contradiction to MV3:
-(1) #1, #2 and #3 are clearly distinct.
+(1) a1, a2 and a3 are clearly distinct.
-(2) #4 - #3 = g1 - g2 != 0, since the scale is a non-trivial AG.
+(2) a4 - a3 = g1 - g2 != 0, since the scale is a non-trivial AG.
-(3) #4 - #1 = 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. By choice of tuning this is very close to the chroma of the mos generated by g0.
-(4) #4 - #2 = g1 - 2 g2 + g3 != 0, by choice of tuning.
+(4) a4 - a2 = g1 - 2 g2 + g3 != 0, by choice of tuning.
 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: