Rank-3 scale theorems: Difference between revisions

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==== AG + unconditionally MV3 implies "ax by bz" and that the scale's ====

==== AG + unconditionally MV3 implies "ax by bz" and that the scale's cardinality is odd ====

'''Assuming both AG and unconditional MV3''', we have two chains of generator g0 (going right). The two cases are:

O-O-...-O (m notes)

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Label the notes (1,k) and (2,k), 1 ≤ k ≤ m or m-1, for notes in the upper and lower chain respectively.

In case 1 (~~singly~~ even scale size), let g1 = (2,1) - (1,1) and g2 = (1,2) - (2,1). 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):

In case 1 (even scale size n = 2^t r where r is odd), let g1 = (2,1) - (1,1) and g2 = (1,2) - (2,1). We have the chain g1 g2 g1 g2... g1 g3. Suppose the k-step is the class generated by r generators (which is an odd number of generator steps):

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

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

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

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

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

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

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

# from g1 (odd) g1 g3 g1 (odd) g1, get a4 = (r+1)/2 g1 + (r-3)/2 g2 + g3.

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

# a1, a2 and a3 are clearly distinct.

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

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(The above holds for any odd n >= 3.)

This proof shows that AG and unconditionally-MV3 scales must have odd or ~~doubly even cardinality~~.

This proof shows that AG and unconditionally-MV3 scales must have cardinality odd or 4.

==== An AG scale is unconditionally MV3 iff its cardinality is odd or 4 ====

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-==== AG + unconditionally MV3 implies "ax by bz" and that the scale's  ====
+==== AG + unconditionally MV3 implies "ax by bz" and that the scale's cardinality is odd  ====
 '''Assuming both AG and unconditional MV3''', we have two chains of generator g0 (going right). The two cases are:
   O-O-...-O (m notes)
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 Label the notes (1,k) and (2,k), 1 ≤ k ≤ m or m-1, for notes in the upper and lower chain respectively.
-In case 1 (singly even scale size), let g1 = (2,1) - (1,1) and g2 = (1,2) - (2,1). 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):
+In case 1 (even scale size n = 2^t r where r is odd), let g1 = (2,1) - (1,1) and g2 = (1,2) - (2,1). We have the chain g1 g2 g1 g2... g1 g3. Suppose the k-step is the class generated by r generators (which is an odd number of generator steps):
-# from g1 ... g1, get a1 = (n/2-1)*g0 + g1 = n/2 g1 + (n/2-1) g2
+# from g1 ... g1, get a1 = (r-1)/2*g0 + g1 = (r+1)/2 g1 + (r-1)/2 g2
-# from g2 ... g2, get a2 =(n/2-1)*g0 + g2 = (n/2-1) g1 + n/2 g2
+# from g2 ... g2, get a2 = (r-1)/2*g0 + g2 = (r-1)/2 g1 + (r+1)/2 g2
-# from g2 (even) g1 g3 g1 (even) g2, get a3 = (n/2-1) g1 + (n/2-1) g2 + g3
+# from g2 (even) g1 g3 g1 (even) g2, get a3 = (r-1)/2 g1 + (r-1)/2 g2 + g3
-# from g1 (odd) g1 g3 g1 (odd) g1, get a4 = n/2 g1 + (n/2-2) g2 + g3.
+# from g1 (odd) g1 g3 g1 (odd) g1, get a4 = (r+1)/2 g1 + (r-3)/2 g2 + g3.
-Choose a tuning where g1 and g2 are both very close to but not exactly 1/2*g0, resulting in a scale very close to the mos generated by 1/2 g0. (i.e. g1 and g2 differ from 1/2*g0 by ε, a quantity much smaller than the chroma of the n/2-note mos generated by g0, which is |g3 - g2|). Assuming n > 4, 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, resulting in a scale very close to the mos generated by 1/2 g0. (i.e. g1 and g2 differ from 1/2*g0 by ε, a quantity much smaller than the chroma of the n/2-note mos generated by g0, which is |g3 - g2|). Assuming n > 4, we have 4 distinct sizes for k-steps, a contradiction to unconditional-MV3:
 # a1, a2 and a3 are clearly distinct.
 # a4 - a3 = g1 - g2 != 0, since the scale is a non-trivial AG.
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 (The above holds for any odd n >= 3.)
-This proof shows that AG and unconditionally-MV3 scales must have odd or doubly even cardinality.
+This proof shows that AG and unconditionally-MV3 scales must have cardinality odd or 4.
 ==== An AG scale is unconditionally MV3 iff its cardinality is odd or 4 ====