Rank-3 scale theorems: Difference between revisions

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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, assume that g0 is a j-step and gcd(j,n) = 1. Let g1 = (2,1)-(1,m) and g2 = (1,1)-(2,m). Then circle of stacked g0 generators is (starting from top left): (m-1 g0's) g1 (m-1 g0's) g2. A scale step is always the same number k (which must be odd) of such generators gi, since the scale is MV3 and thus generically satisfies the constant structure property. Assume (after taking octave complement) that a single step takes less than half of the generators. So a word corresponding to the scale step is formed by one of:

In case 1, assume that g0 is a j-step and gcd(j,n) = 1 (as given by AG). Let g1 = (2,1)-(1,m) and g2 = (1,1)-(2,m). Then circle of stacked g0 generators is (starting from top left): (m-1 g0's) g1 (m-1 g0's) g2. A scale step is always the same number k (which must be odd) of such generators gi, since the scale is MV3 and thus generically satisfies the constant structure property. Assume (after taking octave complement) that a single step takes less than half of the generators. So a word corresponding to the scale step is formed by one of:

* k g0

* (k-1) g0 + g1

* (k-1) g0 + g2.

It is clear that the last two sizes must occur the same number of times.

If on the other hand gcd(j,n) > 1, it must be = 2 because of the sizes of the two g0 chains are >= 1/3*n. The two imperfect generators must close each of the two chains into two circles of size m = n/2: (m-1 g0's) g1 and (m-1 g0's) g1. (By AG, g1 = g2.)

~~Suppose the interval v1 = (2,1) - (1,1) separating the two chains is an r-step for r odd. Suppose v1 = a'x + b'y + c'z. Every even note (that is (1,i) for some i) has a~~

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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 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, assume that g0 is a j-step and gcd(j,n) = 1. Let g1 = (2,1)-(1,m) and g2 = (1,1)-(2,m). Then circle of stacked g0 generators is (starting from top left): (m-1 g0's) g1 (m-1 g0's) g2. A scale step is always the same number k (which must be odd) of such generators gi, since the scale is MV3 and thus generically satisfies the constant structure property. Assume (after taking octave complement) that a single step takes less than half of the generators. So a word corresponding to the scale step is formed by one of:
+In case 1, assume that g0 is a j-step and gcd(j,n) = 1 (as given by AG). Let g1 = (2,1)-(1,m) and g2 = (1,1)-(2,m). Then circle of stacked g0 generators is (starting from top left): (m-1 g0's) g1 (m-1 g0's) g2. A scale step is always the same number k (which must be odd) of such generators gi, since the scale is MV3 and thus generically satisfies the constant structure property. Assume (after taking octave complement) that a single step takes less than half of the generators. So a word corresponding to the scale step is formed by one of:
 * k g0
 * (k-1) g0 + g1
 * (k-1) g0 + g2.
 It is clear that the last two sizes must occur the same number of times.
-If on the other hand gcd(j,n) > 1, it must be = 2 because of the sizes of the two g0 chains are >= 1/3*n. The two imperfect generators must close each of the two chains into two circles of size m = n/2: (m-1 g0's) g1 and (m-1 g0's) g1. (By AG, g1 = g2.)
-Suppose the interval v1 = (2,1) - (1,1) separating the two chains is an r-step for r odd. Suppose v1 = a'x + b'y + c'z. Every even note (that is (1,i) for some i) has a
 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: