Rank-3 scale theorems: Difference between revisions

Line 112:

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 ~~(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:~~

In case 1, a step takes an odd number l of generators gi in the chain, so the other 2 sizes can't come in equal numbers.

* k g0

n/2*g1 + (n/2-1)g2 + g3

* (k-~~1) g0 +~~ g1

But here, the n/2-step comes in 4 sizes:

* (k-~~1) g0~~ + ~~g2.~~

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

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

# g2 ... g2, (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

# g1 (odd) g1 g3 g1 (odd) g1 (inverse of #2) = n/2 g1 + (n/2-2) g2 + g3.

This is a contradiction to MV3: #1, #2 and #3 are clearly distinct. #4 - #3 = g1 - g2 != 0, since the scale is a non-trivial AG. #4 - #1 = g3 - g2 != 0 (shown by "tempering" g1 and g2 together to 1/2*g0 so as to get a mos). #4 - #2 = g1 - 2 g2 + g3 != 0 using the same trick as in #4 - #1.

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:

@@ Line 112: / Line 112: @@
 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 (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:
+In case 1, a step takes an odd number l of generators gi in the chain, so the other 2 sizes can't come in equal numbers.
-* k g0
+ n/2*g1 + (n/2-1)g2 + g3
-* (k-1) g0 + g1
+But here, the n/2-step comes in 4 sizes:
-* (k-1) g0 + g2.
+# g1 ... g1, (n/2-1)*g0 + g1 = n/2 g1 + (n/2-1) g2
-It is clear that the last two sizes must occur the same number of times.
+# g2 ... g2, (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
+# g1 (odd) g1 g3 g1 (odd) g1 (inverse of #2) = n/2 g1 + (n/2-2) g2 + g3.
+This is a contradiction to MV3: #1, #2 and #3 are clearly distinct. #4 - #3 = g1 - g2 != 0, since the scale is a non-trivial AG. #4 - #1 = g3 - g2 != 0 (shown by "tempering" g1 and g2 together to 1/2*g0 so as to get a mos). #4 - #2 = g1 - 2 g2 + g3 != 0 using the same trick as in #4 - #1.
 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: