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

In case 1 (even scale size), consider the sizes of the n/2-step:

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

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

# g1 ... g1, (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

# 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.

We have 4 distinct sizes, 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.

~~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, a step takes an odd number l of generators gi in the chain, so the other 2 sizes can't come in equal numbers.
+In case 1 (even scale size), consider the sizes of the n/2-step:
- n/2*g1 + (n/2-1)g2 + g3
-But here, the n/2-step comes in 4 sizes:
 # g1 ... g1, (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
 # 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.
+We have 4 distinct sizes, 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.
-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: