Rank-3 scale theorems: Difference between revisions

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# 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 very close to 1/2*g0~~). #4 - #2 = g1 - 2 g2 + g3 != 0 ~~using a suitable~~ choice ~~for g1 and g2 (same trick)~~.

Choose a tuning where g1 and g2 are both very close to 1/2*g0. We have 4 distinct sizes for n/2-steps, a contradiction to MV3:

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

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

(3) #4 - #1 = g3 - g2 != 0, by the choice of tuning.

(4) #4 - #2 = g1 - 2 g2 + g3 != 0, by the 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:

@@ Line 117: / Line 117: @@
 # 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 very close to 1/2*g0). #4 - #2 = g1 - 2 g2 + g3 != 0 using a suitable choice for g1 and g2 (same trick).
+Choose a tuning where g1 and g2 are both very close to 1/2*g0. We have 4 distinct sizes for n/2-steps, a contradiction to MV3:
+(1) #1, #2 and #3 are clearly distinct.
+(2) #4 - #3 = g1 - g2 != 0, since the scale is a non-trivial AG.
+(3) #4 - #1 = g3 - g2 != 0, by the choice of tuning.
+(4) #4 - #2 = g1 - 2 g2 + g3 != 0, by the 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: