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 together to 1/2*g0 so as to get a mos). #4 - #2 = g1 - 2 g2 + g3 != 0 using a suitable choice for g1 and g2 (same trick).

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 a suitable choice for g1 and g2 (same trick). This shows that AG and MV3 scales must have odd size.

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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 # 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 a suitable choice for g1 and g2 (same trick).
+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 a suitable choice for g1 and g2 (same trick). This shows that AG and MV3 scales must have odd size.
 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: