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)~~. This shows that AG and MV3 scales must have odd size~~.

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).

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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# (k-1) g1 + (k-1) g2 + g3,

if a step is an odd number of generators (since the scale size is odd, we can always ensure this by taking octave complements of all the generators). The first two sizes must occur the same number of times.

This proof shows that AG and MV3 scales must have odd size.

==== 3-DE implies MV3 (WIP) ====

@@ 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 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.
+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).
 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 124: / Line 124: @@
 # (k-1) g1 + (k-1) g2 + g3,
 if a step is an odd number of generators (since the scale size is odd, we can always ensure this by taking octave complements of all the generators). The first two sizes must occur the same number of times.
+This proof shows that AG and MV3 scales must have odd size.
 ==== 3-DE implies MV3 (WIP) ====