Generator-offset property: Difference between revisions
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Choose a tuning where ''g''<sub>1</sub> and ''g''<sub>2</sub> are both very close to but not exactly 1/2*''g''<sub>0</sub>, resulting in a scale very close to the mos generated by 1/2 ''g''<sub>0</sub>. (i.e. ''g''<sub>1</sub> and ''g''<sub>2</sub> differ from 1/2*''g''<sub>0</sub> by ε, a quantity much smaller than the chroma of the ''n''/2-note mos generated by ''g''<sub>0</sub>, which is |''g''<sub>3</sub> − ''g''<sub>2</sub>|). Thus we have 4 distinct sizes for ''k''-steps: | Choose a tuning where ''g''<sub>1</sub> and ''g''<sub>2</sub> are both very close to but not exactly 1/2*''g''<sub>0</sub>, resulting in a scale very close to the mos generated by 1/2 ''g''<sub>0</sub>. (i.e. ''g''<sub>1</sub> and ''g''<sub>2</sub> differ from 1/2*''g''<sub>0</sub> by ε, a quantity much smaller than the chroma of the ''n''/2-note mos generated by ''g''<sub>0</sub>, which is |''g''<sub>3</sub> − ''g''<sub>2</sub>|). Thus we have 4 distinct sizes for ''k''-steps: | ||
# ''a''<sub>1</sub>, ''a''<sub>2</sub> and ''a''<sub>3</sub> are clearly distinct. | # ''a''<sub>1</sub>, ''a''<sub>2</sub> and ''a''<sub>3</sub> are clearly distinct. | ||
# ''a''<sub>4</sub> − ''a''<sub>3</sub> = ''g''<sub>1</sub> − ''g''<sub>2</sub> | # ''a''<sub>4</sub> − ''a''<sub>3</sub> = ''g''<sub>1</sub> − ''g''<sub>2</sub> ≠ 0, since the scale is a non-degenerate GO scale. | ||
# ''a''<sub>4</sub> − ''a''<sub>1</sub> = ''g''<sub>3</sub> − ''g''<sub>2</sub> = (''g''<sub>3</sub> + ''g''<sub>1</sub>) − (''g''<sub>2</sub> + ''g''<sub>1</sub>) | # ''a''<sub>4</sub> − ''a''<sub>1</sub> = ''g''<sub>3</sub> − ''g''<sub>2</sub> = (''g''<sub>3</sub> + ''g''<sub>1</sub>) − (''g''<sub>2</sub> + ''g''<sub>1</sub>) ≠ 0. This is exactly the chroma of the mos generated by ''g''<sub>0</sub>. | ||
# ''a''<sub>4</sub> − ''a''<sub>2</sub> = ''g''<sub>1</sub> − 2 ''g''<sub>2</sub> + ''g''<sub>3</sub> = (''g''<sub>3</sub> − ''g''<sub>2</sub>) + (''g''<sub>1</sub> − ''g''<sub>2</sub>) = (chroma ± ε) | # ''a''<sub>4</sub> − ''a''<sub>2</sub> = ''g''<sub>1</sub> − 2 ''g''<sub>2</sub> + ''g''<sub>3</sub> = (''g''<sub>3</sub> − ''g''<sub>2</sub>) + (''g''<sub>1</sub> − ''g''<sub>2</sub>) = (chroma ± ε) ≠ 0 by choice of tuning. | ||
By applying this argument to 1-steps, we see that there must be 4 step sizes in some tuning, a contradiction. Thus ''g''<sub>1</sub> and ''g''<sub>2</sub> must themselves be step sizes. Thus we see that an even-length, unconditionally SV3, GO scale must be of the form ''xy...xyxz''. But this pattern is not abstractly SV3 if ''n'' ≥ 6, since 3-steps come in 4 sizes: ''xyx'', ''yxy'', ''yxz'' and ''xzx''. Thus ''n'' = 4 and the scale is ''xyxz''. This proves (3). | By applying this argument to 1-steps, we see that there must be 4 step sizes in some tuning, a contradiction. Thus ''g''<sub>1</sub> and ''g''<sub>2</sub> must themselves be step sizes. Thus we see that an even-length, unconditionally SV3, GO scale must be of the form ''xy...xyxz''. But this pattern is not abstractly SV3 if ''n'' ≥ 6, since 3-steps come in 4 sizes: ''xyx'', ''yxy'', ''yxz'' and ''xzx''. Thus ''n'' = 4 and the scale is ''xyxz''. This proves (3). | ||
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For (1), we now only need to see that SGA + odd length => abstractly SV3. But the argument in case 2 above works for any interval class (abstract SV3 wasn't used), hence any interval class comes in (abstractly) exactly 3 sizes regardless of tuning. | For (1), we now only need to see that SGA + odd length => abstractly SV3. But the argument in case 2 above works for any interval class (abstract SV3 wasn't used), hence any interval class comes in (abstractly) exactly 3 sizes regardless of tuning. | ||
For (4), assume S is aX bY bZ, a odd. If b = 1, there’s nothing to prove. So assume b > 1. If Y’s and Z’s don't alternate perfectly, then (ignoring X's) you have two consecutive Y's somewhere and two consecutive Z's somewhere else. Assume that the perfect generator is iX + jW with j | For (4), assume S is aX bY bZ, a odd. If b = 1, there’s nothing to prove. So assume b > 1. If Y’s and Z’s don't alternate perfectly, then (ignoring X's) you have two consecutive Y's somewhere and two consecutive Z's somewhere else. Assume that the perfect generator is iX + jW with j ≥ 2. (If this is not true, invert the generator, since b > 1.) | ||
In aX bY bZ, consider (i+j)-steps (representing the generator) with: (1) the maximum possible number of Y’s (at least 2 more than the # of Z's), (2) the maximum number of Z’s (at least 2 more than the # of Y's), or (3) an intermediate number of Y’s and Z’s between the two or (4) (the preimage of) the imperfect generator, having a different number of X’s than (1), (2), and (3). Since a + 2b | In aX bY bZ, consider (i+j)-steps (representing the generator) with: (1) the maximum possible number of Y’s (at least 2 more than the # of Z's), (2) the maximum number of Z’s (at least 2 more than the # of Y's), or (3) an intermediate number of Y’s and Z’s between the two or (4) (the preimage of) the imperfect generator, having a different number of X’s than (1), (2), and (3). Since a + 2b ≥ 5, there are at least 4 perfect generators, so there must be at least one of each of (1), (2), and (3), giving a contradiction to SV3. [Whenever the root of the (i+j)-step are moved within S, the numbers of Y's and Z's change one at a time and reach a maximum at some choice of the root and a minimum with another choice, guaranteeing that intermediate values are reached.] | ||
Any generator of aX 2bW must have an odd number of W steps (Otherwise, intervals with an odd number of W steps can't be generated.). This with (4) immediately gives (5). | Any generator of aX 2bW must have an odd number of W steps (Otherwise, intervals with an odd number of W steps can't be generated.). This with (4) immediately gives (5). | ||