Ternary scale theorems: Difference between revisions
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We also know that ''s'' is of the form ''w''('''x''', '''y''', '''z''')''w''('''x''', '''z''', '''y'''). Hence the number of occurrences of '''x''' + '''y''' = the number of occurrences of '''x''' + '''z''', counting all 2-steps in all of ''s''. | We also know that ''s'' is of the form ''w''('''x''', '''y''', '''z''')''w''('''x''', '''z''', '''y'''). Hence the number of occurrences of '''x''' + '''y''' = the number of occurrences of '''x''' + '''z''', counting all 2-steps in all of ''s''. | ||
Write ''s''<sub>1</sub> for the scale word made from stacked 2-steps from the 0-degree, and let ''s''<sub>2</sub> be | Write ''s''<sub>1</sub> for the scale word made from stacked 2-steps from the 0-degree, and let ''s''<sub>2</sub> be as follows: | ||
* In the singly even case, let ''s''<sub>2</sub> be the circular word of 2-steps starting at the (''n''/2)-degree. We know that they differ only by interchanging '''y''' and '''z''', hence that they have the same period. Hence both ''s''<sub>1</sub> and ''s''<sub>2</sub> are primitive. | * In the singly even case, let ''s''<sub>2</sub> be the circular word of 2-steps starting at the (''n''/2)-degree. We know that they differ only by interchanging '''y''' and '''z''', hence that they have the same period. Hence both ''s''<sub>1</sub> and ''s''<sub>2</sub> are primitive. | ||
* In the doubly even case, start from the mode of ''s'' whose template MOS is the brightest mode. Let ''s''<sub>2</sub> be offset at a generator of the even-regular scale, which by Theorem 6 we chooae to have the same dyad class as a bright generator of the MOS ''a'''''x''' 2''k'''''X'''. This is what induces the equality of ''s''<sub>1</sub> and ''s''<sub>2</sub> (in particular, the two scales have the same period, thus they are both primitive): Let ''s''<sub>''t''</sub> be the period of the brightest mode of the template MOS, and let ''g'' be its bright generator class. Then the slice ''s''<sub>''t''</sub>[-''g'' +1 : 1] is the imperfect generator of the MOS. Now when we "darken" the mode by one generator, which is the difference between the template MOSes of ''s''<sub>1</sub> and ''s''<sub>2</sub>, we turn that slice into the bright generator, hence swapping ''s''<sub>''t''</sub>[-''g''] and ''s''<sub>''t''</sub>[-''g'' + 1]. Note that ''g'' must be odd since it generates a 2-period MOS. So (under 1-indexing) the first letter's index is odd and the second letter's index is even, which is what we want since the letters are within a stacked 2-step. While the generator might have to be higher by an (''n''/2)-step, that doesn't affect the parity since ''n''/2 is even. | * In the doubly even case, start from the mode of ''s'' whose template MOS is the brightest mode. Let ''s''<sub>2</sub> be offset at a generator of the even-regular scale, which by Theorem 6 we chooae to have the same dyad class as a bright generator of the MOS ''a'''''x''' 2''k'''''X'''. This is what induces the equality of ''s''<sub>1</sub> and ''s''<sub>2</sub> (in particular, the two scales have the same period, thus they are both primitive): Let ''s''<sub>''t''</sub> be the period of the brightest mode of the template MOS, and let ''g'' be its bright generator class. Then the slice ''s''<sub>''t''</sub>[-''g'' +1 : 1] is the imperfect generator of the MOS. Now when we "darken" the mode by one generator, which is the difference between the template MOSes of ''s''<sub>1</sub> and ''s''<sub>2</sub>, we turn that slice into the bright generator, hence swapping ''s''<sub>''t''</sub>[-''g''] and ''s''<sub>''t''</sub>[-''g'' + 1]. Note that ''g'' must be odd since it generates a 2-period MOS. So (under 1-indexing) the first letter's index is odd and the second letter's index is even, which is what we want since the letters are within a stacked 2-step. While the generator might have to be higher by an (''n''/2)-step, that doesn't affect the parity since ''n''/2 is even. | ||