Ternary scale theorems: Difference between revisions

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Write ''s''1 for the scale word made from stacked 2-steps from the 0-degree, and let ''s''2 be the scale word of stacked 2-steps from the 1-degree.

* In the singly even case, let ''s''2 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''1 and ''s''2 are primitive.

* In the doubly even case, let ''s''2 be offset at the ~~generator~~, which has the same dyad class as ~~some~~ generator of the MOS 2aX 2cZ.

* In the doubly even case, let ''s''2 be offset at a "generator" of the even-regular scale, which by Theorem 6 has the same dyad class as a bright generator of the MOS 2aX 2cZ. This is what induces the equality of ''s''1 and ''s''2: Let ''s''''t'' be the period of the brightest mode of the template MOS, and let ''g'' be its bright generator class. Then the slice ''s''''t''[-''g'' +1 : 1] is the imperfect generator of the MOS. When this mode is "darkened" by one generator, we turn that slice into the bright generator, hence swapping ''s''''t''[-''g''] and s_T[-''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 that's within a stacked 2-step. While the generator might have to be higher by one period of the template MOS, that doesn't affect the parity since the period is even. In particilar, the two scales have the same period, thus they are both primitive.

We prove that ''s''1 and ''s''2 are MOS substitution scales with a filling MOS of period 2. The number the 2-step (1) occurs must be the same in both ''s''1 and ''s''2. The word of stacked 2-steps of the template MOS (which is of the form ''w''('''x''', '''X''', '''X''')''w''('''x''', '''X''', '''X''')), which is itself a MOS word, consists of letters (1) '''x''' + '''X''' and (2) 2'''X''' if more '''X''''s than '''x''''s, 2'''x''' if more '''x''''s than '''X''''s. The word of stacked 2-steps from the n/2-degree is also this same MOS word.

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 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 the scale word of stacked 2-steps from the 1-degree.
 * 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, let ''s''<sub>2</sub> be offset at the generator, which has the same dyad class as some generator of the MOS 2aX 2cZ.
+* In the doubly even case, let ''s''<sub>2</sub> be offset at a "generator" of the even-regular scale, which by Theorem 6 has the same dyad class as a bright generator of the MOS 2aX 2cZ. This is what induces the equality of ''s''<sub>1</sub> and ''s''<sub>2</sub>: 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. When this mode is "darkened" by one generator, we turn that slice into the bright generator, hence swapping ''s''<sub>''t''</sub>[-''g''] and s_T[-''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 that's within a stacked 2-step. While the generator might have to be higher by one period of the template MOS, that doesn't affect the parity since the period is even. In particilar, the two scales have the same period, thus they are both primitive.
 We prove that ''s''<sub>1</sub> and ''s''<sub>2</sub> are MOS substitution scales with a filling MOS of period 2. The number the 2-step (1) occurs must be the same in both ''s''<sub>1</sub> and ''s''<sub>2</sub>. The word of stacked 2-steps of the template MOS (which is of the form ''w''('''x''', '''X''', '''X''')''w''('''x''', '''X''', '''X''')), which is itself a MOS word, consists of letters (1) '''x''' + '''X''' and (2) 2'''X''' if more '''X''''s than '''x''''s, 2'''x''' if more '''x''''s than '''X''''s. The word of stacked 2-steps from the n/2-degree is also this same MOS word.