Ternary scale theorems: Difference between revisions

Line 359:

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, start from the mode of ''s'' whose template MOS is the brightest mode. Let ''s''2 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''1 and ''s''2 (in particular, the two scales have the same period, thus they are both primitive): 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. Now when we "darken" the mode by one generator, which is the difference between the template MOSes of ''s''1 and ''s''2, 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 the letters are 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 the doubly even case, start from the mode of ''s'' whose template MOS is the brightest mode. Let ''s''2 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''1 and ''s''2 (in particular, the two scales have the same period, thus they are both primitive): 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. Now when we "darken" the mode by one generator, which is the difference between the template MOSes of ''s''1 and ''s''2, 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 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.

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 our chosen offset is also this same MOS word.

@@ Line 359: / Line 359: @@
 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, 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 one period of the template MOS, that doesn't affect the parity since the period 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.
 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 our chosen offset is also this same MOS word.