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 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 (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. 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 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 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 (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. 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.

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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First we handle the ''a'' = 2''k'' case. This means that ''k'' = gcd(''a'', ''k'') = 1, and ''a'' = 2. This is the trivial ''n'' = 4 ('''xyxz''') case. Thus it remains to handle the cases (1) and (2) above. The steps of a MOS are distributed maximally evenly. Hence the positions occupied by '''x''' + '''X''' are a maximally even subset of <math>\mathbb{Z}/(n/2)\mathbb{Z},</math> and so are the positions occupied by 2'''X''' resp. 2'''x'''. Whenever the letter '''x''' + '''X''' is encountered, the number of the last letters that are equated to '''X''' that are consumed is 1, which is odd. Whenever the other letter is encountered, that number is even (0 or 2). Hence the letter 2'''X''' resp. 2'''x''' serves as the non-slot letter, and the letters ('''x''' + '''X''') serve as the slot letters where a 2-period filling MOS word (a repetition of ('''x'''+'''y''')('''x'''+'''z''') is substituted.

Now we count the letters that occur in these MOS substitution words of 2-steps~~. We wish to show that ''s'' altogether has doubly evenly many 2-steps that are slot letters~~. Consider the chunk boundaries of the template MOS. For every boundary between chunks, there is one slot letter in the template MOS for ''s''''1'' and one in the template MOS ''s''''2'', due to parity. So it suffices that we have evenly many boundaries between (nonempty) chunks. Equivalently, we have to prove that there are evenly many steps in the template MOS ''a'''''x''' 2''k'''''X''', which is true by assumption (''a'' is even, 2''k'' is even).

Now we count the letters that occur in these MOS substitution words of 2-steps. Consider the chunk boundaries of the template MOS. For every boundary between chunks, there is one slot letter in the template MOS for ''s''''1'' and one in the template MOS ''s''''2'', due to parity. So it suffices that we have evenly many boundaries between (nonempty) chunks. Equivalently, we have to prove that there are evenly many steps in the template MOS ''a'''''x''' 2''k'''''X''', which is true by assumption (''a'' is even, 2''k'' is even).

* In the singly even case, since there are evenly many slot letters in both ''s''1 and ''s''2, there are oddly many non-slot letters in both. Since ''s''1 and ''s''2 differ by interchanging '''y''' and '''z''', they have "opposite" filling letters, '''x''' + '''y''' being the opposite of '''x''' + '''z'''. This makes ''s''1 and ''s''2 opposite chiralities of an odd-regular MV3 scale.

* In the doubly even case, the number of non-slot letters in ''s''1 and ''s''2 is even, and we have a filling MOS of period 2. Since ''s''1 and ''s''2 are both primitive, they are both even-regular scales. {{Qed}}

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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 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> (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. 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 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 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> (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. 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.
 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.
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 First we handle the ''a'' = 2''k'' case. This means that ''k'' = gcd(''a'', ''k'') = 1, and ''a'' = 2. This is the trivial ''n'' = 4 ('''xyxz''') case. Thus it remains to handle the cases (1) and (2) above. The steps of a MOS are distributed maximally evenly. Hence the positions occupied by '''x''' + '''X''' are a maximally even subset of <math>\mathbb{Z}/(n/2)\mathbb{Z},</math> and so are the positions occupied by 2'''X''' resp. 2'''x'''. Whenever the letter '''x''' + '''X''' is encountered, the number of the last letters that are equated to '''X''' that are consumed is 1, which is odd. Whenever the other letter is encountered, that number is even (0 or 2). Hence the letter 2'''X''' resp. 2'''x''' serves as the non-slot letter, and the letters ('''x''' + '''X''') serve as the slot letters where a 2-period filling MOS word (a repetition of ('''x'''+'''y''')('''x'''+'''z''') is substituted.
-Now we count the letters that occur in these MOS substitution words of 2-steps. We wish to show that ''s'' altogether has doubly evenly many 2-steps that are slot letters. Consider the chunk boundaries of the template MOS. For every boundary between chunks, there is one slot letter in the template MOS for ''s''<sub>''1''</sub> and one in the template MOS ''s''<sub>''2''</sub>, due to parity. So it suffices that we have evenly many boundaries between (nonempty) chunks. Equivalently, we have to prove that there are evenly many steps in the template MOS ''a'''''x''' 2''k'''''X''', which is true by assumption (''a'' is even, 2''k'' is even).
+Now we count the letters that occur in these MOS substitution words of 2-steps. Consider the chunk boundaries of the template MOS. For every boundary between chunks, there is one slot letter in the template MOS for ''s''<sub>''1''</sub> and one in the template MOS ''s''<sub>''2''</sub>, due to parity. So it suffices that we have evenly many boundaries between (nonempty) chunks. Equivalently, we have to prove that there are evenly many steps in the template MOS ''a'''''x''' 2''k'''''X''', which is true by assumption (''a'' is even, 2''k'' is even).
 * In the singly even case, since there are evenly many slot letters in both ''s''<sub>1</sub> and ''s''<sub>2</sub>, there are oddly many non-slot letters in both. Since ''s''<sub>1</sub> and ''s''<sub>2</sub> differ by interchanging '''y''' and '''z''', they have "opposite" filling letters, '''x''' + '''y''' being the opposite of '''x''' + '''z'''. This makes ''s''<sub>1</sub> and ''s''<sub>2</sub> opposite chiralities of an odd-regular MV3 scale.
 * In the doubly even case, the number of non-slot letters in ''s''<sub>1</sub> and ''s''<sub>2</sub> is even, and we have a filling MOS of period 2. Since ''s''<sub>1</sub> and ''s''<sub>2</sub> are both primitive, they are both even-regular scales. {{Qed}}

Ternary scale theorems: Difference between revisions

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