Ternary scale theorems: Difference between revisions
Tags: Mobile edit Mobile web edit |
Tags: Mobile edit Mobile web edit |
||
| Line 358: | Line 358: | ||
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. | 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 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 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''<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 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''<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. | ||