Generator-offset property: Difference between revisions

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=== Proposition 1 (Properties of SGA scales) ===

Let ''S'' be a 3-step-size scale word in L, M, and s of length ''n'', and suppose ''S'' is SGA. Then:

# The length of ''S'' is odd, or ''S'' is equivalent to xyxz.

# ''S'' is of the form ''a''x ''b''y ''b''z for some permutation (x, y, z) of (L, M, s).

# If ''n'' is odd, ''S'' is abstractly SV3 (i.e. SV3 for almost all tunings).

~~# ''S'' is of the form ''a''x ''b''y ''b''z for some permutation (x, y, z) of (L, M, s).~~

~~# The length of ''S'' is odd, or ''S'' is equivalent to xyxz.~~

# If ''n'' is odd, ''S'' = ''a''X ''b''Y ''b''Z is obtained from some mode of the (single-period) mos ''a''X 2''b''W by replacing all the W's successively with alternating Y's and Z's (or alternating Z's and Y's for the other chirality, fixing the mode of ''a''X 2''b''W). The two alternants differ by replacing one Y with a Z.

# ''S'' is pairwise-mos. That is, the following operations each result in a mos: setting L = M, setting L = s, and setting M = s.

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# a4 − a2 = g1 − 2 g2 + g3 = (g3 − g2) + (g1 − g2) = (chroma ± ε) ≠ 0 by choice of tuning.

By applying this argument to 1-steps, we see that there must be 4 step sizes in some tuning, a contradiction. Thus g1 and g2 must themselves be step sizes. Thus we see that an even-length, unconditionally SV3, GO scale must be of the form xy...xyxz. But this pattern is not abstractly SV3 if ''n'' ≥ 6, since 3-steps come in 4 sizes: xyx, yxy, yxz and xzx. Thus ''n'' = 4 and the scale is xyxz. (Note that xyxz is not SV3, since it has only two kinds of 2-steps, xy and xz.) This proves (3).

By applying this argument to 1-steps, we see that there must be 4 step sizes in some tuning, a contradiction. Thus g1 and g2 must themselves be step sizes. Thus we see that an even-length, unconditionally SV3, GO scale must be of the form xy...xyxz. But this pattern is not abstractly SV3 if ''n'' ≥ 6, since 3-steps come in 4 sizes: xyx, yxy, yxz and xzx. Thus ''n'' = 4 and the scale is xyxz. (Note that xyxz is not SV3, since it has only two kinds of 2-steps, xy and xz.) This proves (1).

In case 2, let (2, 1) − (1, 1) = g1, (1, 2) − (2, 1) = g2 be the two alternants. Let g3 be the leftover generator after stacking alternating g1 and g2. Then the generator circle looks like g1 g2 g1 g2 ... g1 g2 g3. Then the combinations of alternants corresponding to a step come in exactly 3 sizes:

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(The above holds for any odd ''n'' ≥ 3.)

For (1), we now only need to see that if ''S'' has an odd number of notes and is SGA, ''S'' is abstractly SV3. But the argument in case 2 above works when you substitute any interval class in ''S'' instead of a 1-step (abstract SV3 wasn't used), hence any interval class comes in (abstractly) exactly 3 sizes.

For (3), we now only need to see that if ''S'' has an odd number of notes and is SGA, ''S'' is abstractly SV3. But the argument in case 2 above works when you substitute any interval class in ''S'' instead of a 1-step (abstract SV3 wasn't used), hence any interval class comes in (abstractly) exactly 3 sizes.

For (4), assume ''S'' is ''a''X ''b''Y ''b''Z, a odd. If ''b'' = 1, there's nothing to prove. So assume ''b'' > 1. Suppose for the sake of contradiction that Y′s and Z′s don't alternate perfectly. Assume that the perfect generator of ''a''X 2''b''W is ''i''X + ''j''W with ''j'' ≥ 2. (If ''j'' = 1, we can invert the generator to make ''j'' ≥ 2, since ''b'' > 1.)

@@ Line 30: / Line 30: @@
 === Proposition 1 (Properties of SGA scales) ===
 Let ''S'' be a 3-step-size scale word in L, M, and s of length ''n'', and suppose ''S'' is SGA. Then:
+# The length of ''S'' is odd, or ''S'' is equivalent to xyxz.
+# ''S'' is of the form ''a''x ''b''y ''b''z for some permutation (x, y, z) of (L, M, s).
 # If ''n'' is odd, ''S'' is abstractly SV3 (i.e. SV3 for almost all tunings).
-# ''S'' is of the form ''a''x ''b''y ''b''z for some permutation (x, y, z) of (L, M, s).
-# The length of ''S'' is odd, or ''S'' is equivalent to xyxz.
 # If ''n'' is odd, ''S'' = ''a''X ''b''Y ''b''Z is obtained from some mode of the (single-period) mos ''a''X 2''b''W by replacing all the W's successively with alternating Y's and Z's (or alternating Z's and Y's for the other chirality, fixing the mode of ''a''X 2''b''W). The two alternants differ by replacing one Y with a Z.
 # ''S'' is pairwise-mos. That is, the following operations each result in a mos: setting L = M, setting L = s, and setting M = s.
@@ Line 66: / Line 66: @@
 # a<sub>4</sub> &minus; a<sub>2</sub> = g<sub>1</sub> &minus; 2 g<sub>2</sub> + g<sub>3</sub> = (g<sub>3</sub> &minus; g<sub>2</sub>) + (g<sub>1</sub> &minus; g<sub>2</sub>) = (chroma ± ε) ≠ 0 by choice of tuning.
-By applying this argument to 1-steps, we see that there must be 4 step sizes in some tuning, a contradiction. Thus g<sub>1</sub> and g<sub>2</sub> must themselves be step sizes. Thus we see that an even-length, unconditionally SV3, GO scale must be of the form xy...xyxz. But this pattern is not abstractly SV3 if ''n'' ≥ 6, since 3-steps come in 4 sizes: xyx, yxy, yxz and xzx. Thus ''n'' = 4 and the scale is xyxz. (Note that xyxz is not SV3, since it has only two kinds of 2-steps, xy and xz.) This proves (3).
+By applying this argument to 1-steps, we see that there must be 4 step sizes in some tuning, a contradiction. Thus g<sub>1</sub> and g<sub>2</sub> must themselves be step sizes. Thus we see that an even-length, unconditionally SV3, GO scale must be of the form xy...xyxz. But this pattern is not abstractly SV3 if ''n'' ≥ 6, since 3-steps come in 4 sizes: xyx, yxy, yxz and xzx. Thus ''n'' = 4 and the scale is xyxz. (Note that xyxz is not SV3, since it has only two kinds of 2-steps, xy and xz.) This proves (1).
 In case 2, let (2, 1) &minus; (1, 1) = g<sub>1</sub>, (1, 2) &minus; (2, 1) = g<sub>2</sub> be the two alternants. Let g<sub>3</sub> be the leftover generator after stacking alternating g<sub>1</sub> and g<sub>2</sub>. Then the generator circle looks like g<sub>1</sub> g<sub>2</sub> g<sub>1</sub> g<sub>2</sub> ... g<sub>1</sub> g<sub>2</sub> g<sub>3</sub>. Then the combinations of alternants corresponding to a step come in exactly 3 sizes:
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 (The above holds for any odd ''n'' ≥ 3.)
-For (1), we now only need to see that if ''S'' has an odd number of notes and is SGA, ''S'' is abstractly SV3. But the argument in case 2 above works when you substitute any interval class in ''S'' instead of a 1-step (abstract SV3 wasn't used), hence any interval class comes in (abstractly) exactly 3 sizes.
+For (3), we now only need to see that if ''S'' has an odd number of notes and is SGA, ''S'' is abstractly SV3. But the argument in case 2 above works when you substitute any interval class in ''S'' instead of a 1-step (abstract SV3 wasn't used), hence any interval class comes in (abstractly) exactly 3 sizes.
 For (4), assume ''S'' is ''a''X ''b''Y ''b''Z, a odd. If ''b'' = 1, there's nothing to prove. So assume ''b'' > 1. Suppose for the sake of contradiction that Y′s and Z′s don't alternate perfectly. Assume that the perfect generator of ''a''X 2''b''W is ''i''X + ''j''W with ''j'' ≥ 2. (If ''j'' = 1, we can invert the generator to make ''j'' ≥ 2, since ''b'' > 1.)