Generator-offset property: Difference between revisions

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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, i.e. YX<sup>''t''</sup>Y (for some ''t'' ≥ 0) occurs in ''S''. 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.)

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, i.e. YX<sup>''t''</sup>Y (for some ''t'' ≥ 0) occurs in ''S''.

[correction pending]

<!--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.)

In ''S'', (''i'' + ''j'')-steps (representing the generator) are always one of the following:

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Since ''a'' + 2''b'' ≥ 5, there are at least 4 perfect generators, so there must be at least one of each of (a), (b), and (c), giving a contradiction to SV3. [Whenever the root of the (''i'' + ''j'')-step are moved within ''S'', the numbers of Y's and Z's change one at a time and reach a maximum at some choice of the root and a minimum with another choice, guaranteeing that intermediate values are reached. We can use this "intermediate value theorem" argument because (d) occurs at only one note.]

Any generator of ''a''X 2''b''W must have an odd number of W steps (Otherwise, intervals with an odd number of W steps can't be generated.). We have finished proving (4).

Any generator of ''a''X 2''b''W must have an odd number of W steps (Otherwise, intervals with an odd number of W steps can't be generated.). We have finished proving (4).-->

For (5), odd-numbered SGA scales are [[Fokker block]]s (in the 2-dimensional lattice generated by the generator and the offset). To see this, consider the following lattice depiction of such a scale:

@@ Line 77: / Line 77: @@
 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, i.e. YX<sup>''t''</sup>Y (for some ''t'' &ge; 0) occurs in ''S''. 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.)
+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, i.e. YX<sup>''t''</sup>Y (for some ''t'' &ge; 0) occurs in ''S''.
+[correction pending]
+<!--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.)
 In ''S'', (''i'' + ''j'')-steps (representing the generator) are always one of the following:
@@ Line 86: / Line 89: @@
 Since ''a'' + 2''b'' ≥ 5, there are at least 4 perfect generators, so there must be at least one of each of (a), (b), and (c), giving a contradiction to SV3. [Whenever the root of the (''i'' + ''j'')-step are moved within ''S'', the numbers of Y's and Z's change one at a time and reach a maximum at some choice of the root and a minimum with another choice, guaranteeing that intermediate values are reached. We can use this "intermediate value theorem" argument because (d) occurs at only one note.]
-Any generator of ''a''X 2''b''W must have an odd number of W steps (Otherwise, intervals with an odd number of W steps can't be generated.). We have finished proving (4).
+Any generator of ''a''X 2''b''W must have an odd number of W steps (Otherwise, intervals with an odd number of W steps can't be generated.). We have finished proving (4).-->
 For (5), odd-numbered SGA scales are [[Fokker block]]s (in the 2-dimensional lattice generated by the generator and the offset). To see this, consider the following lattice depiction of such a scale: