Generator-offset property: Difference between revisions

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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]~~

In ''S'', consider ''k''-steps where ''k'' is chosen so one size has two W's. They have the following sizes:

~~<!--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.)~~

# (a) the preimage of the perfect generator with 2 Y's

# (b) the preimage of the perfect generator with 2 Z's

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

# (c) the preimage of the perfect generator with 1 Y and 1 Z

# (a) the preimage of the perfect generator with ~~the maximum number of~~ Y's ~~(at least 2 more than the # of Z's)~~

# (b) the preimage of the perfect generator with ~~the maximum number of~~ Z's ~~(at least 2 more than the # of Y's)~~

# (c) the preimage of the perfect generator with ~~an intermediate number of~~ Y's and Z~~'s between (a) and (b)~~

# (d) (the preimage of) the imperfect generator, having a different number of X's than (a), (b), and (c).

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.]

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.

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.) This implies one of the alternants must have one more Y and one fewer Z than the other. 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:

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 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]
+In ''S'', consider ''k''-steps where ''k'' is chosen so one size has two W's. They have the following sizes:
-<!--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.)
+# (a) the preimage of the perfect generator with 2 Y's
+# (b) the preimage of the perfect generator with 2 Z's
-In ''S'', (''i'' + ''j'')-steps (representing the generator) are always one of the following:
+# (c) the preimage of the perfect generator with 1 Y and 1 Z
-# (a) the preimage of the perfect generator with the maximum number of Y's (at least 2 more than the # of Z's)
-# (b) the preimage of the perfect generator with the maximum number of Z's (at least 2 more than the # of Y's)
-# (c) the preimage of the perfect generator with an intermediate number of Y's and Z's between (a) and (b)
 # (d) (the preimage of) the imperfect generator, having a different number of X's than (a), (b), and (c).
-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.]
+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.
-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.) This implies one of the alternants must have one more Y and one fewer Z than the other. 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: