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''.

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

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

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

# (a) the preimage with 2 Y's

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

# (b) the preimage with 2 Z's

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

# (c) the preimage with 1 Y and 1 Z

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

# (d) the preimage of the interval 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.

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).

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:

@@ Line 79: / Line 79: @@
 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''.
-In ''S'', consider ''k''-steps where ''k'' is chosen so one size has two W's. They have the following sizes:
+In ''S'', consider ''k''-steps where ''k'' is chosen so one size has two W's. [incomplete]<!-- They have either the following sizes:
-# (a) the preimage of the perfect generator with 2 Y's
+# (a) the preimage with 2 Y's
-# (b) the preimage of the perfect generator with 2 Z's
+# (b) the preimage with 2 Z's
-# (c) the preimage of the perfect generator with 1 Y and 1 Z
+# (c) the preimage with 1 Y and 1 Z
-# (d) (the preimage of) the imperfect generator, having a different number of X's than (a), (b), and (c).
+# (d) the preimage of the interval 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.
-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).
+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: