Generator-offset property: Difference between revisions

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# If |S| is odd, then S = aX bY bZ is obtained from the (single-period) mos aX 2bW 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).

# The two alternants differ by replacing one Y with a Z.

# ''S'' is ''pairwise-mos'' (PMOS). That is, the result of identifying any two step sizes of ''S'' is always a mos.

# ''S'' is ''monotone-mos'' (MMOS). That is, equating L = M, equating M = s, and equating s = 0 each results in a mos.

[Note: This is not true with SGA replaced with GO; [[blackdye]] is a counterexample that is MV4.]

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In aX bY bZ, consider (i+j)-steps (representing the generator) with: (1) the maximum possible number of Y’s (at least 2 more than the # of Z's), (2) the maximum maximum number of Z’s (at least 2 more than the # of Y's), or (3) an intermediate number of Y’s and Z’s between the two or (4) (the preimage of) the imperfect generator. Since a + 2b >= 5, are at least 4 perfect generators, so there must be at least one of each of (1), (2), and (3), giving a contradiction to MV3. [Let T be the subword consisting only of Y's and Z's. If T has a substring of length j that's not contained in a perfect generator, you can go somewhere else to find it, since the numbers of Y's and Z's change one at a time and reach a maximum and a minimum somewhere, guaranteeing that intermediate values are reached "on the other side".]

The generator of aX 2bW must have an odd number of W steps; if it had an even number of W steps, it would be generated by stacking the generator of the mos aX bW' with W' = 2W, a contradiction. This with (4) immediately gives (5). <math>\square</math>

The generator of aX 2bW must have an odd number of W steps; if it had an even number of W steps, it would be generated by stacking the generator of the mos aX bW' with W' = 2W, a contradiction. This with (4) immediately gives (5).

For (6) and (7), We know SGA scales are MV3 and that MV3s project to mosses when one removes all instances of one step size, in particular s. Thus it suffices to prove that the result of equating any two step sizes is a mos. Equating Y and Z equates the swung alternants, resulting in the mos aX 2bW (this is a known fact).

It remains to prove that equating X and Y equates the generator with the difference between the start of the 2nd chain with the end of the 1st one or vice versa; similarly for equating X and Z. Consider the swung-generator-alternant chain g1 g2 .... g1 g2 g3. Assume that the offset g1 = δ has one more Y (and one fewer Z) than g2 = g - δ. g3 (which becomes the imperfect generator of aX 2bW) either has one more X or one fewer X than both g1 and g2. If g3 has one more X, then g3 + g1 is equated to g1 + g2 = g, as desired. If g3 has one fewer X, then simply invert the generator class. <math>\square</math>

=== Proposition 2 ===

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# is not of the form ''mx my mz'',

then it is of the form ''W''(''x'', ''y'', ''z'')''W''(''y'', ''x'', ''z'') for some word ''W'' in 3 variables.

~~=== Conjecture 4 ===~~

~~An SGA scale is always ''monotonic-well-formed'' (MWF). That is, the results of equating L with m, equating m with s, and setting s = 0 are all mosses~~

~~This would follow from the fact that MV3s are LQ (thus project to mosses upon eliminating one step size), and Conjecture 5.~~

~~=== Conjecture 5 ===~~

~~An SGA scale is always pairwise-well-formed (PWF). That is, the result of identifying any two step sizes of an SGA scale is always a non-multiperiod mos.~~

~~==== Proof ====~~

Suppose the scale is aX bY bZ. We know SGA scales are MV3. We know that MV3s project to mosses when one removes all instances of one step size, in particular s. Thus it suffices to prove that the result of equating any two step sizes is a mos. Equating Y and Z equates the swung alternants, resulting in the mos aX 2bW (this is a known fact).

~~It remains to prove that equating X and Y equates the generator with the difference between the start of the 2nd chain with the end of the 1st one or vice versa; similarly for equating X and Z.~~

Consider the swung-generator-alternant chain g1 g2 .... g1 g2 g3. Assume that the offset g1 = δ has one more Y (and one fewer Z) than g2 = g - δ. g3 (which becomes the imperfect generator of aX 2bW) either has one more X or one fewer X than both g1 and g2. If g3 has one more X, then g3 + g1 is equated to g1 + g2 = g, as desired. If g3 has one fewer X, then simply invert the generator class.

== Falsified conjectures ==

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 # If |S| is odd, then S = aX bY bZ is obtained from the (single-period) mos aX 2bW 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).
 # The two alternants differ by replacing one Y with a Z.
+# ''S'' is ''pairwise-mos'' (PMOS). That is, the result of identifying any two step sizes of ''S'' is always a mos.
+# ''S'' is ''monotone-mos'' (MMOS). That is, equating L = M, equating M = s, and equating s = 0 each results in a mos.
 [Note: This is not true with SGA replaced with GO; [[blackdye]] is a counterexample that is MV4.]
@@ Line 76: / Line 78: @@
 In aX bY bZ, consider (i+j)-steps (representing the generator) with: (1) the maximum possible number of Y’s (at least 2 more than the # of Z's), (2) the maximum maximum number of Z’s (at least 2 more than the # of Y's), or (3) an intermediate number of Y’s and Z’s between the two or (4) (the preimage of) the imperfect generator. Since a + 2b >= 5, are at least 4 perfect generators, so there must be at least one of each of (1), (2), and (3), giving a contradiction to MV3. [Let T be the subword consisting only of Y's and Z's. If T has a substring of length j that's not contained in a perfect generator, you can go somewhere else to find it, since the numbers of Y's and Z's change one at a time and reach a maximum and a minimum somewhere, guaranteeing that intermediate values are reached "on the other side".]
-The generator of aX 2bW must have an odd number of W steps; if it had an even number of W steps, it would be generated by stacking the generator of the mos aX bW' with W' = 2W, a contradiction. This with (4) immediately gives (5). <math>\square</math>
+The generator of aX 2bW must have an odd number of W steps; if it had an even number of W steps, it would be generated by stacking the generator of the mos aX bW' with W' = 2W, a contradiction. This with (4) immediately gives (5).
+For (6) and (7), We know SGA scales are MV3 and that MV3s project to mosses when one removes all instances of one step size, in particular s. Thus it suffices to prove that the result of equating any two step sizes is a mos. Equating Y and Z equates the swung alternants, resulting in the mos aX 2bW (this is a known fact).
+It remains to prove that equating X and Y equates the generator with the difference between the start of the 2nd chain with the end of the 1st one or vice versa; similarly for equating X and Z. Consider the swung-generator-alternant chain g1 g2 .... g1 g2 g3. Assume that the offset g1 = δ has one more Y (and one fewer Z) than g2 = g - δ. g3 (which becomes the imperfect generator of aX 2bW) either has one more X or one fewer X than both g1 and g2. If g3 has one more X, then g3 + g1 is equated to g1 + g2 = g, as desired. If g3 has one fewer X, then simply invert the generator class. <math>\square</math>
 === Proposition 2 ===
@@ Line 105: / Line 111: @@
 # is not of the form ''mx my mz'',
 then it is of the form ''W''(''x'', ''y'', ''z'')''W''(''y'', ''x'', ''z'') for some word ''W'' in 3 variables.
-=== Conjecture 4 ===
-An SGA scale is always ''monotonic-well-formed'' (MWF). That is, the results of equating L with m, equating m with s, and setting s = 0 are all mosses
-This would follow from the fact that MV3s are LQ (thus project to mosses upon eliminating one step size), and Conjecture 5.
-=== Conjecture 5 ===
-An SGA scale is always pairwise-well-formed (PWF). That is, the result of identifying any two step sizes of an SGA scale is always a non-multiperiod mos.
-==== Proof ====
-Suppose the scale is aX bY bZ. We know SGA scales are MV3. We know that MV3s project to mosses when one removes all instances of one step size, in particular s. Thus it suffices to prove that the result of equating any two step sizes is a mos. Equating Y and Z equates the swung alternants, resulting in the mos aX 2bW (this is a known fact).
-It remains to prove that equating X and Y equates the generator with the difference between the start of the 2nd chain with the end of the 1st one or vice versa; similarly for equating X and Z.
-Consider the swung-generator-alternant chain g1 g2 .... g1 g2 g3. Assume that the offset g1 = δ has one more Y (and one fewer Z) than g2 = g - δ. g3 (which becomes the imperfect generator of aX 2bW) either has one more X or one fewer X than both g1 and g2. If g3 has one more X, then g3 + g1 is equated to g1 + g2 = g, as desired. If g3 has one fewer X, then simply invert the generator class.
 == Falsified conjectures ==