Generator-offset property: Difference between revisions

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===== Statement (3) =====

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.

===== Statement 4 =====

===== Statement (4) =====

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.

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'''Claim 1''': Deleting X's from the alternants of ''S'' gives every ''j''-step in the scale ''E''X(''S'')(Y, Z), the scale word obtained by deleting all X's from X.

Assume that the imperfect generator of ''T'' has ''j'' + 1 W's and the perfect generator has ''j'' W's. Suppose that one ''j''-step word ''R'' on note ''p'' of ''E''X(''S'') is "contained in" the corresponding "imperfect alternant" of ''S'', the unique (''i'' + ''j'')-step, which is ''I'' = ''S''[''p'' : ''p'' + ''i'' + ''j'']. By this we mean that ''E''X(''I'') has ''R'' as a substring. Then ''S''[''p'' - 1: ''p'' - 1 + ''i'' + ''j''] and ''S''[''p'' + 1 : ''p'' + 1 + ''i'' + ''j''] are both perfect, and have one fewer step that is Y or Z. Thus the word ''I'' must both begin and end in a letter that is either Y or Z. Removing all the X's from ''I'' results in a word that is ''j'' + 1 letters long and is the ''j''-step we started with, with just one extra letter. Thus one of the two perfect generators above, namely the one that removes the extra letter, must contain this ''j''-step.

Proof: Assume that the imperfect generator of ''T'' has ''j'' + 1 W's and the perfect generator has ''j'' W's. Suppose that one ''j''-step word ''R'' on note ''p'' of ''E''X(''S'') is "contained in" the corresponding "imperfect alternant" of ''S'', the unique (''i'' + ''j'')-step, which is ''I'' = ''S''[''p'' : ''p'' + ''i'' + ''j'']. By this we mean that ''E''X(''I'') has ''R'' as a substring. Then ''S''[''p'' - 1: ''p'' - 1 + ''i'' + ''j''] and ''S''[''p'' + 1 : ''p'' + 1 + ''i'' + ''j''] are both perfect, and have one fewer step that is Y or Z. Thus the word ''I'' must both begin and end in a letter that is either Y or Z. Removing all the X's from ''I'' results in a word that is ''j'' + 1 letters long and is the ''j''-step we started with, with just one extra letter. Thus one of the two perfect generators above, namely the one that removes the extra letter, must contain this ''j''-step.

~~Proof: Every (''i'' + ''j'')-step in S except one involves *a* j-step in ''E''X(''S'')(Y, Z)~~.

'''Claim 2''': If a scale ''U'' has ''k'' X's and ''k'' Y's, gcd(''j'', 2''k'') = 1, and consecutively stacked ''j''-steps occur in 2 alternating sizes, the ''k'' Y's and ''k'' Z's alternate.

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These two claims prove that ''E''X(S) = (YZ)''k'', which also proves that the two alternants' sizes differ by replacing one Y for a Z.

===== Statement (5) =====

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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 ===== Statement (3) =====
 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.
-===== Statement 4 =====
+===== Statement (4) =====
 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.
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 '''Claim 1''': Deleting X's from the alternants of ''S'' gives every ''j''-step in the scale ''E''<sub>X</sub>(''S'')(Y, Z), the scale word obtained by deleting all X's from X.
-Assume that the imperfect generator of ''T'' has ''j'' + 1 W's and the perfect generator has ''j'' W's. Suppose that one ''j''-step word ''R'' on note ''p'' of ''E''<sub>X</sub>(''S'') is "contained in" the corresponding "imperfect alternant" of ''S'', the unique (''i'' + ''j'')-step, which is ''I'' = ''S''[''p'' : ''p'' + ''i'' + ''j'']. By this we mean that ''E''<sub>X</sub>(''I'') has ''R'' as a substring. Then ''S''[''p'' - 1: ''p'' - 1 + ''i'' + ''j''] and ''S''[''p'' + 1 : ''p'' + 1 + ''i'' + ''j''] are both perfect, and have one fewer step that is Y or Z. Thus the word ''I'' must both begin and end in a letter that is either Y or Z. Removing all the X's from ''I'' results in a word that is ''j'' + 1 letters long and is the ''j''-step we started with, with just one extra letter. Thus one of the two perfect generators above, namely the one that removes the extra letter, must contain this ''j''-step.
+Proof: Assume that the imperfect generator of ''T'' has ''j'' + 1 W's and the perfect generator has ''j'' W's. Suppose that one ''j''-step word ''R'' on note ''p'' of ''E''<sub>X</sub>(''S'') is "contained in" the corresponding "imperfect alternant" of ''S'', the unique (''i'' + ''j'')-step, which is ''I'' = ''S''[''p'' : ''p'' + ''i'' + ''j'']. By this we mean that ''E''<sub>X</sub>(''I'') has ''R'' as a substring. Then ''S''[''p'' - 1: ''p'' - 1 + ''i'' + ''j''] and ''S''[''p'' + 1 : ''p'' + 1 + ''i'' + ''j''] are both perfect, and have one fewer step that is Y or Z. Thus the word ''I'' must both begin and end in a letter that is either Y or Z. Removing all the X's from ''I'' results in a word that is ''j'' + 1 letters long and is the ''j''-step we started with, with just one extra letter. Thus one of the two perfect generators above, namely the one that removes the extra letter, must contain this ''j''-step.
-Proof: Every (''i'' + ''j'')-step in S except one involves *a* j-step in ''E''<sub>X</sub>(''S'')(Y, Z).
 '''Claim 2''': If a scale ''U'' has ''k'' X's and ''k'' Y's, gcd(''j'', 2''k'') = 1, and consecutively stacked ''j''-steps occur in 2 alternating sizes, the ''k'' Y's and ''k'' Z's alternate.
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 These two claims prove that ''E''<sub>X</sub>(S) = (YZ)<sup>''k''</sup>, which also proves that the two alternants' sizes differ by replacing one Y for a Z.
 ===== Statement (5) =====
 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: