Ternary scale theorems: Difference between revisions

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'''Claim 1''': Deleting '''X'''s from the generator subwords of ''s'' gives every ''j''-step subword in the scale ''E''X(''s'')('''Y''', '''Z'''), the scale word obtained by deleting all '''X''''s from ''s''. These ''j''-step subwords are adjacent and alternating.

Proof: ~~Consider~~ the ~~subword for~~ the ~~closing~~ generator of ''s'' ~~on some index~~ ''p'', ~~which~~ is {{nowrap|''I'' {{=}} ''s''[''p'' : ''p'' + ''i'' + ''j'']}}, ~~and suppose~~ the ~~result of deleting all~~ '''X''''~~s from~~ ''I'' ~~has a~~ ''j''~~-step subword~~ ''w''. ~~Shifting~~ ''I'' ~~one step to the left and one step~~ to the right, {{nowrap|''s''[''p'' − 1: ''p'' − 1 + ''i'' + ''j'']~~}} and~~ {{nowrap|''s''[''p'' ~~+ 1~~ : ''p'' + 1 + ''i'' + ''j'']}} ~~are both detemperings of~~ perfect ~~generators~~ of ''T'', and ~~have~~ one ~~fewer non-~~'''X''' ~~step than~~ ''I'' ~~by our assumption~~. ~~Thus~~ the ~~word~~ ''I'' ~~must both begin~~ and ~~end in a~~ non-'''X''' ~~letter~~. ~~Removing all the~~ '''X''''s ~~from~~ ''I'' ~~results in a word~~ that is {{nowrap|''j'' ~~+ 1~~}} ~~letters long and is~~ the ''j''-step ~~word~~ ''w'' ~~with just one extra letter appended~~. ~~Thus one of the two perfect generators above~~, ~~namely the one that removes the extra letter, must contain this~~ ''j''-step~~. The rest of the~~ ''j''-step ~~subwords of~~ ''s'' ~~can all be obtained by deleting~~ '''X'''~~s from detempered perfect generators; take~~ {{nowrap|''q'' ~~≠~~ ''p''}} ~~to be~~ the ~~index of the first letter of one such~~ ''j''-step ~~subword (as contained~~ in ''s''~~) and use~~ {{nowrap|''s''[''q'' : ''q'' + ''i'' + ''j'']}}.

Proof:

* A.1. Say that the generator of T has ''k'' steps.

Also note that we only need to stack {{nowrap|2''b'' ≤ ''n'' − 1}} generators to witness this alternation. Under the ordering induced by this stacking, the 1st ''j''-step subword of ''U'' and the 2''b''-th ''j''-step window differ due to parity. Since {{nowrap|gcd(''j'', 2''b'') {{=}} 1}}, this visits every note of ''U''.

* A.2.i. Each slice/occurrence of the (perfect) generator in the template MOS ''T'' contains a certain number of '''X''' steps, and the imperfect generator occurs only at one position. Call the unique imperfect position ''p''.

* A.2.ii. Say that the number of '''X''' steps in a ''perfect'' generator is ''i'', and the number of '''W''' steps in a ''perfect'' generator is ''j'', we have that {{nowrap|''k'' {{=}} ''i'' + ''j''.}}

* A.2.iii. We know from MOS theory that letter counts in ''k''-steps (for any fixed ''k'') differ by at most 1. Assume, possibly after taking the equave complement, that the imperfect generator has one ''more'' X: the imperfect generator has {{nowrap|(''i'' + 1)-many}} '''X''''s, and {{nowrap|(''j'' - 1)-many}} '''W''''s.

* A.3.i. Recall that ''p'' is the unique bad position, such that the ''k''-letter slice {{nowrap|''I'' {{=}} ''T''[''p'' : ''p'' + ''k'']}} abelianizes to the imperfect generator.

* A.3.ii. Scooting the slice ''I'' to the right yields {{nowrap|''I''''R'' :{{=}} ''T''[''p'' + 1 : ''p'' + 1 + ''k'']}}, also perfect. Since its abelianization is a perfect generator, ''I''''R'' has ''i''-many '''X''''s and j-many '''W''''s.

* A.3.iii. Since ''I''''R'' gains a '''W''' and loses an '''X''' relative to ''I'', the lost letter '''X''' is at the leftmost position of I's window, which is ''p''.

* A.3.iv. Conclusion: ''T''[''p''], the leftmost letter of {{nowrap|''I'' {{=}} ''T[''p'' : ''p'' + ''k''],}} is '''X'''.

* B.1. Now we go back to the original necklace ''w''. Lift each perfect generator window (we have {{nowrap|''n'' − 1}} perfect windows) of ''T'' to ''w''.

* B.2. By the hypothesis that ''w'' has an AS, and since the AS descends to stacking a single generator in the template MOS ''T'', the lifted generators ''g''1 and ''g''2 alternate in their counts of '''Y''' and also alternate in their counts of '''Z'''.

* B.3. For a MOS binary word, the count of a given letter in a generator is coprime to the total count of that letter in one period of the MOS. By this fact applied to ''T'', {{nowrap|gcd(''j'', 2''b'') {{=}} 1}}.

* B.4. Hence, since every instance of the generator in ''T'' has ''j''-many '''W''' letters, every instance of ''g''1 and every instance of ''g''2 has ''j''-many non-'''X''' letters.

* C.1. Importantly, deleting '''X''''s gives windows of length ''j'', such that when you project adjacent lifted generators (by deleting '''X''''s) to the binary necklace {{nowrap|''U'' = ''E'''''X'''(''w'')('''Y''', '''Z''')}}, the resulting ''j''-step windows in ''U'' are adjacent and do not overlap.

* C.2. Moreover, for every ''j''-step window {{nowrap|''U''[''q'' : ''q'' + ''j'']}}, there exists an {{nowrap|(''i'' + ''j'')-step}} window {{nowrap|''w''[''r'' : ''r'' + ''i'' + ''j'']}}, so that {{nowrap|''w''[''r'']}} is a non-'''X'''. Since by subclaim A, the unique imperfect {{nowrap|(''i'' + ''j'')-step}} window in ''w'' begins in an '''X''', we know that {{nowrap|''w''[''r'' : ''r'' + ''i'' + ''j'']}} is perfect.

* C.3. Also note that we only need to stack {{nowrap|2''b'' ≤ ''n'' − 1}} generators to witness this alternation. Under the ordering induced by this stacking, the 1st ''j''-step subword of ''U'' and the 2''b''-th ''j''-step window differ due to parity. Since {{nowrap|gcd(''j'', 2''b'') {{=}} 1}}, this visits every note of ''U''.

'''Claim 2''': If a binary necklace ''U'' has ''b'' '''Y'''s and ''b'' '''Z'''s, {{nowrap|gcd(''j'', 2''b'') {{=}} 1}}, and consecutively stacked ''j''-steps in ''U'' occur in 2 alternating sizes, then {{nowrap|''U'' {{=}} ('''YZ''')''b''}}.

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 '''Claim 1''': Deleting '''X'''s from the generator subwords of ''s'' gives every ''j''-step subword in the scale ''E''<sub>X</sub>(''s'')('''Y''',&nbsp;'''Z'''), the scale word obtained by deleting all '''X''''s from ''s''. These ''j''-step subwords are adjacent and alternating.
-Proof: Consider the subword for the closing generator of ''s'' on some index ''p'', which is {{nowrap|''I'' {{=}} ''s''[''p'' : ''p'' + ''i'' + ''j'']}}, and suppose the result of deleting all '''X''''s from ''I'' has a ''j''-step subword ''w''. Shifting ''I'' one step to the left and one step to the right, {{nowrap|''s''[''p'' − 1: ''p'' − 1 + ''i'' + ''j'']}} and {{nowrap|''s''[''p'' + 1 : ''p'' + 1 + ''i'' + ''j'']}} are both detemperings of perfect generators of ''T'', and have one fewer non-'''X''' step than ''I'' by our assumption. Thus the word ''I'' must both begin and end in a non-'''X''' letter. Removing all the '''X''''s from ''I'' results in a word that is {{nowrap|''j'' + 1}} letters long and is the ''j''-step word ''w'' with just one extra letter appended. Thus one of the two perfect generators above, namely the one that removes the extra letter, must contain this ''j''-step. The rest of the ''j''-step subwords of ''s'' can all be obtained by deleting '''X'''s from detempered perfect generators; take {{nowrap|''q'' &ne; ''p''}} to be the index of the first letter of one such ''j''-step subword (as contained in ''s'') and use {{nowrap|''s''[''q'' : ''q'' + ''i'' + ''j'']}}.
+Proof:
+* A.1. Say that the generator of T has ''k'' steps.
-Also note that we only need to stack {{nowrap|2''b'' &le; ''n'' &minus; 1}} generators to witness this alternation. Under the ordering induced by this stacking, the 1st ''j''-step subword of ''U'' and the 2''b''-th ''j''-step window differ due to parity. Since {{nowrap|gcd(''j'', 2''b'') {{=}} 1}}, this visits every note of ''U''.
+* A.2.i. Each slice/occurrence of the (perfect) generator in the template MOS ''T'' contains a certain number of '''X''' steps, and the imperfect generator occurs only at one position. Call the unique imperfect position ''p''.
+* A.2.ii. Say that the number of '''X''' steps in a ''perfect'' generator is ''i'', and the number of '''W''' steps in a ''perfect'' generator is ''j'', we have that {{nowrap|''k'' {{=}} ''i'' + ''j''.}}
+* A.2.iii. We know from MOS theory that letter counts in ''k''-steps (for any fixed ''k'') differ by at most 1. Assume, possibly after taking the equave complement, that the imperfect generator has one ''more'' X: the imperfect generator has {{nowrap|(''i'' + 1)-many}} '''X''''s, and {{nowrap|(''j'' - 1)-many}} '''W''''s.
+* A.3.i. Recall that ''p'' is the unique bad position, such that the ''k''-letter slice {{nowrap|''I'' {{=}} ''T''[''p'' : ''p'' + ''k'']}} abelianizes to the imperfect generator.
+* A.3.ii. Scooting the slice ''I'' to the right yields {{nowrap|''I''<sub>''R''</sub> :{{=}} ''T''[''p'' + 1 : ''p'' + 1 + ''k'']}}, also perfect. Since its abelianization is a perfect generator, ''I''<sub>''R''</sub> has ''i''-many '''X''''s and j-many '''W''''s.
+* A.3.iii. Since ''I''<sub>''R''</sub> gains a '''W''' and loses an '''X''' relative to ''I'', the lost letter '''X''' is at the leftmost position of <i>I</i>'s window, which is ''p''.
+* A.3.iv. Conclusion: ''T''[''p''], the leftmost letter of {{nowrap|''I'' {{=}} ''T[''p'' : ''p'' + ''k''],}} is '''X'''.
+* B.1. Now we go back to the original necklace ''w''. Lift each perfect generator window (we have {{nowrap|''n'' &minus; 1}} perfect windows) of ''T'' to ''w''.
+* B.2. By the hypothesis that ''w'' has an AS, and since the AS descends to stacking a single generator in the template MOS ''T'', the lifted generators ''g''<sub>1</sub> and ''g''<sub>2</sub> alternate in their counts of '''Y''' and also alternate in their counts of '''Z'''.
+* B.3. For a MOS binary word, the count of a given letter in a generator is coprime to the total count of that letter in one period of the MOS. By this fact applied to ''T'', {{nowrap|gcd(''j'', 2''b'') {{=}} 1}}.
+* B.4. Hence, since every instance of the generator in ''T'' has ''j''-many '''W''' letters, every instance of ''g''<sub>1</sub> and every instance of ''g''<sub>2</sub> has ''j''-many non-'''X''' letters.
+* C.1. Importantly, deleting '''X''''s gives windows of length ''j'', such that when you project adjacent lifted generators (by deleting '''X''''s) to the binary necklace {{nowrap|''U'' = ''E''<sub>'''X'''</sub>(''w'')('''Y''', '''Z''')}}, the resulting ''j''-step windows in ''U'' are adjacent and do not overlap.
+* C.2. Moreover, for every ''j''-step window {{nowrap|''U''[''q'' : ''q'' + ''j'']}}, there exists an {{nowrap|(''i'' + ''j'')-step}} window {{nowrap|''w''[''r'' : ''r'' + ''i'' + ''j'']}}, so that {{nowrap|''w''[''r'']}} is a non-'''X'''. Since by subclaim A, the unique imperfect {{nowrap|(''i'' + ''j'')-step}} window in ''w'' begins in an '''X''', we know that {{nowrap|''w''[''r'' : ''r'' + ''i'' + ''j'']}} is perfect.
+* C.3. Also note that we only need to stack {{nowrap|2''b'' &le; ''n'' &minus; 1}} generators to witness this alternation. Under the ordering induced by this stacking, the 1st ''j''-step subword of ''U'' and the 2''b''-th ''j''-step window differ due to parity. Since {{nowrap|gcd(''j'', 2''b'') {{=}} 1}}, this visits every note of ''U''.
 '''Claim 2''': If a binary necklace ''U'' has ''b'' '''Y'''s and ''b'' '''Z'''s, {{nowrap|gcd(''j'', 2''b'') {{=}} 1}}, and consecutively stacked ''j''-steps in ''U'' occur in 2 alternating sizes, then {{nowrap|''U'' {{=}} ('''YZ''')<sup>''b''</sup>}}.