Ternary scale theorems: Difference between revisions

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* 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.i. Recall that ''p'' is the unique 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'']}}. 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''.

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* 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'']}}, such that {{nowrap|''w''[''r'']}} is the non-'''X''' that corresponds to {{nowrap|''U''[''q'']}} under step deletion. 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.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'']}} such that {{nowrap|''w''[''r'']}} is the non-'''X''' that corresponds to {{nowrap|''U''[''q'']}} under step deletion. 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 {{nowrap|2''b''-th}} ''j''-step window differ due to parity. Since {{nowrap|gcd(''j'', 2''b'') {{=}} 1}}, this visits every note of ''U''.

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 * 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'' &minus; 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.i. Recall that ''p'' is the unique 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'']}}. 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''.
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 * 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'']}}, such that {{nowrap|''w''[''r'']}} is the non-'''X''' that corresponds to {{nowrap|''U''[''q'']}} under step deletion. 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.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'']}} such that {{nowrap|''w''[''r'']}} is the non-'''X''' that corresponds to {{nowrap|''U''[''q'']}} under step deletion. 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 {{nowrap|2''b''-th}} ''j''-step window differ due to parity. Since {{nowrap|gcd(''j'', 2''b'') {{=}} 1}}, this visits every note of ''U''.