Ternary scale theorems: Difference between revisions

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By part (2), we have that ''s'' has step signature {{nowrap|''a'''''X''' ''b'''''Y''' ''b'''''Z'''}}, ''a'' odd. By part (4), we have that {{nowrap|''T''('''X''', '''W''') {{=}} ''s''('''X''', '''W''', '''W''')}} is a MOS scale ''a'''''X'''2''b'''''W'''. If {{nowrap|''b'' {{=}} 1}}, there's nothing to prove, so assume {{nowrap|''b'' > 1}}.

Consider the two generators in the GS of ''s'', which are detemperings of the generator {{nowrap|''i'''''X''' + ''j'''''W'''}} of ''T''('''X''', '''W'''), where {{nowrap|gcd(''j'', 2''k'') {{=}} 1}}. Assume, possibly after inverting the generator, that the imperfect generator of ''T'' has {{nowrap|''j'' + 1}} '''W'''s and the perfect generator has ''j'' '''W'''s.

Consider the two generators in the GS of ''s'', which are detemperings of the generator {{nowrap|''i'''''X''' + ''j'''''W'''}} of ''T''('''X''', '''W'''), where {{nowrap|gcd(''j'', 2''k'') {{=}} 1}}. Assume, possibly after inverting the generator, that the imperfect generator of ''T'' has {{nowrap|''j'' − 1}} '''W'''s and the perfect generator has ''j'' '''W'''s.

'''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.

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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'']}}, 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.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 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 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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 By part (2), we have that ''s'' has step signature {{nowrap|''a'''''X''' ''b'''''Y''' ''b'''''Z'''}}, ''a'' odd. By part (4), we have that {{nowrap|''T''('''X''', '''W''') {{=}} ''s''('''X''', '''W''', '''W''')}} is a MOS scale ''a'''''X'''2''b'''''W'''. If {{nowrap|''b'' {{=}} 1}}, there's nothing to prove, so assume {{nowrap|''b'' &gt; 1}}.
-Consider the two generators in the GS of ''s'', which are detemperings of the generator {{nowrap|''i'''''X''' + ''j'''''W'''}} of ''T''('''X''',&nbsp;'''W'''), where {{nowrap|gcd(''j'', 2''k'') {{=}} 1}}. Assume, possibly after inverting the generator, that the imperfect generator of ''T'' has {{nowrap|''j'' + 1}} '''W'''s and the perfect generator has ''j'' '''W'''s.
+Consider the two generators in the GS of ''s'', which are detemperings of the generator {{nowrap|''i'''''X''' + ''j'''''W'''}} of ''T''('''X''',&nbsp;'''W'''), where {{nowrap|gcd(''j'', 2''k'') {{=}} 1}}. Assume, possibly after inverting the generator, that the imperfect generator of ''T'' has {{nowrap|''j'' &minus; 1}} '''W'''s and the perfect generator has ''j'' '''W'''s.
 '''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.
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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'']}}, 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.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 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 2''b''-th ''j''-step window differ due to parity. Since {{nowrap|gcd(''j'', 2''b'') {{=}} 1}}, this visits every note of ''U''.