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 lifts 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''<sub>X</sub>(''s'')('''Y''', '''Z'''), the scale word obtained by deleting all '''X''''s from ''s''. These ''j''-step subwords are adjacent and alternating under the ordering induced by the AGS stack.

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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'' &minus; 1}} '''W'''s and the perfect generator has ''j'' '''W'''s.
+Consider the two generators in the GS of ''s'', which are lifts 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 under the ordering induced by the AGS stack.