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