Ternary scale theorems: Difference between revisions

Line 92:

* 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.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'']}}. Since its abelianization is a perfect generator, ''I''<sub>''R''</sub> has ''i''-many '''X''''s and j-many '''W''''s.

@@ Line 92: / Line 92: @@
 * 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.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'']}}. Since its abelianization is a perfect generator, ''I''<sub>''R''</sub> has ''i''-many '''X''''s and j-many '''W''''s.