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''''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.

Line 107:

'''Claim 2''': If a binary necklace ''U'' has ''b'' '''Y'''s and ''b'' '''Z'''s, {{nowrap|gcd(''j'', 2''b'') {{=}} 1}}, and consecutively stacked ''j''-steps in ''U'' occur in 2 alternating sizes, then {{nowrap|''U'' {{=}} ('''YZ''')''b''}}.

Proof: Write '''u''' and '''v''' for the two sizes of ''j''-steps. Since {{nowrap|gcd(''j'', 2''b'') {{=}} 1}}, there exists ''m'' such that stacking ''m''-many ''j''-steps yields scale steps of ''U'', and ''m'' is odd because {{nowrap|gcd(''m'', 2''b'') {{=}} 1}}. Hence the scale steps of ''U'' are {{nowrap|('''uv'''){{sfrac|''m'' − 1|2}}'''u''' (mod '''e''')}} and {{nowrap|('''vu'''){{sfrac|''m'' − 1|2}}'''v''' (mod '''e''')}}, and the step sizes alternate because '''u''' and '''v''' do.

These two claims prove that {{nowrap|''E'''''X'''(S) {{=}} ('''YZ''')''b''}} and that the two GS generators' sizes differ by replacing one '''Y''' for a '''Z'''. {{Qed}}

@@ 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'' &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.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 107: / Line 107: @@
 '''Claim 2''': If a binary necklace ''U'' has ''b'' '''Y'''s and ''b'' '''Z'''s, {{nowrap|gcd(''j'', 2''b'') {{=}} 1}}, and consecutively stacked ''j''-steps in ''U'' occur in 2 alternating sizes, then {{nowrap|''U'' {{=}} ('''YZ''')<sup>''b''</sup>}}.
-Proof: Write '''u''' and '''v''' for the two sizes of ''j''-steps. Since {{nowrap|gcd(''j'', 2''b'') {{=}} 1}}, there exists ''m'' such that stacking ''m''-many ''j''-steps yields scale steps of ''U'', and ''m'' is odd because {{nowrap|gcd(''m'', 2''b'') {{=}} 1}}. Hence the scale steps of ''U'' are {{nowrap|('''uv''')<sup>{{sfrac|''m'' − 1|2}}</sup>'''u''' (mod '''e''')}} and {{nowrap|('''vu''')<sup>{{sfrac|''m'' − 1|2}}</sup>'''v''' (mod '''e''')}}, and the step sizes alternate because '''u''' and '''v''' do.
+Proof: Write '''u''' and '''v''' for the two sizes of ''j''-steps. Since {{nowrap|gcd(''j'', 2''b'') {{=}} 1}}, there exists ''m'' such that stacking ''m''-many ''j''-steps yields scale steps of ''U'', and ''m'' is odd because {{nowrap|gcd(''m'', 2''b'') {{=}} 1}}. Hence the scale steps of ''U'' are {{nowrap|('''uv''')<sup>{{sfrac|''m'' &minus; 1|2}}</sup>'''u''' (mod '''e''')}} and {{nowrap|('''vu''')<sup>{{sfrac|''m'' − 1|2}}</sup>'''v''' (mod '''e''')}}, and the step sizes alternate because '''u''' and '''v''' do.
 These two claims prove that {{nowrap|''E''<sub>'''X'''</sub>(S) {{=}} ('''YZ''')<sup>''b''</sup>}} and that the two GS generators' sizes differ by replacing one '''Y''' for a '''Z'''. {{Qed}}