Ternary scale theorems: Difference between revisions

and use the vectors (-1, 2) and ({{ceil|n/2}}, 1) as the Fokker block chromas. A rank-3 Fokker block has the property that tempering out by each of the chromas gives two MOSes. These correspond to two of the temperings '''X''' {{=}} '''Y''', '''Y''' {{=}} '''Z''' and '''X''' {{=}} '''Z'''. The third tempering follows by symmetry (by taking the other chirality).

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

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

Proof: Consider the subword for the closing generator of ''s'' on index ''p'', which is ''I'' {{=}} ''s''[''p'' : ''p'' + ''i'' + ''j''], and suppose the result of deleting all '''X''''s from ''I'' has a ''j''-step subword ''w''. Shifting ''I'' one step to the left and one step to the right, ''s''[''p'' − 1: ''p'' − 1 + ''i'' + ''j''] and ''s''[''p'' + 1 : ''p'' + 1 + ''i'' + ''j''] are both detemperings of perfect generators of ''T'', and have one fewer non-'''X''' step than ''I'' by our assumption. Thus the word ''I'' must both begin and end in a non-'''X''' letter. Removing all the '''X''''s from ''I'' results in a word that is ''j'' + 1 letters long and is the ''j''-step word ''w'' with just one extra letter appended. Thus one of the two perfect generators above, namely the one that removes the extra letter, must contain this ''j''-step. The rest of the ''j''-step subwords of ''s'' can all be obtained by deleting '''X'''s from detempered perfect generators; take ''q'' ≠ ''p'' to be the index of the first letter of one such ''j''-step subword (as contained in ''s'') and use ''s''[''q'' : ''q'' + ''i'' + ''j''].

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

Proof: Write '''u''' and '''v''' for the two sizes of ''j''-steps. Since 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 gcd(''m'', 2''b'') {{=}} 1. Hence the scale steps of ''U'' are ('''uv''')^{(''m''−1)/2}'''u''' mod '''e''' and ('''vu''')^{(''m''−1)/2}'''v''' mod '''e''', and the step sizes alternate because '''u''' and '''v''' do.

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