Ternary scale theorems: Difference between revisions

In case 1, let {{nowrap|'''g'''₁ {{=}} (2, 1) − (1, 1)|'''g'''₂ {{=}} (1, 2) − (2, 1)}}, and {{nowrap|'''g'''₃ {{=}} (1, 1) − ({{frac|''n''|2}}, 2)}} {{nowrap|{{=}} ((−{{frac|''n''|2}} − 1)*'''g'''₁ − {{frac|''n''|2}}*'''g'''₂) (mod '''e''')}}. We assume that '''g'''₁, '''g'''₂ and '''e''' are ℤ-linearly independent. We have the chain '''g'''₁ '''g'''₂ '''g'''₁ '''g'''₂ ... '''g'''₁ '''g'''₃ which visits every note in ''s''.  

# from '''g'''₁ '''g'''₂ ... '''g'''₁, we get {{nowrap|''a''₁ {{=}} {{sfrac|''r'' − 1|2}} * '''g''' + '''g'''₁}} {{nowrap|{{=}} {{ceil|{{frac|''r''|2}}}} '''g'''₁ + {{floor|{{frac|''r''|2}}}} '''g'''₂}}

# from '''g'''₂ '''g'''₁ ... '''g'''₂, we get {{nowrap|''a''₂ {{=}} {{sfrac|''r'' − 1|2}} * '''g''' + '''g'''₂}} {{nowrap|{{=}} {{floor|{{frac|''r''|2}}}} '''g'''₁ + {{ceil|{{frac|''r''|2}}}} '''g'''₂}}

# from '''g'''₂ (...even # of gens...) '''g'''₁ '''g'''₃ '''g'''₁ (...even # of gens...) '''g'''₂, we get {{nowrap|''a''₃ {{=}} {{sfrac|''r'' − 1|2}} '''g'''₁ + {{sfrac|''r'' − 1|2}} '''g'''₂ + '''g'''₃}} {{nowrap|≡ {{sfrac|''r'' − ''n''|2}} − {{sfrac|3|2}}'''g'''₁ + {{sfrac|''r'' − ''n''|2}} − {{sfrac|1|2}}'''g'''₂ (mod '''e''')}}.

# from '''g'''₁ (...odd # of gens...) '''g'''₁ '''g'''₃ '''g'''₁ (...odd # of gens...) '''g'''₁, we get {{nowrap|''a''₄ {{=}} {{sfrac|''r'' + 1|2}} '''g'''₁ + {{sfrac|''r'' − 3|2}} '''g'''₂ + '''g'''₃}} {{nowrap|≡ {{sfrac|''r'' − ''n''|2}} − {{sfrac|1|2}}'''g'''₁ + {{sfrac|''r'' − ''n''|2}} − {{sfrac|3|2}}'''g'''₂ (mod '''e''')}}.

In case 2, let {{nowrap|''n'' ≥ 3}} and let {{nowrap|(2, 1) − (1, 1) {{=}} '''g'''₁|(1, 2) − (2, 1) {{=}} '''g'''₂}} be the two alternants. Let '''g'''₃ be the closing generator after stacking alternating '''g'''₁ and '''g'''₂. Then the generator circle is {{nowrap|('''g'''₁ '''g'''₂)^{{{floor|''n''/2}}}}} '''g'''₃. If a step is formed by stacking ''k'' generators, we may assume that ''k'' is odd, and the combinations of alternants corresponding to a step come in exactly 3 sizes:

(since the scale size is odd, we can always ensure this by taking octave complements of all the generators). By counting the length-''k'' subwords of the (linear) word {{nowrap|('''g'''₁ '''g'''₂)^{{{floor|{{frac|''n''|2}}}}}}}, we see that the first two sizes must both occur {{sfrac|''n'' − ''k''|2}} times. This proves (2).

Proof: Consider the subword for the closing generator of ''s'' on index ''p'', which is {{nowrap|''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, {{nowrap|''s''[''p'' − 1: ''p'' − 1 + ''i'' + ''j'']}} and {{nowrap|''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 {{nowrap|''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 {{nowrap|''q'' ≠ ''p''}} to be the index of the first letter of one such ''j''-step subword (as contained in ''s'') and use {{nowrap|''s''[''q'' : ''q'' + ''i'' + ''j'']}}.

Assume that the generator '''g''' is a ''k''-step and ''k'' is even. (If ''k'' is not even, invert the generator.) On some note ''p'' we have a chain of (''n'' + 1)/2 notes and on ''p′'' {{=}} ''p'' + offset we'll have (''n'' − 1)/2) notes.

Since ''n'' is odd, ''rk'' ≡ ''k''/2 mod ''n'' iff ''r'' ≡ (''n'' + 1)/2 mod ''n''. (Note that both 2 and ''k'' are coprime with ''n'', hence multiplicatively invertible mod ''n''.) This proves that the offset, which must be reached after (''n'' + 1)/2 ''k''-steps, is a ''k''/2-step, as desired. (As [''k''] is a generator of ℤ/''n'', stacking (''n'' − 1)-many ''k''-steps must visit every note exactly once. Thus if the offset wasn't reached in (''n'' + 1)/2 steps, the two generator chains wouldn't have the assumed lengths.) {{qed}}

Suppose {{nowrap|'''Q''' {{=}} (α, β, γ)}} {{nowrap|≠ '''R''' {{=}} (α, β′, γ′)}} are the two ''k''-steps in ''s'' that project to '''P'''. Then {{nowrap|'''T''' {{=}} (α′, β″, γ″)}} projects to '''I'''. Here the values in each component differ by at most 1, and {{nowrap|α ≠ α′}}. Then the circular word Λ₁ formed by the '''a'''-components of the ''k''-steps in '''P''' is α...αα′. Since Σ₂ is a primitive MOS pattern of {{nowrap|β'''b''' + (''n'' − β)('''a'''~'''c''')}} and {{nowrap|β′a + (''n'' − β′)('''a'''~'''c''')}}, the circular word Λ₂ = the pattern of β and β′ must be a primitive MOS. Similarly, Λ₃ = the pattern of γ and γ′ is a primitive MOS.

Either {{nowrap|β″ {{=}} β}} or {{nowrap|β″ {{=}} β′}}. Assume {{nowrap|β″ {{=}} β′}}. Then {{nowrap|γ″ {{=}} γ}}, and {{nowrap|Λ₃ {{=}} (λ + 1)γ (μ − 1)γ′}}. Also assume that the first ''k''-step in Σ is '''Q'''. Then we have:

where {{nowrap|''W'' {{=}} ''W''('''x''', '''y''')}} is a word in two variables '''x''' and '''y''', of length {{nowrap|''n'' − 2}}.

Since, by our assumption, Λ₃ has two γ in a row, Λ₃ must have more γ than γ′, so {{nowrap|μ − 1 < ''n''/2}}. Since Λ₃ is a MOS, {{nowrap|μ − 1 ≥ 1}}. So we have {{nowrap|2 ≤ μ ≤ {{ceil|''n''/2}}}}.

'''Case 1''': {{nowrap|μ {{=}} 2}}, i.e. Λ₂ is the MOS {{nowrap|(''n'' − 2)β 2β′}}.

# the stack has one '''T''' and does not contain any '''R''' (since it's more than {{nowrap|''f'' − 1}} generators away).

In this case ''s'' has two chains of '''Q''', one with {{floor|''n''/2}} notes and one offset by {{nowrap|'''Q'''^{(''f'' − 1)}R}} with {{ceil|''n''/2}} notes. Every instance of Q must be a ''k''-step, since by ℤ-linear independence {{nowrap|'''Q''' {{=}} α'''a''' + β'''b''' + γ'''c'''}} is the only way to write '''Q''' in the basis {{nowrap|('''a''', '''b''', '''c''')}}; so ''s'' is well-formed with respect to '''Q'''. Thus ''s'' also satisfies the generator-offset property with generator '''Q'''.

(The lower bound is reached if Λ₃ has chunks of sizes {{nowrap|{{floor|''n''/μ}} − 1}} and {{floor|''n''/μ}}, and the upper bound is reached if Λ₃ has chunks of sizes {{floor|''n''/μ}} and {{ceil|''n''/μ}}.)

The difference between the chunk sizes of Λ₃ is {{nowrap|''y'' − ''x''}}, which must be 1 since Λ₃ is pairwise well-formed. We thus have the following subcases: (In the following, chunk of Λ₂ means chunk of β, and chunk of Λ₃ means chunk of γ.)

'''Case 3.1:''' {{nowrap|(''x'', ''y'') {{=}} ({{floor|''n''/μ}}, 2*{{floor|''n''/μ}} − 1)}}.

Since {{nowrap|''y'' − ''x'' {{=}} {{floor|''n''/μ}} − 1}}, we have {{nowrap|''x'' {{=}} {{floor|''n''/μ}} {{=}} 2}} and {{nowrap|''y'' {{=}} 3}}. The chunk in Λ₃ whose size was defined to be ''y'' is made from two consecutive chunks in Λ₂ of size 1. (So Λ₂ has chunks of size 1 and 2, and Λ₃ has chunks of size 2 and 3.) Since chunk sizes of a MOS themselves form a MOS, Λ₂ has more chunks of size 1 than it has chunks of size 2.

Λ₂ has only two chunks of size 1, {{nowrap|Λ₂[''n'' − 1]}} and Λ₂[1], since otherwise Λ₃ would have a chunk of size 1 within {{nowrap|Λ₃[1 : ''n'']}}. Thus Λ₂ has exactly one chunk of size 2. Thus {{nowrap|Λ₂ {{=}} ββ′βββ′ββ′}} and {{nowrap|Λ₃ {{=}} γγ′γγγ′γγ}}. Thus we have:

'''Case 3.2''': {{nowrap|(''x'', ''y'') {{=}}}} {{nowrap|({{floor|''n''/μ}} + 1, 2*{{floor|''n''/μ}} − 1)}} is impossible: Here {{nowrap|(''x'', ''y'') {{=}} (4, 5)}}. But then Λ₂ has a chunk of size < 3 because of the β′ at index ''n'', contradicting that ''x'' is one of the chunk sizes of Λ₂.

The remaining cases are all impossible because they imply {{nowrap|''y'' − ''x'' ≥ 2}}:

* the perfect generator {{nowrap|'''g''' {{=}} ''t'''''X''' + (''k'' − ''t'')'''W'''}} (note that {{nowrap|(''k'' − ''t'')}} is odd by a previous proof), and

* the imperfect generator '''i''' {{=}} (''t'' + 1)'''X''' + (''k'' − ''t'' − 1)'''W'''.

We have {{nowrap|π('''g'''₁) {{=}} π('''g'''₂)}} = '''g'''. Let '''h''' be the size in ''s'' such that {{nowrap|π('''h''') {{=}} '''i'''}}. Hence only one ''k''-step subword ''h'' has this size in ''s''. By Theorem 1, we also may assume {{nowrap|{{abs|'''g'''₁}}_'''Y''' {{=}} {{abs|'''g'''₂}}_'''Y''' + 1}} and {{nowrap|{{abs|'''g'''₁}}_'''Z''' {{=}} {{abs|'''g'''₂}}_'''Z''' − 1}} (the other case corresponds to the opposite chirality).

* {{nowrap|{{abs|'''u'''₁}}_'''X''' {{=}} {{abs|'''u'''₂}}_'''X'''}} {{nowrap|{{=}} {{abs|v}}_'''X''' − 1}}

* {{nowrap|{{abs|'''u'''₁}}_'''Z''' {{=}} {{abs|'''u'''₂}}_'''Z''' − 1}}