Ternary scale theorems: Difference between revisions

Let ''s''('''a''', '''b''', '''c''') be a scale word in three ℤ-linearly independent step sizes '''a''', '''b''', '''c'''. Suppose ''s'' is pairwise well-formed (equivalently, all its projections are primitive MOSes). Then ''s'' is SV3 and has an odd number of notes. Moreover, ''s'' is either generator-offset or equivalent to the scale word '''abacaba'''.

Suppose ''s'' has ''n'' notes (after dealing with small cases, we may assume ''n'' ≥ 7) and ''s'' projects to primitive MOSes ''s''₁ (via identifying '''b''' with '''c'''), ''s''₂ (via identifying '''a''' with '''c'''), and ''s''₃ (via identifying '''a''' with '''b'''). Suppose ''s''₁'s generator is a ''k''-step, which comes in two sizes: '''P''', the perfect ''k''-step, and '''I''', the imperfect ''k''-step. By stacking ''n''-many ''k''-steps, we get two words of length ''n'' of ''k''-steps of ''s''₂ and ''s''₃, respectively. These binary words, which we call Σ₂ and Σ₃, must be MOSes, since ''m''-steps in the new words correspond to ''mk''-steps in the MOS words ''s''₁ and ''s''₂, which come in at most two sizes. Since ''s''₁ is a primitive MOS, gcd(''k'', ''n'') {{=}} 1. Hence when 0 < ''m'' < ''n'', ''mk'' is ''not'' divisible by ''n'' and ''mk''-steps come in ''exactly'' two sizes; hence both Σ₂ and Σ₃ are primitive MOSes.

We can write sizes of dyads in ''s'' as vectors (''p'', ''q'', ''r'') using the basis ('''a''', '''b''', '''c''').  

Suppose for sake of contradiction that only one size of ''k''-step ('''α''', '''β''', '''γ''') in ''s'' projects to '''P''' in ''s''₁. Then projecting to ''s''₂ shows that ''s''₂'s generator is the ''k''-step (α + γ)*('''a'''~'''c''') + β'''b''', and Σ₂'s imperfect generator is located at index ''n'', like Σ₁'s imperfect generator is. Then ''s''₁ and ''s''₂ are the same mode of the same MOS pattern (up to knowing which step size is the bigger one). Assume the '''L''' of ''s''₁ (it could be '''s''', but it doesn't matter) is the result of identifying '''b''' and '''c''', and all '''s''' steps in ''s''₁ come from '''a'''. Then the steps of ''s''₂ corresponding to the '''L''' of ''s''₁ must be either all '''b''''s or all '''a'''~'''c''''s, thus these steps are all '''b''''s in ''s'' (otherwise they would be identified with the '''a''', against the assumption that ''s''₁ and ''s''₂ are the same MOS pattern and mode). So ''s'' has only two step sizes (a and b), contradicting the assumption that ''s'' is ternary.

Only two sizes of ''k''-steps of ''s'' can project to P in ''s''₁, for if there are three sizes of ''k''-steps (α, β, γ), (α, β′, γ′), (α, β′′, γ′′) in ''s'' that project to P, then β, β′ and β′′ are three distinct values. Thus these would project to three different ''k''-steps in ''s''₃, contradicting the MOS property of ''s''₃.

Suppose '''Q''' {{=}} (α, β, γ) ≠ '''R''' {{=}} (α, β′, γ′) are the two ''k''-steps in ''s'' that project to '''P'''. Then '''T''' {{=}} (α′, β′′, γ′′) projects to '''I'''. Here the values in each component differ by at most 1, and 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 β'''b''' + (''n'' − β)('''a'''~'''c''') and 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.

Suppose Λ₂ is the MOS λβ μβ′. Then Λ₃ is the MOS (λ ± 1)γ (μ ∓ 1)γ′. Since both Λ₂ and Λ₃ are primitive, and at least one of μ and (μ ∓ 1) are even, it is now immediate that ''n'' is odd.

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

  Σ {{=}} Q ''W''(Q, R)  T

  Λ₁ {{=}} α …      α α′

  Λ₂ {{=}} β ''W''(β, β′) β′

  Λ₃ {{=}} γ ''W''(γ, γ′) γ

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

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

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

For Λ₂ to be a MOS, the first, and only, occurrence of '''R''' must be at either ''f'' {{=}} {{floor|''n''/2}} or {{ceil|''n''/2}}. We may assume that it is at ''f''; otherwise reverse the chain and reindex the words to start at 2''f''.

  Λ₁ {{=}} α … α α  α … α  α′

  Λ₂ {{=}} β … β β′ β … β  β′

  Λ₃ {{=}} γ … γ γ′ γ … γ  γ

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

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

'''Case 2:''' μ ≥ {{ceil|''n''/2}}, i.e. Λ₂ has fewer β than β′.

Since Λ₃ has more β than β′, Λ₂ is {{floor|''n''/2}}β {{ceil|''n''/2}}β′, and Λ₃ is {{ceil|''n''/2}}γ {{floor|''n''/2}}γ′. There is a unique mode of {{ceil|''n''/2}}γ {{floor|''n''/2}}γ′ that both begins and ends with γ, namely γγ′γγ′…γγ′γ. Thus Λ₂ is ββ′ββ′…ββ′β′. It is now easy to see that if the number of ''k''-steps stacked is odd, then there are two sizes that do not contain '''T''' and one size that contains '''T'''; if the number of ''k''-steps stacked is even, then there is one size that does not contain '''T''' and two sizes that contain T. Hence ''s'' is SV3.

In this case we have Σ {{=}} '''QRQR'''...'''QRT''', and ''s'' is well-formed with respect to the generator '''Q''' + '''R''', thus ''s'' satisfies the generator-offset property. By Proposition 1, ''s'' is SV3.

'''Case 3:''' 3 ≤ μ ≤ {{floor|''n''/2}}.

Λ₂ has a chunk of β (after the first β′) of size ''x'' where ''x'' {{=}} {{floor|''n''/μ}} ≥ {{floor|''n''/{{floor|''n''/2}}}} {{=}} 2 or ''x'' {{=}} {{ceil|''n''/μ}} {{=}} {{floor|''n''/μ}} + 1. Hence Λ₃ has a chunk of γ of size ''x''. Λ₃ also has a chunk that contains Λ₃[''n'' : 2] as a subword. This chunk must be of size ''y'', where  

(The lower bound is reached if Λ₃ has chunks of sizes {{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 ''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:''' (''x'', ''y'') {{=}} ({{floor|''n''/μ}}, 2*{{floor|''n''/μ}} − 1).

Since ''y'' − ''x'' {{=}} {{floor|''n''/μ}} − 1, we have ''x'' {{=}} {{floor|''n''/μ}} {{=}} 2 and ''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, Λ₂[''n'' − 1] and Λ₂[1], since otherwise Λ₃ would have a chunk of size 1 within Λ₃[1 : ''n'']. Thus Λ₂ has exactly one chunk of size 2. Thus Λ₂ {{=}} ββ′βββ′ββ′ and Λ₃ {{=}} γγ′γγγ′γγ. Thus we have:

  Σ {{=}} Q R  Q Q R  Q T

  Λ₁ {{=}} α α  α α α  α α′

  Λ₂ {{=}} β β′ β β β′ β β′

  Λ₃ {{=}} γ γ′ γ γ γ′ γ γ

# ''t'' {{=}} 1: ''s'' is equivalent to '''abacaba'''.

# ''t'' {{=}} 2: ''s'' is '''QR QQ RQ TQ RQ QR QT''' => ''s'' is equivalent to '''abacaba'''.

# ''t'' {{=}} 3: ''s'' is '''QRQ QRQ TQR QQR QTQ RQQ RQT''' => ''s'' is equivalent to '''abacaba'''.

'''Case 3.2''': (''x'', ''y'') {{=}} ({{floor|''n''/μ}} + 1, 2*{{floor|''n''/μ}} − 1) is impossible: Here (''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.

'''Case 3.3''': (''x'', ''y'') {{=}} ({{floor|''n''/μ}} + 1, 2*{{floor|''n''/μ}}) is impossible: Here (''x'', ''y'') {{=}} (3, 4). But then Λ₂ has a chunk of size 1 because of the β' at index ''n'', and another chunk of size 0 or 2, contradicting that ''x'' is one of the chunk sizes of Λ₂.

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

* '''Case 3.4''': (''x'', ''y'') {{=}} ({{floor|''n''/μ}} + 1, 2*{{floor|''n''/μ}} + 1)

* '''Case 3.5''': (''x'', ''y'') {{=}} ({{floor|''n''/μ}} + 1, 2*{{floor|''n''/μ}} + 2)

* '''Case 3.6''': (''x'', ''y'') {{=}} ({{floor|''n''/μ}} + 1, 2*{{floor|''n''/μ}} + 3)

* '''Case 3.7''': (''x'', ''y'') {{=}} ({{floor|''n''/μ}}, 2*{{floor|''n''/μ}})

* '''Case 3.8''': (''x'', ''y'') {{=}} ({{floor|''n''/μ}}, 2*{{floor|''n''/μ}} + 1)

* '''Case 3.9''': (''x'', ''y'') {{=}} ({{floor|''n''/μ}}, 2*{{floor|''n''/μ}} + 2)

* '''Case 3.10''': (''x'', ''y'') {{=}} ({{floor|''n''/μ}}, 2*{{floor|''n''/μ}} + 3)