Generator-offset property: 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 PWF. Then S is SV3 and has an odd number of notes. Moreover, S is either GO or equivalent to the scale word abacaba.

==== Proof ====

Suppose S has n notes (after dealing with small cases, we may assume n ≥ 7) and S projects to single-period mosses S₁ (via identifying b ~ c), S₂ (via identifying a ~ c) and S₃ (via identifying a ~ b). Suppose S₁'s generator is a k-step, which comes in two sizes: P, the perfect k-step, and I, the imperfect k-step. By stacking k-steps, we get two words of length n of k-steps of S₂ and S₃, respectively. These two-"step-size" words, which we call Σ₂ and Σ₃, must be mosses, since m-steps in the new words correspond to mk-steps in the mos words S₁ and S₂, which come in at most two sizes. Since S₁ is a single-period mos, gcd(k, n) = 1. Hence when 0 < m < n, km is ''not'' divisible by n and km-steps come in ''exactly'' two sizes; hence both Σ₂ and Σ₃ are single-period mosses.  

(Note: We say “assume” as shorthand for either “assume without loss of generality” or “assume after excluding other possibilities”.)

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

Only two sizes of k-steps of S can project to P in S₁, 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 S₃, contradicting the mos property of S₃.

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Since, by our assumption, Λ₃ has two γ’s in a row, Λ₃ must have more γ than γ’, so μ &minus; 1 < n/2. Since Λ₃ is a mos, μ &minus; 1 ≥ 1. So we have 2 ≤ μ ≤ ceil(n/2).

We have three cases to consider:

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 flip the chain and reindex the words to start at 2f.

In this case S has two chains of Qs, one with floor(n/2) notes and and one with ceil(n/2) notes. Every instance of Q must be a k-step, since by '''Z'''-linear independence Q = αa + βb + γc 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.

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, thus S satisfies the generator-offset property.

Since y &minus; x = floor(n/μ) &minus; 1 and floor(n/μ) ≥ 2, we have: x = floor(n/μ) = 2 and y &minus; x = 1; hence y = 2*floor(n/μ) &minus; 1 = 3. The chunk in Λ₃ whose size is 3 is made from two chunks in Λ₂ of size 1. (So Λ₂ has chunks of size 1 and 2, and Λ₃ has chunks of size 2 and 3.)

(This also implies S is SV3.)

The remaining cases are all impossible because they imply y &minus; x ≥ 2:

== Open conjectures ==