Ternary scale theorems: Difference between revisions

* Non-italicized Latin variables refer to interval sizes, for example step sizes.

* Indices for all words are 1-indexed.

* The notation ''S''(X₁, ..., X_''r'') is used for an ''r''-ary scale word with variables X₁, ..., X_''r'' possibly standing in for any sizes. If ''S''(X, Y) = XXY then ''S''(A, B) = AAB.

* We leave the distinction between linear and cyclic words up to context. We usually also elide the distinction between subwords and the interval sizes that subtend them.

* A cyclic word ''S'' (representing the steps of a [[periodic scale]]) of size ''n'' is '''generator-offset''' if it satisfies the following properties. The following conditions do not imply that g₁ and g₂ are the same number of scale steps. For example, 5-limit [[blackdye]] has g₁ = 9/5 (a 9-step)  and g₂ = 5/3 (a 7-step).

*# ''S'' is generated by two chains of stacked generators g separated by a fixed offset δ; either both chains are of size ''n''/2, or one chain has size (''n'' + 1)/2 and the second has size (''n'' − 1)/2. Equivalently, ''S'' can be built by stacking a single chain of alternants g₁ and g₂, resulting in a circle of the form either g₁ g₂ ... g₁ g₂ g₁ g₃ or g₁ g₂ ... g₁ g₂ g₃.

* A ternary scale is ''pairwise-well-formed'' if all its projections are well-formed (i.e. single-period mosses).

Let ''S'' be a ternary scale word in L, M, and s of length ''n'', and suppose ''S'' is SGA. Then:

# The length of ''S'' is odd, or ''S'' is equivalent to (xy)^''r''xz for some integer ''r'' ≥ 1.

[Note: This is not true with SGA replaced with generator-offset; [[blackdye]] is a counterexample that is MV4.]

Let e be the equave of ''S''.

Label the notes (1, ''j'') and (2, ''j''), 1 ≤ ''j'' ≤ (number of notes in the chain), for notes in the upper and lower chain respectively.

In case 1, let g₁ = (2, 1) − (1, 1), g₂ = (1, 2) − (2, 1), and g₃ = (1, 1) − (''n''/2, 2) = (−''n''/2*g₁ − g₁ − ''n''/2*g₂) mod e. We assume that g₁, g₂ and e are '''Z'''-linearly independent. We have the chain g₁ g₂ g₁ g₂ ... g₁ g₃ which visits every note in ''S''.  

These are all distinct by '''Z'''-linear independence. By applying this argument to 1-steps, we see that there must be 4 step sizes in some tuning, a contradiction. Thus g₁ and g₂ must themselves be step sizes. Thus we see that an even-length, abstractly SV3, generator-offset scale must be of the form (xy)^''r''xz. (Note that (xy)^''r''xz is not SV3, since it has only two kinds of 2-steps, xy and xz.) This proves (1).

In case 2, let (2, 1) − (1, 1) = g₁, (1, 2) − (2, 1) = g₂ be the two alternants. Let g₃ be the leftover generator after stacking alternating g₁ and g₂. Then the generator circle looks like g₁ g₂ g₁ g₂ ... g₁ g₂ g₃. Assuming that a step is an odd number of generators, the combinations of alternants corresponding to a step come in exactly 3 sizes:

# ''k''g₁ + (''k'' − 1)g₂

(The above holds for any odd ''n'' ≥ 3.)

For (3), we now only need to see that if len(''S'') is odd and ''S'' is SGA, ''S'' is abstractly SV3. But the argument in case 2 above works when you substitute any interval class in ''S'' instead of a 1-step (abstract SV3 wasn't used), hence any interval class comes in (abstractly) exactly 3 sizes.  

For (4), assume ''S'' is ''a''X ''b''Y ''b''Z, ''a'' odd. If ''b'' = 1, there's nothing to prove, so assume ''b'' > 1.

These two claims prove that ''E''_X(S) = (YZ)^''b'' and that the two alternants' sizes differ by replacing one Y for a Z.

For (5), odd-numbered SGA scales are [[Fokker block]]s (in the 2-dimensional lattice generated by the generator and the offset). To see this, consider the following lattice depiction of such a scale:

  x x x ... x  

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 mosses. 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).

For (6), consider the mos ''a''X 2''b''W as chunks of X separated by W (tempering Y and Z together into W). Eliminating every other W (corresponding to either Y or Z) turns it into a mos:

# The chunk sizes of X form a mos, and taking every ''k''th note of an ''n''-note mos yields a mos.

Lastly, ''E''_X(''S'') is the mos ''b''Y ''b''Z; hence ''S'' is elimination-mos.

Suppose that a periodic scale satisfies the following:

* is generator-offset

Then the scale is SGA.

Assume that the generator 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.

By modular arithmetic we have ''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. (If the offset wasn't reached in (''n'' + 1)/2 steps, the two generator chains either wouldn't be disjoint or wouldn't have the assumed lengths.)

A primitive generator-offset ternary scale of even size where the generator g is an even-step (i.e. g subtends an even number of steps) has the following properties:

# It is a union of two primitive mosses of size ''n''/2 generated by g

# It is ''not'' chiral

# It is a primitive mos word of two of the letters interleaved with the third letter, for example: XYXZXYXZXY, made by interleaving X's in the mos word YZYZY.

(1) and (2) were proved in the proof of Proposition 1 (the part that we appeal to, from "all multiples of the generator g must be even-steps ..." to "These are all distinct by Z-linear independence", does not rely on ''S'' having the SGA property). (3) and (4) are easy to check using (1).

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 single-period mosses). 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 single-period mosses ''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 mosses, 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 single-period 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 single-period mosses.

Below we write step sizes resulting from identification as a~b, b~c, and a~c.

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 ''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 cyclic word Λ₁ formed by the a-components of the ''k''-steps in P is α...αα′. Since Σ₂ is a single-period mos pattern of βb + (''n'' &minus; β)(a~c) and β′a + (''n'' &minus; β′)(a~c), the cyclic word Λ₂ = the pattern of β and β′ must be a single-period mos. Similarly, Λ₃ = the pattern of γ and γ′ is a single-period mos.

where ''W'' = ''W''(''x'', ''y'') is a word in two variables ''x'' and ''y'', of length ''n'' &minus; 2.

Since, by our assumption, Λ₃ has two γ 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).

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

# A single-period MV3 is either (1) pairwise well-formed, (2) equivalent to XYZYX, or (3) a "twisted" word constructed as follows:

## Start with a power of a multimos word ''w''(X, Z) = ''ka''X ''kb''Z such that ''a'' is even and each ''a''X ''b''Z subword of ''w'' is of the form X''P''(X, Z)Z where ''P''(X, Z) is a palindrome.

## Replace every other X with Y in ''w''.  

# Single-period MV3 scales not of type (3) are always SV3.

Proven by Bulgakova, Buzhinsky and Goncharov (2023), "[https://www.sciencedirect.com/science/article/pii/S0304397522006417 On balanced and abelian properties of circular words over a ternary alphabet]" (and Theorem 4).