Rank-3 scale theorems

Theorems

• Every triple Fokker block is max variety 3.
• Every max variety 3 block is a triple Fokker block. (However, not every max-variety 3 scale, in general, need be a Fokker block.)
• Triple Fokker blocks form a trihexagonal tiling on the lattice.
• A scale imprint is that of a Fokker block if and only if it is the product word of two DE scale imprints with the same number of notes. See Introduction to Scale Theory over Words in Two Dimensions | SpringerLink
• If the step sizes for a rank-3 Fokker block are L, m, n, and s, where L > m > n > s, then the following identity must hold: (n-s) + (m-s) = (L-s), hence n+m=L+s
• Any convex object on the lattice can be converted into a hexagon.
• Any convex scale with 3 step sizes is a hexagon on the lattice, in which each set of parallel lines corresponds to one of the steps.

Unproven Conjectures

• Every rank-3 Fokker block has mean-variety < 4, meaning that some interval class will come in less than 4 sizes.

MV3 proofs

Under construction

Definitions and theorems

Throughout, let S be a scale word in steps x, y, z (and assume all three of these letters are used).

Definition: Unconditionally MV3

S is unconditionally MV3 or intrinsically MV3 if S is MV3 for all possible choices of step ratio x:y:z.

Definition: EMOS

S is elimination-MOS (EMOS) if the result of removing (all instances of) any one of the step sizes is a MOS.

Definition: PMOS

S is pairwise MOS (PMOS) if the result of equating any two of the step sizes is a MOS.

Definition: AG

S satisfies the alternating generator property (AG) if it satisfies the following equivalent properties:

1. S can be built by stacking a single chain of alternating generators g1 and g2, resulting in a circle of the form either g1 g2 ... g1 g2 g1 g3 or g1 g2 ... g1 g2 g3.
2. S is generated by two chains of generators separated by a fixed interval; either both chains are of size m, or one chain has size m and the second has size m-1.

These are equivalent, since the separating interval can be taken to be g1 and the generator of each chain = g1 + g2.

Definitions: LQ

Let n = a_1 + ... + a_r be the scale size, w a scale word with signature a_1 X_1, ..., a_r X_r, let L be a line of the form L(t) = (a_1, ..., a_r)t + v_0, where v_0 is a constant vector in R^r. We say that L is in generic position if L contains a point (0, α_1, α_2, ... α_{r-1}) where α_i and α_i/α_j for i != j are irrational.

• Assume S is a 2-step scale. Then S is slope-LQ if the slope between any two pair of points (representing a k-mosstep) is one of the two nearest possible slopes (in the set {k/0,...,0/k}) to b/a.
• Say that a 2-step scale S is floor-LQ if some mode M of S satisfies that γ(M) = the graph of floor(b/a*x).
• Say that an r-step scale S is LQ if any appropriate line in generic position, (a_1, ..., a_r)t + v_0, has intersections with coordinate level planes x_i = k that spell out the scale as you move in the positive t direction.

AG + unconditionally MV3 implies "ax by bz" and that the scale's cardinality is odd

Assuming both AG and unconditional MV3, we have two chains of generator g0 (going right). The two cases are:

O-O-...-O (m notes)
O-O-...-O (m notes)

and

O-O-O-...-O (m notes)
O-O-...-O (m-1 notes).

Label the notes (1,k) and (2,k), 1 ≤ k ≤ m or m-1, for notes in the upper and lower chain respectively.

In case 1 (even scale size n = 2^t r where r is odd), let g1 = (2,1) - (1,1) and g2 = (1,2) - (2,1). We have the chain g1 g2 g1 g2... g1 g3. Suppose the k-step is the class generated by r generators (which is an odd number of generator steps):

1. from g1 ... g1, get a1 = (r-1)/2*g0 + g1 = (r+1)/2 g1 + (r-1)/2 g2
2. from g2 ... g2, get a2 = (r-1)/2*g0 + g2 = (r-1)/2 g1 + (r+1)/2 g2
3. from g2 (even) g1 g3 g1 (even) g2, get a3 = (r-1)/2 g1 + (r-1)/2 g2 + g3
4. from g1 (odd) g1 g3 g1 (odd) g1, get a4 = (r+1)/2 g1 + (r-3)/2 g2 + g3.

Choose a tuning where g1 and g2 are both very close to but not exactly 1/2*g0, resulting in a scale very close to the mos generated by 1/2 g0. (i.e. g1 and g2 differ from 1/2*g0 by ε, a quantity much smaller than the chroma of the n/2-note mos generated by g0, which is |g3 - g2|). Assuming n > 4, we have 4 distinct sizes for k-steps, a contradiction to unconditional-MV3:

1. a1, a2 and a3 are clearly distinct.
2. a4 - a3 = g1 - g2 != 0, since the scale is a non-trivial AG.
3. a4 - a1 = g3 - g2 = (g3 + g1) - (g2 + g1) != 0. This is exactly the chroma of the mos generated by g0.
4. a4 - a2 = g1 - 2 g2 + g3 = (g3 - g2) + (g1 - g2) = (chroma ± ε) != 0 by choice of tuning.

(For n = 4, the above argument doesn't work because a3 = a4, and xyxz is a counterexample.)

In case 2, let (2,1)-(1,1) = g1, (1,2)-(2,1) = g2 be the two alternating generators. 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. Then the generators corresponding to a step are:

1. k g1 + (k-1) g2
2. (k-1) g1 + k g2
3. (k-1) g1 + (k-1) g2 + g3,

if a step is an odd number of generators (since the scale size is odd, we can always ensure this by taking octave complements of all the generators). The first two sizes must occur the same number of times.

(The above holds for any odd n >= 3.)

This proof shows that AG and unconditionally-MV3 scales must have cardinality odd or 4.

An AG scale is unconditionally MV3 iff its cardinality is odd or 4

We only need to see that AG + odd cardinality => MV3. But the argument in case 2 above works for any interval class (MV3 wasn't used), hence any interval class comes in at most 3 sizes regardless of tuning.

3-DE implies MV3 (WIP)

We prove that 3-DE + not abcba implies PMOS, which is known to imply MV3.