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: 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 alternating generators, 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.

Definitions: LQ

Let n = a+b+c be the scale size, w = aX bY cZ be the scale word, let L be a line of the form L(t) = (a, b, c)t ÷ v_0, where v_0 is a constant vector. We say that L is in generic position if L intersects the yz-plane at (0, α, β) where α, β, and β/α 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 a k-step scale S is LQ if any appropriate line in generic position, L(t) = (a, b, c)t + v_0, has intersections with coordinate level planes x = k, y = k or z = k that spell out the scale as you move in the positive t direction.

MV2 is equivalent to floor-LQ in 2-step scales (WIP)

Assume wlog there are more L's than s's.

Take the graph of the brightest mode of the mos, M_b(x) (right = L, up = s). We claim that this is the required graph of F(x) = floor(b/a*x).

M_b <= F: Prove that F(x) describes a mos.

Say F has #s s's and #L L's across interval [m, m']. Say there is #s+t small steps and #L-t large steps on some k step [r, r'], t >= 2. This implies that the slope of the line b/a* x itself satisfies

(F(m')-F(m)-1)/(m'-m) <= b/a <= (F(m')-F(m)+1)/(m'-m).

(bounded by "floor minus ceiling" and "ceiling minus floor" slopes; this is because x-x' <= x-floor(x') <= floor(x)+1-floor(x').)

Rearranging, F(m') - F(m) - 1 <= b/a(m'-m) <= F(m') - F(m) + 1

But F(m') -F(m) = #s and m'-m = #L. So #s -1 <= b/a*#L <= #s+1. Do the same thing for the "bad" interval [r', r] and you get #s+t-1 <= b/a(#L-t) <= #s+t+1.

Thus b/a#L <= b/a(#L-t), a contradiction.

M_b >= F: (bc it's a mos) Suppose there is an x-value n_0 where M_b(n_0) <= F(n_0) - 1. n_0 > 1 since otherwise, M_b(1) < 0. Let k = min(n_0, n-n_0), n = scale size. Then find three different k-mossteps/average slopes by taking the interval [n_0-k, n_0] before n_0, one interval containing n_0 and one interval after n_0. (We already know that mosses are slope-LQ.)

Since M_b is a mos mode, there is a k-step within [0, n_0] that has the slope which is just smaller than (F(n_0)-1)/n_0 (1). Similarly, there is a k-step within [n_0, n] that has the slope which is just bigger than (F(n_0)+1)/(n-n_0). These slopes are "two or more steps away" from each other, which is a contradiction. (State this more formally)

LQ is equivalent to floor-LQ in case of 2-step scales

A floor-LQ scale S is LQ since the graph of F(x) = floor(b/a*x) has the desired lattice points: the lattice points are

floor({(a,b)t : 0 <= t <= 1}) = floor_x(floor_y({(a,b)t : 0 <= t <= 1})) = floor_x([graph of floor(b/a*x)]) (*).

Conversely, if a 2-step scale S is LQ, floor({(a,b)t : 0 <= t <= 1}) gives you the graph of floor(b/a*x) (plus the vertical lines) when you connect the dots. This follows from the same equation (*).

MV3 Theorem 1 (WIP)

The following are equivalent for a non-multiperiod scale word S with steps x, y, z:

1. S is MV3.
2. S is PMOS, or S is of the form x'y'z'y'x' or its repetitions.
3. S is AG, or S is of the form x'y'z'y'x' or its repetitions, or x'y'x'z'x'y'x' or its repetitions.

MV3 + LQ implies EMOS (WIP)

Proof sketch:

Let L = L(t) = (a, b, c)t + (0, α, β) be a line in generic position corresponding to the signature aX bY cZ. The projection matrices

[math]\displaystyle{ P_1 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}, \ P_2 = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}, \ P_3 = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, }[/math]

(which map (x, y, z) to (x, y), (y, z) and (z, x), respectively) map L to lines in R^2 that are in generic position (i.e. they intersect the x- and y-axes at irrational points). The projections record intersections with two of the planes to intersections with x- and y- axes, and these intersections must spell out the result of removing one of the step sizes; hence the resulting scales must be mosses.

(MV3 has not been used yet)

PMOS implies AG (except in the case xyxzxyx) (WIP)

We now prove that except in the case xyxzxyx, if the scale is pairwise MOS, then it is AG.

To eliminate xyxzxyx we manually check all words up to length 7... (todo)

Now assume len(S) >= 8.

PMOS -> Consider mos temperings

• in x, ξ (ξ = y or z), with gen g1 -> g1g1...g1g1' (g1' = imperfect gen)
• in y, η (η = x or z), with gen g2
• in z, ζ (ζ = x or y), with gen g3.

Denote their detemperings as G11, G12, G13, G21, G22, G23, G31, G32, G33.

To be continued...

AG implies "ax by bz"

Assuming the alternating generator property, 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, let g1 = (2,1)-(1,m) and g2 = (1,1)-(2,m). The circle of stacked g0 generators is (starting from top left): (m-1 g0's) g1 (m-1 g0's) g2. A scale step is always a same number k (which must be odd) of such generators gi. Assume (after taking octave complement) that a single step takes less than half of the generators. So a word corresponding to the scale step is formed by one of:

• k g0
• (k-1) g0 + g1
• (k-1) g0 + g2.

It is clear that the last two sizes must occur the same number of times.

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:

• k g1 + (k-1) g2
• (k-1) g1 + k g2
• (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. QED.

3-DE implies MV3 (WIP)

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