Ternary parallelogram scales are MOS substitution

This article proves the following theorem:

Ternary parallelogram scale words are MOS substitution scale words, where the period count of the template MOS is the number of rows of the parallelogram parallel to the unique step size parallel to a side of the parallelogram.

Definitions

Pitch-class group

The pitch-class group of a scale word w in letters x₁, ..., x_r with step signature e ∈ ℤ^r⟨x₁, ..., x_r⟩ is the abelian group C(w) := ℤ^r⟨x₁, ..., x_r⟩/⟨e⟩. The pitch-class group is associated with a canonical map π that sends every step vector to its pitch class.

Parallelogram scale

A scale word w is a parallelogram scale word if C(w) is torsion-free (equiv. a free abelian group) and there exists integers m, n > 1 and linearly independent elements v and w in C(w) such that the π-image of

[math]\displaystyle{ \mathcal{I}_w := \{\mathrm{ab}(\epsilon), \mathrm{ab}(w[0:1]), ..., \mathrm{ab}(w[0:|w|-1])\} }[/math]

is of the form

[math]\displaystyle{ \{i\mathbf{v} + j\mathbf{w} : i \in [0:m], j \in [0:n]\}. }[/math]

MOS substitution scale

See MOS substitution.

Proof

Step 1: Get a surjective homomorphism [math]\displaystyle{ \mathbb{Z}^2 \to \mathbb{Z}/mn\mathbb{Z} }[/math]

The π-image of any k-step interval (abelianized slice) ab(w[i : i + k]) is identical to that of ab(w[i : i + k + mn]). Hence there is a well-defined map from the pitch classes of intervals of w to ℤ/mnℤ. Traversing w step by step gives a traversal of [0 : m] × [0 : n] where we label each grid point with the index of the current note in w. We also recall that the pitch-class vector v has a representative that is a k_v-step interval in w, 0 < k_v < mn, and similarly for w.

We thus wish to constrain ways of labeling [0 : m] × [0 : n] with ℤ/mnℤ elements such that

• v = (1, 0) is consistently the π-image of a k_v-step interval of w, 0 < k_v < mn, so traveling one step east (while staying within the grid) always shifts the label by k_v
• w = (0, 1) is consistently the π-image of a k_w-step interval, 0 < k_w < mn, so traveling one step north (while staying within the grid) always shifts the label by k_w
• every element of ℤ/mnℤ is used exactly once in the labeling.

After rotating w, we may assume that (0, 0) is labeled 0. The labeling now extends to a surjective homomorphism [math]\displaystyle{ \varphi: \mathbb{Z}^2\langle \mathbf{v},\mathbf{w}\rangle \to \mathbb{Z}/mn\mathbb{Z}, }[/math] where φ(v) = k_v and φ(w) = k_w. φ has [0 : m] × [0 : n] as a fundamental domain.

Step 2: Any m × n window in ℤ² has the same cyclic ordering of elements under the φ-labeling

For any integers i₀ and j₀, if φ((i₀, j₀)) = a, then for any (i, j) in [0 : m] × [0 : n], we have

[math]\displaystyle{ \begin{align*}\varphi((i_0 + i, j_0 + j)) &= \varphi((i_0, j_0)) + \varphi((i, j)) \\ &= a + \varphi((i, j)),\end{align*} }[/math]

as φ is a homomorphism. Hence corresponding elements of any two m × n windows get the same ℤ/mnℤ labeling under φ modulo a shift.

Step 3: By ternarity, exactly one of the 1-step vectors is parallel to a coordinate axis

Consider the following four "quadrants":

1. Q₁ = [0 : m] × [0 : n]
2. Q₂ = [-m + 1 : 1] × [0 : n]
3. Q₃ = [-m + 1 : 1] × [-n + 1 : 1]
4. Q₄ = [0 : m] × [0 : n]

By the previous step, φ restricted to any m × n window in ℤ² is surjective, hence all of the four windows Q₁, ..., Q₄ have 1 somewhere in them. Call these positions u₁, ..., u₄ (note that none of them are the zero vector). Since φ((0, 0)) = 0 by another application of the previous step we have u₁, ..., u₄ as the images of 1-step vectors of w. Since w is ternary, exactly two of these vectors will be pairwise equal, say u_k = u_l. These four "quadrants" intersect in [math]\displaystyle{ [-m + 1 : m] \times \{0\} \cup \{0\} \times [-n + 1 : n], }[/math] entailing that u_k = u_l is on a coordinate axis (either the v-coordinate is 0 or the w-coordinate is 0 but not both).

A technical lemma

Statement: If m > 1, n > 2, a has order > n in ℤ/mnℤ, and ℤ/mnℤ is partitioned into bins where one bin consists of ≤ 2m - 1 adjacent elements and the rest of the bins consist of 2m - 1, then {0, a, 2a, ..., (n - 1)a} meets some bin at least twice.

Proof: Apply the pigeonhole principle: the number of bins is

[math]\displaystyle{ \begin{align*} \bigg\lceil \frac{mn}{2m-1} \bigg\rceil &=\bigg\lceil \frac{n}{2-1/m} \bigg\rceil \\ &\le \bigg\lceil \frac{2n}{3} \bigg\rceil < n. \end{align*} }[/math]

The bin where two multiples of a coincide must have two distinct multiples of a, because the order of a is > n.

Step 4: The axial step is a MOS substitution slot letter

Relabel the axial vector as u_x (= π(x)) and the two nonaxial vectors as u_y = π(y) and u_z = π(z). We shall now show that x is the slot letter of a MOS substitution scale.

Assume without loss of generality that u_x = (t, 0), t > 0 (parallel to v).

The two non-axial step vectors differ by (0, n) if the axial step is parallel to v and by (m, 0) otherwise

Under the above assumption, since φ(tv) = 1 we have that φ(v) = k_v is a cyclic generator of the group ℤ/mnℤ, such that tk_v = 1 mod mn. Multiplication by t is a group automorphism ψ of ℤ/mnℤ such that ψ(k_v) = 1, so the first row [0 : m] × {0} is mapped as ψφ((i, 0)) = i; so the image of the first row under ψφ is {0, ..., m - 1}.

Claim: k_w has order n in ℤ/mnℤ.

Proof: The order cannot be less than n, lest we have φ((0, uk_w)) = 0 for some 0 < u < n, contradicting injectivity of φ within a fundamental domain (following from Step 2). If the order is N > n, we have two cases. If n = 2, then clearly the image of the (0, 1) translation of [0 : m] must be [m : 2m], forcing k_w = m which is of order 2. If m > 3, then by the lemma, [0 : m], [0 : m] + a, ..., [0 : m] + (n - 1)a, where a = ψ(k_w), are not disjoint, contradicting injectivity.

The claim establishes that

1. the two other 1-step vectors are not found on the axis orthogonal to the first vector, as that would imply order mn, nor on the axis opposite to the first
2. φ is minimally periodic under translation by nw. Hence the two non-axial 1's on the grid from the 0 at the origin are translated by nw.
3. Since m is of order n, m must be found on the w-axis as well.

Template word is MOS

Since the two non-axial steps differ by nw, to identify the two we wrap ℤ² with the vector nw, identifying lattice points separated by nw. This makes the space ℤ × ℤ/nℤ. Now the image of [math]\displaystyle{ \pi(\mathcal{I}_w) }[/math] under this wrap is of the form [0 : m] × ℤ/nℤ. Note that m (its order being n) corresponds to a generator of the ℤ/nℤ factor, establishing (by minimality) that the period of the template word is m. Now do another wrap, identifying m with 0, and we're left with [0 : m] in ℤ. We're now in period-equivalent pitch-class space with one generator chain. Since the template word is binary we have a MOS with period m and period count n, which is "the number of rows of the parallelogram parallel to the unique step size parallel to a side of the parallelogram."

Filling word is MOS

Delete all instances of the axial step x and consider what the two remaining step sizes do to the w-coordinate.

Without loss of generality assume that u_y = (b, c), c > 0, and u_z = (b, c - n). As the v-coordinates of both vectors are equal, we only need to look at the w-coordinate. Since the w-coordinate of a point must stay within [0 : n], at any point it must follow the rule: "If the current w-coordinate + c ≥ n, then move by c - n units (using the letter z). Otherwise, move by c units (using the letter y)."

This pattern of movements is in fact the same as the one produced by taking the circular word "1 1 1 ... 1 (1 - n)" and stacking (abelianized) c-step subwords. As there is only one bad position per period, the filling word can easily be seen to be MOS by stacking kc-step subwords of the latter word for 2 ≤ k ≤ length - 1. [math]\displaystyle{ \square }[/math]