Ternary parallelogram scales are MOS substitution

This is an expert page. It is written to allow experienced readers to learn more about the advanced elements of the topic.

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 exist 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

A ternary scale word w(x₁, x₂, x₃) is a MOS substitution scale word if there exists a permutation [math]\displaystyle{ \pi \in S_3 }[/math] such that the following holds:

• identifying x_π(1) and x_π(2) results in a MOS scale word (called the template MOS) and
• deleting all instances of x_π(3) (called the slot letter) results in a MOS scale word (called the filling MOS).

Proof

Step 1: Get a surjective homomorphism [math]\displaystyle{ \varphi:\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. Because w is a parallelogram scale, 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, 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).

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: ψ(k_w) (thus k_w as well) 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 by disjointness the image of the (0, 1) translation of [0 : m] must be [m : 2m], forcing ψ(k_w) = m which is of order 2.
• If n ≥ 3, assume by way of contradiction that [0 : m], [0 : m] + a, ..., [0 : m] + (n - 1)a, where a = ψ(k_w), are disjoint. Then [0 : m], [0 : m] + a, ..., [0 : m] + (n - 2)a are disjoint. We ask: where do we place [0 : m] + (n - 1)a?
  • The number of remaining slots on all of ℤ/mnℤ is m.
  • There is no contiguous region of m unoccupied slots. If there were, then the (n - 1) windows that have already been placed must have no gap between them. This implies that a generates ⟨m⟩ and has order n. Hence there is no way to place [0 : m] + (n - 1)a without it overlapping with one of the other windows.

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 for k_w, nor on [-m + 1 : 0] × {0} as that contradicts k_v having order mn
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, and the cyclic group ℤ/mnℤ has at most one subgroup of any given order, 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 v = "1 1 1 ... 1 (1 - n)" ((n - 1)-many 1's) and stacking sums [math]\displaystyle{ \sum_{i=0}^{c-1} v[i_0+i] }[/math] of c-step subwords. As the resulting word has only one bad position per period, the filling word can easily be seen to be MOS by stacking kc-step subwords of v for 2 ≤ k ≤ length - 1. [math]\displaystyle{ \square }[/math]

Open problems

1. Conjecture: Ternary scales that are parallelogram substrings with full row length m, full column length n, and cardinality mn - 1 are MOS substitution.
2. Conjecture: Ternary scales that are parallelogram substrings with full row length m, full column length n, and cardinality mn - 2 are MOS substitution.
3. Prove a converse to this theorem.
   • If the template MOS of a MOS substitution scale has m > 1 periods and the step signature's gcd = 1 => m × n parallelogram scale. This is false as stated, since no 5L(3m7s) scale is a parallelogram scale.