Ternary parallelogram scales are MOS substitution: Difference between revisions

''[[Ternary]] parallelogram scale [[word]]s 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.''

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

* 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'').

 

=== Step 1: Get a surjective homomorphism <math>\varphi:\mathbb{Z}^2 \to \mathbb{Z}/mn\mathbb{Z}</math> ===

The π-image of any ''k''-step interval (abelianized slice) {{nowrap|ab(''w''[''i'' : ''i'' + ''k''])}} is identical to that of {{nowrap|ab(''w''[''i'' : ''i'' + ''k'' + ''mn'']).}} Hence there is a well-defined map from the pitch classes of intervals of ''w'' to {{nowrap|ℤ/''mn''ℤ.}} Because ''w'' is a parallelogram scale, traversing ''w'' step by step gives a traversal of {{nowrap|[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'', {{nowrap|0 < ''k''_'''v''' < ''mn'',}} and similarly for '''w'''.  

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

By the previous step, φ restricted to any ''m'' × ''n'' window in ℤ² is surjective, hence all of the four windows ''Q''₁, ..., ''Q''₄ have 1 somewhere in them. Call these positions '''u'''₁, ..., '''u'''₄ (note that none of them are the zero vector). Since {{Nowrap|φ((0, 0)) {{=}} 0,}} by another application of the previous step we have '''u'''₁, ..., '''u'''₄ 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>[-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).

Claim: ψ(''k''_'''w''') (thus ''k''_'''w''' as well) has order ''n'' in {{nowrap|ℤ/''mn''ℤ.}}

Proof: The order cannot be less than ''n'', lest we have {{nowrap|φ((0, ''uk''_'''w''')) {{=}} 0}} for some {{nowrap|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'' : 2''m''], forcing ψ(''k''_'''w''') = ''m'' which is of order 2.

* If ''n'' &ge; 3, assume by way of contradiction that {{nowrap|[0 : ''m''], [0 : ''m''] + ''a'', ..., [0 : ''m''] + (''n'' - 1)''a''}}, where ''a'' = ψ(''k''_'''w'''), are disjoint. Then {{nowrap|[0 : ''m''], [0 : ''m''] + ''a'', ..., [0 : ''m''] + (''n'' - 2)''a''}} are disjoint. We ask: where do we place {{nowrap|[0 : ''m''] + (''n'' - 1)''a''?}}

** 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 {{angbr|''m''}} and has order ''n''. Hence there is no way to place {{nowrap|[0 : ''m''] + (''n'' - 1)''a''}} without it overlapping with one of the other windows.

# 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 {{nowrap|[-''m'' + 1 : 0] × {0}}} as that contradicts ''k''_'''v''' having order ''mn''

# Since ''m'' is of order ''n'', and the cyclic group {{nowrap|ℤ/''mn''ℤ}} has at most one subgroup of any given order, ''m'' must be found on the '''w'''-axis as well.

Since the two non-axial steps differ by ''n'''''w''', to identify the two we wrap ℤ² with the vector ''n'''''w''', identifying lattice points separated by ''n'''''w'''. This makes the space ℤ × ℤ/''n''ℤ. Now the image of <math>\pi(\mathcal{I}_w)</math> under this wrap is of the form [0 : ''m''] × ℤ/''n''ℤ. Note that ''m'' (its order being ''n'') corresponds to a generator of the ℤ/''n''ℤ 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."

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 {{nowrap|'''u'''_'''y''' {{=}} (''b'', ''c''), ''c'' > 0,}} and {{nowrap|'''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 {{nowrap|[0 : ''n''],}} at any point it must follow the rule: "If the current '''w'''-coordinate + c &ge; ''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 {{nowrap|''v'' {{=}} "1 1 1 ... 1 (1 - ''n'')"}} ((''n'' - 1)-many 1's) and stacking sums <math>\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 {{nowrap|2 &le; ''k'' &le; length - 1.}} {{Qed}}

# Conjecture: Ternary scales that are parallelogram substrings with full row length ''m'', full column length ''n'', and cardinality {{nowrap|''mn'' - 1}} are MOS substitution.

# Conjecture: Ternary scales that are parallelogram substrings with full row length ''m'', full column length ''n'', and cardinality {{nowrap|''mn'' - 2}} are MOS substitution.

#* <s>If the template MOS of a MOS substitution scale has m > 1 periods and the step signature's gcd = 1 => m × n parallelogram scale.</s> This is false as stated, since no 5L(3m7s) scale is a parallelogram scale.

#* Conjecture: If the template MOS of a MOS substitution scale has m > 1 periods, the step signature's gcd = 1, and the filling MOS is a multiMOS, then the scale is a parallelogram scale. (But the filling MOS need not be a multiMOS for a parallelogram scale: LLmLLsLLmLLsLLs is of type 10L(2m3s))

#* Conjecture: If a MOS substitution scale's template MOS is a multiMOS, and its step signature's gcd = 1, then its generator structure is a (possibly) *shifted* full parallelogram, i.e. either a full parallelogram or the first and last rows' lengths add up to the full row length.