Ternary parallelogram scales are MOS substitution: Difference between revisions
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After rotating ''w'', we may assume that (0, 0) is labeled 0. The labeling now extends to a surjective homomorphism <math>\varphi: \mathbb{Z}^2\langle \mathbf{v},\mathbf{w}\rangle \to \mathbb{Z}/mn\mathbb{Z},</math> where {{nowrap|φ('''v''') {{=}} ''k''<sub>'''v'''</sub>}} and {{nowrap|φ('''w''') {{=}} ''k''<sub>'''w'''</sub>.}} φ has {{nowrap|[0 : ''m''] × [0 : ''n'']}} as a fundamental domain. | After rotating ''w'', we may assume that (0, 0) is labeled 0. The labeling now extends to a surjective homomorphism <math>\varphi: \mathbb{Z}^2\langle \mathbf{v},\mathbf{w}\rangle \to \mathbb{Z}/mn\mathbb{Z},</math> where {{nowrap|φ('''v''') {{=}} ''k''<sub>'''v'''</sub>}} and {{nowrap|φ('''w''') {{=}} ''k''<sub>'''w'''</sub>.}} φ has {{nowrap|[0 : ''m''] × [0 : ''n'']}} as a fundamental domain. | ||
=== | === Step 2: Any ''m'' × ''n'' window in ℤ<sup>2</sup> has the same cyclic ordering of elements under the φ-labeling === | ||
For any integers ''i''<sub>0</sub> and ''j''<sub>0</sub>, if {{nowrap|φ((''i''<sub>0</sub>, ''j''<sub>0</sub>)) {{=}} ''a'',}} then for any (''i'', ''j'') in [0 : ''m''] × [0 : ''n''], we have | For any integers ''i''<sub>0</sub> and ''j''<sub>0</sub>, if {{nowrap|φ((''i''<sub>0</sub>, ''j''<sub>0</sub>)) {{=}} ''a'',}} then for any (''i'', ''j'') in [0 : ''m''] × [0 : ''n''], we have | ||
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as φ is a homomorphism. Hence corresponding elements of any two ''m'' × ''n'' windows get the same {{nowrap|ℤ/''mn''ℤ}} labeling under φ modulo a shift. | as φ is a homomorphism. Hence corresponding elements of any two ''m'' × ''n'' windows get the same {{nowrap|ℤ/''mn''ℤ}} labeling under φ modulo a shift. | ||
=== Step | === Step 3: By ternarity, exactly one of the 1-step vectors is parallel to a coordinate axis === | ||
Consider the following four "quadrants": | Consider the following four "quadrants": | ||
# ''Q''<sub>1</sub> = [0 : ''m''] × [0 : ''n''] | # ''Q''<sub>1</sub> = [0 : ''m''] × [0 : ''n''] | ||
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# ''Q''<sub>3</sub> = [-''m'' + 1 : 1] × [-''n'' + 1 : 1] | # ''Q''<sub>3</sub> = [-''m'' + 1 : 1] × [-''n'' + 1 : 1] | ||
# ''Q''<sub>4</sub> = [0 : ''m''] × [0 : ''n''] | # ''Q''<sub>4</sub> = [0 : ''m''] × [0 : ''n''] | ||
By the previous | By the previous step, φ restricted to any ''m'' × ''n'' window in ℤ<sup>2</sup> is surjective, hence all of the four windows ''Q''<sub>1</sub>, ..., ''Q''<sub>4</sub> have 1 somewhere in them. Call these positions '''u'''<sub>1</sub>, ..., '''u'''<sub>4</sub> (note that none of them are the zero vector). Since {{Nowrap|φ((0, 0)) {{=}} 0}} by another application of the lemma we have '''u'''<sub>1</sub>, ..., '''u'''<sub>4</sub> as the images of 1-step vectors of ''w''. Since ''w'' is ternary, exactly two of these vectors will be pairwise equal, say '''u'''<sub>''k''</sub> = '''u'''<sub>''l''</sub>. These four "quadrants" intersect in <math>[-m + 1 : m] \times \{0\} \cup \{0\} \times [-n + 1 : n],</math> entailing that '''u'''<sub>''k''</sub> = '''u'''<sub>''l''</sub> is on a coordinate axis (either the '''v'''-coordinate is 0 or the '''w'''-coordinate is 0 but not both). | ||
=== Step | === Lemma: If ''a'' has order > ''n'' in {{nowrap|ℤ/''mn''ℤ}}, then {{nowrap|{''a'', 2''a'', ..., (''n'' - 1)''a''}}} and {{nowrap|[-''m'' + 1 : ''m'']}} are not disjoint === | ||
This is a technical lemma about ℤ/''mn''ℤ dynamics. | |||
=== Step 4: The axial step is a MOS substitution slot letter === | |||
Relabel the axial vector as '''u'''<sub>'''x'''</sub> (= π('''x''')) and the two nonaxial vectors as '''u'''<sub>'''y'''</sub> = π('''y''') and '''u'''<sub>'''z'''</sub> = π('''z'''). We shall now show that '''x''' is the slot letter of a MOS substitution scale. | Relabel the axial vector as '''u'''<sub>'''x'''</sub> (= π('''x''')) and the two nonaxial vectors as '''u'''<sub>'''y'''</sub> = π('''y''') and '''u'''<sub>'''z'''</sub> = π('''z'''). We shall now show that '''x''' is the slot letter of a MOS substitution scale. | ||
Assume without loss of generality that | Assume without loss of generality that | ||
'''u'''<sub>'''x'''</sub> = (''t'', 0), ''t'' > 0 (parallel to '''v'''). | '''u'''<sub>'''x'''</sub> = (''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 ==== | ==== The two non-axial step vectors differ by (0, ''n'') if the axial step is parallel to '''v''' and by (''m'', 0) otherwise ==== | ||