Ternary scale theorems: Difference between revisions

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== Theorem 6 (Generator-offset structure of even-regular scales) ==

=== Definition (Even-regular scale) ===

A primitive ternary scale ''s'' is ''even-regular'' if len(''s'') is even and ''s'' is equivalent to a word constructed from taking the MOS 2''a'''''X'''2''c'''''Z''' with ''a'' odd and gcd(''a'', ''c'') {{=}} 1, and replacing every other '''X''' with '''Y'''. In particular, ''s'' has [[step signature]] equivalent to ''a'''''X'''''a'''''Y'''''b'''''Z''' with ''a'' odd and ''b'' even. For example, '''LsLsLmsLsLsm''' (achiral [[diachrome]], 5'''L'''2'''m'''5'''s''') is an even-regular scale.

A primitive ternary scale ''s'' is ''even-regular'' if len(''s'') is even and ''s'' is equivalent to a word constructed from taking the MOS 2''a'''''X''' 2''c'''''Z''' with ''a'' odd and {{nowrap|gcd(''a'', ''c'') {{=}} 1}}, and replacing every other '''X''' with '''Y'''. In particular, ''s'' has [[step signature]] equivalent to ''a'''''X''' ''a'''''Y''' ''b'''''Z''' with ''a'' odd and ''b'' even. For example, '''LsLsLmsLsLsm''' (achiral [[diachrome]], 5'''L''' 2'''m''' 5'''s''') is an even-regular scale.

=== Theorem ===

If ''s'' {{=}} ''s''('''X''', '''Y''', '''Z''') is even-regular, then:

If {{nowrap|''s'' {{=}} ''s''('''X''', '''Y''', '''Z''')}} is even-regular, then:

# ''s'' consists of two generator chains, each with len(''s'')/2 notes;

# the generator has the same interval class as some generator of the MOS 2''a'''''W'''2''c'''''Z''';

# the generator has the same interval class as some generator of the MOS 2''a'''''W''' 2''c'''''Z''';

# the two generator chains are offset by a len(''s'')/2-step interval;

# ''s'' is balanced.

=== Proof ===

The result of substituting '''Y''' with '''X''' (let us call this map ''p'') is the MOS ''M'' {{=}} 2''a'''''X'''2''c'''''Z''', which has exactly 2 periods since gcd(''a'', ''c'') {{=}} 1. ''M'' thus consists of two generator chains separated by the period of ''M'', which has ''a'' + ''c'' {{=}} len(''s'') steps. It thus suffices for there to exist ''k'', 0 < ''k'' < ''a'' + ''c'', such that every perfect ''k''-step generator has the same preimage in ''s'', which will be our desired generator. Suppose that the perfect ''k''-step of ''M'' is ''i'''''W''' + ''j'''''Z''' where 0 < ''i'' < ''a''. Since ''a'' is odd, possibly after taking the period-complement we may assume that ''i'' is even. Hence each subword ''w'' of ''s'' such that its projection ''p''(''w'') subtends a perfect ''k''-step satisfies {{abs|''w''}}'''X''' {{=}} {{abs|''w''}}'''Y''' (= ''i''/2). It plainly follows that every such ''w'' satisfies {{abs|''w''}}'''X''' {{=}} {{abs|''w''}}'''Y''' {{~~=}}~~ ''i''/2 and {{abs|''w''}}'''Z''' {{=}} ''j''.

The result of substituting '''Y''' with '''X''' (let us call this map ''p'') is the MOS {{nowrap|''M'' {{=}} 2''a'''''X''' 2''c'''''Z'''}}, which has exactly 2 periods since {{nowrap|gcd(''a'', ''c'') {{=}} 1}}. ''M'' thus consists of two generator chains separated by the period of ''M'', which has {{nowrap|''a'' + ''c'' {{=}} len(''s'')}} steps. It thus suffices for there to exist ''k'', {{nowrap|0 < ''k'' < ''a'' + ''c''}}, such that every perfect ''k''-step generator has the same preimage in ''s'', which will be our desired generator. Suppose that the perfect ''k''-step of ''M'' is {{nowrap|''i'''''W''' + ''j'''''Z'''}} where {{nowrap|0 < ''i'' < ''a''}}. Since ''a'' is odd, possibly after taking the period-complement we may assume that ''i'' is even. Hence each subword ''w'' of ''s'' such that its projection ''p''(''w'') subtends a perfect ''k''-step satisfies {{nowrap|{{abs|''w''}}'''X''' {{=}} {{abs|''w''}}'''Y''' {{=}} ''i''/2}}. It plainly follows that every such ''w'' satisfies {{nowrap|{{abs|''w''}}'''X''' {{=}} {{abs|''w''}}'''Y'''}} = {{sfrac|''i''|2}} and {{nowrap|{{abs|''w''}}'''Z''' {{=}} ''j''}}.

It remains to show that ''s'' is balanced...

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 == Theorem 6 (Generator-offset structure of even-regular scales) ==
 === Definition (Even-regular scale) ===
-A primitive ternary scale ''s'' is ''even-regular'' if len(''s'') is even and ''s'' is equivalent to a word constructed from taking the MOS 2''a'''''X'''2''c'''''Z''' with ''a'' odd and gcd(''a'', ''c'') {{=}} 1, and replacing every other '''X''' with '''Y'''. In particular,  ''s'' has [[step signature]] equivalent to ''a'''''X'''''a'''''Y'''''b'''''Z''' with ''a'' odd and ''b'' even. For example, '''LsLsLmsLsLsm''' (achiral [[diachrome]], 5'''L'''2'''m'''5'''s''') is an even-regular scale.
+A primitive ternary scale ''s'' is ''even-regular'' if len(''s'') is even and ''s'' is equivalent to a word constructed from taking the MOS 2''a'''''X'''&nbsp;2''c'''''Z''' with ''a'' odd and {{nowrap|gcd(''a'', ''c'') {{=}} 1}}, and replacing every other '''X''' with '''Y'''. In particular,  ''s'' has [[step signature]] equivalent to ''a'''''X'''&nbsp;''a'''''Y'''&nbsp;''b'''''Z''' with ''a'' odd and ''b'' even. For example, '''LsLsLmsLsLsm''' (achiral [[diachrome]], 5'''L'''&nbsp;2'''m'''&nbsp;5'''s''') is an even-regular scale.
 === Theorem ===
-If ''s'' {{=}} ''s''('''X''', '''Y''', '''Z''') is even-regular, then:
+If {{nowrap|''s'' {{=}} ''s''('''X''', '''Y''', '''Z''')}} is even-regular, then:
 # ''s'' consists of two generator chains, each with len(''s'')/2 notes;
-# the generator has the same interval class as some generator of the MOS 2''a'''''W'''2''c'''''Z''';
+# the generator has the same interval class as some generator of the MOS 2''a'''''W'''&nbsp;2''c'''''Z''';
 # the two generator chains are offset by a len(''s'')/2-step interval;
 # ''s'' is balanced.
 === Proof ===
-The result of substituting '''Y''' with '''X''' (let us call this map ''p'') is the MOS ''M'' {{=}} 2''a'''''X'''2''c'''''Z''', which has exactly 2 periods since gcd(''a'', ''c'') {{=}} 1. ''M'' thus consists of two generator chains separated by the period of ''M'', which has ''a'' + ''c'' {{=}} len(''s'') steps. It thus suffices for there to exist ''k'', 0 &lt; ''k'' &lt; ''a'' + ''c'', such that every perfect ''k''-step generator has the same preimage in ''s'', which will be our desired generator. Suppose that the perfect ''k''-step of ''M'' is ''i'''''W''' + ''j'''''Z''' where 0 &lt; ''i'' &lt; ''a''. Since ''a'' is odd, possibly after taking the period-complement we may assume that ''i'' is even. Hence each subword ''w'' of ''s'' such that its projection ''p''(''w'') subtends a perfect ''k''-step satisfies {{abs|''w''}}<sub>'''X'''</sub> {{=}} {{abs|''w''}}<sub>'''Y'''</sub> (= ''i''/2). It plainly follows that every such ''w'' satisfies {{abs|''w''}}<sub>'''X'''</sub> {{=}} {{abs|''w''}}<sub>'''Y'''</sub> {{=}} ''i''/2 and {{abs|''w''}}<sub>'''Z'''</sub> {{=}} ''j''.
+The result of substituting '''Y''' with '''X''' (let us call this map ''p'') is the MOS {{nowrap|''M'' {{=}} 2''a'''''X''' 2''c'''''Z'''}}, which has exactly 2 periods since {{nowrap|gcd(''a'', ''c'') {{=}} 1}}. ''M'' thus consists of two generator chains separated by the period of ''M'', which has {{nowrap|''a'' + ''c'' {{=}} len(''s'')}} steps. It thus suffices for there to exist ''k'', {{nowrap|0 &lt; ''k'' &lt; ''a'' + ''c''}}, such that every perfect ''k''-step generator has the same preimage in ''s'', which will be our desired generator. Suppose that the perfect ''k''-step of ''M'' is {{nowrap|''i'''''W''' + ''j'''''Z'''}} where {{nowrap|0 &lt; ''i'' &lt; ''a''}}. Since ''a'' is odd, possibly after taking the period-complement we may assume that ''i'' is even. Hence each subword ''w'' of ''s'' such that its projection ''p''(''w'') subtends a perfect ''k''-step satisfies {{nowrap|{{abs|''w''}}<sub>'''X'''</sub> {{=}} {{abs|''w''}}<sub>'''Y'''</sub> {{=}} ''i''/2}}. It plainly follows that every such ''w'' satisfies {{nowrap|{{abs|''w''}}<sub>'''X'''</sub> {{=}} {{abs|''w''}}<sub>'''Y'''</sub>}} =&nbsp;{{sfrac|''i''|2}} and {{nowrap|{{abs|''w''}}<sub>'''Z'''</sub> {{=}} ''j''}}.
 It remains to show that ''s'' is balanced... <!--{{qed}}-->