Ternary scale theorems: Difference between revisions
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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''}}<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>}} = {{sfrac|''i''|2}} and {{nowrap|{{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 < ''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''}}<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>}} = {{sfrac|''i''|2}} and {{nowrap|{{abs|''w''}}<sub>'''Z'''</sub> {{=}} ''j''}}. | ||
It remains to show that ''s'' is balanced... <!--{{qed}}--> | It remains to show that ''s'' is balanced. Any ''k''-step subword has either ''j'' or ''j'' + 1 '''Z'''s for some ''j'' the result of conflating '''X''' and '''Y''' is a MOS. If the number of non-'''Z''' letters in a ''k''-step subword is even, then there is only one possibility for the number of '''X''' and the number of '''Y'''. If the number of non-'''Z''' letters in a ''k''-step subwrod is odd, then both the number of '''X'''s and the number of '''Y'''s differ by at most 1. <!--{{qed}}--> | ||
== Theorem 7 (Classification of MV3 scales) == | == Theorem 7 (Classification of MV3 scales) == | ||