Recursive structure of MOS scales: Difference between revisions
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s = ''eg'' + ''fp'' | s = ''eg'' + ''fp'' | ||
for appropriate integers ''c, d, e, f'', where |''c''| < ''n'' and |''e''| < ''n''. We must have that ''c'' = ±''b'' and ''e'' = ∓''a'', as by assumption ''a''L + ''b''s = (''ac'' + ''be'')''g'' + (''ad'' + ''bf'')''p'' = ''p''. In fact, [L, s] is another valid basis for the group | for appropriate integers ''c, d, e, f'', where |''c''| < ''n'' and |''e''| < ''n''. We must have that ''c'' = ±''b'' and ''e'' = ∓''a'', as by assumption ''a''L + ''b''s = (''ac'' + ''be'')''g'' + (''ad'' + ''bf'')''p'' = ''p''. In fact, [L, s] is another valid basis for the abelian group with basis {''p'', ''g''}, since by binarity we have ''p, g'' ∈ span(L, s). | ||
Assume ''c'' = ''b'' and ''e'' = −''a''. [This corresponds to assuming that ''g'' is the "bright" generator.] Let χ = L − s > 0; then χ is ''p''-equivalent to ''+ng''. Now by generatedness and binarity, any interval class that has at least two sizes must have sizes separated by ''ng'' (the separation corresponding to changing an L step to an s step). Since ''g'' and ''p'' are linearly independent, for each ''j'' ∈ {1, ..., ''n'' − 1} there exists at most one ''k'' = ''k''(''j'') ∈ {1, ..., ''n'' − 1}</sub> such that ''jg'' is ''p''-equivalent to one size of ''k''-step. Hence if an interval class that has two sizes must have the sizes ''jg'' and (''j'' − ''n'')''g''; any other size must leave the range −(''n'' − 1)''g'', ..., 0, ..., (''n'' − 1)''g''. Thus the class of ''k''-steps has at most two sizes for 1 ≤ ''k'' ≤ (''n'' − 1). Each non-''p''-equivalent class must have ''exactly'' two sizes, since the inverse of the ''k''-step that is equivalent to ''jg'' is an (''n'' − ''k'')-step equivalent to ''−jg'', which must be distinct from an (''n'' − ''k'')-step equivalent to a positive number of ''g'' generators; note that the latter (''n'' − ''k'')-step does occur in the "brightest" (most ''g'' generators stacked ''up'' rather than ''down'') mode of ''S''. | Assume ''c'' = ''b'' and ''e'' = −''a''. [This corresponds to assuming that ''g'' is the "bright" generator.] Let χ = L − s > 0; then χ is ''p''-equivalent to ''+ng''. Now by generatedness and binarity, any interval class that has at least two sizes must have sizes separated by ''ng'' (the separation corresponding to changing an L step to an s step). Since ''g'' and ''p'' are linearly independent, for each ''j'' ∈ {1, ..., ''n'' − 1} there exists at most one ''k'' = ''k''(''j'') ∈ {1, ..., ''n'' − 1}</sub> such that ''jg'' is ''p''-equivalent to one size of ''k''-step. Hence if an interval class that has two sizes must have the sizes ''jg'' and (''j'' − ''n'')''g''; any other size must leave the range −(''n'' − 1)''g'', ..., 0, ..., (''n'' − 1)''g''. Thus the class of ''k''-steps has at most two sizes for 1 ≤ ''k'' ≤ (''n'' − 1). Each non-''p''-equivalent class must have ''exactly'' two sizes, since the inverse of the ''k''-step that is equivalent to ''jg'' is an (''n'' − ''k'')-step equivalent to ''−jg'', which must be distinct from an (''n'' − ''k'')-step equivalent to a positive number of ''g'' generators; note that the latter (''n'' − ''k'')-step does occur in the "brightest" (most ''g'' generators stacked ''up'' rather than ''down'') mode of ''S''. | ||