Wedgie/Archived version: Difference between revisions

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Consider the 2.''q''1.(…).q''n'' [[JI subgroup]], with basis '''2''', '''q'''1, ..., '''q'''''n''.

The period '''p''' (fraction of octave) and generator '''g''' form a basis for all the intervals of a rank-2 temperament. For example, '''p''' = 2/1 and '''g''' = 3/2 form a basis for meantone. But from a purely linear-algebra perspective, there's nothing special about the basis {'''p''', '''g'''}; I could have chosen another basis, for example '''p'''' = 3/1 for my "period" and '''g'''' = 2/1 for my "generator". What makes the wedgie a unique identifier for a temperament is that rather than specify a basis directly, the wedgie specifies a ''constraint'' that any basis for the temperament must satisfy: namely, that a basis '''e'''1, '''e'''2 must satisfy W('''e'''1, '''e'''2) = ±1.

The period '''p''' (fraction of octave) and generator '''g''' form a basis for all the intervals of a rank-2 temperament. For example, '''p''' = '''2''' and '''g''' = '''3''' - '''2''' (representing 3/2) form a basis for meantone. But from a purely linear-algebra perspective, there's nothing special about the basis {'''p''', '''g'''}; I could have chosen another basis, for example '''p'''' = '''3''' for my "period" and '''g'''' = '''2''' for my "generator". What makes the wedgie a unique identifier for a temperament is that rather than specify a basis directly, the wedgie specifies a ''constraint'' that any basis for the temperament must satisfy: namely, that a basis '''e'''1, '''e'''2 must satisfy W('''e'''1, '''e'''2) = ±1.

In the language of linear algebra, the wedgie is an "alternating bilinear form" on the appropriate JI group ''M''; this means that (ignoring sign) it acts like the operation of finding the determinant of two vectors on the appropriate quotient group ''M' '' = ''M''/''K'' of ''M'', where ''K'' is the kernel of the bilinear form W. Using the fact that W = a&b where a and b are two edos (properly, rank-1 [[val]]s), you can verify that ''K'' is exactly the kernel of the rank-2 temperament, as follows. (Hence ''M''/''K' '' is a rank-2 lattice on which W is an alternating non-degenerate bilinear form, which justifies the intuition of viewing W as a determinant-like function.)

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 Consider the 2.''q''<sub>1</sub>.(…).q<sub>''n''</sub> [[JI subgroup]], with basis '''2''', '''q'''<sub>1</sub>, ..., '''q'''<sub>''n''</sub>.
-The period '''p''' (fraction of octave) and generator '''g''' form a basis for all the intervals of a rank-2 temperament. For example, '''p''' = 2/1 and '''g''' = 3/2 form a basis for meantone. But from a purely linear-algebra perspective, there's nothing special about the basis {'''p''', '''g'''}; I could have chosen another basis, for example '''p'''' = 3/1 for my "period" and '''g'''' = 2/1 for my "generator". What makes the wedgie a unique identifier for a temperament is that rather than specify a basis directly, the wedgie specifies a ''constraint'' that any basis for the temperament must satisfy: namely, that a basis '''e'''<sub>1</sub>, '''e'''<sub>2</sub> must satisfy W('''e'''<sub>1</sub>, '''e'''<sub>2</sub>) = ±1.
+The period '''p''' (fraction of octave) and generator '''g''' form a basis for all the intervals of a rank-2 temperament. For example, '''p''' = '''2''' and '''g''' = '''3''' - '''2''' (representing 3/2) form a basis for meantone. But from a purely linear-algebra perspective, there's nothing special about the basis {'''p''', '''g'''}; I could have chosen another basis, for example '''p'''' = '''3''' for my "period" and '''g'''' = '''2''' for my "generator". What makes the wedgie a unique identifier for a temperament is that rather than specify a basis directly, the wedgie specifies a ''constraint'' that any basis for the temperament must satisfy: namely, that a basis '''e'''<sub>1</sub>, '''e'''<sub>2</sub> must satisfy W('''e'''<sub>1</sub>, '''e'''<sub>2</sub>) = ±1.
 In the language of linear algebra, the wedgie is an "alternating bilinear form" on the appropriate JI group ''M''; this means that (ignoring sign) it acts like the operation of finding the determinant of two vectors on the appropriate quotient group ''M' '' = ''M''/''K'' of ''M'', where ''K'' is the kernel of the bilinear form W. Using the fact that W = a&b where a and b are two edos (properly, rank-1 [[val]]s), you can verify that ''K'' is exactly the kernel of the rank-2 temperament, as follows. (Hence ''M''/''K' '' is a rank-2 lattice on which W is an alternating non-degenerate bilinear form, which justifies the intuition of viewing W as a determinant-like function.)