Wedgie/Archived version: Difference between revisions

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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.

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.)

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.)

Let ''K''1 = the kernel of the temperament (i.e. the set of commas tempered out by the temperament), and ''K''2 = ker W = {'''v''' ∈ ''M'' : W('''v''', '''w''') = 0 ∀'''w''' ∈ ''M''}. If '''v''' ∈ ''K''1, then '''v''' is tempered out by both a and b, so W('''v''', '''w''') = a('''v''')b('''w''') − a('''w''')b('''v''') = 0, and '''v''' ∈ ''K''2. Conversely, if '''v''' ∈ ''K''2, then W('''v''', '''w''') = a('''v''')b('''w''') − a('''w''')b('''v''') = 0 for all '''w''', which implies a('''v''')b('''w''') = a('''w''')b('''v''') (*) for all '''w'''. Since a and b both have rank 1 but a&b has rank 2, a and b are linearly independent as vals; so we can choose '''w''' such that a('''w''') = 0 but b('''w''') ≠ 0. Then (*) shows a('''v''') = 0. By the same argument, b('''v''') = 0. So '''v''' is in ''K''1 and ''K''1 = ''K''2; the kernel of the temperament is exactly the intervals that the wedgie "treats as zero".

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 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.
-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.)
+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.)
 Let ''K''<sub>1</sub> = the kernel of the temperament (i.e. the set of commas tempered out by the temperament), and ''K''<sub>2</sub> = ker W = {'''v''' ∈ ''M'' : W('''v''', '''w''') = 0 ∀'''w''' ∈ ''M''}. If '''v''' ∈ ''K''<sub>1</sub>, then '''v''' is tempered out by both a and b, so W('''v''', '''w''') = a('''v''')b('''w''') &minus; a('''w''')b('''v''') = 0, and '''v''' ∈ ''K''<sub>2</sub>. Conversely, if '''v''' ∈ ''K''<sub>2</sub>, then W('''v''', '''w''') = a('''v''')b('''w''') &minus; a('''w''')b('''v''') = 0 for all '''w''', which implies a('''v''')b('''w''') = a('''w''')b('''v''') (*) for all '''w'''. Since a and b both have rank 1 but a&b has rank 2, a and b are linearly independent as vals; so we can choose '''w''' such that a('''w''') = 0 but b('''w''') ≠ 0. Then (*) shows a('''v''') = 0. By the same argument, b('''v''') = 0. So '''v''' is in ''K''<sub>1</sub> and ''K''<sub>1</sub> = ''K''<sub>2</sub>; the kernel of the temperament is exactly the intervals that the wedgie "treats as zero".