Wedgie/Archived version: Difference between revisions

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The following additionally assumes that you know what the words "basis", "linear map", and "determinant" mean.

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 acts ~~more like~~ a ~~set of constraints~~ that any basis for the temperament must satisfy.

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 acts as a constraint that any basis for the temperament must satisfy.

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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 The following additionally assumes that you know what the words "basis", "linear map", and "determinant" mean.
-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 acts more like a set of constraints that any basis for the temperament must satisfy.
+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 acts as a constraint that any basis for the temperament must satisfy.
 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.)