The wedgie: 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 linear algebra perspective, there's nothing special about the basis {p, g}; I could have chosen the basis p' = 3/1 and g' = 2/1. 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.

In the language of linear algebra, the wedgie is an "alternating bilinear form" on the appropriate JI group M; this means that 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 ~~biliear~~ 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:

In the language of linear algebra, the wedgie is an "alternating bilinear form" on the appropriate JI group M; this means that 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:

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 ∈ K1, 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, 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, as claimed.

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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 linear algebra perspective, there's nothing special about the basis {p, g}; I could have chosen the basis p' = 3/1 and g' = 2/1. 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.
-In the language of linear algebra, the wedgie is an "alternating bilinear form" on the appropriate JI group M; this means that 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 biliear 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:
+In the language of linear algebra, the wedgie is an "alternating bilinear form" on the appropriate JI group M; this means that 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:
 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 ∈ K1, 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, 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, as claimed.