Mathematical theory of saturation: Difference between revisions

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Because unsaturated subgroups of <math>\mathbb{Z}^n</math> are for these reasons problematic, it is useful to have a means to saturate them; that is, to find the minimal saturated subgroup of <math>\mathbb{Z}^n</math> containing the given subgroup. We may do this by inverting the right-reducing matrix which in part converts a matrix of basis elements for the subgroup <math>V</math> to [http://en.wikipedia.org/wiki/Smith_normal_form Smith normal form]. If <math>A</math> is a matrix with <math>r</math> (the rank) rows of dimension <math>n</math> whose rows form a basis for <math>V</math>, then there are two square matrices <math>L</math> and <math>R</math>, such that <math>S = LAR</math>, where <math>S</math> is the Smith normal form. The right-reducing matrix is <math>R</math>, the matrix multiplying <math>A</math> on the right. The first <math>r</math> rows of <math>R</math> generate the saturation of <math>V</math>. This procedure is only useful if there is a routine for finding the Smith normal form available, so we will assume there is one and not concern ourselves with the Smith form as such.

To give an example, consider the matrix {{~~vector~~|{{map|12 19 28 34}} {{map|26 41 60 72}}}} whose rows are the two vals we considered above. The Smith form itself is the 2×4 matrix {{~~vector~~|{{map|1 0 0 0}} {{map|0 2 0 0}}}}; this does not concern us. The left-reducing matrix does not concern us either; our interest lies in the right reducing matrix, which is an invertible square integral matrix, {{~~vector~~|{{map|-11 19 4 13}} {{map|7 -12 -4 -10}} {{map|0 0 1 0}} {{map|0 0 0 1}}}}. Inverting this matrix gives another square integral matrix, {{~~vector~~|{{map|12 19 28 34}} {{map|7 11 16 19}} {{map|0 0 1 0}} {{map|0 0 0 1}}}}. The rank of <math>V</math> is two, so to find a basis for the saturation of <math>V</math>, we take the first two rows, which gives us the group generated by {{~~vector~~|{{map|12 19 28 34}} {{map|7 11 16 19}}}}. The [[Normal_lists|normal val list]] for this is {{~~vector~~|{{map|1 0 -4 -13}} {{map|0 1 4 10}}}}, which are period and generator maps for septimal meantone, which is the saturated temperament corresponding to the contorted <math>V</math>.

To give an example, consider the matrix {{ket|{{map|12 19 28 34}} {{map|26 41 60 72}}}} whose rows are the two vals we considered above. The Smith form itself is the 2×4 matrix {{ket|{{map|1 0 0 0}} {{map|0 2 0 0}}}}; this does not concern us. The left-reducing matrix does not concern us either; our interest lies in the right reducing matrix, which is an invertible square integral matrix, {{ket|{{map|-11 19 4 13}} {{map|7 -12 -4 -10}} {{map|0 0 1 0}} {{map|0 0 0 1}}}}. Inverting this matrix gives another square integral matrix, {{ket|{{map|12 19 28 34}} {{map|7 11 16 19}} {{map|0 0 1 0}} {{map|0 0 0 1}}}}. The rank of <math>V</math> is two, so to find a basis for the saturation of <math>V</math>, we take the first two rows, which gives us the group generated by {{ket|{{map|12 19 28 34}} {{map|7 11 16 19}}}}. The [[Normal_lists|normal val list]] for this is {{ket|{{map|1 0 -4 -13}} {{map|0 1 4 10}}}}, which are period and generator maps for septimal meantone, which is the saturated temperament corresponding to the contorted <math>V</math>.

To test for saturation, we may take the wedge product of the generators. Wedging {{map|26 41 60 72}} with {{map|12 19 28 34}} gives us {{multimap|2 8 20 8 26 24}}; this is not zero, so the rank of the group these generate is two. However the coefficients have a gcd of two, and hence the group is not saturated; for saturation, the coefficients must be relatively prime, with a gcd of one.

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 Because unsaturated subgroups of <span><math>\mathbb{Z}^n</math></span> are for these reasons problematic, it is useful to have a means to saturate them; that is, to find the minimal saturated subgroup of <span><math>\mathbb{Z}^n</math></span> containing the given subgroup. We may do this by inverting the right-reducing matrix which in part converts a matrix of basis elements for the subgroup <span><math>V</math></span> to [http://en.wikipedia.org/wiki/Smith_normal_form Smith normal form]. If <span><math>A</math></span> is a matrix with <span><math>r</math></span> (the rank) rows of dimension <span><math>n</math></span> whose rows form a basis for <span><math>V</math></span>, then there are two square matrices <span><math>L</math></span> and <span><math>R</math></span>, such that <span><math>S = LAR</math></span>, where <span><math>S</math></span> is the Smith normal form. The right-reducing matrix is <span><math>R</math></span>, the matrix multiplying <span><math>A</math></span> on the right. The first <span><math>r</math></span> rows of <span><math>R</math></span> generate the saturation of <span><math>V</math></span>. This procedure is only useful if there is a routine for finding the Smith normal form available, so we will assume there is one and not concern ourselves with the Smith form as such.
-To give an example, consider the matrix {{vector|{{map|12 19 28 34}} {{map|26 41 60 72}}}} whose rows are the two vals we considered above. The Smith form itself is the 2×4 matrix {{vector|{{map|1 0 0 0}} {{map|0 2 0 0}}}}; this does not concern us. The left-reducing matrix does not concern us either; our interest lies in the right reducing matrix, which is an invertible square integral matrix, {{vector|{{map|-11 19 4 13}} {{map|7 -12 -4 -10}} {{map|0 0 1 0}} {{map|0 0 0 1}}}}. Inverting this matrix gives another square integral matrix, {{vector|{{map|12 19 28 34}} {{map|7 11 16 19}} {{map|0 0 1 0}} {{map|0 0 0 1}}}}. The rank of <span><math>V</math></span> is two, so to find a basis for the saturation of <span><math>V</math></span>, we take the first two rows, which gives us the group generated by {{vector|{{map|12 19 28 34}} {{map|7 11 16 19}}}}. The [[Normal_lists|normal val list]] for this is {{vector|{{map|1 0 -4 -13}} {{map|0 1 4 10}}}}, which are period and generator maps for septimal meantone, which is the saturated temperament corresponding to the contorted <span><math>V</math></span>.
+To give an example, consider the matrix {{ket|{{map|12 19 28 34}} {{map|26 41 60 72}}}} whose rows are the two vals we considered above. The Smith form itself is the 2×4 matrix {{ket|{{map|1 0 0 0}} {{map|0 2 0 0}}}}; this does not concern us. The left-reducing matrix does not concern us either; our interest lies in the right reducing matrix, which is an invertible square integral matrix, {{ket|{{map|-11 19 4 13}} {{map|7 -12 -4 -10}} {{map|0 0 1 0}} {{map|0 0 0 1}}}}. Inverting this matrix gives another square integral matrix, {{ket|{{map|12 19 28 34}} {{map|7 11 16 19}} {{map|0 0 1 0}} {{map|0 0 0 1}}}}. The rank of <span><math>V</math></span> is two, so to find a basis for the saturation of <span><math>V</math></span>, we take the first two rows, which gives us the group generated by {{ket|{{map|12 19 28 34}} {{map|7 11 16 19}}}}. The [[Normal_lists|normal val list]] for this is {{ket|{{map|1 0 -4 -13}} {{map|0 1 4 10}}}}, which are period and generator maps for septimal meantone, which is the saturated temperament corresponding to the contorted <span><math>V</math></span>.
 To test for saturation, we may take the wedge product of the generators. Wedging {{map|26 41 60 72}} with {{map|12 19 28 34}} gives us {{multimap|2 8 20 8 26 24}}; this is not zero, so the rank of the group these generate is two. However the coefficients have a gcd of two, and hence the group is not saturated; for saturation, the coefficients must be relatively prime, with a gcd of one.