Mathematical theory of saturation: Difference between revisions
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Because unsaturated subgroups of Z^n are problematic, it is useful to have a means to saturate them; that is, to find the minimal saturated subgroup of Z^n 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 V to [[http://en.wikipedia.org/wiki/Smith_normal_form|Smith normal form]]. If A is a matrix with r (the rank) rows of dimension n whose rows form a basis for V, then there are two square matricies L and R, such that S = LAR, where S is the Smith normal form. The right-reducing matrix is R, the matrix multiplying A on the right. The first r rows of R generate the saturation of V. 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. | Because unsaturated subgroups of Z^n are problematic, it is useful to have a means to saturate them; that is, to find the minimal saturated subgroup of Z^n 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 V to [[http://en.wikipedia.org/wiki/Smith_normal_form|Smith normal form]]. If A is a matrix with r (the rank) rows of dimension n whose rows form a basis for V, then there are two square matricies L and R, such that S = LAR, where S is the Smith normal form. The right-reducing matrix is R, the matrix multiplying A on the right. The first r rows of R generate the saturation of V. 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 [<12 19 28 34|, <26 41 60 72|] whose rows are the two vals we considered above. The Smith form itself is the 2x4 matrix [[1 0 0 0], [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, [[-11 19 4 13], [7 -12 -4 10], [0 0 1 0], [0 0 0 1]]. Inverting this matrix gives another square integral matrix, [<12 19 28 34|, <7 11 16 19|, <0 0 1 0|, <0 0 0 1|]. The rank of V is two, so to find a basis for the saturation of V, we take the first two rows, which gives us the group generated by [<12 19 28 34|, <7 11 16 19|]. The normal val list for this is [<1 0 -4 -13|, <0 1 4 10|], which are period and generator maps for septimal meantone, which is the saturated temperament corresponding to the contorted V. | To give an example, consider the matrix [<12 19 28 34|, <26 41 60 72|] whose rows are the two vals we considered above. The Smith form itself is the 2x4 matrix [[1 0 0 0], [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, [[-11 19 4 13], [7 -12 -4 10], [0 0 1 0], [0 0 0 1]]. Inverting this matrix gives another square integral matrix, [<12 19 28 34|, <7 11 16 19|, <0 0 1 0|, <0 0 0 1|]. The rank of V is two, so to find a basis for the saturation of V, we take the first two rows, which gives us the group generated by [<12 19 28 34|, <7 11 16 19|]. The [[Normal lists|normal val list]] for this is [<1 0 -4 -13|, <0 1 4 10|], which are period and generator maps for septimal meantone, which is the saturated temperament corresponding to the contorted V. | ||
To test for saturation, we may take the wedge product of the generators. Wedging <26 41 60 72| with <12 19 28 34| gives us <<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.</pre></div> | To test for saturation, we may take the wedge product of the generators. Wedging <26 41 60 72| with <12 19 28 34| gives us <<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.</pre></div> | ||
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Because unsaturated subgroups of Z^n are problematic, it is useful to have a means to saturate them; that is, to find the minimal saturated subgroup of Z^n 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 V to <a class="wiki_link_ext" href="http://en.wikipedia.org/wiki/Smith_normal_form" rel="nofollow">Smith normal form</a>. If A is a matrix with r (the rank) rows of dimension n whose rows form a basis for V, then there are two square matricies L and R, such that S = LAR, where S is the Smith normal form. The right-reducing matrix is R, the matrix multiplying A on the right. The first r rows of R generate the saturation of V. 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.<br /> | Because unsaturated subgroups of Z^n are problematic, it is useful to have a means to saturate them; that is, to find the minimal saturated subgroup of Z^n 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 V to <a class="wiki_link_ext" href="http://en.wikipedia.org/wiki/Smith_normal_form" rel="nofollow">Smith normal form</a>. If A is a matrix with r (the rank) rows of dimension n whose rows form a basis for V, then there are two square matricies L and R, such that S = LAR, where S is the Smith normal form. The right-reducing matrix is R, the matrix multiplying A on the right. The first r rows of R generate the saturation of V. 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.<br /> | ||
<br /> | <br /> | ||
To give an example, consider the matrix [&lt;12 19 28 34|, &lt;26 41 60 72|] whose rows are the two vals we considered above. The Smith form itself is the 2x4 matrix [[1 0 0 0], [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, [[-11 19 4 13], [7 -12 -4 10], [0 0 1 0], [0 0 0 1]]. Inverting this matrix gives another square integral matrix, [&lt;12 19 28 34|, &lt;7 11 16 19|, &lt;0 0 1 0|, &lt;0 0 0 1|]. The rank of V is two, so to find a basis for the saturation of V, we take the first two rows, which gives us the group generated by [&lt;12 19 28 34|, &lt;7 11 16 19|]. The normal val list for this is [&lt;1 0 -4 -13|, &lt;0 1 4 10|], which are period and generator maps for septimal meantone, which is the saturated temperament corresponding to the contorted V.<br /> | To give an example, consider the matrix [&lt;12 19 28 34|, &lt;26 41 60 72|] whose rows are the two vals we considered above. The Smith form itself is the 2x4 matrix [[1 0 0 0], [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, [[-11 19 4 13], [7 -12 -4 10], [0 0 1 0], [0 0 0 1]]. Inverting this matrix gives another square integral matrix, [&lt;12 19 28 34|, &lt;7 11 16 19|, &lt;0 0 1 0|, &lt;0 0 0 1|]. The rank of V is two, so to find a basis for the saturation of V, we take the first two rows, which gives us the group generated by [&lt;12 19 28 34|, &lt;7 11 16 19|]. The <a class="wiki_link" href="/Normal%20lists">normal val list</a> for this is [&lt;1 0 -4 -13|, &lt;0 1 4 10|], which are period and generator maps for septimal meantone, which is the saturated temperament corresponding to the contorted V.<br /> | ||
<br /> | <br /> | ||
To test for saturation, we may take the wedge product of the generators. Wedging &lt;26 41 60 72| with &lt;12 19 28 34| gives us &lt;&lt;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.</body></html></pre></div> | To test for saturation, we may take the wedge product of the generators. Wedging &lt;26 41 60 72| with &lt;12 19 28 34| gives us &lt;&lt;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.</body></html></pre></div> | ||