Mathematical theory of saturation: Difference between revisions
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The set of n-tuples of integers Z^n such that two n-tuples can be added coordinatewise is the [http://en.wikipedia.org/wiki/Free_abelian_group free abelian group] of rank n. Its subgroups have the property of '''saturation''' if for any element a of Z^n, if an integer multiple | The set of n-tuples of integers <span><math>\mathbb{Z}^n</math></span> such that two <span><math>n</math></span>-tuples can be added coordinatewise is the [http://en.wikipedia.org/wiki/Free_abelian_group free abelian group] of rank <span><math>n</math></span>. Its subgroups have the property of '''saturation''' if for any element <span><math>a</math></span> of <span><math>\mathbb{Z}^n</math></span>, if an integer multiple <span><math>m·a</math></span> of <span><math>a</math></span> belongs to a sublattice <span><math>V</math></span>, then <span><math>a</math></span> already belongs to <span><math>V</math></span>. Another way to put it is that if some linear combination with rational coefficients <span><math>q_1v_1 + \dots + q_kv_k</math></span> of elements of <span><math>V</math></span> belongs to <span><math>\mathbb{Z}^n</math></span>, then it belongs to <span><math>V</math></span>. For the latter definition we consider <span><math>\mathbb{Z}^n</math></span> to be contained in the <span><math>n</math></span>-dimensional [http://en.wikipedia.org/wiki/Vector_space real vector space] <span><math>\mathbb{R}^n</math></span>, in which case <span><math>\mathbb{Z}^n</math></span> is often called the [http://en.wikipedia.org/wiki/Integer_lattice integer lattice], or grid lattice. | ||
If C represents the commas (nullspace or kernel) of a supposed regular temperament, i.e. the intervals it tempers out, then if C isn't saturated the supposed temperament it defines may be regarded as pathological, as it has notes with no clear interpretation (some JI intervals cannot be reached by a generator in the tempered lattice). For example, if (81/80) | If <span><math>C</math></span> represents the commas (nullspace or kernel) of a supposed regular temperament, i.e. the intervals it tempers out, then if <span><math>C</math></span> isn't saturated the supposed temperament it defines may be regarded as pathological, as it has notes with no clear interpretation (some JI intervals cannot be reached by a generator in the tempered lattice). For example, if (81/80)² = 6561/6400 is tempered out but 81/80 is not, then it is not clear how the tempered versions of 5/4 and 81/64 are related, as they are not the same note yet two of them in succession ''are'' the same note. This is called a ''torsion'' problem. Similarly, if <span><math>V</math></span> is the subgroup of vals of the temperament, and is not saturated, then we obtain a temperament of sorts in which all of the notes cannot be reached by tempered intervals (cannot be reached by tempering a JI interval); this at least is an actual system of musical intervals, but disconnected. This has been called a '''contorsion''' problem. | ||
For example, consider the "temperament" with commas generated by 126/125 and 3645/3584. The group generated by the [[monzo|monzos]] |1 2 -3 1 | For example, consider the "temperament" with commas generated by 126/125 and 3645/3584. The group generated by the [[monzo|monzos]] {{vector|1 2 -3 1}} and {{vector|-9 6 1 -1}} is not saturated, since (126/125)*(3645/3584) = (81/80)², but 81/80 does not belong to the group. Hence (81/80)² is tempered out, but 81/80 is not, and we have torsion. If we take the two vals {{map|12 19 28 34}} and {{map|26 41 60 72}} we similarly get contorsion. However, this 5- and 7-limit contorsion can be fixed in a way by extending to the 11-limit, and interpreting the "unobtainable" notes as notes reached in the 11-limit. Adding 245/242 to the commas (81/80 and 126/125) of septimal meantone is one way of reinterpreting the situation. | ||
Because unsaturated subgroups of Z^n are for these reasons 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 matrices 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 <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 | 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 test for saturation, we may take the wedge product of the generators. Wedging | 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. | ||
= Wolfram Language implementation = | = Wolfram Language implementation = | ||