Saturation, torsion, and contorsion: Difference between revisions

Line 4:

== Saturation ==

A temperament matrix is '''saturated''' when it represents a temperament without any redundancies due to a common factor. A [[mapping]] is saturated when no common factor is found in its rows (i.e. generator ~~mappings~~). A [[comma basis]], the dual of a mapping, is saturated when no common factor is found in its columns (i.e. comma vectors).

A temperament matrix is '''saturated''' when it represents a temperament without any redundancies due to a common factor. A [[mapping]] is saturated when no common factor is found in its rows (i.e. generator maps). A [[comma basis]], the dual of a mapping, is saturated when no common factor is found in its columns (i.e. comma vectors).

To be more specific, a mapping is saturated if no linear combination of its rows can produce another row whose entries have a common factor (other than 1). For example, {{ket|{{map|3 0 -1}} {{map|0 3 5}}}} is ''not'' saturated, because {{map|3 0 -1}} - {{map|0 3 5}} = {{map|3 -3 6}}, which has a common factor of 3<ref>If the multiples used on the linear combinations themselves have a GCD>1, the resulting row will always have a GCD>1, and such a linear combination therefore can not be used to demonstrate unsaturation. For example, consider the matrix {{ket|{{map|1 0 -4}} {{map|0 1 4}}}}; we can find the linear combination of rows 2×{{map|1 0 -4}} + 4×{{map|0 1 4}} = {{map|2 4 8}}, which has a GCD of 2, but that's clearly a result of the fact that we used 2× and 4× of the original rows, and 2 and 4 have a GCD of 2. So this linear combination doesn't prove that the matrix is unsaturated. It's still possible that another linear combination might prove it, but this one does not.</ref>. A mapping which consists of a single row with a common factor, such as {{ket|{{map|24 38 56}}}} with a visible common factor of 2, is also not saturated.

@@ Line 4: / Line 4: @@
 == Saturation ==
-A temperament matrix is '''saturated''' when it represents a temperament without any redundancies due to a common factor. A [[mapping]] is saturated when no common factor is found in its rows (i.e. generator mappings). A [[comma basis]], the dual of a mapping, is saturated when no common factor is found in its columns (i.e. comma vectors).
+A temperament matrix is '''saturated''' when it represents a temperament without any redundancies due to a common factor. A [[mapping]] is saturated when no common factor is found in its rows (i.e. generator maps). A [[comma basis]], the dual of a mapping, is saturated when no common factor is found in its columns (i.e. comma vectors).
 To be more specific, a mapping is saturated if no linear combination of its rows can produce another row whose entries have a common factor (other than 1). For example, {{ket|{{map|3 0 -1}} {{map|0 3 5}}}} is ''not'' saturated, because {{map|3 0 -1}} - {{map|0 3 5}} = {{map|3 -3 6}}, which has a common factor of 3<ref>If the multiples used on the linear combinations themselves have a GCD>1, the resulting row will always have a GCD>1, and such a linear combination therefore can not be used to demonstrate unsaturation. For example, consider the matrix {{ket|{{map|1 0 -4}} {{map|0 1 4}}}}; we can find the linear combination of rows 2×{{map|1 0 -4}} + 4×{{map|0 1 4}} = {{map|2 4 8}}, which has a GCD of 2, but that's clearly a result of the fact that we used 2× and 4× of the original rows, and 2 and 4 have a GCD of 2. So this linear combination doesn't prove that the matrix is unsaturated. It's still possible that another linear combination might prove it, but this one does not.</ref>. A mapping which consists of a single row with a common factor, such as {{ket|{{map|24 38 56}}}} with a visible common factor of 2, is also not saturated.