Rank and codimension: Difference between revisions

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Looking only at the number of independent generators of a tuning can obscure its real nature, at least as it is being applied. For instance, a 31et tuning of meantone temperament, with a meantone fifth of 18\31 octaves, is of rank one in the sense that all the intervals in the tuning are generated from 1\31; however, it is being used as a rank two tuning. This issue can be gotten around by means of [[Abstract_regular_temperament|abstract regular temperaments]]; an abstract regular temperament is of rank r if it is defined by a [[Normal_lists|normal val list]] of r vals, or equivalently by an r-multival. The abstractly characterized intervals of the abstract temperament can then be mapped to a tuning; if the mapping is to a rank one tuning such as 31et, that does not affect the rank of the temperament.

Although the term "rank" as used here is exactly the same as used in group theory and linear algebra, it is important to note that the term "co-rank" is being used slightly differently. In both cases, the co-rank is the dimension of the cokernel (the quotient of codomain by image), and hence can be thought of as measuring the degree to which a homomorphism fails to be surjective. However, for any so-called temperament, if the group-theoretic co-rank isn't 0, it isn't a temperament at all - it is contorted. And if the linear-algebraic co-rank isn't 0, that's even worse - it means you have a completely free generator with no mapping specified at any point along the chain. So the both the group-theoretic co-rank and the linear-algebraic co-rank are useless pieces of information for a temperament - they are always 0.

Although the term "rank" as used here is exactly the same as used in group theory and linear algebra, it is important to note that the term "co-rank" is being used slightly differently. In both cases, the co-rank is the dimension of the cokernel (the quotient of codomain by image), and hence can be thought of as measuring the degree to which a homomorphism fails to be surjective. However, for any so-called temperament, if the group-theoretic co-rank isn't 0, it isn't a temperament at all - it is [[contorted]] (or [[enfactored]]). And if the linear-algebraic co-rank isn't 0, that's even worse - it means you have a completely free generator with no mapping specified at any point along the chain. So the both the group-theoretic co-rank and the linear-algebraic co-rank are useless pieces of information for a temperament - they are always 0.

The thing called "codimension" above can be interpreted in linear algebra terms as the codimension of the subspace of supporting vals, relative to the ambient space of all vals. For any mapping matrix mapping from monzos to tmonzos, it's also the co-rank of the dual transformation from tvals back to vals.

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 Looking only at the number of independent generators of a tuning can obscure its real nature, at least as it is being applied. For instance, a 31et tuning of meantone temperament, with a meantone fifth of 18\31 octaves, is of rank one in the sense that all the intervals in the tuning are generated from 1\31; however, it is being used as a rank two tuning. This issue can be gotten around by means of [[Abstract_regular_temperament|abstract regular temperaments]]; an abstract regular temperament is of rank r if it is defined by a [[Normal_lists|normal val list]] of r vals, or equivalently by an r-multival. The abstractly characterized intervals of the abstract temperament can then be mapped to a tuning; if the mapping is to a rank one tuning such as 31et, that does not affect the rank of the temperament.
-Although the term "rank" as used here is exactly the same as used in group theory and linear algebra, it is important to note that the term "co-rank" is being used slightly differently. In both cases, the co-rank is the dimension of the cokernel (the quotient of codomain by image), and hence can be thought of as measuring the degree to which a homomorphism fails to be surjective. However, for any so-called temperament, if the group-theoretic co-rank isn't 0, it isn't a temperament at all - it is contorted. And if the linear-algebraic co-rank isn't 0, that's even worse - it means you have a completely free generator with no mapping specified at any point along the chain. So the both the group-theoretic co-rank and the linear-algebraic co-rank are useless pieces of information for a temperament - they are always 0.
+Although the term "rank" as used here is exactly the same as used in group theory and linear algebra, it is important to note that the term "co-rank" is being used slightly differently. In both cases, the co-rank is the dimension of the cokernel (the quotient of codomain by image), and hence can be thought of as measuring the degree to which a homomorphism fails to be surjective. However, for any so-called temperament, if the group-theoretic co-rank isn't 0, it isn't a temperament at all - it is [[contorted]] (or [[enfactored]]). And if the linear-algebraic co-rank isn't 0, that's even worse - it means you have a completely free generator with no mapping specified at any point along the chain. So the both the group-theoretic co-rank and the linear-algebraic co-rank are useless pieces of information for a temperament - they are always 0.
 The thing called "codimension" above can be interpreted in linear algebra terms as the codimension of the subspace of supporting vals, relative to the ambient space of all vals. For any mapping matrix mapping from monzos to tmonzos, it's also the co-rank of the dual transformation from tvals back to vals.