Equivalence continuum

An equivalence continuum is the space of all rank-k temperaments on a specified JI subgroup that are supported by a specified temperament of a lower rank on the same subgroup (such as an edo viewed as a temperament on said subgroup).

Examples:

• The syntonic-chromatic equivalence continuum is the 5-limit rank-2 equivalence continuum of 7edo.

Mathematical theory

Mathematically, the rank-k equivalence continuum C(k, T, S) associated with a rank-r temperament T on a rank-n subgroup S is the space of saturated n-k-dimensional sublattices of the kernel of T, the rank-(n-r) lattice of commas tempered out by T. This is a set of rational points on the Grassmannian G = Gr(n-k, n-r) = Gr(n-k, ker(T) ⊗ R).

This has a particularly simple description when r = 1 (i.e. when T is an edo), n = 3 (for example, when S is the 5-limit, 2.3.7 or 2.5.7) and k = 2 (so that we're considering the equivalence continua of rank-2 temperaments associated with an edo), as then G = Gr(1, 2) = RP¹ (the real projective line), which can be viewed as a circle. Then the continuum corresponds to the set of lines with rational slope passing through the origin on the Cartesian plane R² where the lattice of ker(T) lives. The lattice of ker(T) is generated by a basis of some choice of two commas u and v in S tempered out by the edo; view the plane as having two perpendicular axes corresponding to u and v directions. A rational point, i.e. a temperament on the continuum, then corresponds to a rational ratio p/q, where u^p/v^q is tempered out by the temperament.

A higher-dimensional example: Say that r = 1, n = 4 (e.g. when S is the 7-limit), and k = 2, for example the set of rank-2 7-limit temperaments supported by 31edo. Then our Grassmannian G becomes Gr(2, 3). Define a coordinate system (x, y, z) for ker(T) using some fixed comma basis u_x, u_y, u_z for ker(T). Then our Grassmannian can be identified with RP² (the real projective plane, the space of lines through the origin in 3-dimensional space) by taking the unique line Rv perpendicular (according to the dot product given by the given coordinates) to the plane of commas tempered out for each temperament. Say that the vector v defining the unique line has components (v₁, v₂, v₃), so that the plane associated with the rank-2 temperament has equation v₁x + v₂y + v₃z = 0. [We may further assume that v₁, v₂, v₃ are integers with gcd 1, since the condition of being perpendicular to two integer vectors is defined by a system of linear equations with integer coefficients, thus has a unique rational solution up to scaling.] One coordinate v_i is always guaranteed to be nonzero. If v₁ ≠ 0 and we scale v by 1/v₁, then the resulting vector v/v₁ = (1, v₂/v₁, v₃/v₁) = (1, s, t) points in the same direction as v and describes two rational (or infinite) parameters s and t which defines any temperament with v₁ ≠ 0 on 31edo's 7-limit rank-2 continuum uniquely. Similarly for every temperament with v₂ ≠ 0 and every temperament with v₃ ≠ 0. Note that this continuum is actually part of a mathematical manifold with a more complicated topology and needs to be described using more than one local chart (coordinate system) constructed like this; unlike for the k - r = 1 case, a single circle won't define every point on this 2-dimensional continuum, just like a single circle won't define every point on a 2-dimensional sphere.

NB: Not all tuning continua are projective spaces, because Gr(2, 4) is not a projective space.

Example (7-limit rank-2 temperaments in 31edo)

Let ux, uy, uz = 81/80, 126/125, 1029/1024 be the basis for the kernel of 7-limit 31edo. Then:

• septimal meantone tempers out 81/80 = ux = (1, 0, 0) and 126/125 = uy = (0,1,0), thus corresponds to the plane z = 0. This corresponds to v = (0, 0, 1).
• hemithirds tempers out 1029/1024 = uz = (0, 0, 1) and 3136/3125 = 2ux + uy = (2, 1, 0). This corresponds to v = (1, -2, 0).