User:Frostburn/Geometric algebra for regular temperaments: Difference between revisions

The simplest kinds of tunings only have a finite set of notes like our western [[12edo|12-tone equal temperament]]. To understand such scales harmonically we need to know how to translate musically meaningful fractions such as [[3/2]] or [[5/3]] into numbers of steps in the scale of interest. Let's start with 5 equal divisions of the octave (the octave is the fraction 2/1). How should we represent the next prime 3/1? If 5 steps is the octave then the closest we can get to 3/1 is with 8 steps. The next prime 5/1 takes 12 steps of our 5-tone scale. These step counts are collected into a vector <math>< 5, 8, 12 ]</math> known as a [[Val|val]] where the prime factorization of intervals of interest are collected into vectors known as [[Monzo|monzos]]: <math>3/2 \mapsto [ -1, 1, 0 ></math> and <math>5/3 \mapsto [ 0, -1, 1 ></math>. We can now use the dot product to work out how many steps we need to represent 3/2 (known as the fifth) <math>< 5, 8, 12 | -1, 1, 0 > = 8 - 5 = 3</math> steps of 5-tone equal temperament. Similarly 5/3 (known as the major sixth) equals <math>< 5, 8, 12 | 0, -1, 1 > = 12 - 8 = 4</math> steps.

Notice that it doesn't matter if we scale our val. <math>< 10, 16, 24 ]</math> represents only 5 unique steps within 10-tone equal temperament. There is no integral monzo (that is, no rational number) that would map to an odd number of steps in this rescaled version. Because the sizes don't matter when we add vals together we're producing a sort of average. Let's take the val for 7-tone equal temperament <math>< 7, 11, 16 ]</math> and add it to <math>< 5, 8, 12 ]</math>. The result is <math>< 12, 19, 28 ]</math> which just happens to line up with the val for our familiar 12-tone equal temperament. If we add the vals for 7-tone and 12-tone together we get <math>< 17, 27, 40 ]</math> which is different from the optimal ([[Patent val|patent]]) val for 17-tone equal temperament <math>< 17, 27, \mathbf{39} ]</math>. By insisting that the first non-zero component of a val is positive and that all of the components are in lowest terms (GCD = 1) we get 1-[[Wedgies and multivals|wedgies]] which are unique identifiers of tunings.

:<math> < 5, 8, 12 ] \mapsto 5e_1 + 8e_2 + 12e_3 =: \overleftarrow{5} </math>

In higher dimensions with algebra <math>\mathcal{G}(n,0)</math> pseudovals can be decomposed into a wedge of <math>(n-1)</math> vals and vals into a vee of <math>(n-1)</math> pseudovals.

To find the '''period''' (in n dimensions). Calculate <math>d = gcd(\mathbf{W}_{12}, \mathbf{W}_{13}, \ldots, \mathbf{W}_{1n})</math>, where <math>\mathbf{W}_{1n}</math> are all of the bivector components associated with <math>e_1</math> (i.e. with prime 2) of the 2-wedgie for the temperament. Your period will be 1\d in backslash notation i.e. one step of d equal divisions of the octave.

To find the '''generator'''. Solve the equation <math>c_2\mathbf{W}_{12} + c_3\mathbf{W}_{13} + \ldots + c_n\mathbf{W}_{1n} = d</math> for the coefficients <math>c_2,\ldots,c_n</math> (using some algorithm such as the [[Wikipedia:extended Euclidean algorithm|extended Euclidean algorithm]]). Then one valid generator for the temperament is <math>\overrightarrow{g}=c_2e_2 + c_3e_3 + \ldots + c_ne_n</math>. Thus (the tempered version of) the generator represents <math>q_2^{c_2} \cdots q_n^{c_n}</math> where 2.''q''₂.(…).''q''_n is the [[JI subgroup]].

Canonize rank-2 temperament <math>\mathbf{T}</math> to its wedgie form <math>\mathbf{W}</math> and find integral <math>\overrightarrow{g}</math> such that

:<math>\overrightarrow{g} \cdot (e_1 \cdot \mathbf{W}) = d</math>

where <math>d = gcd(...(e_1 \cdot \mathbf{W}))</math>. The period can be expressed (non-integrally) as <math>\overrightarrow{p} = e_1 / d</math> and a valid generator is <math>\overrightarrow{g}</math>. Other valid generators are of the form