Generator preimage

Definition

Given a reduced list of p-limit vals V, we may define a set of transversal generators for V as a set of p-limit intervals q such that one of the vals of V maps q to 1 and the rest map it to 0. By reduced is meant that the gcd of the elements of each of the vals is 1--or in other words, none of the vals are contorted--and that they are linearly independent, so that if there are r vals, the rank of V as a matrix is r (or in other words, that the set is defactored).

If v1, v2, ... vr are the vals of V and t1, t2, ... tr are the corresponding transversal generators, then for any p-limit q we have, modulo the regular temperament defined by V

q ≅ t1^v1(q) * t2^v2(q) * ... * tr^vr(q)

In this way the transversal generators provide a transversal of the p-limit, and hence the name.

Examples

Suppose V consists of the 7-limit patent vals for 12 and 19; that is, V = [<12 19 28 34|, <19 30 44 53|]. Then a corresponding list of transversal generators is [49/48, 36/35]. 49/48 corresponds to one step of 12et, and zero steps of 19et, whereas 36/35 is zero steps of 12et, and one step of 19et. This gives us a septimal meantone transversal of the 7-limit where 3/2 is represented by (49/48)^7 * (36/35)^11, and 2 is represented by (49/48)^12 * (36/35)^19. A more familiar septimal meantone transversal starts from the normal val list, [<1 0 -4 -13|, <0 1 4 10|], which corresponds to the transversal generators [2, 3].

Given a list of transversal generators, we may append a comma basis for V and obtain a basis for the entire p-limit. For instance, we may extend [49/48, 36/35] to [49/48, 36/35, 81/80, 126/125]. Taking the corresponding matrix of monzos, whose rows are monzos for this list, inverting it and then transposing, we obtain

[<12 19 28 34|, <19 30 44 53|, <-4 -6 -9 -11|, <-5 -8 -12 -14|]

This is a unimodular matrix defining a change of basis for the p-limit.

For another example, consider [<1 1 1 2|, <0 2 1 1|, <0 0 2 1|] which is the normal val list for breed temperament, the temperament tempering out 2401/2400. A corresponding list of transversal generators is [2, 49/40, 10/7].

Finding the transversal generators

We can find transveral generators for V by the following procedure:

• Take the transpose of the pseudoinverse of V, call that U
• Find a basis for the commas of V
• For each row U[i] of U, clear denominators and append the monzos of the comma basis for V
• Saturate (or defactor) the result to a list of monzos, call that S
• Apply the ith val V[i] (dot product) to each element of S
• Insert V[i].S[j] in front of the elements of S[j] as the first element, obtaining the jth element T[j] of a modified list T
• Hermite reduce the modified list T, take the first row, and remove the first element (which should be a 1.)
• Consider the rest to be a monzo and convert it to a rational number
• This is a corresponding transveral generator to the ith val V[i] of V; it may be reduced to an equivalent generator of minimal Tenney height by multiplying by the commas of V

Example

Note: I've followed the algorithm as described above. But clearly I am transposing things here way more often than is necessary than if the algorithm was superficially revised.

For 5-limit meantone, [math]\displaystyle{ V }[/math] is

[math]\displaystyle{ \begin{bmatrix} 1 & 2 & 4 \\ 0 & -1 & -4 \\ \end{bmatrix} }[/math]

The pseudoinverse [math]\displaystyle{ V⁺ }[/math] is

[math]\displaystyle{ \frac{1}{33} \begin{bmatrix} 17 & 18 \\ 16 & 15 \\ -4 & -12 \\ \end{bmatrix} }[/math]

And that transposed is [math]\displaystyle{ (V⁺)ᵀ }[/math], which we'll call [math]\displaystyle{ U }[/math]:

[math]\displaystyle{ \frac{1}{33} \begin{bmatrix} 17 & 16 & -4 \\ 18 & 15 & -12 \\ \end{bmatrix} }[/math]

And here's the comma basis for [math]\displaystyle{ V }[/math]:

[math]\displaystyle{ \begin{bmatrix} -4 \\ 4 \\ -1 \\ \end{bmatrix} }[/math]

Beginning with [math]\displaystyle{ i = 1 }[/math], we'll create a matrix out of [math]\displaystyle{ U[i] }[/math] with denominators cleared and the comma basis appended (transposed so each comma is a row):

[math]\displaystyle{ \begin{bmatrix} 17 & 16 & -4 \\ -4 & 4 & -1 \\ \end{bmatrix} }[/math]

Defactor and call this [math]\displaystyle{ S }[/math]:

[math]\displaystyle{ \begin{bmatrix} 1 & 32 & -8 \\ 0 & 4 & -1 \\ \end{bmatrix} }[/math]

Now take the dot product of [math]\displaystyle{ V[1] }[/math] to each element of [math]\displaystyle{ S }[/math], or in other words, left multiply [math]\displaystyle{ Sᵀ }[/math] by [math]\displaystyle{ V[1] }[/math]:

[math]\displaystyle{ \begin{bmatrix} 1 & 2 & 4 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 32 & 4 \\ -8 & -1 \\ \end{bmatrix} = \begin{bmatrix} 33 & 4 \\ \end{bmatrix} }[/math]

Now prepend that result, transposed, to [math]\displaystyle{ S }[/math]:

[math]\displaystyle{ \begin{bmatrix} 33 & 1 & 32 & -8 \\ 4 & 0 & 4 & -1 \\ \end{bmatrix} }[/math]

Take the Hermite Normal Form:

[math]\displaystyle{ \begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 4 & -4 & 1 \\ \end{bmatrix} }[/math]

Take the first row:

[math]\displaystyle{ \begin{bmatrix} 1 & 1 & 0 & 0 \\ \end{bmatrix} }[/math]

Remove the first element (which should and indeed is a 1):

[math]\displaystyle{ \begin{bmatrix} 1 & 0 & 0 \\ \end{bmatrix} }[/math]

So indeed that gives the period for meantone, 2/1.

Let's repeat the latter steps but now with [math]\displaystyle{ i = 2 }[/math]. Here's [math]\displaystyle{ U[2] }[/math] with denominators cleared and the comma basis appended:

[math]\displaystyle{ \begin{bmatrix} 18 & 15 & -12 \\ -4 & 4 & -1 \\ \end{bmatrix} }[/math]

Defactor to get our new [math]\displaystyle{ S }[/math]:

[math]\displaystyle{ \begin{bmatrix} 2 & 31 & -16 \\ 0 & 2 & -1 \\ \end{bmatrix} }[/math]

Now take the dot product of [math]\displaystyle{ V[2] }[/math] to each element of [math]\displaystyle{ S }[/math], or in other words, left multiply [math]\displaystyle{ Sᵀ }[/math] by [math]\displaystyle{ V[2] }[/math]:

[math]\displaystyle{ \begin{bmatrix} 0 & -1 & -4 \\ \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 31 & 2 \\ -16 & -1 \\ \end{bmatrix} = \begin{bmatrix} 33 & 2 \\ \end{bmatrix} }[/math]

Now prepend that result, transposed, to [math]\displaystyle{ S }[/math]:

[math]\displaystyle{ \begin{bmatrix} 33 & 2 & 31 & -16 \\ 2 & 0 & 2 & -1 \\ \end{bmatrix} }[/math]

Take the Hermite Normal Form:

[math]\displaystyle{ \begin{bmatrix} 1 & 2 & -1 & 0 \\ 0 & 4 & -4 & 1 \\ \end{bmatrix} }[/math]

Take the first row:

[math]\displaystyle{ \begin{bmatrix} 1 & 2 & -1 & 0 \\ \end{bmatrix} }[/math]

Remove the first element (which should and indeed is a 1):

[math]\displaystyle{ \begin{bmatrix} 2 & -1 & 0 \\ \end{bmatrix} }[/math]

So indeed that gives the generator for meantone, 4/3. We're done!