Contributions to module-theoretic classification of musical motifs

Invariant: Range

As one step towards the algorithmic solution of the classification in Z12 x Zn, we now describe an invariant of the isomorphy class of a subset in any abelian group.

Definition:

Let G be an abelian group and m = {v0, v1, ..., vt} be a set in G (it has t+1 elements). Then we say

Rm := <v1-v0, v2-v0, ..., vt-v0> (the subgroup generated by the elements vk-v0)

is the range of the set m.

The name "range" was chosen because Rm actually is the range of a certain module homomorphism; we will treat this in a later chapter.

Rm has the following properties:

• Rm is independent of the numbering of m since vk-vl = (vk-v0) - (vl-v0).
• Rm is invariant under translations of m since (vk+w)-(v0-w) = vk-v0.
• If f is an automorphism of G, then Rf(m) = f(Rm) and is isomorphic to Rm.

From this follows:

Application to motifs with 2 elements

If the motif has 2 elements: m = {v0, v1}, then Rm has exactly one generator. If G is finite, then Rm is a cyclical group of the order o(v1-v0). Since the isomorphy class of a cyclical group is determined by its order, we conclude that o(v1-v0) is an invariant of the isomorphy class of the motif {v0, v1}.

Applying this to the already known classification in Z12 x Z12, we see that this invariant is sort of "complementary" to the other invariant, gcd of the coordinates of v1-v0: If the gcd of the coordinates of v1-v0 is 1, then its order is 12; if the gcd is 2, then the order is 6; if the gcd is 3, then the order is 4 and so on.

Hence in Z12 x Z12 Rm is not only an invariant but determines the isomorphy class uniquely. We can not rely on this to hold in arbitrary measures.

Recursive classification

As described in one of the previous chapters, the invariant Rm can be used to construct invariants for motifs of any size via taking all 2-element subsets and calculating their ranges.

Splitting into p-components: the transformation

As I said some chapters before, any finite abelian group can be written as the direct sum of its p-components. In this chapter I will construct an actual isomorphism for the case of cyclical groups Zn.

A remark about the notation I use here:
Elements of Zn I denote with uppercase letters, integers with lowercase letters. Hence:

A = a + nZ = { a + n*x/x an arbitrary integer}

Let n and m be integers with gcd(n,m)=1. Then the cyclical group Z(m*n) is isomorphic to Zm x Zn. Now, how to find an isomorphism between Z(m*n) and Zm x Zn?

Let A=a+mZ be a generator element of Zm and B=b+nZ a generator element of Zn - which means that o(A)=m and o(B)=n, and this is the case if gcd(a,m)=1 and gcd(b,n)=1. (For example, in Z4 the elements 1+4Z and 3+4Z are generators whereas 2+4Z is not.)

Then the element (A,B) in Zm x Zn has the order lcm(m,n) = m*n, i.e. the subgroup generated by (A,B) is the whole group Zm x Zn. Since the subgroup generated by (A,B) is isomorphic to Z(m*n), the following function f is an isomorphism:

f: Z(m*n) ® Zm x Zn, x + (m*n)Z ® x*(A,B) = (x*a + mZ, x*b + nZ)

Inversion of f (isomorphism in the opposite direction):

Let (X,Y) be in Zm x Zn where X = x + mZ, Y = y + nZ. Now, since (A,B) is a generator element of Zm x Zn, there exist integers q, r such that

q * (A,B) = ( 1 + mZ, 0 + nZ ), r * (A,B) = ( 0 + mZ, 1 + nZ)

Now

Hence the inversion of the isomorphism f is:

g: Zm x Zn ® Z(m*n), (x + mZ, y + nZ) ® q*x+r*y + (m*n)Z

Example: ismorphism between Z12 and Z3 x Z4

This one is particularly important since it is needed to decompose the pitch component.

A generating element of Z3 x Z4 is ( 1 + 3Z, -1 + 4Z ). Then

( 1 + 3Z, 0 + 4Z ) = 4 * ( 1 + 3Z, -1 + 4Z )
( 0 + 3Z, 1 + 4Z ) = 3 * ( 1 + 3Z, -1 + 4Z ) (since 3*(-1) = -3 = 1 mod 4)

Hence we have the following pair of isomorphisms:

f: x + 12Z ® ( x + 3Z,-x + 4Z )
g: ( x + 3Z, y + 4Z ) ® 4x+3*y + 12Z

Note that the choice of the generating element is arbitrary (for each choice we get another pair of isomorphisms). I chose this particular one because the resulting formula for g is expecially "suggestive".