Contributions to module-theoretic classification of musical motifs

Module-theoretic approach for the complete classification of arbitrary motifs

In this chapter I will describe another invariant that will enable us to do the complete isomorphy classification of arbitrary motifs in (Z₁₂)^2 and some other measures (not in all, though).

Let again be R the ring Z or Z₁₂ and M be an R-module. let m = {v₀, v₁, ..., v_t} be a set in M with t+1 elements. Now we define the mapping

f: R^t ® M, e_i ® v_i-v₀

where e₁, e₂, ... e_t form a basis for the R-module R^t. f is an R-module homomorphism from R^t to M. Its range is our previously described "Rm" (so you know now why we called it "range"). The kernel of f (the elements of R^t whose function value is zero) is a submodule of R^t. Now, what happens to the kernel if we apply an affine transformation to the motif m?

If we apply a translation, the mapping f will be identical since f'(e_i) = (v_i+w)-(v₀+w) = v_i-v₀ = f(e_i)

If we apply an automorphism g of M the transformed mapping will look like this:

f'(e_i) = g(v_i-g(v₀) = g(v_i-v₀) = g(f(e_i))

i.e. f = g f (the concatenation of f and g). As we showed in the chapter where we introduced Rm, the range of gf is isomorphic to the range of f; the kernel of gf, in contrary, is even IDENTICAL to the kernel of f!

However, what the kernel does depend on is the ordering of the elements of m. So we will now have to examine the effect of a permutation of the elements.

Effect of permutations

Let now s be a permutation of the set {0, 1, ...t }. Then we get the mapping

f'(e_i) = v_s(i)-v_s(0)

How do the mappings f and f' hang together?

What we are looking for is a linear mapping from R^t to R^t - which we will designate with Ñ_s - for which holds: f' = f Ñ_s (making the diagram commutative); i.e.:

If s(i) is 0 (this is the case for exactly one i) e_s(i) is not defined. But in this case v_s(i)-v₀ is 0, and we can save us a separate formula for this special case if we define (as a pure convention):

e₀:= 0 (zero vector in R^t).

Then f(e₀) vanishes, too, and for the mapping Ñ_s we get the following formula:

Ñ_s(e_i) = e_s(i) - e_s(0).

For the simple case where the permutation s just exchanges the element 0 with another element k (in cycle notation: (0 k)), we get:

Hence Ñ_s is described by the following (t x t)-matrix:

(the -1 in the k-th row)

This matrix's determinant is -1, i.e. the matrix is regular, the mapping has an inverse, and

Ker f' = Ker (f Ñ_s) = {v in R^t, such that Ñ_s(v) is in Ker f}= (Ñ_s)^-1(Ker f).

If s is an arbitrary permutation, it can be composed of simple permutations of the form (0 k), and there holds

1. Ñ_{(s1 s2)} = Ñ_s1Ñ_s2
2. Ñ_s = 1 (the unit matrix) exactly if s = 1 (the trivial permutation)

The proof for this is straightforward (I have done it in detail in my original work, but I will skip it here). So what we have here is a representation of the permutation group as a matrix group in GL ( t, R ).

Completion of the classification

We have now established a connection between subsets of the R-module M with t+1 elements and orbits of submodules of R^t with respect to a certain matrix group. So far, this is just another invariant - but it turns out that for the case R = Z₁₂ and M = (Z₁₂)^2 this invariant determines the isomorphy class for any t COMPLETELY! This is the essence of the following theorem.

Theorem:

Let R be Z₁₂. Let t be an integer and K₁ = {x₀, x₁, ..., x_t}, K₂ = {y₀, y₁, ..., y_t} two sets of R^2 with t+1 elements. As in the previous section, we define linear mappings:

f₁: R^t ® M, e_i ® x_i-x₀
f₂: R^t ® M, e_i ® y_i-y₀

Then holds:

Proof:

(As throughout this website, I merely sketch what I did in detail in my original work. Since this is a central theorem of the work, I give some more details.)

The first part of the proof ("Þ") was outlined in the first section of this chapter. Now I sketch the second part ("Ü") .

We write F_k := Ker f_k (submodules of R^t), RK_k := Range f_k (submodules of R^2). Let s be a permutation such that Ñ_s(Ker f₁) = Ker f₂. Since Ñ_s is in Gl(t,R), F₁ is isomorphic to F₂.

		fk
	R^t	®	RKk
p	¯
	R^t/Fk

(p is the canonical projection.)

The dashed line defines a function f'_k via: f'_k([x]) := f_k(x). This function is well-defined and an isomorphism from R^t/F_k to RK_k. There holds: f'₁([e_i]) = x_i-x₀, f'₂([e_i]) = y_i-y₀.

			p1
F1	Ì	R^t	®	R^t/F1
¯	Ñs	¯			Ñs'
F2	Ì	R^t	®	R^t/F2
			p2

We want a mapping Ñ_s' such that the diagram commutes.Claim: Ñ_s' is well-defined and an isomorhpism from R^t/F₁ to R^t/F₂.

Proof:
Ñ_s'([x]) = Ñ_s'(x + F₁) = Ñ_s(x + F₁) = Ñ_s(x) + F₂ = [Ñ_s(x)].
Ñ_s' is a homomorhihsm; it is surjective since Ñ_s and p₂ are.
[x] = 0 Û x Î F₁ Û Ñ_s(x) Î F₂ Û [Ñ_s(x)] = Ñ_s'([x]) = 0. Hence Ñ_s' is injective.
(Since R is finite in our case, injectivity implies surjectivity.)

		f'1
	R^t/F1	@	RK1	Ì	R^2
Ñs'	¯			?
	R^t/F2	@	RK2	Ì	R^2
		f'2

Both the mappings f'_k as Ñ_s' are isomorphisms and hence have inverses. We make the diagram commutative with the mapping

f'₂ Ñ_s' (f'₁)^-1=: a

a is an isomorphism from RK₁ to RK₂, and

a(x_i-x₀) = f'₂ Ñ_s' (f'₁)^-1(e_i) = f'₂(e_s(i)-e_s(0)) = y_s(i) - y_s(0)

Now we are nearly finished. Observe that the mapping a, combined with appropriate translations, yields an affine transformation that indeed maps K₁ to K₂! a has but one single flaw: it is not an automorphism of R^2 but an isomorphism between 2 submodules of it. The remaining task is now to find an automorphism a' of R^2 whose restriction to K₁ is a (or a' is an extension of a).

On first sight, this may seem trivial - actually, it were close to trivial if R^2 were a vector space. But for general modules or groups, it is a priori not clear whether such an automorphism exists at all! This question, BTW, is addressed in a theorem which is known as the Fitting Lemma. (Look in standard literature for more information.)

Here (at least for now) I just state that in the special case of abelian groups all whose p-components have the form Z_n^k with n a power of p, any isomorphism between 2 subgroups can always be extended to a group automorphism. (Actually, I have proved this in detail in my original work - but again, at least for now, I skip the details here...)

So, with this theorem at hand, the classification of musical motifs in Z₁₂^2 up to isomorphy is completed!