Composing with automorphisms

Can you compose with automorphisms ? Well, this is a question which is very vague, so let me be more specific.



If you followed the latest posts on musical networks and graphs, you’ll surely have recognized in the picture above the ‘Cube Dance’ graph of Douthett and Steinbach. In red, I’ve highlighted the opening chords of Muse’s pop song Take a Bow.

As I mentionned at the end of this post, the whole progression of chords in the first part of this song can be described by the iterated application of a so-called complete isography on the musical network corresponding to this three-chord cell. Let’s do a quick recap: in that previous post, we defined a monoid M_{\mathcal{U}\mathcal{P}\mathcal{L}} of relations on the set X of the major, minor, and augmented triads. It is generated by the three relations \mathcal{U}, \mathcal{P}, and \mathcal{L} which are pictured below.


Equivalently, we can say that we have defined a functor S \colon M_{\mathcal{U}\mathcal{P}\mathcal{L}} \to \mathbf{Rel}. We have seen that we can use category theory to define musical networks with values in S. We can then transform these networks into other networks by complete isographies: these are automorphisms of the functor S. More precisely, they are defined by the data of an automorphism N of M_{\mathcal{U}\mathcal{P}\mathcal{L}}, and a natural transformation \eta \colon S \to SN, which in this case will correspond to the data of a bijection of the set X onto itself.

So, in the song Take a Bow, Muse used one automorphism of S and applied it repeatedly to get the next chords. But are there others ? In this post, I propose to (informally) compute the automorphism group of the functor  S \colon M_{\mathcal{U}\mathcal{P}\mathcal{L}} \to \mathbf{Rel} and to see how it may be used for composition.


The automorphisms N of M_{\mathcal{U}\mathcal{P}\mathcal{L}}

First, we need to compute the automorphisms of the M_{\mathcal{U}\mathcal{P}\mathcal{L}} monoid. This is actually quite easy ! The automorphisms are defined by the image of the generators. Notice that the \mathcal{P} and \mathcal{L} relations are in fact bijections of the set X, whereas the relation \mathcal{U} is not. Therefore, they can only be sent to inversible elements of M_{\mathcal{U}\mathcal{P}\mathcal{L}}. Since \mathcal{P} and \mathcal{L} generate the dihedral group D_6, their image induce an automorphism of this subgroup. From known results, the automorphism group of D_6 is isomorphic to D_6. Thus, we obtain a homomorphism from \text{Aut}(M_{\mathcal{U}\mathcal{P}\mathcal{L}}) to D_6, and its kernel corresponds to the subgroup generated by the automorphisms in which \mathcal{P} and \mathcal{L} are left unchanged, thus reducing to the question of the image of \mathcal{U}. An exhaustive search yields that only \mathcal{U} and \mathcal{PUP} are valid candidates, and we can check that the kernel is thus isomorphic to \mathbb{Z}_2. So \text{Aut}(M_{\mathcal{U}\mathcal{P}\mathcal{L}}) is an extension of \mathbb{Z}_2 by D_6. It is a matter of technical details (which I’m leaving out here) to check that \text{Aut}(M_{\mathcal{U}\mathcal{P}\mathcal{L}}) is in fact a direct product D_6 \times \mathbb{Z}_2.


The natural transformations \eta

Now this part is a bit harder. Given an automorphism N in \text{Aut}(M_{\mathcal{U}\mathcal{P}\mathcal{L}}), we need to find a bijection \eta \colon X \to X such that for any x \in X and any m \in M_{\mathcal{U}\mathcal{P}\mathcal{L}}, we have \eta(m(x)) = N(m)(\eta(x)) (this is the definition for a natural transformation \eta \colon S \to SN; in an abuse of notation, since there is only one set here, I’m writing \eta for both the natural transformation and its only component on X).

Before we go any further, a little bit of additional notation. I’ve represented below the ‘Cube Dance’ graph, and I’ve identified and named in red some subsets of X.

Now, assume that we have chosen an automorphism N of M_{\mathcal{U}\mathcal{P}\mathcal{L}}. Let’s assume, without loss of generality, that N(\mathcal{U})=\mathcal{U}. We will now look for the possible bijections \eta of X. It is quite obvious that the bijection \eta will map the subset 0_a \cup 3_a \cup 6_a \cup 9_a onto itself. Take the element A\flat_{\text{aug}} in the singleton 0_a (from now on, by another abuse of notation, I will write x_a for both the singleton and the single element it contains). At this point, we can freely choose its image by \eta; let’s write it \sigma(0)_a, where \sigma is a permutation of 0_a \cup 3_a \cup 6_a \cup 9_a. We know that 0_a is related to the elements in 0_+ \cup 9_- by the relation \mathcal{U}, so if we write the naturality condition of \eta in this case, we will get that 0_+ \cup 9_- should be bijectively mapped by \eta to the subset \sigma(0)_+ \cup (\sigma(0)-3 \mod 12)_-.

(Another abuse of notation: I will now consider that arithmetic operations on x_+, x_-, and x_a are to be understood \mod 12.)

Notice that any subset x_+ or x_- is an orbit of one of its element by the subgroup of D_6 isomorphic to \mathbb{Z}_3 generated by \mathcal{L}\mathcal{P}. So, using the naturality condition on \eta, we conclude that we have only two choices: either 0_+ is mapped to \sigma(0)_+ and 9_- to (\sigma(0)-3)_-, or the other way around.

Let’s decide that we want to bijectively map 0_+ to (\sigma(0)-3)_-, and 9_- to \sigma(0)_+. Then, 0_- is bijectively mapped to (\sigma(0)-3)_+, and 9_+ to \sigma(0)_-. We now use the naturality condition on \eta with the morphism \mathcal{U} to conclude that 3_a should be mapped to (\sigma(0)-3)_a and that 9_a should be mapped to (\sigma(0)+3)_a.

I won’t detail the rest, but if you continue chasing the images of the subsets by using the naturality condition, you will end up with a complete determination of \eta by the image of the subsets x_a and x_+ (the image of the subsets x_- are determined automatically by the naturality condition).

However, we need to determine how the subsets x_+ are mapped bijectively to their images. Since any subset x_+ is an orbit of one of its element by the subgroup of D_6 isomorphic to \mathbb{Z}_3 generated by \mathcal{L}\mathcal{P}, we can conclude that we only need to pick a representative of x_+ and choose to send it either to an element in its image subsets. In fact, if we pick representatives in each subset x_+, this is equivalent to choosing for each subset x_+ an element of the subgroup isomorphic to \mathbb{Z}_3.

By enumerating all the possibilities, we obtain the following table of the possible automorphisms of the functor S \colon M_{\mathcal{U}\mathcal{P}\mathcal{L}} \to \mathbf{Rel}. I’ve represented for an automorphism N the possible natural transformations \eta via the image of the subsets x_a and x_+. The g_i are elements of the subgroup isomorphic to \mathbb{Z}_3.

Composition of these automorphisms can be obtained by juxtaposing the diagrams, as shown in the picture below, taking care that the automorphism N_2 applies on the group elements h_i.

All in all, the group of automorphisms of the functor S \colon M_{\mathcal{U}\mathcal{P}\mathcal{L}} \to \mathbf{Rel} is of order 7776 and, if I’m not mistaken, is isomorphic to (({\mathbb{Z}_3}^4 \rtimes \mathbb{Z}_2) \rtimes D_8) \rtimes D_6.

In Muse’s Take a Bow, one element of this group of automorphisms is applied repeatedly on the cell D_MD_{\text{aug}}G_m to obtain the chords in the opening progression. Can we “compose” something else by using other automorphisms ? I’ve chosen two different elements (one such that N(\mathcal{U})=\mathcal{P}\mathcal{U}\mathcal{P} and the second one such that N(\mathcal{U})=\mathcal{U}), each of order 12, and applied them repeatedly on the same initial cell. You can listen to the results in the two videos below.

To be honest, this is not really composition (more some sort of “enumerative composition” à la Tom Johnson), but who knows ? It might be used as a creative tool for something more elaborate…

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