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 of relations on the set of the major, minor, and augmented triads. It is generated by the three relations , , and which are pictured below.

Equivalently, we can say that we have defined a functor . We have seen that we can use category theory to define musical networks with values in . We can then transform these networks into other networks by complete isographies: these are automorphisms of the functor . More precisely, they are defined by the data of an automorphism of , and a natural transformation , which in this case will correspond to the data of a bijection of the set onto itself.

So, in the song *Take a Bow*, Muse used one automorphism of 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 and to see how it may be used for composition.

**The automorphisms of **

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

**The natural transformations **

Now this part is a bit harder. Given an automorphism in , we need to find a bijection such that for any and any , we have (this is the definition for a natural transformation ; in an abuse of notation, since there is only one set here, I’m writing for both the natural transformation and its only component on ).

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 .

Now, assume that we have chosen an automorphism of . Let’s assume, without loss of generality, that . We will now look for the possible bijections of . It is quite obvious that the bijection will map the subset onto itself. Take the element in the singleton (from now on, by another abuse of notation, I will write for both the singleton and the single element it contains). At this point, we can freely choose its image by ; let’s write it , where is a permutation of . We know that is related to the elements in by the relation , so if we write the naturality condition of in this case, we will get that should be bijectively mapped by to the subset .

(Another abuse of notation: I will now consider that arithmetic operations on , , and are to be understood .)

Notice that any subset or is an orbit of one of its element by the subgroup of isomorphic to generated by . So, using the naturality condition on , we conclude that we have only two choices: either is mapped to and to , or the other way around.

Let’s decide that we want to bijectively map to , and to . Then, is bijectively mapped to , and to . We now use the naturality condition on with the morphism to conclude that should be mapped to and that should be mapped to .

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 by the image of the subsets and (the image of the subsets are determined automatically by the naturality condition).

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

By enumerating all the possibilities, we obtain the following table of the possible automorphisms of the functor . I’ve represented for an automorphism the possible natural transformations *via* the image of the subsets and . The are elements of the subgroup isomorphic to .

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

All in all, the group of automorphisms of the functor is of order 7776 and, if I’m not mistaken, is isomorphic to .

In Muse’s Take a Bow, one element of this group of automorphisms is applied repeatedly on the cell –– 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 and the second one such that ), 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…