Emmanuel Amiot Modèles algébriques et algorithmes pour la ...

24 autosimilar melodies
But in Z15 for instance, X 2 = 1 has other solutions (eg 4); taking a = 2, n = 15 and filling in the ordered
orbits (1 2 4 8), (3 6 12 9), (5 10), (7 14 13 11) with alternate opposite values of one note (0 is C, 1 is C♯,
2 is B, a.s.o.), we get by construction M f(k) = −Mk ∀k, close to Feldman’s example.
7.5.2 Proof of Thm. 7.2. Proof Consider an autosimilar melody generated by f : x ↦→ a x + b and let
r = −a −1 , which is coprime with n; hence the sequence
M0, Mr, M2r, . . .
contains all the notes of the sequence M0, M1, . . . though in a different order. We know by construction
that
Hence (putting y = a x + b)
∀x, Ma x+b = Mx.
∀y, My = M−r y+c ⇐⇒ Mr y = Md−y
for some offsets c, d. This is what we endeavoured to prove: that some augmentation of the melody is one
of its retrogrades.
7.5.3 Inverse-retrograde symmetry. The last situation is about melodies whose inverse IS the retrograde
(like in Tom Johnson’s conjecture).
For instance, with f : x ↦→ 3x + 1 mod 26: see fig. 14.
Figure 14. Autosimilar melody (ratio 3) with inverse-retrograde symmetry
It can be seen, and even better, heard, that O0 (the B flat’s) and O8 (the C’s) (resp. O2 and O3)
are retrogrades one of another. This allows a pretty rendition of the melody, setting opposite notes for
symmetric orbits: then the retrograde of the melody will be its inversion, as seen on figure 14 (the symmetry
axis for pitches is around F).
It is somewhat difficult in fact, to construct an example of an autosimilar (primitive) structure without
such a symmetry (this is akin to Johnson’s conjecture). For one thing, if f is a homothety (recall this
happens whenever a − 1 | b in Zn, not only when b = 0), then x ↦→ −x permutes the orbits, as any other
homothety does. Also if some power of a is equal to c − 1, we get directly a palindrom.
Still, an autosimilar melody built from 4x + 1 mod 21 does the trick, as its orbits
(0, 1, 5), (2, 9, 16), (3, 11, 13), (4, 6, 17), (7, 8, 12), (10, 18, 20), (14, 15, 19)
exhibit no inversional symmetry whatsoever. We have given in the main text a condition ensuring that
such retrogradation symmetries between orbits exist: Thm. 7.3. Here is its proof.
Proof We assume that 2b = c(a − 1) and consider symmetry S : x ↦→ −c − x. The theorem states that S
acts on the set of all orbits, i.e. the symmetric of any orbit is some orbit, i.e. for any x, S(Ox) = O S(x).
Recall Ox = {x, f(x), f 2 (x), . . . } where f(x) = a x + b; it is sufficient to notice that f ◦ S = S ◦ f, i.e.
f(S(x)) = a (−x − c) + b = −a x + b − a c = −a x − b − c = −c − (a x + b) = S(f(x))