Evanston, oct. 24th
Rhythmic Canons, Galois Theory, Spectral Conjecture by Emmanuel Amiot Perpignan, France
Rhythmic Canons • What is a rhythmic canon ? • Mathematical tools • Canons modulo p • Transformation, reduction, conservation
Rhythmic Canons •
A canon is a number of voices playing the same tune at different onsets.
•
A rhythmic canon is a number of voices playing repeatedly the same rhythmic pattern at different onsets.
A rhythmic canon is periodic (here modulo 16)
Rhythmic Canons •
The rhythmic pattern is called inner voice,
•
The set of onsets is the outer voice
•
Together they tile a cyclic group
A is the rhythmic pattern, the sequence of onsets. nto F2 [X], there is noBThis good map back into Z[X]. For instance one c e mentioned today. is related elsewhere ([?],[?]). thmic canons and tiling cyclic groups ollowing be seen that the simplification of assuming integer beats (see [?], [?]) and the periodi
Maths for canons
strations are given during the talk. factors nces) of the whole canon are not arbitrary but necessary. 2 3 canon the play+of on different voices, with diffe (1 + X)(1 Xthe + Xsame )possible = rhythmic 1+X ∈ pattern F2 5X stemming e other is hand, there are other specifications, from musical consider condition. Thethere exponentiation ofplaying awhere subset of Nwhich means hatwith on augmentations, every beat is a single note atinstrumental a time, makes the pro ns retrogradations) which in gathering impor for the studyisof:canons but will not be mentioned today. This is related elsewhere ([?], .rminology
mpty) subset.and Then we set factors Polynomials cyclotomic nt to (T ) in Z[X]. The best one was 0 {0, a1 , . . . ak−1 } is 1. If A = a subset of N, it is the inner voice of a rhythmic ca ! ponentiation an equivalentr b1 , . . . b!−1 } one andgets period n iff k condition. The exponentiation of a subset of N me A(x) sum ⇐⇒ •it Aisdirect true in=any Fxp [X] A ⊕ B = Z/nZ
ITION 2. Let A ⊂ N be a finite (non empty) subset. Then we set
k∈A
! e rhythmic pattern, B the sequence of onsets. meaning that the number of notes on each beat is 1 iff it is equal t Exponentiation A(x) = xk sible). k∈A(see [?], [?]) and the per en that the simplification of assuming integer beats
•
A ×Condition B % (a,are b) )&→ arbitrary a + b is but onenecessary. to one) iff ofmap the whole canon (T0not ed thatThe thesum above is different assertio OSITION + Bstatement is possible direct (i.e.specifications, map A × B % (a,from b) &→ athe + b following isfrom one to one) iff cons r hand,. there areAother stemming musical n−1 n A(x) × B(x) = (A ⊕ B)(x) ] ⊂ Z[X], n such that A(X) × B(X) ≡ 1 + x + . . . x (mod X − 1) inimZ h augmentations, retrogradations) which where instrumental in gathering A(x) × B(x) = (A ⊕ B)(x) Bstudy , n such that T yields in F [X] » of canons but will not be mentioned today. This is related elsewhere ( p p 0 p
•
!0, 1, 3, 6" ! !0, 8, 12, 4"
nomials and we for any canon thefactors condition (T0 ) like : forhave some time the reciprocal, in condition (T0rhythmic ) cyclotomic :to prove 6 3 12 Y ONEDA 8 philosophy 4
!X !surprising. Xcondition. ! X !The 1" !Xrhythm!BXand ! ! 1" iation one an equivalentr exponentiation of aXsubset cOSITION forms. result was . AThe isgets a pattern for a rhytmic canon with outer period n iff of N
canon with outer rhythm B period iff212 10n−1 18 15and1314 11 9 10 8n 15 14 1213 n 11 X ! X ! X ! X ! X ! X X! X! X ! X ! X ! X ! X ! X X (mod 2. Let A ⊂ N be a finite(T(non empty) subset. Then we set A(x) × B(x) = 1 + x + x + . . . x ! ! x !− 1) 0) te fields 68 n−1 57 46 3! 5n 2 4 3 279! X X ! X ! X ! X ! X ! X ! 1! X ! 1 X ! X ! X ! X ! X ! X ! X A(x) × B(x) = 1 + x + x + . . . x (mod x − k 1)
+ X9 + X11 + X12 + X13 + X14 + X15 + X18 + X20 + X21 + X23 ith {0, 1, 3, 6} ⊕ {0, 8, 12, 15 20} one gets polynomials (1 + X + X3 + − 1 yields 1 + x + . . . x .
Some Rings
11 12 13 14 15 1 + X + X3 + X6 + X8 +0-1 X9 + X + X + X + X + X + X polynomials
16 ring k[X], indeedXA[X] A.. Of course it is uction modulo − 1inasmuch yields 1 +as x +0,.1. .∈x15 is notthis a ring ements of the set {0, 1}[X]. But set is not a ring : it
rfluchtes Ring
2)(1+X2)=??? (1+X+X makes sense initso any ring k[X], A[X] asof 0, th 1 big (there are many polynomials not inasmuch 0-1) and indeed inasmuch as contains only indeed 0-1 polynomials, and all
that is allowing to say elements the setit{0,tiles 1}[X]. this sea lynomial, to knowofwhether or But not (more ps,
is weird F2many [X] ent that Z[X] is too big (there are so polynomials n n for a given 0-1 polynomial, Pallowing to know whether it (mod 2)
(1+X+X2)(1+X2)= (1+X+X2)(1+X2)= 2+2X3map [X], there is no good For 4instance one 1+X+2X +X4 back into Z[X]. 1+X+X
ng
(1 + X)(1 + X + X2 ) = 1 + X3 ∈ F2 5X
Where does (T0) make sense ? • 0 and 1 are elements of any field • 'Tiling modulo p' means '(T ) holds in F [X]' 0
p
Chinese rhythmic canon theorem (2002): If A(x) B(x) =1+x+…xn-1 mod xn - 1 in all Fp[X], then it holds in Z[X].
Galois theory in Fq • First occurence : Johnson's problem • {0 1 4} and its augmentations tile with period a multiple of 15, because 1+X+X4 splits in F16 5
Theorem 1 (december 2001)
4 3 2 1 0
0
2
4
6
8
10
12
14
Several other cases suggested following question: Is there a 'local to global' approach for the general tiling problem ?
computer program itA.emerged that solutions had mposer andA. mathematician T ANGIANUniversity fromall Hanover Universi ematician T ANGIANand from Hanover quickly cam 4 ned by the order ofsolutions the roots of J(X) =a1multiple +which X + X was in15. the field it emerged that had a length a multi all solutions hadall a length which was of This 4 p oots of X + X in the field with 16 elements 4 J(X) = 1 + of sical illustration Johnson’s problem can be at [?]). http X + X in the field with 16 elements (see [?]). heard(see son’s problem canproductive be heard aton http ://canonsrythmiques.free pproach proved other generalized rhythmic n be heard at http ://canonsrythmiques.free.fr/Midi/. ivetry on to other generalized rhythmic canon problems too. he find a local characterization of condition (T0 ) led to ralized rhythmic canon problems too. acterization of condition (T0 ) led to a hidden result. ndition (Tmodulo a 2004) hidden result. 0 ) led Tiling p Theorem 2 to (april
Galois theory in F
NO !
` REME 2. For any finite (non empty) subset A ⊂ N, for any p
non empty) subset A ⊂ N, for any prime p, there exists B ⊂ N, n ∗ t A ⊂ N, for any prime p, there exists B ⊂ N, n ∈ N with 2 A(X) × B(X) ≡ 1 + X + X + .. 2 n−1 n A(X) × 2B(X) ≡ 1n−1 + X + X + . . n−1 .X (mod X − 1, p) ) ≡ 1 + X + X + ...X (mod X , p)
that is to say in Fp [X].
«Anythat rhythmic pattern ally this means on each beatmakes there aiscanon a number of not each beat there is a number of notes which is 1, up to a multi — modulo p» al illustration in http ://canonsrythmiques.free.fr/Midi/. s a number of notes which is 1, up to a multiple of p. /canonsrythmiques.free.fr/Midi/. ¨ M roof isintothe be Graz published shortlyfur inMathematik. the Graz Zentralblatt fur es.free.fr/Midi/. ¨ hortly Zentralblatt The main tool i Example with 0 1 4 : n olynomial in n 0 big divides some Xis − 1this for bigFe ¨vanishing Zentralblatt fur Mathematik. Theenough. main tool that inn any in 0 dividesnot some Xn − 1 for Though lemma
•
Conditions (T1) and (T2)
Remember A(X).B(X)=1+X+…Xn-1 mod Xn - 1.
• Cyclotomic factors : irreducible factors of
1+X+…Xn-1 must divide A(X) or B(X). They are the Φd, d | n.
•
Let RA = {d ; Φd| A(X)}, SA = {pα ε RA }. !
! p and (T1 ) : A(1) =pα ∈SA p and
(T1 ) : A(1) =
α βp ∈SA
β (T2 ) : if p α, q β, · · · ∈ SA then pα .q · · · ∈ R . A α β (T ) : if p , q , · · · ∈ S then p .q · · · ∈ R . 2 A A Then : –Then If A :tiles then (T1 ) is true. α
Conditions (T1) and (T2) Theorems (1998, Coven-Meyerowitz)
• If A tiles, then (T ) is true • If (T ) and (T ) are true, then A tiles • If A tiles and |A|=p q , then (T ) and (T ) 1
1
2
α
β
1
2
are true.
Also a very special case with 3 prime factors in 2000 (Lagarias-Wang)
Transformations • • Other transformations • duality : A B = B A ! dilatation • dilatation • affine transform • Useful for classifying and building up new Concatenation Concatenation
canons (cf.Vuza canons, in a minute)
Conservation Theorem 3 (2004):
All usual transformations preserve conditions (T1) and (T2) Basic lemma :factorizing the metronome 1+Xk+X2k+…X(p-1)k is the product of the Φd whence d is a divisor of n=p k, but not a divisor of k All this is Galois theory (in cyclotomic fields)
Vuza canons • Definition: no internal period
,
•
( unlike (say) {0,1,4,5} + {0,2,8,10} )
• Hajòs groups (M. A.) good/bad • Rather scarse • Popular with composers
Vuza canons How do we find them ?
• Difficult to get them all • Algorithms exist that give a few solutions • Transformations allow to find much more • Exhaustive search achieved for n=72 and n=108 (january 2004, H. Fripertinger)
Fuglede's conjecture Conjecture (Fuglede 1974)
• A set A tiles by translations iff it is spectral (meaning L2(A) admits a Hilbert basis)
• True in a number of cases (A convex, set of translations a group…)
• False in high dimension (T. Tao, 2003)
Fuglede's conjecture • A link with (T ) and (T ) 1
•
2
Theorem : (Isabella Laba 2000)
If A verifies (T1) and (T2) then A is spectral. If A is spectral then (T1) is true.
Last step • If A tiles but (T ) is false, • If A is not Vuza, then either inner rhythm A 2
or outer rhythm B reduces to a smaller canon ((T2) still false by theorem 3)
• the process cannot end with the trivial canon ({0}
{0}) - but (T2) is true here !
• Hence it ends up with a Vuza canon.
Latest news Theorem 4 (may 2004)
• A canon with (T ) false can only occur in a 2
non-Hajòs group (and reduces to a Vuza canon)
• Any tiling of a Hajòs group is spectral • This means n = p , p q, p q , p qr, pqrs α
α
From Laba + Coven-Meyerowitz
2 2
2
New
(T2) is true for a tiling of an interval; checked also by computer in Z72 and Z108
The end ? • Are all rhythmic canons spectral sets ? this should be found out via cooperation between different fields (musica, algebra, perhaps topology…)
• Both sides ot the Atlantic will be needed. Emmanuel Amiot
[email protected] http://canonsrythmiques.free.fr
