Discrete planes 1D case Generalized substitutions Continued fractions Multidimensional continued fractions, numeration and discrete geometry V. Berth´ e, T. Fernique LIRMM-CNRS-Montpellier-France berthe@lirmm.fr http://www.lirmm.fr/˜berthe Journ´ ees de num´ eration Graz 2007
Discrete planes 1D case Generalized substitutions Continued fractions Recognition of arithmetic discrete planes Problem Given a set of points in Z d , does there exist a standard arithmetic discrete plane that contains them?
Discrete planes 1D case Generalized substitutions Continued fractions Arithmetic discrete planes x , 1 ∗ ), ( � x , 2 ∗ ), ( � x , 3 ∗ ) be the following faces: Let ( � e 3 e 3 e 3 x x x e 2 e 2 e 2 e 1 e 1 e 1 Definition A standard arithmetic discrete plane or stepped plane is defined as x , i ∗ ) | 0 ≤ � � P � α,ρ = { ( � x , � α � + ρ < � � e i , � α �} .
Discrete planes 1D case Generalized substitutions Continued fractions Discrete lines and Sturmian words One can code a standard arithmetic discrete line (Freeman code) over the two-letter alphabet { 0 , 1 } . One gets a Stumian word ( u n ) n ∈ N ∈ { 0 , 1 } N 0100101001001010010100100101 Question How to recognize that a finite word is Sturmian?
Discrete planes 1D case Generalized substitutions Continued fractions Finite Sturmian words 0110110101101101 We consider the substitutions σ 0 : 0 �→ 0 , σ 0 : 1 �→ 10 σ 1 : 0 �→ 01 , σ 1 : 1 �→ 1 One has 01 1 01 1 01 01 1 01 1 01 = σ 1 (0101001010) 0 10 10 0 10 10 = σ 0 (011011) 01 1 01 1 = σ 1 (0101) 01 01 = σ 1 (00) � Continued fractions and Ostrowski numeration system
Discrete planes 1D case Generalized substitutions Continued fractions A classical recognition problem in discrete geometry in connection with • Geometric representations, Rauzy fractals, tilings • Multidimensional continued fractions and S -adic systems
Discrete planes 1D case Generalized substitutions Continued fractions A classical recognition problem in discrete geometry in connection with • Geometric representations, Rauzy fractals, tilings • Multidimensional continued fractions and S -adic systems We need a multidimensional notion of • Substitutions � Arnoux-Ito-Ei’s formalism for unimodular morphisms of the free group • Continued fraction algorithm � Brun’s algorithm • Words � Stepped surfaces
Discrete planes 1D case Generalized substitutions Continued fractions Substitution Let σ be a substitution over A .
Discrete planes 1D case Generalized substitutions Continued fractions Substitution Let σ be a substitution over A . Example: σ (1) = 12 , σ (2) = 13 , σ (3) = 1 .
Discrete planes 1D case Generalized substitutions Continued fractions Substitution Let σ be a substitution over A . The incidence matrix M σ of σ is the matrix defined by: M σ = ( | σ ( j ) | i ) ( i , j ) ∈A 2 , where | σ ( j ) | i is the number of occurrences of the letter i in σ ( j ).
Discrete planes 1D case Generalized substitutions Continued fractions Substitution Let σ be a substitution over A . Unimodular substitution A substitution σ is unimodular if det M σ = ± 1. Abelianization l : A ⋆ → N d be the Parikh mapping: Let d stand for the cardinality of A . Let � � l ( w ) = t ( | w | 1 , | w | 2 , · · · , | w | d ) .
Discrete planes 1D case Generalized substitutions Continued fractions Generalized substitution Generalized substitution [Arnoux-Ito] Let σ be a unimodular substitution. We call generalized substitution the following x , i ∗ ) defined by: tranformation acting on the faces ( � “ ” x − � Θ ∗ x , i ∗ ) = [ [ ( M − 1 , k ∗ ) . σ ( � � l ( P ) σ k ∈A P , σ ( k )= PiS ∗ Θ σ ∗ ∗ Θ σ Θ σ ∗ Θ σ
Discrete planes 1D case Generalized substitutions Continued fractions Stepped surface Definition A stepped surface (also called functional discrete surface) is defined as a union of pointed faces such that the orthogonal projection onto the diagonal plane x + y + z = 0 induces an homeomorphism from the stepped surface onto the diagonal plane.
Discrete planes 1D case Generalized substitutions Continued fractions Stepped surface Definition A stepped surface (also called functional discrete surface) is defined as a union of pointed faces such that the orthogonal projection onto the diagonal plane x + y + z = 0 induces an homeomorphism from the stepped surface onto the diagonal plane.
Discrete planes 1D case Generalized substitutions Continued fractions Stepped surface Definition A stepped surface (also called functional discrete surface) is defined as a union of pointed faces such that the orthogonal projection onto the diagonal plane x + y + z = 0 induces an homeomorphism from the stepped surface onto the diagonal plane. Recognition [Jamet] It is possible to recognize whether a set of points in Z d is contained in a stepped surface by considering a finite neighbour of each point.
Discrete planes 1D case Generalized substitutions Continued fractions Action on a plane Theorem [Arnoux-Ito, Fernique] α ∈ R d Let σ be a unimodular substitution. Let � + be a nonzero vector. The generalized substitution Θ ∗ σ maps without overlaps the stepped plane P � α,ρ onto P t M σ � α,ρ . ∗ Θ σ
Discrete planes 1D case Generalized substitutions Continued fractions Theorem [Arnoux-B.-Fernique-Jamet 2007] Let σ be a unimodular subtitution. The generalized substitution Θ ∗ σ acts without overlaps on stepped surfaces. ∗ Θ σ
Discrete planes 1D case Generalized substitutions Continued fractions Tiling Definition Let σ be a unimodular substitution. A stepped surface is said to be σ -tilable if it is a union of translates of Θ ∗ σ ( � 0 , i ∗ ). Question: Can we desubstitute a σ -tilable stepped surface?
Discrete planes 1D case Generalized substitutions Continued fractions Desubstitution We want to desubstitute a stepped surface according to Θ ∗ σ . Theorem Let σ be an invertible substitution. Let S be a σ -tilable stepped surface. There exists a unique stepped surface S ′ such that Θ ∗ σ ( S ′ ) = S .
Discrete planes 1D case Generalized substitutions Continued fractions Desubstitution We want to desubstitute a stepped surface according to Θ ∗ σ . Theorem Let σ be an invertible substitution. Let S be a σ -tilable stepped surface. There exists a unique stepped surface S ′ such that Θ ∗ σ ( S ′ ) = S . Idea of the proof: Θ ∗ σ ◦ µ = Θ ∗ µ ◦ Θ ∗ 1 σ Θ ∗ σ − 1 ( S ) is a stepped surface. 2 Θ ∗ σ − 1 ( S ) is thus an antecedent of S under the action of Θ ∗ σ . 3
Discrete planes 1D case Generalized substitutions Continued fractions Desubstitution Property Let σ be a unimodular morphism of the free group. Let S be a σ -tilable stepped surface. Then Θ ∗ σ − 1 ( S ) is a stepped surface. We use the following fact: Fact α ∈ R d Let σ be a unimodular morphism of the free group. Let � + be a nonzero vector such that t M σ � α ≥ 0 . Then, Θ ∗ σ maps without overlaps the stepped plane P � α,ρ onto P t M σ � α,ρ .
Discrete planes 1D case Generalized substitutions Continued fractions Desubstitution Property Let σ be a unimodular morphism of the free group. Let S be a σ -tilable stepped surface. Then Θ ∗ σ − 1 ( S ) is a stepped surface. We use the following fact: Fact α ∈ R d Let σ be a unimodular morphism of the free group. Let � + be a nonzero vector such that t M σ � α ≥ 0 . Then, Θ ∗ σ maps without overlaps the stepped plane P � α,ρ onto P t M σ � α,ρ . The stepped plane P � α,ρ is σ -tilable iff t M σ − 1 � α ≥ 0 .
Discrete planes 1D case Generalized substitutions Continued fractions Brun’s algorithm ( d = 2) We consider the following transformation acting on [0 , 1] 2 ( ( { 1 /α } , β α ) if α ≥ β T ( α, β ) = ( α β , { 1 /β } ) otherwise. For all n ∈ N , we set ( α n , β n ) = T n ( α, β ). One has (1 , α n , β n ) ∝ B a n ,ε n (1 , α n +1 , β n +1 ) with ( ⌊ 1 /α n ⌋ , 1) if α n ≥ β n ( a n , ε n ) = ( ⌊ 1 /β n ⌋ , 2) otherwise, with 0 1 0 1 a 1 0 a 0 1 B a , 1 = 1 0 0 and B a , 2 = 0 1 0 @ A @ A 0 0 1 1 0 0
Discrete planes 1D case Generalized substitutions Continued fractions Continued fraction algorithm One thus gets B n − 1 B 0 B 1 B n � α = � α 0 − → � α 1 − → . . . − → � α n − → . . . where B n ∈ GL ( d + 1 , N ). Convergents (1 , � α ) ∝ B 0 × . . . × B n (1 , � α n +1 ) B 0 × . . . × B n (1 ,� ( q n ,� p n ) ∝ 0) .
Discrete planes 1D case Generalized substitutions Continued fractions Continued fraction algorithm One thus gets B n − 1 B 0 B 1 B n α = � � α 0 − → � α 1 − → . . . − → � α n − → . . . where B n ∈ GL ( d + 1 , N ). Convergents (1 , � α ) ∝ B 0 × . . . × B n (1 , � α n +1 ) B 0 × . . . × B n (1 ,� ( q n ,� p n ) ∝ 0) . • Unimodular algorithm • Weak convergence (convergence of the type | α − p n / q n | )
Discrete planes 1D case Generalized substitutions Continued fractions Arithmetics Geometry α ∈ [0 , 1] d d -uple � stepped plane P (1 ,� α ) α n ) = Θ ∗ (1 , � α n ) = B n (1 , � α n +1 ) P (1 ,� σ n ( P (1 ,� α n +1 ) ) with t Bn incidence matrice of σ n
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