Non-Unimodularity. Bernd Sing. CIRM, 24 March Department of Mathematical Sciences. Numeration: Mathematics and Computer Science

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1 Department of Mathematical Sciences CIRM, 24 March 2009 Numeration: Mathematics and Computer Science

2 β-substitutions Let β > 1 be a PV-number (Pisot-number) and consider the greedy expansion to base β. Any (nonnegative) number x has such an expansion n=m a n β n with coefficients a i {0, 1,..., β 1}. One might be interested in characterising the set of β-integers { } 0 Z β = x x = a n β n n=m Z β can also be obtained using a substitution There are essentially two classes of substitutions depending on the β-expansion of 1 being periodic or eventually periodic. a a d 1 b, b a d 2 c,..., z a dn

3 Earlier This Afternoon we saw the polynomial p(x) = x 4 x 3 x 2 x 1, which has dominant (Pisot!) root β Alg. conj. are λ r and λ c, λ c ± i Associated substitution: a ab, b ac, c ad, d a. Geometrically, we can describe this as a tile substitution: Choose l a = 1, l b = β 1, l c = β 2 β 1 and l d = β 3 β 2 β 1 (the components of the PF-eigenvector l of the substitution/abelianization matrix). Then inflate and subdivide, and the set of left endpoints of the intervals yields Z β. l a β l a l b β l a l b l a l c β...

4 Geometric Realization in C R We now replace β k in the expansion of x Z β by (Re(λ k c ), Im(λ k c ), λ k r ) T to get an object in C R = R 3. Calling this the -map, the closure of Z β is now of interest. Let Ω i is the closure of the set of i-type endpoints; then cl Z β = Ω a Ω b Ω c Ω d, The substitution rule yields an iterated function system. Using the notations Re λ c Im λ c 0 1 λ = Im λ c Re λ c 0 and t = 0, we get: 0 0 λ r 1

5 Geometric Realization in C R We now replace β k in the expansion of x Z β by (Re(λ k c ), Im(λ k c ), λ k r ) T to get an object in C R = R 3. Calling this the -map, the closure of Z β is now of interest. Let Ω i is the closure of the set of i-type endpoints; then cl Z β = Ω a Ω b Ω c Ω d, The substitution rule yields an iterated function system. Ω a = λ Ω a λ Ω b λ Ω c λ Ω d Ω b = λ Ω a + t Ω c = λ Ω b + t Ω d = λ Ω c + t

6 A Measure Calculation The sets Ω i are compact and for their Lebesgue-measure we find: µ (Ω a ) = µ (λ Ω a λ Ω b λ Ω c λ Ω d ) µ (Ω j ) = µ (λ Ω j 1 + t)

7 A Measure Calculation The sets Ω i are compact and for their Lebesgue-measure we find: µ (Ω a ) µ (λ Ω a ) + µ (λ Ω b ) + µ (λ Ω c ) + µ (λ Ω d ) µ (Ω j ) = µ (λ Ω j 1 )

8 A Measure Calculation The sets Ω i are compact and for their Lebesgue-measure we find: µ (Ω a ) λ c 2 λ r (µ (Ω a ) + µ (Ω b ) + µ (Ω c ) + µ (Ω d )) µ (Ω j ) = λ c 2 λ r µ (Ω j 1 )

9 A Measure Calculation The sets Ω i are compact and for their Lebesgue-measure we find: µ (Ω a ) λ c 2 λ r (µ (Ω a ) + µ (Ω b ) + µ (Ω c ) + µ (Ω d )) µ (Ω j ) = λ c 2 λ r µ (Ω j 1 ) But λ c 2 λ r = 1/β and we have componentwise µ (Ω a ) µ (Ω a ) µ (Ω b ) µ (Ω b ) β µ (Ω c ) µ (Ω c ) µ (Ω d ) µ (Ω d ) Perron-Frobenius: equality holds and all unions are µ-disjoint.

10 A Measure Calculation The sets Ω i are compact and for their Lebesgue-measure we find: µ (Ω a ) = µ (λ Ω a λ Ω b λ Ω c λ Ω d ) µ (Ω j ) = µ (λ Ω j 1 + t) But λ c 2 λ r = 1/β and we have componentwise µ (Ω a ) µ (Ω a ) µ (Ω b ) µ (Ω b ) β µ (Ω c ) µ (Ω c ) µ (Ω d ) µ (Ω d ) Perron-Frobenius: equality holds and all unions are µ-disjoint.

11 Ω d

12 Ω d

13 Ω d

14 Ω d

15 Ω d

16 Ω d

17 Ω d

18 Ω d

19 Ω d

20 Ω d

21 Ω d

22 Ω d

23 Ω d

24 Ω d

25 Ω d

26 Ω d

27 A Non-Unimodular Example Consider the polynomial p(x) = x 3 3 x 2 x 2, which has dominant (Pisot!) root β Algebraic conjugates are λ c, λ c ± i Associated substitution: a aaab, b ac, c aa. Geometrically, we can describe this as a tile substitution: Choose l a = 1, l b = β 3 and l c = β 2 3 β 1 etc. However, the measure calculation goes wrong (β λ c 2 = 2): µ (Ω a ) µ (Ω a ) 1 2 β µ (Ω b ) µ (Ω b ) µ (Ω c ) µ (Ω c )

28 p-adic Fields Q p The p-adic integers Z p are a complete discrete valuation ring. An element x Z p can be written as Taylor series in powers of p, i.e., x = s n p n =.s 0 s 1... with s n {0,..., p 1}. n=0 The p-adic numbers Q p are the field of fractions of Z p. An element x Q p can be written as Laurent series x = s n p n = s m... s 1.s 0 s 1... with s n {0,..., p 1}. n=m Some 3-adic integers: 0 =.0, 1 =.10, 8 =.220, 1 =.2, 13 =.2112, 3 4 =.0120, 7 = and/or

29 p-adic Fields Q p The p-adic integers Z p are a complete discrete valuation ring. An element x Z p can be written as Taylor series in powers of p, i.e., x = s n p n =.s 0 s 1... with s n {0,..., p 1}. n=0 The p-adic numbers Q p are the field of fractions of Z p. An element x Q p can be written as Laurent series x = s n p n = s m... s 1.s 0 s 1... with s n {0,..., p 1}. n=m Some 3-adic integers: 0 =.0, 1 =.10, 8 =.220, 1 =.2, 13 =.2112, 3 4 =.0120, 7 = and/or

30 Visualisizing Q B }{{}}{{} 7 17 (0) = B 1 (9) = B 3 1 ( 3 2 ) B (2)

31 Geometric Realization in C Q 2 We consider the polynomial p(x) = x 3 3 x 2 x 2 in Q 2 : it has a 2-adic root λ of absolute value 1 2! (It also has two roots in Q 2( 3), but they are of absolute value 1.) the Haar measure µ H on Q 2 has the property µ H (λ 2 A) = 1 2 µ H(A). we here define the -map Z β C Q 2 by replacing each power β k by (Re(λ k c ), Im(λ k c ), λ k 2 )T. now the measure calculation (Haar measure on C Q 2 ) works out!

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83 Remark: Coincidences et al. One is often/usually interested whether the sets Ω j are the prototiles of a specific aperiodic tiling! pure point dynamical system One can check this (by an algorithm) by making use of the measure-disjointness on the right-hand side of the iterated function system. This yields so-called coincidence conditions, finiteness conditions etc.

84 Remark: Cut-and-Project Schemes π 1 π 2 π 1 Ω π 1 π 2 H π 2 π 2 π 1 R

85 Further Example I 1 2 Z 3 Substitution a aaaab, b ab R

86 Further Example II Q 2 1 Substitution a aaba, b aa 3 Z 2 1 Ω a 2 0 Ω b R

87 Further Example II The aperiodic tiling of R Q 2 with prototiles Ω a and Ω b. Q 2 Z Z Z Z 2 R