MacWilliams Extension Theorem for the Lee Weight

Transcription

1 MacWilliams Extension Theorem for the Lee Weight Noncommutative rings and their applications V Lens June 2017 Philippe Langevin IMATH, Toulon last revision June 11, A serie of joint works with Serhii Dyshko and Jay Wood. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

2 Sommaire 1 Isometry and MacWilliams Extension Theorem 2 Extension property 3 Wood criterion 4 Lee metric (Finite field case) 5 Dyshko criterion Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

3 Sommaire Isometry and MacWilliams Extension Theorem 1 Isometry and MacWilliams Extension Theorem 2 Extension property 3 Wood criterion 4 Lee metric (Finite field case) 5 Dyshko criterion Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

4 Isometry Isometry and MacWilliams Extension Theorem Let K be { a finite field 0, x = 0; h(x) = 1, else. n a positive integer C a subspace of K n Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

5 Isometry Isometry and MacWilliams Extension Theorem Let K be { a finite field 0, x = 0; h(x) = 1, else. The Hamming weight of x K n n a positive integer C a subspace of K n w h (x) = {i x i 0} = n h(x i ), i=1 Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

6 Isometry Isometry and MacWilliams Extension Theorem Let K be { a finite field 0, x = 0; h(x) = 1, else. The Hamming weight of x K n n a positive integer C a subspace of K n n w h (x) = {i x i 0} = h(x i ), i=1 A linear map f : C K n preserving the Hamming weight x C, w h (x) = w h (f (x)) is called a (linear) isometry over C. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

7 Isometry and MacWilliams Extension Theorem Monomial transformation consider (e i ) 1 i n the canonical basis of K n. An isometry over the ambiant space K n permutes the vectors of weight one. where λ i K π permutes {1, 2,..., n}. e i λ i e π(i) Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

8 Isometry and MacWilliams Extension Theorem Monomial transformation consider (e i ) 1 i n the canonical basis of K n. An isometry over the ambiant space K n permutes the vectors of weight one. where λ i K π permutes {1, 2,..., n}. e i λ i e π(i) x (λ 1 x π(1), λ 2 x π(2),..., λ n x π(n) ) often called a monomial transformation of K n. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

9 Isometry and MacWilliams Extension Theorem Monomial transformation consider (e i ) 1 i n the canonical basis of K n. An isometry over the ambiant space K n permutes the vectors of weight one. where λ i K π permutes {1, 2,..., n}. e i λ i e π(i) x (λ 1 x π(1), λ 2 x π(2),..., λ n x π(n) ) often called a monomial transformation of K n. S n K n Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

10 Isometry and MacWilliams Extension Theorem MacWilliams Extension Theorem Theorem (MacWilliams, 1962) An isometry over C K n extends to an isometry over K n. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

11 Isometry and MacWilliams Extension Theorem MacWilliams Extension Theorem Theorem (MacWilliams, 1962) An isometry over C K n extends to an isometry over K n. In other words, for an isometry f : C K n there exists a permutation π and scalars λ i s such that x C, f (x) = (λ 1 x π(1), λ 2 x π(2),..., λ n x π(n) ) S n K n res Isom(C) 0 Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

12 Isometry and MacWilliams Extension Theorem Frobenius ring case From the character theorycal proof of Ward & Wood, one deduces that MacWilliams extension theorem works for the Hamming space over any finite Frobenius rings. H. N. Ward, J. A. Wood, Characters and the Equivalence of Codes, J. Comb. Theory, Ser. A, (1996). Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

13 Isometry and MacWilliams Extension Theorem Homogeneous weight The same holds for any homogeneous weight on a finite Frobenius ring : ω(0) = 0; If x and y are associate then ω(x) = ω(y); There exists a constant c such that for all principal ideal I, ω(y) = c I. y I M. Greferath and S. E. Schmidt, Finite-ring combinatorics and MacWilliams s equivalence theorem, J. Combin. Theory Ser. A, (2000). Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

14 Isometry and MacWilliams Extension Theorem Homogeneous weight The same holds for any homogeneous weight on a finite Frobenius ring : ω(0) = 0; If x and y are associate then ω(x) = ω(y); There exists a constant c such that for all principal ideal I, ω(y) = c I. y I M. Greferath and S. E. Schmidt, Finite-ring combinatorics and MacWilliams s equivalence theorem, J. Combin. Theory Ser. A, (2000). Of course, MacWilliams extension works over the Z/(4) with its Lee weight l(0) = 0, l(1) = l(3) = 1, l(2) = 2. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

15 Isometry and MacWilliams Extension Theorem MacWilliams for Lee weight q a positive integer l the Lee weight over Z/(q). l(r) = { r, 0 r q/2; q r, q/2 < r < q. Remark Lee weight is not homogeneous for q > 4. Do we have a MacWilliams extension statement for the Lee weight? A. Barra, Equivalence Theorems and the Local-Global Property, ProQuest LLC, Ann Arbor, MI, 2012, Thesis (Ph.D.) University of Kentucky. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

16 Isometry and MacWilliams Extension Theorem Known results, new results In the last NCRA IV proceedings : q = 2p + 1, p prime (Folklore). q = 4p + 1 (Barra, 2012) q = 2 r or q = 3 r (Lens, 2015) Despite all this progress, there are glaring gaps in our knowledge : does extension theorem holds for linear codes over Z/(q)? Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

17 Isometry and MacWilliams Extension Theorem Known results, new results In the last NCRA IV proceedings : q = 2p + 1, p prime (Folklore). q = 4p + 1 (Barra, 2012) q = 2 r or q = 3 r (Lens, 2015) Despite all this progress, there are glaring gaps in our knowledge : does extension theorem holds for linear codes over Z/(q)? YES! Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

18 Isometry and MacWilliams Extension Theorem Connection with classical tools We have two ways to prove MacWilliams extension Theorem for the Lee weight using classical results of 1 Number Theory 2 Harmonic Analysis Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

19 Isometry and MacWilliams Extension Theorem Connection with classical tools We have two ways to prove MacWilliams extension Theorem for the Lee weight using classical results of 1 Number Theory 2 Harmonic Analysis The first works when the module q is primary, the second due to Sergey Dyshko works for a general module. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

20 Isometry and MacWilliams Extension Theorem Connection with classical tools We have two ways to prove MacWilliams extension Theorem for the Lee weight using classical results of 1 Number Theory 2 Harmonic Analysis The first works when the module q is primary, the second due to Sergey Dyshko works for a general module. I will sketch the proofs in the case of prime fields. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

21 Isometry and MacWilliams Extension Theorem Extension property holds for Lee weight Deux analogues au déterminant de Maillet C. R. Acad. Sci. Paris vol. Ser. I, 2016 Ph. Langevin, J. Wood: The extension problem for Lee and Euclidean weights Journal of Algebra Combinatorics Discrete Structures and Applications Vol. 4 2 pp , Ph. Langevin, J. Wood: The extension theorem for the Lee and Euclidean Weight over Z/p k Z Journal of Pure and Applied Algebra, submitted S. Dyshko: The Extension Theorem for the Lee weight Code, Design and Cryptography, submitted Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

22 Sommaire Extension property 1 Isometry and MacWilliams Extension Theorem 2 Extension property 3 Wood criterion 4 Lee metric (Finite field case) 5 Dyshko criterion Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

23 Extension property Isometry in general Let R be a finite ring ω a weight function on R n a positive integer M a submodule of R n Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

24 Extension property Isometry in general Let R be a finite ring ω a weight function on R n a positive integer M a submodule of R n ω is a real function such that ω(r) = 0 iff r = 0. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

25 Extension property Isometry in general Let R be a finite ring ω a weight function on R n a positive integer M a submodule of R n The ω-weight of x R n ω is a real function such that ω(r) = 0 iff r = 0. w ω (x) = n ω(x i ) i=1 Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

26 Extension property Isometry in general Let R be a finite ring ω a weight function on R n a positive integer M a submodule of R n ω is a real function such that ω(r) = 0 iff r = 0. The ω-weight of x R n n w ω (x) = ω(x i ) i=1 A linear map f : M K n preserving the ω-weight x M, ω(x) = ω(f (x)) is called a (linear) ω-isometry over M. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

27 U-monomial map Extension property e i the canonical basis of R n. Again, an isometry over R n maps e i on λ i e π(i) where λ i R and π permutes {1, 2,..., n}, moreover : t R, ω(t) = w ω (te i ) = w ω (tλ i e π(i) ) = ω(tλ i ) thus λ i lies in the symmetry group of ω U(ω) := {λ R t R, ω(λt) = ω(t)} Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

28 U-monomial map Extension property e i the canonical basis of R n. Again, an isometry over R n maps e i on λ i e π(i) where λ i R and π permutes {1, 2,..., n}, moreover : t R, ω(t) = w ω (te i ) = w ω (tλ i e π(i) ) = ω(tλ i ) thus λ i lies in the symmetry group of ω U(ω) := {λ R t R, ω(λt) = ω(t)} Definition (U-monomial transformation) Given U a subgroup of R, a monomial transformation with scalars in U. S n U n Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

29 Extension property Extension Property Definition (extension property) We say that Extension Property holds for the pair (R, ω) when each ω-isometry over M R n extends to a U(ω)-monomial transformation. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

30 Extension property Extension Property Definition (extension property) We say that Extension Property holds for the pair (R, ω) when each ω-isometry over M R n extends to a U(ω)-monomial transformation. EP holds for Hamming weight on Frobenius ring EP holds for Homogeneous weight on Frobenius ring Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

31 Extension property Extension Property Definition (extension property) We say that Extension Property holds for the pair (R, ω) when each ω-isometry over M R n extends to a U(ω)-monomial transformation. EP holds for Hamming weight on Frobenius ring EP holds for Homogeneous weight on Frobenius ring It looks difficult to decide if EP holds for an arbitrary weight function! Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

32 Sommaire Wood criterion 1 Isometry and MacWilliams Extension Theorem 2 Extension property 3 Wood criterion 4 Lee metric (Finite field case) 5 Dyshko criterion Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

33 Wood criterion Preserving map U be a subgroup of R r s if and only if s ru Ω a set of representatives of R \ {0} c r (x) := {i x i = r} cr U (x) := {i x i r} Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

34 Wood criterion Preserving map U be a subgroup of R r s if and only if s ru Ω a set of representatives of R \ {0} c r (x) := {i x i = r} cr U (x) := {i x i r} A linear map f : M R n such that x C, r Ω c U r (x) = c U r (f (x)) is called a U-preserving map over M. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

35 Wood criterion Goldberg Extension Theorem preserving map over K n The U-preserving maps over K n are precisely the U-monomial transformations. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

36 Wood criterion Goldberg Extension Theorem preserving map over K n The U-preserving maps over K n are precisely the U-monomial transformations. Theorem (Goldberg, 1980) A linear U-preserving map extends to U-monomial transformation. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

37 Wood criterion Goldberg Extension Theorem preserving map over K n The U-preserving maps over K n are precisely the U-monomial transformations. Theorem (Goldberg, 1980) A linear U-preserving map extends to U-monomial transformation. The same holds modular rings : Constantinescu, Heise, Honold (1996). J. A. Wood. Weight functions and the extension theorem for linear codes over finite rings. In R. C. Mullin and G. L. Mullen, editors, Finite fields: theory, applications, and algorithms (Waterloo, ON, 1997), volume 225 of Contemp. Math., pages Amer. Math. Soc., Providence, RI, Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

38 Wood criterion Extensibility Property (recall) The symmetry group of ω. U(ω) = {λ K x K, ω(λx) = ω(x)} K Extension Property We say the extension property holds for the weight ω when each ω-isometry of K n is the restriction of a U(ω)-monomial map. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

39 Wood criterion Extensibility Property (recall) The symmetry group of ω. U(ω) = {λ K x K, ω(λx) = ω(x)} K Extension Property We say the extension property holds for the weight ω when each ω-isometry of K n is the restriction of a U(ω)-monomial map. From Goldberg Theorem, one gets a criterion. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

40 Wood criterion A sufficient condition for Extension Property w ω (x) = n i=1 ω(x i ) = r R ω(r)c r (x) = r Ω ω(r)c U r (x). Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

41 Wood criterion A sufficient condition for Extension Property For all s Ω, w ω (x) = n i=1 ω(x i ) = r R ω(r)c r (x) = r Ω ω(r)c U r (x). w ω (xs) = r R ω(rs)c r (x) = r Ω ω(rs)c U r (x) Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

42 Wood criterion A sufficient condition for Extension Property For all s Ω, w ω (x) = n ω(x i ) = ω(r)c r (x) = ω(r)cr U (x). i=1 r R r Ω idem, w ω (xs) = r R ω(rs)c r (x) = r Ω ω(rs)c U r (x) w ω (f (xs)) = w ω (f (x)s) = r Ω ω(rs)c U r (f (x)) Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

43 Wood criterion A sufficient condition for Extension Property For all s Ω, w ω (x) = n ω(x i ) = ω(r)c r (x) = ω(r)cr U (x). i=1 r R r Ω idem, w ω (xs) = r R ω(rs)c r (x) = r Ω ω(rs)c U r (x) w ω (f (xs)) = w ω (f (x)s) = r Ω ω(rs)c U r (f (x)) Lemma The invertibility of ( ω(rs) ) implies the U-preservation of ω whence r,s Ω Extension Property. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

44 determinantal criterion Wood criterion Let Ω a set of repretentatives for the action of U := U(ω).. W ω :=... ω(rs)... ω := det(w ω ). r,s Ω Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

45 determinantal criterion Wood criterion Let Ω a set of repretentatives for the action of U := U(ω).. W ω :=... ω(rs)... ω := det(w ω ). r,s Ω Proposition (Wood) If ω 0 then Extension Property holds for the weight ω. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

46 determinantal criterion Wood criterion Let Ω a set of repretentatives for the action of U := U(ω).. W ω :=... ω(rs)... ω := det(w ω ). r,s Ω Proposition (Wood) If ω 0 then Extension Property holds for the weight ω. Remark One has an analogue criterion non commutative case. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

47 determinantal criterion Wood criterion Let Ω a set of repretentatives for the action of U := U(ω).. W ω :=... ω(rs)... ω := det(w ω ). r,s Ω Proposition (Wood) If ω 0 then Extension Property holds for the weight ω. Remark One has an analogue criterion non commutative case. Numerical evidence for the Lee weight! Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

48 Sommaire Lee metric (Finite field case) 1 Isometry and MacWilliams Extension Theorem 2 Extension property 3 Wood criterion 4 Lee metric (Finite field case) 5 Dyshko criterion Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

49 Fourier coefficient Lee metric (Finite field case) l an odd prime Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

50 Fourier coefficient Lee metric (Finite field case) l an odd prime l the Lee metric of F l Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

51 Fourier coefficient Lee metric (Finite field case) l an odd prime l the Lee metric of F l U(l) = { 1, +1} Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

52 Lee metric (Finite field case) Fourier coefficient l an odd prime l the Lee metric of F l U(l) = { 1, +1} G := Ω = F l /{ 1, +1} is cyclic of order l 1 2. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

53 Lee metric (Finite field case) Fourier coefficient l an odd prime l the Lee metric of F l U(l) = { 1, +1} G := Ω = F l /{ 1, +1} is cyclic of order l 1 2. Ĝ is the group of even (multiplicative) characters i.e. χ( 1) = 1. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

54 Lee metric (Finite field case) Fourier coefficient l an odd prime l the Lee metric of F l U(l) = { 1, +1} G := Ω = F l /{ 1, +1} is cyclic of order l 1 2. Ĝ is the group of even (multiplicative) characters i.e. χ( 1) = 1. By the Dedekind determinant formula l = ± χ b G l(χ) Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

55 Lee metric (Finite field case) Fourier coefficient l an odd prime l the Lee metric of F l U(l) = { 1, +1} G := Ω = F l /{ 1, +1} is cyclic of order l 1 2. Ĝ is the group of even (multiplicative) characters i.e. χ( 1) = 1. By the Dedekind determinant formula l = ± χ b G l(χ) where l(χ) = s G l(s)χ(s) is the Fourier coefficient of l at χ. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

56 Lee metric (Finite field case) Sophie Germain case Proposition Certainly, Extension Property holds for the Lee weight in the case of sure prime module i.e. l = 2p + 1 with p prime. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

57 Lee metric (Finite field case) Sophie Germain case Proposition Certainly, Extension Property holds for the Lee weight in the case of sure prime module i.e. l = 2p + 1 with p prime. β a generator of G Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

58 Lee metric (Finite field case) Sophie Germain case Proposition Certainly, Extension Property holds for the Lee weight in the case of sure prime module i.e. l = 2p + 1 with p prime. β a generator of G χ a non trivial character Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

59 Lee metric (Finite field case) Sophie Germain case Proposition Certainly, Extension Property holds for the Lee weight in the case of sure prime module i.e. l = 2p + 1 with p prime. β a generator of G χ a non trivial character ζ := χ(β) is a primitive p-th root of unity. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

60 Lee metric (Finite field case) Sophie Germain case Proposition Certainly, Extension Property holds for the Lee weight in the case of sure prime module i.e. l = 2p + 1 with p prime. β a generator of G χ a non trivial character ζ := χ(β) is a primitive p-th root of unity. The minimal polynomial of ζ is Φ p (T ) = T p T 1 + T 0 Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

61 Sophie Germain case Lee metric (Finite field case) Proposition Certainly, Extension Property holds for the Lee weight in the case of sure prime module i.e. l = 2p + 1 with p prime. β a generator of G χ a non trivial character ζ := χ(β) is a primitive p-th root of unity. The minimal polynomial of ζ is thus Φ p (T ) = T p T 1 + T 0 p 1 l(χ) = l(β k )ζ k k=0 does not vanish simply because l is not constant on G. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

62 Two in one Lee metric (Finite field case) We consider the Lee and Euclidean weights : { t, 0 t l/2; l(t) = l t, l/2 < t < l; e(t) = l(t) 2. they share the same symmetry group U := U(l) = { 1, +1} = U(e). Theorem If l is an odd prime then l 0 and e 0. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

63 Lee metric (Finite field case) Fourier coefficient of the Lee map The quotient group is cyclic of order n := (l 1)/2. we want to prove : G := F l /{±1} = {1, 2,..., (l 1)/2} χ Ĝ, 0 l(χ) = s G l(s)χ(s). It is trivial when l = 2p + 1, p prime. Barra proved the case l = 4p + 1. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

64 Fourier analysis Lee metric (Finite field case) We identify Ĝ with the group of even characters of F l : Ĝ = {χ F l χ( 1) = 1} The Fourier coefficients of l and e are given by l(χ) = l(x)χ(x) = l(k)χ(k) = kχ(k) x G ê(χ) = x G k<l/2 k<l/2 e(x)χ(x) = e(k)χ(k) = k 2 χ(k) k<l/2 k<l/2 Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

65 Lee metric (Finite field case) Links between the determinants It is easy to verify the following quadratic relation holds l(2x) 2 4l(x) 2 = ( l(2x) 2l(x) ) l. In other words On spectra e(2x) 4e(x) = ( l(2x) 2l(x) ) l. ( χ(2) 4) ê(χ) = ( χ(2) 2) l(χ) l. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

66 Lee metric (Finite field case) Links between the determinants It is easy to verify the following quadratic relation holds l(2x) 2 4l(x) 2 = ( l(2x) 2l(x) ) l. In other words On spectra e(2x) 4e(x) = ( l(2x) 2l(x) ) l. ( χ(2) 4) ê(χ) = ( χ(2) 2) l(χ) l. Scholie Let r be the smallest positive integer such that 2 r ±1 mod l. (2 r + 1) l 1 2r e = l l 1 2 l. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

67 Lee metric (Finite field case) basic fact for non trivial even characters 1 χ even and not trivial 1(χ) = 2 χ(k) = 0. k<l/2 Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

68 Lee metric (Finite field case) basic fact for non trivial even characters 1 χ even and not trivial 1(χ) = 2 χ(k) = 0. k<l/2 The first generalized Bernoulli s number vanishes too B 1 (χ) = 1 l l kχ(k) = 0 k=1 Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

69 Lee metric (Finite field case) basic fact for non trivial even characters 1 χ even and not trivial 1(χ) = 2 χ(k) = 0. k<l/2 The first generalized Bernoulli s number vanishes too We want to prove that B 1 (χ) = 1 l l kχ(k) = 0 k=1 0 1 kχ(k) = l(χ) l k<l/2 Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

70 Lee metric (Finite field case) Consequence of l(χ) = 0 on the 2nd Bernoulli s number Let us observe the consequence of l(χ) = 0= ê(χ), 1 χ, χ( 1) = 1, on the second generalized Bernoulli s number B 2 (χ) = 1 2l l (k 2 lk)χ(k). k=1 2lB 2 (χ) = 2ê(χ) 2 l(χ)l + 1(χ)l 2 = zero.

71 Lee metric (Finite field case) Consequence of l(χ) = 0 on the 2nd Bernoulli s number Let us observe the consequence of l(χ) = 0= ê(χ), 1 χ, χ( 1) = 1, on the second generalized Bernoulli s number B 2 (χ) = 1 2l l (k 2 lk)χ(k). k=1 2lB 2 (χ) = 2ê(χ) 2 l(χ)l + 1(χ)l 2 = zero. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

72 Lee metric (Finite field case) Contradiction with classical fact from number theory In number theory, there is a long story concerning the analytic continuation of the Dirichlet serie L(s, χ) = n=1 χ(n) n s Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

73 Lee metric (Finite field case) Contradiction with classical fact from number theory In number theory, there is a long story concerning the analytic continuation of the Dirichlet serie On the one hand L(s, χ) = n=1 χ(n) n s B 2 (χ)/2 = L( 1, χ) Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

74 Lee metric (Finite field case) Contradiction with classical fact from number theory In number theory, there is a long story concerning the analytic continuation of the Dirichlet serie On the one hand On the other hand L(s, χ) = n=1 χ(n) n s B 2 (χ)/2 = L( 1, χ) L( 1, χ) = 0 if and only if χ is odd. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

75 Lee metric (Finite field case) Contradiction with classical fact from number theory In number theory, there is a long story concerning the analytic continuation of the Dirichlet serie On the one hand On the other hand L(s, χ) = n=1 χ(n) n s B 2 (χ)/2 = L( 1, χ) L( 1, χ) = 0 if and only if χ is odd. whence the determinants l and e do not vanish. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

76 Lee metric (Finite field case) Contradiction with classical fact from number theory In number theory, there is a long story concerning the analytic continuation of the Dirichlet serie On the one hand On the other hand L(s, χ) = n=1 χ(n) n s B 2 (χ)/2 = L( 1, χ) L( 1, χ) = 0 if and only if χ is odd. whence the determinants l and e do not vanish. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

77 Lee metric (Finite field case) primary case Corollary (extension property) The Lee and Euclidean isometries are the restriction of { 1, +1}-monomial transformations. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

78 Lee metric (Finite field case) primary case Corollary (extension property) The Lee and Euclidean isometries are the restriction of { 1, +1}-monomial transformations. The same approach works in the case of a primary module Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

79 Lee metric (Finite field case) primary case Corollary (extension property) The Lee and Euclidean isometries are the restriction of { 1, +1}-monomial transformations. The same approach works in the case of a primary module but not for a composite module! Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

80 Sommaire Dyshko criterion 1 Isometry and MacWilliams Extension Theorem 2 Extension property 3 Wood criterion 4 Lee metric (Finite field case) 5 Dyshko criterion Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

81 Dyshko criterion Additive Fourier coefficient The additive Fourier coefficient of ω : ω (a) = x F l ω(x)µ(ax) where µ is the standard additive character of F l. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

82 Dyshko criterion Additive Fourier coefficient The additive Fourier coefficient of ω : ω (a) = x F l ω(x)µ(ax) where µ is the standard additive character of F l. Note that U(ω ) = U(ω) and a F l ω (a) = l ω(0) = 0 Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

83 change of determinant Dyshko criterion Since ω(0) = 0, ω (χ) = τ(χ) ω( χ) where τ(χ) is a Gauss sum Wω. =... ω (rs).... r,s Fl /±1 Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

84 change of determinant Dyshko criterion Since ω(0) = 0, ω (χ) = τ(χ) ω( χ) where τ(χ) is a Gauss sum Wω. =... ω (rs).... r,s Fl /±1 ω = 0 det(w ω) = 0 Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

85 Dyshko criterion Levy-Desplanques dominant criterion A strictly diagonally dominant n n-matrix (a ij ) i.e. i, a ii > i j a ij is not singular. Corollary If r 0, ω (r) < 0 and ω (0) < 2 U(ω) ω (1) then ω 0. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

86 sketch Dyshko criterion We consider the matrices Wω. =... ω (rs).... r,s Fl /±1.... ω (r/s).... r,s Fl /±1 ω (1) is on the diagonal Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

87 sketch Dyshko criterion We consider the matrices Wω. =... ω (rs).... r,s Fl /±1.... ω (r/s).... r,s Fl /±1 ω (1) is on the diagonal ω (1) ω (r) = ω (1) + ω (r) 1 r Ω 1 r Ω Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

88 sketch Dyshko criterion We consider the matrices Wω. =... ω (rs).... r,s Fl /±1.... ω (r/s).... r,s Fl /±1 ω (1) is on the diagonal ω (1) ω (r) = ω (1) + ω (r) 1 r Ω 1 r Ω The sum of the Fourier coefficients ω (0) + U(ω) r Ω ω (r) vanishes. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

89 sketch Dyshko criterion We consider the matrices Wω. =... ω (rs).... r,s Fl /±1.... ω (r/s).... r,s Fl /±1 ω (1) is on the diagonal ω (1) ω (r) = ω (1) + ω (r) 1 r Ω 1 r Ω The sum of the Fourier coefficients ω (0) + U(ω) r Ω ω (r) vanishes. ω (1) 1 r Ω ω (r) = 2ω (1) + ω (0) U(ω) Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

90 Dyshko criterion Additive Fourier coefficient of the Lee map 0 r < l n := l 1 2 t := 2πr l Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

91 Dyshko criterion Additive Fourier coefficient of the Lee map 0 r < l n := l 1 2 t := 2πr l l 1 l (r) = l(k)e it = 2 k=0 where D n is the Dirichlet kernel n cos kt = 2n (D n (t) F n (t)) k=1 Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

92 Dyshko criterion Additive Fourier coefficient of the Lee map 0 r < l n := l 1 2 t := 2πr l l 1 l (r) = l(k)e it = 2 k=0 where D n is the Dirichlet kernel D n (t) := and F n the Fejér kernel n k= n n cos kt = 2n (D n (t) F n (t)) k=1 e ikt = n k=1 cos kt = sin(n )t 2 sin 1 2 t = 0 Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

93 Dyshko criterion Additive Fourier coefficient of the Lee map 0 r < l n := l 1 2 t := 2πr l l 1 l (r) = l(k)e it = 2 k=0 where D n is the Dirichlet kernel D n (t) := and F n the Fejér kernel n k= n n cos kt = 2n (D n (t) F n (t)) k=1 e ikt = n F n (t) := 1 n 1 D k (t) = 1 n n k=0 k=1 k=1 cos kt = sin(n )t 2 sin 1 2 t = 0 ( n (1 k 1 sin n 2 ) cos kt = t n 2n sin 1 2 t ) 2 Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

94 Dyshko criterion Lee weight satisfies the two conditions 0 r < l n := l 1 2 First condition : Second condition : and l (r) = 2nF n ( 2πr l )<0 sin l l (1) = 4 sin 1 2 2π l 2π l 2 l (0) = 2 l 1 2 k = 1 4 (l2 1) k=1 Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

95 second condition Dyshko criterion We have to prove 4l (1) > l (0) Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

96 second condition Dyshko criterion We have to prove and now it is very easy! 4l (1) > l (0) Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

97 second condition Dyshko criterion We have to prove and now it is very easy! 4l (1) > l (0) Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

98 Dyshko criterion second condition We have to prove and now it is very easy! 4l (1) > l (0) Indeed, 4 π 2 l2 4l (1) and l (0) 1 4 l2 Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

99 Dyshko criterion Dyshko criterion for modular ring consider the ring Z/(q) ω a weight function b a divisor of q. write q = ab Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

100 Dyshko criterion Dyshko criterion for modular ring consider the ring Z/(q) ω a weight function b a divisor of q. write q = ab Consider the additive Fourier coefficients of the map x ω(bx) F a (t) = x Z/(a) ω(bx)ζ tx a Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

101 Dyshko criterion Dyshko criterion for modular ring consider the ring Z/(q) ω a weight function b a divisor of q. write q = ab Consider the additive Fourier coefficients of the map x ω(bx) F a (t) = x Z/(a). W a (ω) =... ω (rs).... ω(bx)ζ tx a r,s Z/(a) /G a(ω) where G a (ω) = {h Z/(a) t Z/(a) wt(bht) = ω(bt)}. Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

102 Dyshko criterion Dyshko criterion for modular ring Theorem (Dyshko) Let ω : Z/(q) C be a weight function. If for all 1 a q the matrix W a (ω) is non singular and h G a (ω) g G q (ω) g h mod a then Extension Property holds for the weight ω. Corollary For every integer q 2 the Extension Property of the Lee weight holds over the ring Z/(q). Philippe Langevin (IMATH, Toulon) last revision June 11, / 44

103 Dyshko criterion Philippe Langevin (IMATH, Toulon) last revision June 11, / 44