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Characters and the Macwilliams Identities

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Characters and the Macwilliams Identities Characters and the MacWilliams identities Jay A. Wood Department of Mathematics Western Michigan University http://homepages.wmich.edu/∼jwood/ Algebra for Secure and Reliable Communications Modeling Morelia, Michoac´an,Mexico October 8, 2012 Characters I Let G be a finite abelian group (usually additive). I A character of G is a group homomorphism π : G ! C×, where C× is the multiplicative group of nonzero complex numbers. I Example. Let G = Z=mZ and let a 2 Z=mZ. Then × πa : Z=mZ ! C , πa(g) = exp(2πiag=m) is a character. Every character of Z=mZ is of this form. JW (WMU) MacWilliams identities October 8, 2012 2 / 21 Character group I Let Gb be the set of all characters of G. I Then Gb is itself a finite abelian group under pointwise multiplication of characters. That is, (π1π2)(g) := π1(g)π2(g) for g 2 G. I Gb is the character group of G. ∼ I jGbj = jGj; in fact, Gb = G (but not naturally). ∼ I Gb = G, naturally: g 7! [π 7! π(g)]. JW (WMU) MacWilliams identities October 8, 2012 3 / 21 Products I If G1 and G2 are two finite abelian groups, then ∼ (G1 × G2)b = Gc1 × Gc2. I (π1; π2)(g1; g2) := π1(g1)π2(g2). JW (WMU) MacWilliams identities October 8, 2012 4 / 21 Vanishing formulas × I The character sending every g 2 G to 1 2 C is written 1, the principal character. ( X jGj; π = 1; π(g) = 0; π 6= 1: g2G ( X jGj; g = 0; π(g) = 0; g 6= 0: π2Gb JW (WMU) MacWilliams identities October 8, 2012 5 / 21 Linear independence of characters I Let F (G; C) = ff : G ! Cg be the set of all functions from G to C. I F (G; C) is a vector space over C of dimension jGj. I The characters of G are linearly independent in F (G; C); in fact, they form a basis. I The characters are orthonormal under 1 X hf ; f i = f (g)f (g): 1 2 jGj 1 2 g2G JW (WMU) MacWilliams identities October 8, 2012 6 / 21 Fourier transform I Let V be a complex vector space. I The Fourier transform b : F (G; V ) ! F (Gb; V ) is a linear transformation defined for f 2 F (G; V ) by: X f^(π) := π(g)f (g): g2G I Fourier inversion: b is invertible 1 X f (g) = π(−g)f^(π): jGj π2Gb JW (WMU) MacWilliams identities October 8, 2012 7 / 21 Annihilators I For H a subgroup of G, define the annihilator (Gb : H) := fπ 2 Gb : π(H) = 1g: I (Gb : H) is a subgroup of Gb. ∼ I (Gb : H) = (G=H)b, so j(Gb : H)j = jGj=jHj. I 0 ! H ! G ! G=H ! 0 induces 1 ! (Gb : H) ! Gb ! Hb ! 1. I Double annihilator: (G :(Gb : H)) = H. JW (WMU) MacWilliams identities October 8, 2012 8 / 21 Poisson summation formula I For H a subgroup of G and f : G ! V , X 1 X f (h) = f^(π): j(Gb : H)j h2H π2(Gb:H) JW (WMU) MacWilliams identities October 8, 2012 9 / 21 Additive codes I Let the alphabet A be a finite abelian group. I An additive code of length n over A is a subgroup C ⊂ An. I The Hamming weight wt(a) = 1 for a 6= 0 in A. n P I Extend to vectors a 2 A by wt(a) = wt(ai ). JW (WMU) MacWilliams identities October 8, 2012 10 / 21 Hamming weight enumerator n I For an additive code C ⊂ A , the Hamming weight enumerator is the generating function n X n−wt(c) wt(c) X n−i i WC (X ; Y ) := X Y = Ai X Y ; c2C i=0 where Ai equals the number of codewords in C of Hamming weight i. JW (WMU) MacWilliams identities October 8, 2012 11 / 21 The MacWilliams identities Theorem For an additive code C ⊂ An, with annihilator (Abn : C), 1 WC (X ; Y ) = W(An:C)(X + (jAj − 1)Y ; X − Y ): j(Abn : C)j b n n I Because jAbj = jAj and (A :(Ab : C)) = C, roles of C and (Abn : C) can be reversed. I This version is due to Delsarte, 1972. JW (WMU) MacWilliams identities October 8, 2012 12 / 21 Strategy of proof I The idea of the proof is due to Gleason. I Apply the Poisson summation formula X 1 X f (h) = f^(π); j(Gb : H)j h2H π2(Gb:H) with G = An, H = C,(Gb : H) = (Abn : C), and f (h) = X n−wt(c)Y wt(c). JW (WMU) MacWilliams identities October 8, 2012 13 / 21 Form of f^, n = 1 n−wt(c) wt(c) ^ I For f (c) = X Y , what is f ? Case n = 1: X f^(π) = π(a)f (a) a2A X = π(a)X 1−wt(a)Y wt(a) a2A X = X + π(a)Y a6=0 ( X + (jAj − 1)Y ; π = 1; = X − Y ; π 6= 1: JW (WMU) MacWilliams identities October 8, 2012 14 / 21 Form of f^, general n I For general n, we take the product of 1-dimensional results. n I For π = (π1; π2; : : : ; πn) 2 Ab , f^(π) = f^(π1)f^(π2) ··· f^(πn) = (X + (jAj − 1)Y )n−wt(π)(X − Y )wt(π): I Here, wt(π) counts the number of πi 6= 1. JW (WMU) MacWilliams identities October 8, 2012 15 / 21 Comments I The format of the MacWilliams identities for additive codes was 1 WC (X ; Y ) = W(An:C)(X + (jAj − 1)Y ; X − Y ): j(Abn : C)j b n n I Can we identify (Ab : C) with a subgroup of A ? In a natural way? I What about additional structure: if A is a field, a ring, or a module? JW (WMU) MacWilliams identities October 8, 2012 16 / 21 Finite fields (a) ∼ I Prime fields: for Fp = Z=pZ, then θp(a) = exp(2πia=p) is a character of Fp. I Every character of Fp is of the form a 7! θp(ba), for some b 2 Fp. ` I In general, consider Fq, q = p , with trace function Trq=p : Fq ! Fp, p p2 p`−1 Trq=p(a) = a + a + a + ··· + a . Define θq = θp ◦ Trq=p, a character of Fq. I Every character of Fq is of the form a 7! θq(ba), for some b 2 Fq. JW (WMU) MacWilliams identities October 8, 2012 17 / 21 Finite fields (b) ∼ n ∼ n I Fcq = Fq as vector spaces, and Fcq = Fq as vector n spaces: y 2 Fq 7! (a 7! θq(a · y)), where · is the n dot product on Fq. n I Under this isomorphism, (Fcq : C) corresponds to ? n C := fy 2 Fq : c · y = 0; c 2 Cg: JW (WMU) MacWilliams identities October 8, 2012 18 / 21 MacWilliams identities for finite fields I This leads to the original version of the MacWilliams identities, due to MacWilliams, 1961{1962, for A = Fq: 1 W (X ; Y ) = W ? (X + (q − 1)Y ; X − Y ): C jC ?j C JW (WMU) MacWilliams identities October 8, 2012 19 / 21 Finite rings I Suppose A = R, a finite ring. The same identifications that worked for finite fields will work here, provided Rb ∼= R as one-sided R-modules. n ∼ n n I Then Rb = R , and (Rb : C) can be identified with l(C) := fb 2 Rn : b · c = 0; c 2 Cg; r(C) := fb 2 Rn : c · b = 0; c 2 Cg: JW (WMU) MacWilliams identities October 8, 2012 20 / 21 MacWilliams identities for finite rings ∼ I Assume R is a finite ring satisfying Rb = R as one-sided R-modules. Then: 1 W (X ; Y ) = W (X + (jRj − 1)Y ; X − Y ): C jl(C)j l(C) I Similarly for r(C) in place of l(C). I In the next lecture we discuss rings satisfying Rb ∼= R: the finite Frobenius rings. JW (WMU) MacWilliams identities October 8, 2012 21 / 21.
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