Foundational Aspects of Linear Codes: 1. Characters and Frobenius rings

Transcription

1 Foundational Aspects of Linear Codes: 1. Characters and Frobenius rings Jay A. Wood Department of Mathematics Western Michigan University jwood/ On the Algebraic and Geometric Classifications of Projective Varieties University of Messina June 20, 2016

2 Acknowledgments Thanks to Gaetana Restuccia and Mustapha Lahyane for organizing the workshop, for inviting me to speak, and for their kind hospitality. Foundational Aspects June 20, / 28

3 Overview of lectures 0. Overview of lectures Basic terminology Duality and weight enumerators Extension problem Foundational Aspects June 20, / 28

4 Basic vocabulary Overview of lectures Let R be a finite associative ring with 1, not necessarily commutative. Let A be a finite unital left R-module; A will be the alphabet. A left R-linear code over A of length n is a left R-submodule C A n. Right linear codes are defined similarly. Module alphabets are due to Nechaev and collaborators, Foundational Aspects June 20, / 28

5 Weights Overview of lectures A weight on A is any function w : A C with w(0) = 0. Extend to w : A n C by w(a 1, a 2,..., a n ) = n w(a i ). i=1 Restrict w to linear code C A n. Foundational Aspects June 20, / 28

6 Examples Overview of lectures For any alphabet A, define the Hamming weight: { 1, a 0, wt(a) = 0, a = 0. Lee weight: for R = A = Z/NZ, restrict N/2 < a N/2 and set w L (a) = a (ordinary absolute value). Homogeneous weight (later). Foundational Aspects June 20, / 28

7 Overview of lectures Hamming weight enumerator For a linear code C A n, define the Hamming weight enumerator of C by hwe C (X, Y ) = x C X n wt(x) Y wt(x). hwe C (X, Y ) = n i=0 A ix n i Y i, where A i is the number of codewords in C of Hamming weight i. Foundational Aspects June 20, / 28

8 Dual codes Overview of lectures Let A = R itself. Define the dot product on R n by x y = n x i y i, i=1 where x = (x 1, x 2,..., x n ) and y = (y 1, y 2,..., y n ). For a linear code C R n, define annihilators l(c) = {y R n : y C = 0}, r(c) = {y R n : C y = 0}. Foundational Aspects June 20, / 28

9 Questions Overview of lectures Are the annihilators well-behaved? Is there a nice relationship between the Hamming weight enumerators of C and its annihilators? Yes: the MacWilliams identities. We will discuss these questions in the second lecture. Foundational Aspects June 20, / 28

10 Isometries Overview of lectures Let C 1, C 2 A n be two linear codes. An R-linear isomorphism f : C 1 C 2 is a linear isometry with respect to a weight w if w(xf ) = w(x) for all x C 1. I will usually write homomorphisms of left modules M on the right side, so that (rx)f = r(xf ) for r R, x M. Foundational Aspects June 20, / 28

11 Overview of lectures Symmetry groups Suppose the alphabet A has weight w. Define symmetry groups by G lt = {u U(R) : w(ua) = w(a), a A}, G rt = {φ GL R (A) : w(aφ) = w(a), a A}. Here, U(R) is the group of units of R, and GL R (A) is the group of invertible R-linear homomorphisms of A to itself. Foundational Aspects June 20, / 28

12 Overview of lectures Monomial transformations Let G GL R (A) be a subgroup. A G-monomial transformation of A n is an invertible R-linear homomorphism T : A n A n of the form (a 1, a 2,..., a n )T = (a σ(1) φ 1, a σ(2) φ 2,..., a σ(n) φ n ), for (a 1, a 2,..., a n ) A n. Here, σ is a permutation of {1, 2,..., n} and φ i G for i = 1, 2,..., n. Foundational Aspects June 20, / 28

13 Overview of lectures Monomial transformations are isometries Easy: G rt -monomial transformations are isometries of A n with respect to the weight w. Let C 1 A n be a linear code and let T be a G rt -monomial transformation of A n. Set C 2 = C 1 T. Then the restriction of T to C 1 is a linear isometry from C 1 to C 2. Is the converse true? That is, does every linear isometry between linear codes extend to a G rt -monomial transformation? Call this the Extension Problem. More on this in the third and fourth lecture and in the research seminar. Foundational Aspects June 20, / 28

14 Characters and Frobenius rings 1. Characters and Frobenius rings Definitions Properties Fourier transform Character modules Generating characters Foundational Aspects June 20, / 28

15 Definitions Characters and Frobenius rings Let A be a finite abelian group (additive notation); A will be a module later. A character of A is a group homomorphism π : A C, where C is the multiplicative group of nonzero complex numbers. The set  of all characters of A is a multiplicative abelian group under pointwise multiplication. Additive version:  = Hom Z (A, Q/Z). Foundational Aspects June 20, / 28

16 Duality functor Characters and Frobenius rings Pontryagin duality: A Â Â = A, naturally. Â = A, but not naturally. (A B) = Â B. Exact contravariant functor: exact sequence 0 A 1 A 2 A 3 0 induces exact sequence 0 Â3 Â2 Â1 0. Foundational Aspects June 20, / 28

17 Characters and Frobenius rings Summation formulas For a A, For π Â, π(a) = π Â π(a) = a A { A, a = 0, 0, a 0. { A, π = 1, 0, π 1. Foundational Aspects June 20, / 28

18 Characters and Frobenius rings Linear independence Let F (A, C) = {f : A C}, a complex vector space of dimension A. The elements of  form a basis for F (A, C). In particular, characters are linearly independent over C. Need multiplicative form of characters for linear independence and for the summation formulas. Foundational Aspects June 20, / 28

19 Annihilators Characters and Frobenius rings Let B A be any subgroup. Define the annihilator (Â : B): (Â : B) = {π Â : π(b) = 1}. (Â : B) = (A/B). B (Â : B) = A. Double annihilator: (A : (Â : B)) = B. Foundational Aspects June 20, / 28

20 Characters and Frobenius rings Fourier transform Given a function f : A V, V a complex vector space. Define its Fourier transform ˆf : Â V by ˆf (π) = a A π(a)f (a), π Â. : F (A, V ) F ( Â, V ). Invert: f (a) = 1 A π( a)ˆf (π), a A. π Â Foundational Aspects June 20, / 28

21 Characters and Frobenius rings Poisson summation formula Let B be any subgroup of A, and let f : A V. Then for any a A, 1 f (a + b) = b B (Â : B) π( a)ˆf (π). π (Â:B) If a = 0, then f (b) = b B 1 ˆf (Â : B) (π). π (Â:B) Foundational Aspects June 20, / 28

22 Characters and Frobenius rings Character modules Now suppose R is a finite ring with 1 and A is a finite unital left R-module. Then  becomes a right R-module by π r (a) = π(ra), a A. ( r π(a) = π(ar) for right to left case.) is an exact contravariant functor of R-modules. For left R-submodule B A, ( : B) is a right R-submodule of Â. Foundational Aspects June 20, / 28

23 Characters and Frobenius rings Top-bottom duality An R-module is simple if it has no nontrivial proper submodules. The Jacobson radical Rad(R) is the intersection of all maximal left ideals of R; a two-sided ideal. For a left R-module A, the socle Soc(A) is the left R-submodule generated by all simple left R-submodules of A. (A/ Rad(R)A) = Soc(Â) Foundational Aspects June 20, / 28

24 Characters and Frobenius rings Generating characters Left R-module A. A character ρ Â is a generating character if ker ρ contains no nonzero left R-submodules. Not every module admits a generating character. Foundational Aspects June 20, / 28

25 Embedding Characters and Frobenius rings Suppose ρ is a generating character for A. Define α : A R by (aα)(r) = ρ(ra), a A, r R. α is an injective homomorphism of left R-modules. Dual map R Â, r ρr, is surjective homomorphism of right R-modules. ρ generates Â. Conversely: if  is cyclic, or A embeds in R, then A has a generating character. Foundational Aspects June 20, / 28

26 Frobenius rings Characters and Frobenius rings Recall: R = R. Consider R as a module over itself. If R has a generating character, then R = R as left and as right R-modules. Soc(R) = Soc( R) = (R/ Rad(R)) = R/ Rad(R). Such a finite ring is a Frobenius ring. Foundational Aspects June 20, / 28

27 Characters and Frobenius rings Some examples of generating characters Z/NZ admits θ N (a) = exp(2πia/n), a Z/NZ. exp is the standard complex exponential function. F q admits θ q (a) = θ p (Tr q p (a)), a F q. R = M k k (F q ) admits ρ(p) = θ q (Tr P), P R. A = M k l (F q ), k > l, admits ρ A. When k < l, A does not admit a generating character. π Q (P) = θ q (Tr(PQ)), for Q M l k (F q ). Find nonzero X M k l (F q ) with XQ = 0, as k < l. Then RX ker π Q. Foundational Aspects June 20, / 28

28 Cyclic socle Characters and Frobenius rings Suppose A has cyclic socle Soc(A). Soc(A) is a sum of matrix modules with k l. Soc(A) admits a generating character: multiply together those from matrix modules. Extension exists, by exactness. Any extension is a generating character for A. Theorem R-module A has a generating character if and only if Soc(A) is cyclic. Foundational Aspects June 20, / 28