On self-dual MRD codes Wolfgang Willems DARNEC15, Istanbul, Nov. - PowerPoint PPT Presentation

On self-dual MRD codes Wolfgang Willems DARNEC’15, Istanbul, Nov. 4-6, 2015

set up: C ≤ k m × n , linear of dimension ℓ , k = F q . • ( m ≥ n ) d( A, B ) = rank ( A − B ) for A, B ∈ k m × n . • � A, B � = trace ( AB t ). • If C = C ⊥ , then C is called self-dual. • C is called MRD if d( C ) = d = n − ℓ • m + 1. If C is a self-dual MRD code, then ℓ = mn • and 2 d = n 2 + 1 ≥ 2. 1

Problem. What can we say about self-dual MRD codes? • Do they exist? If so, are they of interest? • Joint work with G. Nebe, RWTH Aachen, Germany. 2

Disappointing: They do not exist in characteristic 2 . Theorem 1. Assume that char k = 2 and C ⊆ C ⊥ ≤ k m × n . Then the all-ones matrix J is in C ⊥ . In particular, d( C ⊥ ) = 1 . Proof: • A = ( a ij ) ∈ C . j =1 a ij ) 2 = � A, J � 2 . 0 = � A, A � = � m � n j =1 a 2 ij = ( � m � n • i =1 i =1 d( C ⊥ ) ≤ d( J, 0) = rank J = 1. • 3

Example. Let C ≤ F 2 × 2 be an MRD code of dimension 2. Then q � � � � 1 0 0 1 C = � A, B � with A = and B = . a b c d C is a self-dual MRD code if and only if the Lemma 1. following holds true: (i) − 1 �∈ F 2 q , i.e. q ≡ 3 mod 4 and (ii) a 2 + b 2 = − 1 and ( c, d ) ∈ { ( − b, a ) , ( b, − a ) } . Remark. All codes in Lemma 1 are pairwise equivalent and equivalent to Gabidulin codes of full length. 4

Theorem 2. (Hua, Wan; ∼ ’50, ’60) If ϕ is a linear isometry of k m × n ( m, n ≥ 2) w.r.t. d( · , · ), then there exist X ∈ GL( m, k ) and Y ∈ GL( n, k ) s.t. for all A ∈ k m × n ϕ ( A ) = κ X,Y ( A ) = XAY (proper isometry) or, but only in case m = n , ϕ ( A ) = τ X,Y = XA t Y for all A ∈ k n × n Remark. If ϕ also preserves �· , ·� , then XX t = aI m and Y Y t = a − 1 I n . 5

Proposition. C ≤ k m × n with char k � = 2 is properly equivalent to a self- dual code if and only if the following holds: X = X t ∈ GL( m, k ) , Y = Y t ∈ GL( n, k ) (i) det X, det Y ∈ ( k × ) 2 (ii) C ⊥ = X C Y . (iii) 6

Proof. Suppose that X 0 C Y 0 = D = D ⊥ . 0 = trace ( X 0 C 1 Y 0 ( X 0 C 2 Y 0 ) t ) = trace ( X 0 C 1 Y 0 Y t 0 C t 2 X t 0 ) = trace ( X t 0 X 0 C 1 Y 0 Y t 0 C t 2 ) Put X := X t 0 X 0 and Y := Y 0 Y t 0 . Then X and X are sym- metric of square determinant and C ⊥ = X C Y . Conversely, (i) and (ii) imply X = X t 0 X 0 and Y = Y 0 Y t 0 (due to the classification of regular quadratic forms). 7

Main Theorem. 2 , Γ ≤ k n × n be a Gabidulin code of dimension n 2 Let C = G n 2 . Then C is equivalent to a self-dual Gabidulin code if and only if n ≡ 2 mod 4 and q ≡ 3 mod 4. (compare the result with Lemma 1) 8

To prove the main theorem we mainly need Theorem 3. For 0 < ℓ < n and k = F q we have. a) The group of proper automorphisms of G ℓ, Γ ≤ k n × n is Aut ( p ) ( G ℓ, Γ ) = { κ X,Y | ( X, Y ) ∈ ( A j G × 1 , Γ × A − j G × 1 , Γ ) , 0 ≤ j ≤ n − 1 } b) Aut( G ℓ, Γ ) = � Aut ( p ) ( G ℓ, Γ ) , τ T − 1 ,TA ℓ − 1 � c) | Aut( G ℓ, Γ ) | = 2 n ( q n − 1) q n − 1 q − 1 . (Note: G × Singer cycle, det S �∈ ( k × ) 2 ) 1 , Γ = � S � , 9

Lemma 4. 2 , Γ = TA n/ 2 G n 2 , Γ T − 1 G ⊥ n , where Γ = ( γ, γ q , . . . , γ q n − 1 ) • T = ( t ij ) where t ij = trace F qn / F q ( γ q i + j ) •   0 . . . 0 1 1 0 . . . 0   • A =  .  .  ... ... . 0 .    0 . . . 1 0 (Essentially in Berger ’02 and Ravagnani ’15) 10

Proof of the main theorem. • Suppose that C = G n 2 , Γ is equiv. to a self-dual one. (1) C is properly equiv. to a self-dual code: • C �→ XC t Y ∈ D = D ⊥ . • Y t CX t ∈ D t = ( D t ) ⊥ . C ⊥ = TA n/ 2 C T − 1 (2) (by Lemma 4) C ⊥ = X C Y with X, Y sym. and det X, det Y ∈ ( k × ) 2 (3) (by Proposition). 11

( A − n/ 2 T − 1 X, Y T ) = ( A j S i , A − j S h ) ∈ Aut( C ) (4) (by Theorem 3) (5) What are the conditions that there exist triples ( i, j, h ) such that X i,j = TA n/ 2+ j S i Y h,j = A − j S h T − 1 and are symmetric and have a square determinant. ... is equivalent to n ≡ 2 mod 4 and q ≡ 3 mod 4. 12

Final remarks. • If q ≡ 1 mod 4 or 4 | n we do not know any example of a self-dual MRD code in F n × n . q • Is there a self-dual MRD code in F 4 × 4 ? 3 • (Morrison) In F 4 × 2 there are 5 equivalence classes of self- 5 dual MRD codes. • Are there interesting automorphism groups in the class of self-dual MRD codes? ( t � = 1 = t 2 ) Aut( G ℓ, Γ ) = (( C q n − 1 Y q − 1 C q n − 1 ) ⋊ C n ) ⋊ � t � , 13