Intertwined results on linear codes and Galois geometries
Peter Vandendriessche · Ghent University Academic Bibliography (Ghent University) · 2014
The fact that C m,t,2 h = χ π | π ∈ B ⊥ is a cyclic code follows directly from the fact that P m,t,2 h = χ π | π ∈ B is cyclic as a linear code.Proposition 2.2.5.P m,t,2 h = C ⊥ m,t,2 h , 1 , where 1 ∈ C ⊥ m,t,2 h .Proof.Because all generators χ π of C ⊥ m,t,2 h are of even weight, all code words of C ⊥ m,t,2 h are also of even weight.Since the length 2 h(m+1) -1 2 h -1 of this cyclic code is odd, the all-one vector is not a code word.Recall that P m,t,2 h is the vector space spanned by the incidence vectors of t-dimensional subspaces in PG(m, 2 h ).Because the number of t-dimensional subspaces that contain a given point in PG(m, 2 h ) is always odd, we have π∈B χ π = 1, which implies that 1 ∈ P m,t,2 h .Because χ π = χ π + 1, it follows that C ⊥ m,t,2 h ⊂ P m,t,2 h .Thus, we have C ⊥ m,t,2 h , 1 ⊆ P m,t,2 h .Now the fact that χ π = χ π + 1 is equivalent to the relation that χ π = χ π + 1, which implies that P m,t,2 h ⊆ C ⊥ m,t,2 h , 1 .Thus, P m,t,2 h = C ⊥ m,t,2 h , 1 as desired.Proposition 2.2.6.C m,t,2 h = P ⊥ m,t,2 h , 1 , where 1 ∈ P ⊥ m,t,2 h .Proof.Because the generators of P m,t,2 h are all of odd weight, the inner product between 1 and any of the generators is nonzero, which implies that 1 ∈ P ⊥ m,t,2 h .By the same token, because the generators of C ⊥ m,t,2 h are of even weight, we have 1 ∈ C m,t,2 h .By Proposition 2.2.5, C ⊥ m,t,2 h ⊂ P m,t,2 h , which implies that P ⊥ m,t,2 h ⊂ C m,t,2 h .Again by Proposition 2.2.5, the dimensions of P m,t,2 h and C ⊥ m,t,2 h satisfy the equation that dim P m,t,2 h = dim C ⊥ m,t,2 h + 1.Hence, C m,t,2 h = P ⊥ m,t,2 h , 1 as desired.