User:Will Orrick/Cocyclic Hadamard matrix

In mathematics, a cocyclic Hadamard matrix is a Hadamard matrix generated by a binary cocycle, ψ, of a finite group G. A binary cocycle is a map ψ:G×G → ⟨−1⟩, where ⟨−1⟩ is the two-element multiplicative group and ψ satisfies ψ(a,b)ψ(ab,c) = ψ(b,c)ψ(a,bc) for all a,b,c ∈ G. It follows from this definition that ψ(1,a) and ψ(b,1) are equal, constant functions. If their value is +1, then the cocycle is normalized. The cocyclic matrix Mψ is the matrix whose (i,j)th element is ψ(gi,gj) where g1 = 1, g2, ..., gn is a particular listing of the group elements.

A cocyclic matrix is a Hadamard matrix if and only if ∑g∈G ψ(a,g) = 0 for all a ≠ 1.

Proof: The "only if" part follows from the fact that the first row of Mψ, corresponding to the identity element of the group, is constant. For the "if" part, we first prove the general statement that the inner product of rows i and j of a cocylic matrix equals the sum of the elements of the row corresponding to the group element gigj−1, up to a sign:


 * $$\begin{align}\sum_{k=1}^n \psi(g_i,g_k)\psi(g_j,g_k)&=\sum_{k=1}^n \psi(g_i,g_j^{-1}g_k)\psi(g_j,g_j^{-1}g_k)\\ &=\sum_{k=1}^n \psi(g_i,g_j^{-1})\psi(g_ig_j^{-1},g_k)\psi(g_j^{-1},g_k)\psi(g_j,g_j^{-1}g_k).\end{align}$$

In the second line the defining property of cocycles was used with a=gi, b=gj−1, c=gk to expand the first factor in the sum. Using the defining property a second time with a=gj, b=gj−1, c=gk, we rewrite the last two factors as


 * $$\psi(g_j^{-1},g_k)\psi(g_j,g_j^{-1}g_k)=\psi(g_j,g_j^{-1})\psi(g_jg_j^{-1},g_k)=\psi(g_j,g_j^{-1})\psi(1,g_k)=\pm \psi(g_j,g_j^{-1})$$

with the plus sign taken if the cocycle is normalized and the minus sign taken otherwise. The inner product of rows i and j therefore equals


 * $$\pm\psi(g_i,g_j^{-1})\psi(g_j,g_j^{-1})\sum_{k=1}^n \psi(g_ig_j^{-1},g_k)=\psi(g_ig_j^{-1},g_j)\sum_{k=1}^n \psi(g_ig_j^{-1},g_k),$$

where in the last step we used the defining property a third time, with a=gigj−1, b=gj, c=gj−1.

The statement now follows, since the inner product evaluates to 0 if i ≠ j and to n otherwise. Therefore the rows of Mψ are pairwise orthogonal.