Question inspired by statement in Lecture 6 of notes https://www2.perimeterinstitute.ca/personal/dgottesman/CO639-2004/Lecture6.pdf
The statement that I am a bit stuck on is: "For reasons that we will see below, we actually require the two row vectors $u_{1}$ and $u_{2}$ to be orthogonal to each other and to themselves: that is, to have $u_{i}·\bar{u_{j}} = 0$ for $i, j ∈ ${$1, 2$}. I would like to know why it is REQUIRED that the row vectors be orthogonal.
Background info and notation:
Let $S$ be an abelian subgroup in $GF(4)$ and let $u_{i}$ be vectors in $S$.
Commutativity of vectors $u,v \in GF(4)$ is specified as being when the (symplectic) inner product $$(u|v)=0$$
Where $(u|v)$ is defined by $$(u|v) = tr(u.\bar{v}) = \sum_{j} tr(u_{j}\bar{v_{j}})$$ Note:
- $\bar{v} = v^{2}$,
- $v=${$v_{1}, \dots, v_{n}$}
- $\bar{v} =${$v_{1}^{2} \dots, v_{n}^{2}$}
- $tr(x) = x + x^{2}$
Additionally, vectors $u_{i}$ and $u_{j}$ are orthogonal if: $$u_{i}.\bar{u_{j} }=0$$
Why would orthogonality be a necessary criteria in order for each of the elements of $u_{i} i \in $ {$1,2,3,4$} to commute?
I understand how orthogonality implies commutation but I don't understand the vice versa.
There is a note that explains : "because the stabilizer $S$ is itself a subset of $N(S)$, we can deduce that all the vectors of the parity check matrix must be orthogonal to every other row, and orthogonal to themselves: this is where that condition came from in producing the parity check matrix earlier"
I don't understand why this would require orthogonality. As $S$ is an abelian group, then $N_{E}(S)=C_{E}(S)$ which should imply that every element in $S$ needs to commute with each other. However, I don't understand how commutation, as it is defined here, requires orthogonality.
$tr(0)=tr(1)=0$, so if $u_{1}.\bar{u_{2}}=1$ then $(u|v)=0$. So it is my understanding that this shows vectors can commute even if not necessarily orthogonal?