How to represent elements from $\bar{E}=E/Z(E)$ in the form $(a|b)$

Question

From https://arxiv.org/abs/quant-ph/9608006

Background

The group $E$ of tensor products $\pm w_{1} \otimes \dots \otimes w_{n}$ and $\pm i w_{1} \otimes \dots \otimes w_{n}$, where each $w_{j}$ is one of $I, X, Y, Z$ describes the possible errors in $n$ qubits. $E$ has order $2^{2n+2}$.

$E$ has centre $Z(E)=\langle iI \rangle $

$\bar{E}$ is the quotient group $\bar{E}=E/Z(E)$. $\bar{E}$ is a binary vector space of order $2^{2n}$.

Elements of $\bar{E}$ are written $(a|b)=(a_{1} \dots a_{n}|b_{1} \dots b_{n})$ and which is equipped with the inner product $((a|b),(a'|b'))=a.b' + a'.b$

My Question

It is my understanding that elements of the quotient group $\bar{E}$ must have the form $\bar{e_{i}}=e_{i}\langle iI \rangle=${$iw_{1}\otimes \dots \otimes w_{n},-iw_{1}\otimes \dots \otimes w_{n}, w_{1}\otimes \dots \otimes w_{n}, -w_{1}\otimes \dots \otimes w_{n}$ }

However, I am not seeing how this corresponds to the form that elements are supposed to take in $\bar{E}$, which is $(a|b)$?

In the coset notation, each element appears to have 4 terms, but in the concatenated notation, it appears to be two $n$ length vectors concatenated together?

Answer 1

For understanding how they relate, there are two main points to make. Point 1 is technical, making a correction about the elements of the quotient group. Point 2 somewhat intuitive rather than technical, by digging into what the representations mean.

1. The center consists of four elements, which correspond to multiplication by the constants $1,-1,i,-i$. Considering $E / \Xi (E)$ is basically just eliminating those constants out front, so we don’t have to keep considering them. So we just have to think about elements that are essentially of the form $$w_1 \otimes \dots \otimes w_n.$$ (Looking towards the next part, it is in this way that we start only thinking about bit-flip and phase-flip errors.)

2. To relate $w_1 \otimes \dots \otimes w_n$ to $(a | b)$, consider what each is keeping track of.

• The term $w_1 \otimes \dots \otimes w_n$ is of length $n$, and each $w_i$ can be one of four things (specifically, one of $I,\sigma_x, \sigma_y, \sigma_z$). So this is a set of $(2^2)^n = 2^{2n}$ elements keeping track of the possible bit-flip and phase-flip errors.

• The term $(a|b)$ is keeping track of the possible $\sigma_x$ errors through the binary vector $a$, and the possible $\sigma_z$ errors through the binary vector $b$. Even though its length $2n$ instead of $n$, we can see that since all entries of $a$ and $b$ are either $0$ or $1$ that there are $2^{(2n)} = 2^{2n}$ of them, again representing all possible bit-flip and phase-flip errors.

In summary, $(a|b)$ is a way of representing the elements of $E/\Xi(E)$ that separates the bit-flip and phase-flip errors, as opposed to compacting them together in each $w_i$ of the tensor product representation.