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I am working from page $8$ of this paper.

The generator for the coset representatives of $C^{\perp}$ in $C$ for the $[6,5,2]$ classical code is given by $$G_{C/C^{\perp}} = \begin{bmatrix} 1 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 & 0 \end{bmatrix}$$

The rows of this matrix, $\underline{g_{i}}, i \in 1,2,3,4$, generate all coset representatives for $C^{\perp}$ in $C$.

CSS construction provides a $[[6,4]]$ stabilizer code $Q$ spanned by the set of basis vectors {$| \psi_{x} \rangle | \underline{x} \in \mathbb{F}_{2}^{4}$}, where $\underline{x} = [x_{1}, x_{2}, x_{3}, x_{4}]$ and $$|\psi_{x} \rangle = \frac{1}{\sqrt{2}}| (000000) + \sum_{j=1}^{4}x_{j}\underline{g_{j}} \rangle + \frac{1}{\sqrt{2}}| (111111) + \sum_{j=1}^{4}x_{j}\underline{g_{j}} \rangle$$

However, I am very confused by this. If $\underline{x} \in \mathbb{F}_{2}^{4}$, then how are we multiplying it by $\underline{g_{j}}$, which has length $5$ (if it is referring to the rows of the generator of $C/C^{\perp}$, however, if it is referring to the columns then $\underline{g_{j}}$ has length $4$, but then it doesn't make sense to add the resulting vector $x_{j}\underline{g_{j}}$ of length $4$ to the vectors $(000000)$ and $(111111)$ of length $6$.

am567
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1 Answers1

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$\underline{x} \in \mathbb{F}_{2}^{4}$ as you say, but $x_j \in \mathbb{F}_{2}$ is a scalar. $\underline{g_j}$ is a row of $G_{C/C^{\perp}}$, thus has length 6. Therefore, $\sum_{j=1}^{4}x_{j}\underline{g_{j}}$ is an $\mathbb{F}_2$ linear combination of length 6 row vectors, and thus has length 6. You can also write this as $\underline{x} G_{C/C^{\perp}}$ (a $1 \times 4$ row vector times a $4 \times 6$ matrix, so a $1 \times 6$ row vector).

user2533488
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