Is the Steane code the only $ [\![7,1,3]\!] $ CSS code?

Question

Is the Steane code the only $ [\![7,1,3]\!] $ CSS code?

This paper claims there are 10 non-equivalent $ [\![7,1,3]\!] $ stabilizer codes. How many of these are CSS codes? Is it just the Steane code?

Answer 1

The CSS construction takes a classical linear $[n,k_1]$ code $C_1$ and a classical linear $[n,k_2]$ code $C_2\subset C_1$ such that $C_1$ and $C_2^\perp$ both have code distance at least $d$, and yields a quantum stabilizer code $[\![n, k_1-k_2, d]\!]$. Thus, the question can be answered by counting suitable pairs of classical codes. By the classical Hamming bound $k_1,k_2\leqslant 4$, so there are only three cases to consider.

Case one: $[7,2]+[7,1]$

A classical $[7,1]$ code consists of two code words $a$ and $b$. By a simple counting argument, we can find two distinct $i,j\in\{0,1,\dots,6\}$ such that $a_i=a_j$ and $b_i=b_j$. But then the $7$-bit string $e_{i,j}$ with $1$ on positions $i$ and $j$ and zeroes everywhere else belongs to $C_2^\perp$. However, its Hamming weight is $2$, so $C_2^\perp$ cannot have distance $3$. Therefore, there are no suitable classical codes with $k_1=2$ and $k_2=1$.

Case two: $[7,3]+[7,2]$

The above argument applies in this case, too, albeit we now have to use the fact that the code is linear. In more detail, a classical linear $[7,2]$ code $C_2$ is a subspace of $\mathbb{Z}_2^7$ spanned by two linearly independent $7$-bit strings $a$ and $b$. Once again, we can find two distinct positions $i,j\in\{0,1,\dots,6\}$ such that $a_i=a_j$ and $b_i=b_j$. But then $e_{i,j}\in C_2^\perp$, so $C_2^\perp$ has distance at most $2$ and once again we find no suitable classical codes with $k_1=3$ and $k_2=2$.

Case three: $[7,4]+[7,3]$

Let $G\in\mathbb{Z}_2^{4\times 7}$ be the generator matrix of a $[7,4]$ code $C_1$. By adding and rearranging rows we can transform the matrix into the standard form $$ G=[I_4|P]\tag1 $$ where $I_4$ is the $4\times 4$ identity matrix and $P\in\mathbb{Z}_2^{4\times 3}$. But $C_1$ has distance at least $3$, so every row of $P$ must have Hamming weight at least two. For the same reason, the rows of $P$ must be distinct. But there are only four possible $3$-bit strings that satisfy these requirements: $111$, $110$, $101$, and $011$. Therefore, up to permutations of rows and columns, the generator matrix is $$ G=\begin{bmatrix} 1 & 0 & 0 & 0 & 1 & 1 & 1\\ 0 & 1 & 0 & 0 & 1 & 1 & 0\\ 0 & 0 & 1 & 0 & 1 & 0 & 1\\ 0 & 0 & 0 & 1 & 0 & 1 & 1\\ \end{bmatrix}.\tag2 $$ This fixes the $C_1$ code. Our task is then reduced to the task of enumerating all $3$-dimensional subspaces of $C_1$ whose dual is a linear code of distance at least three. However, $C_1$ is a set of mere sixteen elements, so there are fewer than ${15 \choose 3}=455$ eligible subspaces. The exhaustive search is left as a simple coding exercise for the reader.