1
$\begingroup$

I'm going through pymatching tutorial on constructing a toric code using hypergraph product of two repetition codes. The hypergraph product code construction $H G P\left(H_1, H_2\right)$ takes as input the parity check matrices of two linear codes $C_1:=\operatorname{ker} H_1$ and $C_2:=\operatorname{ker} H_2$. The code $H G P\left(H_1, H_2\right)$ is a CSS code with the parity check matrix $H_X$ corresponding to the $X$-type stabilisers \begin{equation} H_X=\left[H_1 \otimes I_{n_2}, I_{r_1} \otimes H_2^T\right] \end{equation} and the parity check matrix $H_Z$ corresponding to the $Z$-type stabilisers is \begin{equation} H_Z=\left[I_{n_1} \otimes H_2, H_1^T \otimes I_{r_2}\right] \end{equation} where $H_1$ has dimensions $r_1 \times n_1, H_2$ has dimensions $r_2 \times n_2$ and $I_l$ denotes the $l \times l$ identity matrix.

So the full parity check matrix for the toric code is \begin{equation} H=\left(\begin{array}{cc} H_X & 0 \\ 0 & H_Z \end{array}\right) \end{equation}

From the Künneth theorem, the logical $X$ operators of the toric code are given by \begin{equation} L_X=\left(\begin{array}{cc} \mathcal{H}^1 \otimes \mathcal{H}^0 & 0 \\ 0 & \mathcal{H}^0 \otimes \mathcal{H}^1 \end{array}\right) \end{equation} where $\mathcal{H}^0$ and $\mathcal{H}^1$ are the zeroth and first cohomology groups of the length-one chain complex that has the repetition code parity check matrix as its boundary operator.

Question: Is $L_Z$ simply \begin{equation} L_Z=\left(\begin{array}{cc} \mathcal{H}_1 \otimes \mathcal{H}_0 & 0 \\ 0 & \mathcal{H}_0 \otimes \mathcal{H}_1 \end{array}\right) \end{equation} where $\mathcal{H}_0$ and $\mathcal{H}_1$ are the zeroth and first homology groups? How can I compute them? The cohomology groups for $L_X$ in the tutorial are: $\mathcal{H}^0$ is a vector of all $1$'s and $\mathcal{H}^1$ is a vector of all $0$'s but the first entry which is $1$ but I don't underastand why and how to get $\mathcal{H}_0$ and $\mathcal{H}_1$.

$\endgroup$

1 Answer 1

2
$\begingroup$

I don't understand why and how to get $\mathcal{H}_0$ and $\mathcal{H}_1$

The repetition code can be written as a chain complex $$ C_1 \xrightarrow{\partial_1} C_0, $$ For example for the distance 4 repetition code $$\partial_1= \pmatrix{1 & 1 & 0 & 0 \cr 0 & 1 & 1 & 0 \cr 0 & 0 & 1 & 1 \cr 1 & 0 & 0 & 1} ,$$ and $C_1$ is a vector space over $\mathbb{Z}_2^{4\times 1}$ and $C_0$ is a vector space over $\mathbb{Z}_2^{4\times 1}$

Elements in $C_1$ can be understood as a set of edges and elements in $C_0$ as sets of vertexes. The boundary operator $d_1$ is a map from a set of edges to a set of vertexes. For example for edges $c_1 =\pmatrix{1 0 1 0}$, $\partial_1 c_1 = \pmatrix{1,1,1,1}$. The graphical interpretation of this is: enter image description here

The group $H_n$ is defined as the quotient group obtained by partitioning n-cycles $Z_n$ into equivalence classes under composition with elements of n-boundaries $B_n$.

Let's try to figure out what $Z_0, Z_1, B_0$ and $B_1$ are. $B_1$ is the trivial one-element group, because we're considering a length 1 chain complex. Every element in $C_0$ is in $Z_0$, because any set of vertexes is a cycle. A handwavy explanation of this is that if you take the boundary of a vertex you get nothing.

The boundary of every 1-chain is a 0-chain with an even number of non-zero vertices. Therefore $H_0$ can be represented as any vector with an odd number of $1$s (For example first entry one followed by all zeros). Because $B_1$ is trivial $H_1 = Z_1$. You can form a cycle using all edges, so $H_1$ can be represented using the all ones vector.

Is $𝐿_𝑍$ simply \begin{equation} L_Z=\left(\begin{array}{cc} \mathcal{H}_1 \otimes \mathcal{H}_0 & 0 \\ 0 & \mathcal{H}_0 \otimes \mathcal{H}_1 \end{array}\right) \end{equation}

Yes this looks correct. You can check it by testing if it commutes with the parity check matrix and anticommutes with $L_X$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.