How to compute Z logical operators of a toric code using Kunneth theorem?

Question

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 $\mathcal{H}_1$.

Answer 1

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:

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 $L_X$