For
$$
K_1=\begin{pmatrix}0&2\\2&0\end{pmatrix},
\qquad
K_1^{-1}=
\begin{pmatrix}0&1/2\\1/2&0\end{pmatrix},
$$
the two basic particles have bosonic self-statistics and full mutual-braid phase $-1$. Moreover $|\det K_1|=4$. These are the electric and magnetic particles of the <surface code>.

For
$$
K_2=\begin{pmatrix}0&2\\2&4\end{pmatrix},
\qquad
K_2^{-1}=
\begin{pmatrix}-1&1/2\\1/2&0\end{pmatrix},
$$
the first basic particle is a fermion, the second is a boson, and they are <mutual semions>; again $|\det K_2|=4$. They can be identified with $f=e\times m$ and $m$, so this is merely another integral basis for the same surface-code phase.

Finally,
$$
|\det K_3|
=\det\begin{pmatrix}4&2\\2&4\end{pmatrix}
=12,
$$
which already rules it out. Therefore
$$
\boxed{K_1\ \text{and}\ K_2\ \text{describe the surface code};\quad K_3\ \text{does not}.}
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 342 / 1 / f / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-342/1/f/solution