Since $\det M=-3$, the condensed top-sector anyons modulo local particles form
$$
\mathcal L/(\mathcal L\cap K\mathbb Z^4)
\simeq\mathbb Z^2/M\mathbb Z^2,
$$
which has order $|\det M|=3$. The same lower bound follows directly from Wilson-operator algebra. Take
$$
q=(1,0,0,0)^T,
\qquad
q'=(0,0,1,0)^T.
$$
Because
$$
M^{-1}=\frac{-1}{3}\begin{pmatrix}1&-2\\-2&1\end{pmatrix},
$$
their crossing operators obey
$$
W_qW_{q'}=e^{-2\pi i/3}W_{q'}W_q.
$$
Acting repeatedly with one operator on an eigenstate of the other produces three states with distinct eigenvalues; they are linearly independent and have the same energy. Hence
$$
\boxed{\operatorname{GSD}\geq3.}
$$
For these identical maximal boundaries the bound is saturated, although only the lower bound was requested.


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 342 / 1 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-342/1/d/solution