Past exam of the mathematics course of the University of Cambridge 2024 iii Paper 342 2 d Solution 2026-09-28
Choose both circular ends of the cylinder to be electric, or rough, boundaries. In the convention of part 1, retain plaquette generatorswith their boundary truncations, and use star generatorsonly at vertices not lying on an electric boundary. Omitting the endpoint star checks allows a string to terminate at either boundary, which is precisely anyon condensation at a boundary for .
Relative homology now has one nontrivial class represented by a primal path joining the two boundaries. The dual absolute homology has one class represented by an loop around the cylinder. Their strings intersect once and anticommute, so they form one logical pair . Every other closed or boundary-ending string is a product of stabilizers, so the code has exactly one logical qubit.
In the Random-bond Ising model mapping, there is no Ising spin for an omitted boundary star. An edge joining an interior vertex to an electric boundary is toggled by only , so its bond factor is replaced by the boundary-field factor . Equivalently, attach exterior spins fixed to and retain the same bond formula. Edges wholly on a boundary contribute fixed constants. The bulk Hamiltonian is therefore supplemented bywith the signs still determined by . This is the required boundary modification.