 Solution

With identical $m$-condensing boundaries, the cylinder supports one logical qubit and hence hasOne logical operator is an electric $Z$ string around the circumference; its conjugate is a magnetic $X$ string joining the two boundaries. The perturbation $h{\sum }_j{X}_j$ can generate the latter only after a virtual magnetic anyon traverses the length of the cylinder. Degenerate perturbation theory therefore gives the ground-state splitting of a surface-code cylinderwhere the factor $C$ counts translated shortest paths and nonuniversal order-one factors have been suppressed.

If both boundaries instead condense $e$, the unperturbed degeneracy remains two. The logical operator made solely from $X$ is now a magnetic loop winding around the circumference, so the same perturbation first acts nontrivially at order $C$:Thus exchanging the condensed anyon exchanges the geometrical length controlling this perturbative splitting.