 Solution

Cut one nonseparating circle on each handle of the closed orientable surface ${\Sigma }_g$ of genus $g$. The cut surface is a sphere with $2g$ boundary circles. Gluing a pair back together requires an anyon label $a_i$ on one boundary and the antianyon label ${\bar{a}}_i$ on the oppositely oriented boundary. Sum over these labels independently for the $g$ handles.

For a fixed labeling, the dimension of the resulting anyon fusion space on the sphere, with total vacuum anyon charge, isThe gluing rule of the topological quantum field theory therefore gives the handle-cut formula for anyon ground-state degeneracy:Here $1$ denotes the vacuum anyon, and the anyon fusion matrices have entries $(N_b{)}_{ac}=N_{ab}^c$, as in the source. At $g=0$, the empty product is the identity matrix, giving one ground state. At $g=1$, $(N_a{N}_{\bar{a}}{)}_{11}=1$ for each simple charge, so the ground-state degeneracy of the torus is the number of anyon types.