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The dual charge to an anyon $a$, characterized by $N_{a\bar{a}}^1=1$. Reversing the orientation of an anyon strand replaces its label by its antianyon. Its quantum dimension satisfies $d_{\bar{a}}=d_a$.

An anyon fusion rule specifies which total anyon charges $c$ can result from combining charges $a,b$. The nonnegative integers $N_{ab}^c$ are anyon fusion multiplicities. The vacuum anyon is the identity for this product, and $N_{ab}^1={\delta }_{b,\bar{a}}$ identifies the antianyon.

Write $c_{u,v}$ for braiding and $M_{u,v}=c_{v,u}{c}_{u,v}$ for anyon monodromy. Transparency says $M_{a,x}=1$ for every $x$. The hexagon identities for braiding express braiding a composite $a\otimes b$ past $x$ as successive braidings of its two factors past $x$. If $a,b$ are both transparent anyons, move $x$ around $b$ and then around $a$; both full moves are identities, henceThis is the fusion closure of transparent anyons at the level of the composite object.

For completeness, let $i:c\to a\otimes b$ and $p:a\otimes b\to c$ be an inclusion and projection for an allowed anyon fusion summand, with $pi=1_c$. Naturality of braiding givesThusThe vacuum anyon belongs to this collection, and taking duals preserves transparency, so it is also closed under antianyons.

There is a further restriction on the spin of a transparent anyon. Apply the anyon balancing identity to the vacuum anyon channel of $a\times \bar{a}$:Here ribbon duality gives ${\theta }_{\bar{a}}={\theta }_a$ and ${\theta }_1=1$. ConsequentlyTheir topological spins are those of bosons or fermions. This does not imply $d_a=1$: a transparent anyon can have several anyon fusion channels, with nontrivial exchange action within them, even though every full anyon monodromy is trivial. If the theory is additionally a modular tensor category, the only transparent anyon is the vacuum anyon; Question 1 need not impose that extra hypothesis.

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.

The positive square root of the sum of the squared quantum dimensions of all simple anyons. The vacuum anyon has $d_1=1$ and antianyons have equal quantum dimensions.