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A matrix encoding multiplication by charge $b$ in the anyon fusion rule. These matrices commute in a braided monoidal category. On a sphere, the dimension of the anyon fusion space with puncture charges $a_1,\dots ,a_n$ and total vacuum anyon charge is $(N_{a_1}\cdots N_{a_n}{)}_{11}$.

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.

In a modular tensor category, a loop with the normalized Kirby color encircling a charge-$x$ strand projects onto the vacuum anyon charge and multiplies it by $\mathcal{D}$. Indeed, the encircling scalar is ${\mathcal{D}}^{-1}{\sum }_a{d}_a{S}_{ax}/S_{1x}$, which equals $\mathcal{D}{\delta }_{x1}$ by unitary matrix orthogonality for the modular S-matrix.

A finite semisimple ribbon braided monoidal category with nondegenerate modular S-matrix. For a unitary anyon theory, this means that the only transparent anyon is the vacuum anyon. Such data define a three-dimensional topological quantum field theory.

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.

Let $B_c^{ab}$ denote the split-fuse diagram with a charge-$c$ edge joining the lower fusion vertex to the upper splitting vertex, and outer strands $a,b$. Completeness in the isotopy normalization of anyon diagrams readsResolve the two strands inside the encircling normalized Kirby color loop in this way. Move that loop onto the intermediate charge-$c$ edge. The Kirby killing property multiplies it by $\mathcal{D}{\delta }_{c1}$, leavingThe vacuum anyon channel is allowed only for $b=\bar{a}$, and then $d_b=d_a$ and $d_1=1$. Erasing the vacuum edge turns $B_1^{a\bar{a}}$ into the cup-cap diagram $E_a$, which connects the incoming pair by a cap and the outgoing pair by a cup. Thus the requested two-strand Kirby projector isHere $P_1^{a\bar{a}}=E_a/d_a$ is the orthogonal projection onto total vacuum anyon charge; for $b\ne \bar{a}$ it is zero. Both strands in the source point upwards, so the condition is $b=\bar{a}$, not $b=a$ unless $a$ is self-dual.

The coefficient can be checked without a convention guess: stacking two cup-cap diagrams creates one charge-$a$ loop, so $E_a^2=d_a{E}_a$. Consequently $(E_a/d_a{)}^2=E_a/d_a$, as required of an orthogonal projection. Omitting $1/d_a$ would fail this check.

Apply the anyon balancing identity to the vacuum anyon channel in $a\times \bar{a}$. Transparency gives $1=({\theta }_a{\theta }_{\bar{a}}{)}^{-1}={\theta }_a^{-2}$. Thus a transparent anyon has bosonic or fermionic topological spin. This does not require its quantum dimension to equal one.

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.

An anyon whose full anyon monodromy with every other charge is the identity in every allowed anyon fusion channel. The vacuum anyon is transparent. Transparent charges and their anyon fusion summands form a symmetric subcategory of the braided monoidal category. They can have quantum dimension greater than one.

An encircling normalized Kirby color loop around two equally oriented anyon strands is $\mathcal{D}$ times the orthogonal projection onto their total vacuum anyon charge. In the isotopy normalization of anyon diagrams, erasing the vacuum edge writes this as ${\delta }_{b,\bar{a}}(\mathcal{D}/d_a)$ times the cup-cap diagram. The cup-cap squared is $d_a$ times the cup-cap, so $d_a^{-1}$ is essential.