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For transparent anyons $a,b$, the hexagon identities for braiding for braiding express double exchange of $a\otimes b$ with any charge $x$ as the successive double exchanges of $a$ and $b$ with $x$. Both are identities. Naturality restricts the resulting identity to every simple anyon fusion summand $c$ of $a\otimes b$, proving that $c$ is transparent.

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.

The two crossings of $j,h$ make their full anyon monodromy. For a transparent anyon $h$, this is the identity, so that pair of crossings can be removed together. No single-crossing exchange phase, and no topological spin from a twist, is introduced.

After this removal the upper and lower anyon fusion trees differ only in their intermediate labels. In the physics normalization of anyon diagrams, successive split-fuse bubbles give the Kronecker deltas ${\delta }_{bf}$ and ${\delta }_{cg}$. The remaining closed normalized state has norm one. Hence, for the admissible labelings of a,In particular, every admissible diagram satisfying the label matching in a evaluates to $1$. There is no factor $d_a$ here: the source explicitly permits the physics normalization of anyon diagrams, in which a closed normalized state has value $1$.

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.