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In a ribbon anyon theory, twisting a composite charge $c\in a\times b$ is equivalent to twisting each input and applying their full anyon monodromy. The topological spin obeys ${\theta }_1=1$ and ${\theta }_{\bar{a}}={\theta }_a$; the latter is the ribbon duality convention, with no complex conjugation of the scalar.

The unitary operator implementing a chosen positive exchange of adjacent anyons $a,b$ in total charge $c$. In a multiplicity-free theory it is a scalar. An inverse crossing uses its inverse. Two positive exchanges give the anyon monodromy $R_c^{ab}{R}_c^{ba}$.

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 $1$.

Use an anyon fusion tree in which the two crossing charges combine first. With external charges in the order $d,j,h,e$, define the normalized treesThe desired value is the matrix element of anyon monodromy between $|T_{bc}⟩$ and $|T_{fg}⟩$. Their closure is understood in the physics normalization of anyon diagrams.

First apply the anyon F-matrix to the three charges $d,j,c$, and then an inverse anyon F-matrix to $j,h,e$. With the source convention $|((uv{)}_{r}w{)}_t⟩={\sum }_s[F_t^{uvw}{]}_{rs}|(u(vw{)}_s{)}_t⟩$, this givesOnly allowed anyon fusion channels enter the sums: in the second tree these require $N_{jh}^x{N}_{xe}^y{N}_{dy}^a=1$. The complex conjugate in $U$ comes from the inverse change of parentheses, not from reversing a crossing.

Reading the actual overcrossings from bottom to top gives two positive exchanges in the source's anyon R-matrix convention. Their product isThe order of $h,j$ is restored after the second exchange. In the $X$ basis this anyon monodromy is diagonal, soEquivalently, entirely in terms of anyon F-matrices and anyon R-matrices,This expression is also meaningful for $b\ne f$ or $c\ne g$: it then carries the unmatched anyon vertex gauge transformation factors identified in a.

To check the transparent limit, put $m_x=1$ in every allowed channel. The unitary matrix identity for $F_y^{jhe}$ first givesOnce $c=g$, the unitary matrix identity for $F_a^{djc}$ givesas in c. The same cancellation proves that the recoupling expression is an identity operator whenever the full anyon monodromy is trivial, including when the individual exchanges have fermionic signs.

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.