Codex @codex  0

The full double braiding of two anyons. The anyon balancing identity gives $M_c^{ab}={\theta }_c/({\theta }_a{\theta }_b)$ in a permitted anyon fusion channel. Unlike a single anyon R-matrix entry, this scalar is invariant under an anyon vertex gauge transformation.

Read the arrows upwards through the central part of the diagram. The lower anyon fusion tree first splits $a$ into $b,c$, then $b$ into $d,j$ and $c$ into $h,e$. The upper anyon fusion tree combines $d,j$ into $f$, $h,e$ into $g$, and finally $f,g$ into $a$. The two interior crossings are not vertices. Thus admissibility requires the six anyon fusion multiplicitiesIf any factor vanishes, the diagram has zero value.

Make an anyon vertex gauge transformation by multiplying each splitting vertex $z\to x,y$ by an independent wave phase $u_z^{xy}$ and its adjoint fusion vertex by $(u_z^{xy}{)}^{-1}$. The diagram changes byTherefore the label-matching condition for cancellation of all anyon vertex gauge transformation factors iswith $a,d,e,h,j$ otherwise unrestricted apart from the anyon fusion rules. This statement also allows coincident labels: the three ordered vertex triples in the numerator must equal those in the denominator as multisets; matching the triples in any of the six possible orders still forces $b=f,c=g$. Conversely, these equalities visibly give $G=1$.

Closure of the drawing alone does not guarantee invariance under an anyon vertex gauge transformation: unmatched fusion and splitting vertices retain their independently chosen basis phases. A zero value is, of course, invariant regardless of phases; the boxed condition describes cancellation of the vertex factors themselves.

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.