Local anyonic channel projector

A semisimple decomposition of an object $X$ has inclusions $i_{z,\mu }:z\to X$ and retractions $p_{z,\mu }:X\to z$, with $p_{z,\mu }{i}_{z,\nu }={\delta }_{\mu \nu }{\mathrm{id}}_z$, zero cross-composites for distinct representatives, and ${\sum }_{z,\mu }{i}_{z,\mu }{p}_{z,\mu }={\mathrm{id}}_X$. By [F2], its inclusions for each $z$ form a basis of $\mathrm{Hom}(z,X)$. Any other basis is obtained by an invertible column change; multiplying the retractions by the inverse matrix retains both identities. Thus any specified split basis can be the inclusion basis of such a decomposition. A simple leaf itself has the one-term decomposition with identity inclusion, even when it is an external object different from its representative.

Build up a finite binary tree from its leaves. Suppose its child objects $T_L,T_R$ have decompositions with inclusion bases $(i_{x,\mu })$ and $(j_{y,\nu })$ as in step 1.1. Tensoring their decompositions gives inclusions $i_{x,\mu }\otimes j_{y,\nu }$ of $x\otimes y$ into $T_L\otimes T_R$, with tensor retractions whose composites and sum satisfy the same identities by bifunctoriality. Therefore, for any simple output $z$, composition gives the isomorphism ${⨁}_{x,y,\mu ,\nu }\mathrm{Hom}(z,x\otimes y)\to \mathrm{Hom}(z,T_L\otimes T_R)$; its inverse applies those tensor retractions. Inserting the specified vertex basis in each summand produces exactly all split-tree composites for output $z$, and proves they are a basis. For a nonroot subtree take $z$ once over the internal representative set and apply step 1.1 to obtain its decomposition. At the root use the actual external output $c$ and its specified vertex bases directly. This finite recursion proves the basis assertion for arbitrary external simple objects, without requiring arbitrary endpoint bases to be transported copies of representative bases. Every internal simple label and every multiplicity index is retained.

Postcomposition by $A_i$ is linear and invertible, with inverse postcomposition by $A_i^{-1}$. Step 2.1 supplies bases on both sides, hence its matrix $F_i$ is invertible. The pair-first basis is partitioned by the pair output $s$, retaining every coordinate within that channel, so ${\mathrm{\Pi }}_s^2={\mathrm{\Pi }}_s$. Consequently $P_{i,s}^2=F_i^{-1}{\mathrm{\Pi }}_s^2{F}_i=P_{i,s}$, proving the general algebraic assertion without unitarity.

In the unitary case choose orthonormal split vertex bases. Start step 1.1 with an orthogonal decomposition; any orthonormal change of its inclusion basis is unitary and retains $p_{z,\mu }=i_{z,\mu }^{†}$. Tensor compatibility and composition make the decompositions in step 2.1 orthogonal. Thus its tree inclusions are orthonormal for $g^{†}h=⟨h,g⟩{\mathrm{id}}_c$. For unitary $A_i$ one has $(A_{i}g{)}^{†}(A_{i}h)=g^{†}{A}_i^{†}{A}_{i}h=g^{†}h$. Its matrix $F_i$ is therefore unitary. The coordinate projection ${\mathrm{\Pi }}_s$ is self-adjoint in the orthonormal pair-first basis, so $P_{i,s}=F_i^{†}{\mathrm{\Pi }}_s{F}_i$ is self-adjoint as well as idempotent.

To compare Hom conventions, take daggers in [F5]. Unitarity of the associator gives $\alpha \circ f_j^{†}={\sum }_k\overline{F_{kj}}{g}_k^{†}$. The maps $f_j^{†},g_k^{†}$ are exactly the corresponding adjoint split-tree bases. Hence the split postcomposition matrix for this local move is $\bar{F}$, while the QC25 fusion Hom matrix is $F$. This proves the claimed convention conversion and completes the well-definedness and unitary assertions, including zero-dimensional spaces with their unique empty coordinate maps.