Noninvertible Symmetries and Anyonic Chains

For $n=0$, $T_0=b$ and [F4] gives $\mathrm{Hom}(c,b)=0$ when $c\cong ̸b$. If $c\cong b$, composition with the chosen isomorphism $u_{c,b}:c\to b$ identifies $\mathrm{End}(c)=\mathbb{C}{\mathrm{id}}_c$ with $\mathrm{Hom}(c,b)$, so the latter has basis $u_{c,b}$. This is exactly the typed empty-path convention of [F1]. For $n=1$, the path vectors are directly the chosen basis $B_{b,a}^c$ of $\mathrm{Hom}(c,b\otimes a)$, since the starting map is ${\mathrm{id}}_b$. This separately proves the positive-length base case without requiring the bases for isomorphic but unequal external objects to coincide.

Fix $n\ge 1$ and assume for every simple endpoint $x$ that the length-$n$ path composites form a basis of $\mathrm{Hom}(x,T_n)$.

By finite semisimplicity, the induction hypothesis decomposes $T_n\cong {⨁}_{x,\gamma }x$, with one summand for each length-$n$ path $\gamma$ ending at $x$. Here $x$ ranges once over the fixed simple representatives, even when the external endpoints are different isomorphic objects. For each such $x$, finite semisimplicity gives $x\otimes a\cong {⨁}_y{y}^{\oplus N_{xa}^y}$. Applying $\mathrm{Hom}(y,-)$, all nonisomorphic simple summands vanish and each of the $N_{xa}^y$ copies of $y$ contributes $\mathrm{End}(y)=\mathbb{C}{\mathrm{id}}_y$ by [F4]. Hence $\dim W_{x,a}^y=N_{xa}^y$ and the specified inclusion basis has exactly that many vectors.

Choose retractions $p_{x,\gamma }:T_n\to x$ dual to the path inclusions $\gamma :x\to T_n$ in step 2.1. They exist by taking any semisimple decomposition and changing its inclusion basis to the given basis for each simple $x$; the inverse basis matrix changes its retractions. Thus ${\sum }_{x,\gamma }\gamma p_{x,\gamma }={\mathrm{id}}_{T_n}$ and $p_{x,\gamma }\gamma \mathrm{\prime }$ is the appropriate Kronecker identity. Tensoring gives a direct-sum decomposition of $T_n\otimes a$ with inclusions $\gamma \otimes {\mathrm{id}}_a$ and retractions $p_{x,\gamma }\otimes {\mathrm{id}}_a$. Applying $\mathrm{Hom}(c,-)$ therefore gives the isomorphism ${⨁}_{x,\gamma }\mathrm{Hom}(c,x\otimes a)\to \mathrm{Hom}(c,T_{n+1})$ sending $(h_{x,\gamma })$ to ${\sum }_{x,\gamma }(\gamma \otimes {\mathrm{id}}_a)h_{x,\gamma }$. Its inverse applies each displayed retraction. Inserting the specified basis $B_{x,a}^c$ in each summand gives exactly the length-$(n+1)$ path composites of [F1], so they form a basis, including when the external $c$ differs from its representative.

For $n\ge 1$, step 3.1 shows that each admissible path contributes one basis vector for every tuple of vertex indices, hence contributes ${\prod }_{j=1}^n\dim W_{x_{j-1},a}^{x_j}={\prod }_{j=1}^n{N}_{x_{j-1}a}^{x_j}$ vectors. Summing over the finitely many internal simple labels gives the displayed dimension; every sum is finite because the category has finitely many simple isomorphism classes and $n$ is finite. Step 1.1 gives the separate $n=0$ count. Induction completes the basis proof and count.

For $n=0$, $T_0=b$. If $b\cong ̸c$, simplicity gives $\mathrm{Hom}(c,b)=0$ and there are no paths. If $b\cong c$, the only empty-path vector is the chosen unitary isomorphism $u_{c,b}:c\to b$, and $u_{c,b}^{†}{u}_{c,b}={\mathrm{id}}_c$ shows it has norm one. Such a choice exists: for any isomorphism $v:c\to b$, [F3] gives $\mathrm{End}(c)=\mathbb{C}{\mathrm{id}}_c$, so $v^{†}v=\lambda {\mathrm{id}}_c$. Positivity in [F2] gives $\lambda \ge 0$, and $v^{†}v=0$ would imply $v=0$ by [F2]; hence $\lambda >0$. The rescaled map $u_{c,b}={\lambda }^{-1\mathrm{/}2}v$ satisfies $u_{c,b}^{†}{u}_{c,b}={\mathrm{id}}_c$. Since it is invertible, $u_{c,b}^{†}=u_{c,b}^{-1}$, so it is unitary. Thus the length-zero path family is orthonormal. For $n=1$, the path family is the given orthonormal split vertex basis $B_{b,a}^c$ itself, so the positive-length base case holds independently of the empty-path choice.

Fix $n\ge 1$ and assume that all length-$n$ path composites, for every simple endpoint, are orthonormal in their Hom spaces.

If $f_{\mu }:x\otimes a\to y$ are the chosen orthonormal vertex maps, their adjoints $i_{\mu }=f_{\mu }^{†}:y\to x\otimes a$ satisfy $i_{\mu }^{†}{i}_{\nu }=f_{\mu }{f}_{\nu }^{†}={\delta }_{\mu \nu }{\mathrm{id}}_y$ by [F3]. Distinct simple outputs are orthogonal, and the orthogonal semisimple decomposition has projections $i_{\mu }{i}_{\mu }^{†}$ summing to the identity.

Tensoring the orthonormal length-$n$ path inclusions by ${\mathrm{id}}_a$ preserves their adjoint products because $(f\otimes {\mathrm{id}}_a{)}^{†}=f^{†}\otimes {\mathrm{id}}_a$ by [F2]. Composing with the isometric inclusions of step 1.3 therefore gives orthonormal path inclusions at length $n+1$. Induction from step 1.1 proves orthonormality for every finite $n$.

By induction, every path composite is orthonormal. The basis lemma [F4] says the same finite family spans $H_{a,n;b,c}$; if there are no admissible paths, that space is zero and the empty family is its basis. Thus the path composites are an orthonormal basis for each $n$, completing induction from step 1.1.

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.

Let $L_{\mathbf{1}}$ be the left-tree basis map obtained by applying $v^{†}$ to the first pair and then the unit unitor, and let $R_{\mathbf{1}}$ be the right-tree basis map obtained by applying $v^{†}$ to the last pair and then the other unit unitor. Both have norm one because $v^{†}v={\mathrm{id}}_{\mathbf{1}}$ and the unit constraints are unitary.

By the F-symbol definition, $L_{\mathbf{1}}\circ {\alpha }^{-1}={\sum }_f(F_a^{aaa}{)}_{f,\mathbf{1}}{R}_f$. Composing on the right with $R_{\mathbf{1}}^{†}$ kills every $f\mathrm{\ne }\mathbf{1}$ by orthogonality, so $(L_{\mathbf{1}}\circ {\alpha }^{-1})R_{\mathbf{1}}^{†}=(F_a^{aaa}{)}_{\mathbf{1},\mathbf{1}}{\mathrm{id}}_a$. In the chosen basis maps this left side is $(v^{†}\otimes {\mathrm{id}}_a){\alpha }^{-1}({\mathrm{id}}_a\otimes v)$, with unitors inserted at the two ends.

The first duality zig-zag is $(\mathrm{ev}\otimes {\mathrm{id}}_a){\alpha }^{-1}({\mathrm{id}}_a\otimes \mathrm{coev})={\mathrm{id}}_a$. Substituting [F1] gives ${\kappa }_a{d}_a(v^{†}\otimes {\mathrm{id}}_a){\alpha }^{-1}({\mathrm{id}}_a\otimes v)={\mathrm{id}}_a$. Step 2.1 therefore yields $(F_a^{aaa}{)}_{\mathbf{1},\mathbf{1}}=1\mathrm{/}({\kappa }_a{d}_a)={\kappa }_a\mathrm{/}{d}_a$, since ${\kappa }_a^2=1$.

The entry in step 3.1 is real. Unitarity [F3] gives $((F_a^{aaa}{)}^{-1}{)}_{\mathbf{1},\mathbf{1}}=\overline{(F_a^{aaa}{)}_{\mathbf{1},\mathbf{1}}}={\kappa }_a\mathrm{/}{d}_a$. Its reciprocal is ${\delta }_a={\kappa }_a{d}_a$, proving the normalization with the actual FS sign.

For a finite direct sum $A={⨁}_{s,p}s$, let $I_{s,p}:s\to A$ be its fixed copy inclusions. Set $A^{\vee }=A$ and $h_A={\sum }_{s,p}(I_{s,p}\otimes I_{s,p})h_s$, $e_A=h_A^{†}$, where $h_{\mathbf{1}}=e_{\mathbf{1}}={\mathrm{id}}_{\mathbf{1}}$. Each cross-copy term in a snake vanishes, and each diagonal term is the simple snake from [F1], so these are duality maps on every finite sum. In the evaluation mate identity, the coefficient for copies $p,q$ of simple $s$ is $\sqrt{d_s}(f_s{)}_{pq}$ on one side and $\sqrt{d_s}(f_s^{\vee }{)}_{qp}$ on the other, where $d_{\mathbf{1}}=1$ and $d_{\tau }=\varphi$ denote the cup/cap normalization constants at this stage. Thus $f_s^{\vee }=f_s^{\mathsf{T}}$, without complex conjugation, by uniqueness of mates. The double dual is therefore literally the identity on objects and morphisms. This alone does not yet establish its tensor comparison. The empty sum uses the unique zero maps and satisfies the same identities.

Define ${\theta }_A$ to be the identity on every unit copy and $x{\mathrm{id}}_{\tau }$ on every $\tau$ copy. These central scalar blocks are natural for arbitrary morphism matrices. Of the four ordered simple pairs, the three containing a unit satisfy the twist law by the strict unit gauge. On $\tau \otimes \tau =\mathbf{1}\oplus \tau$ its two channel equations are $x^2(x^{-1}{)}^2=1,x^2(y^{-1}{)}^2=x,$ the latter because $x=y^2$. Tensoring copy inclusions and then inserting the simple channel inclusions exhausts every finite direct sum, so these equations prove the twist law on all objects, including zero. Also ${\theta }_{\mathbf{1}}=1$. Duality is block transpose by step 1.1, so ${\theta }_A^{\vee }={\theta }_{A^{\vee }}$; hence this twist is ribbon.

In [F2]'s typed formula for ${\mathfrak{u}}_{\tau }$, the map ${\mathrm{id}}_{\tau }\otimes h_{\tau }$ starts in the right tree $\tau \otimes (\tau \otimes \tau )$, with its last pair in the vacuum channel. Apply ${\alpha }^{-1}$ to reach $(\tau \otimes \tau )\otimes \tau$, then $c_{\tau ,\tau }\otimes {\mathrm{id}}_{\tau }$, and finally $e_{\tau }\otimes {\mathrm{id}}_{\tau }$. Let $R_{\mathbf{1}}={\mathrm{id}}_{\tau }\otimes w$ be the right-vacuum inclusion, and let $L_{\mathbf{1}}=w\otimes {\mathrm{id}}_{\tau },L_{\tau }$ be the orthonormal left-tree inclusions. The object block is $F^{-1}=F$, so ${\alpha }^{-1}{R}_{\mathbf{1}}=u{L}_{\mathbf{1}}+v{L}_{\tau }$. The middle braiding gives $u{x}^{-1}{L}_{\mathbf{1}}+v{y}^{-1}{L}_{\tau }$. The final cap is on this same first pair: $L_{\mathbf{1}}^{†}{L}_{\mathbf{1}}=1$ and $L_{\mathbf{1}}^{†}{L}_{\tau }=0$. With $e_0=(1,0{)}^{\mathsf{T}}$, the complete two-channel compression is therefore ${\mathfrak{u}}_{\tau }=\varphi e_0^{\mathsf{T}}\mathrm{diag}(x^{-1},y^{-1})F{e}_0{\mathrm{id}}_{\tau }=\varphi (u{x}^{-1}+0\cdot v{y}^{-1}){\mathrm{id}}_{\tau }=x^{-1}{\mathrm{id}}_{\tau },{\mathfrak{u}}_{\mathbf{1}}={\mathrm{id}}_{\mathbf{1}}\mathrm{.}$ Naturality at all copy inclusions gives precisely these scalar blocks on every finite sum. No dagger conjugation of $x$ or extra F move is inserted.

By steps 2.1–2.2 and [F3], $j=\mathfrak{u}\theta$ is a monoidal natural isomorphism, and its blocks are all identity. To check the tensor comparison explicitly, the three unit-input cases in [F2]'s relation give $d_{a,b}=1$. In the two $\tau \tau$ channels that relation gives $d_{\tau ,\tau }{\mathrm{\mid }}_{\mathbf{1}}=\frac{1\cdot x^{-2}}{x^{-2}}=1,d_{\tau ,\tau }{\mathrm{\mid }}_{\tau }=\frac{x^{-1}{y}^{-2}}{x^{-2}}=x{y}^{-2}=1.$ Thus the four ordered simple pairs, including both nonunit root channels, have identity comparison. Naturality at all tensor products of copy inclusions extends this to every finite sum; the unit comparison is identity in the strict unit gauge. This proves the required pivotal monoidality rather than merely objectwise double-dual isomorphisms.

Since $A^{\vee }=A$ and $j_A=1$, the left closure is $e_A(f\otimes 1)h_A$ and the right closure is $e_A(1\otimes f)h_A$. Orthogonal copy inclusions kill all off-diagonal terms; the diagonal term for a copy of $s$ is $d_s(f_s{)}_{pp}$ on either side. Summing proves both displayed trace formulas. In particular ${\dim }_j(A)=m_1+\varphi m_{\tau }={\dim }_j(A^{\vee })$, so [F4] makes $j$ spherical with simple dimensions $1,\varphi >0$. For $f=g^{†}g$, the formula is a sum of these positive weights times squared absolute matrix entries, and vanishes only for $g=0$. The zero object has trace and dimension zero.

If $j^{\mathrm{\prime }}$ is another pivotal structure, $\gamma =j^{-1}{j}^{\mathrm{\prime }}$ is a monoidal natural automorphism of the identity. Naturality at copy inclusions makes it scalar $1$ on unit copies and $t\mathrm{\ne }0$ on $\tau$ copies. Monoidality on the unit and $\tau$ summands of $\tau \otimes \tau$ gives respectively $t^2=1$ and $t^2=t$. Hence $t=1$, and all components of $\gamma$ are identity. This proves uniqueness and the asserted canonical positive spherical structure, with every trace and ribbon convention already established.

In $K_0(\mathcal{C})$, [F1] gives $[\tau {]}^2=[\mathbf{1}]+[\tau ]$. Hence $[\tau ]([\tau ]-[\mathbf{1}])=[\mathbf{1}]$ and $([\tau ]-[\mathbf{1}])[\tau ]=[\mathbf{1}]$ by the same polynomial calculation. Thus $[\tau ]$ is a fusion-ring unit, with inverse the virtual class $[\tau ]-[\mathbf{1}]$.

If an object $Y$ were a right tensor inverse, [F3] would give $[Y]=m[\mathbf{1}]+n[\tau ]$ for some integers $m,n\ge 0$. Then $[\tau ][Y]=n[\mathbf{1}]+(m+n)[\tau ]$. Equality with $[\mathbf{1}]$ would force $n=1$ and $m+n=0$, impossible for $m,n\ge 0$. Therefore $\tau$ has no right tensor inverse and cannot be tensor-invertible.

For a canonical associator map $A:T\to T^{\mathrm{\prime }}$, [F1] defines the split Hom map $A_{\ast }:h\mapsto A\circ h$. If $B:T^{\mathrm{\prime }}\to T^{\mathrm{\prime }\mathrm{\prime }}$, then $B_{\ast }(A_{\ast }(h))=(B\circ A)\circ h$. Thus each finite F-move sequence is the matrix of postcomposition by its canonical composite, including tensor products with identities and inverse associators. This argument applies in any monoidal category with the fixed coordinate spaces and bases posited in the Given data.

The two sequences in the Statement have the same source and target parenthesized words, so their canonical morphisms are equal by [F2]. Coordinates of equal morphisms in the fixed fusion-tree bases are equal; hence the two matrix composites agree.

Conjugating the same channel projection by equal coordinate maps gives equal operators, since matrix multiplication preserves equality and the maps are invertible. This is exactly the definition in [F3].

Orthogonal simple summand inclusions satisfy $v_{\mu }^{†}{v}_{\nu }={\delta }_{\mu \nu }\mathrm{id}$ and ${\sum }_{\mu }{v}_{\mu }{v}_{\mu }^{†}=\mathrm{id}$, with zero cross-composites for distinct simple labels. Tensoring and composing these identities up a finite tree proves that its split composites form an orthonormal basis, for the arbitrary simple inputs in the Statement. Each associator and its inverse is unitary, so $A_i$ is unitary and $(A_{i}g{)}^{†}(A_{i}h)=g^{†}h$; hence $F_i^{-1}=F_i^{†}$. In the pair-first basis the coordinates are partitioned by pair output. Thus ${\mathrm{\Pi }}_s^{†}={\mathrm{\Pi }}_s$, ${\mathrm{\Pi }}_s^2={\mathrm{\Pi }}_s$, ${\mathrm{\Pi }}_s{\mathrm{\Pi }}_t=0$ for $s\mathrm{\ne }t$, and ${\sum }_s{\mathrm{\Pi }}_s=I$, including every multiplicity coordinate. The same identities hold on a zero-dimensional block with its unique empty maps.

By step 1.1, $P_{i,s}=F_i^{†}{\mathrm{\Pi }}_s{F}_i$, so $P_{i,s}^{†}=P_{i,s}$ and $P_{i,s}^2=F_i^{†}{\mathrm{\Pi }}_s^2{F}_i=P_{i,s}$. For $s\mathrm{\ne }t$, $P_{i,s}{P}_{i,t}=F_i^{†}{\mathrm{\Pi }}_s{\mathrm{\Pi }}_t{F}_i=0$, and ${\sum }_s{P}_{i,s}=F_i^{†}I{F}_i=I$. When all site charges coincide this is exactly [F1]'s family.

Restrict three adjacent $a$ sites to total charge $a$. The subspace where the first pair fuses to $\mathbf{1}$ is one-dimensional, since $N_{aa}^{\mathbf{1}}=1$ and $\mathbf{1}\otimes a\cong a$. The same holds for the subspace where the last pair fuses to $\mathbf{1}$. In the two orthonormal tree bases their unit vectors have overlap $(F_a^{aaa}{)}_{\mathbf{1},\mathbf{1}}$. The product of rank-one projections onto unit vectors $e,f$ is $P_e{P}_f{P}_e=\mathrm{\mid }⟨e,f⟩{\mathrm{\mid }}^2{P}_e$, so [F3] gives $P_i{P}_{i+1}{P}_i=d_a^{-2}{P}_i$ on this total-$a$ block. Outside total $a$, both vacuum-pair projections vanish because the remaining site is $a$; hence the identity holds on the full three-site space and, after tensoring with spectator identities, on the path space.

By [F2], $P_i^2=P_i$, so $U_i^2={\delta }_a^2{P}_i={\delta }_a{U}_i$. The adjacent relation follows from step 1.1: $U_i{U}_{i+1}{U}_i={\delta }_a^3{d}_a^{-2}{P}_i={\delta }_a{P}_i=U_i$, since ${\delta }_a^2=d_a^2$.