Unitary Fusion Spaces, F/R Symbols, and Anyon Models

Let $\mathcal{C}$ be a fusion category over $\mathbb{C}$ in the sense of Fusion and multifusion categories. A dagger structure assigns to every morphism $f:X\to Y$ a morphism $f^{†}:Y\to X$, conjugate-linearly, with $(g\circ f{)}^{†}=f^{†}\circ g^{†}$ and $(f^{†}{)}^{†}=f$. A unitary fusion category is such a fusion category equipped with this dagger so that every morphism space is a finite-dimensional normed space, composition is submultiplicative, $\mathrm{\parallel }{f}^{†}\mathrm{\parallel }=\mathrm{\parallel }f\mathrm{\parallel }$, and $\mathrm{\parallel }{f}^{†}f\mathrm{\parallel }=\mathrm{\parallel }f{\mathrm{\parallel }}^2$. Thus each ${\mathrm{End}}_{\mathcal{C}}(X)$ is a finite-dimensional $C^{\ast }$-algebra. Positive endomorphisms are the elements of its usual $C^{\ast }$-positive cone (equivalently $u^{†}u$ for an endomorphism $u$); in particular, if $c$ is simple and $\mathrm{End}(c)=\mathbb{C}{\mathrm{id}}_c$, then $\lambda {\mathrm{id}}_c$ is positive exactly when $\lambda \ge 0$. For any morphism $f:X\to Y$, $f^{†}f$ is positive in $\mathrm{End}(X)$; also require $f^{†}f=0$ to imply $f=0$. Semisimple direct-sum decompositions can be chosen orthogonally: their inclusions $i_j$ satisfy $i_j^{†}{i}_k=0$ for $j\mathrm{\ne }k$, $i_j^{†}{i}_j=\mathrm{id}$, and ${\sum }_j{i}_j{i}_j^{†}=\mathrm{id}$. Tensor product preserves daggers: $(f\otimes g{)}^{†}=f^{†}\otimes g^{†}$. Its associator and unit constraints are unitary, meaning ${\alpha }^{†}={\alpha }^{-1}$, ${\lambda }^{†}={\lambda }^{-1}$, and ${\rho }^{†}={\rho }^{-1}$.

The underlying fusion category is rigid, so its evaluation and coevaluation morphisms satisfy the duality zig-zag identities. Those morphisms are not required to be unitary isomorphisms. A braiding is additional structure; when it is specified as a unitary braiding, each component satisfies $c_{X,Y}^{†}=c_{X,Y}^{-1}$. Positivity and unitarity are extra structure, not consequences of the fusion rules or of the braid relations alone.

Let $a,b,c$ be simple objects of a unitary fusion category $\mathcal{C}$ (Unitary fusion category). The fusion multiplicity is

$N_{ab}^c:={\dim }_{\mathbb{C}}{\mathrm{Hom}}_{\mathcal{C}}(a\otimes b,c),$

and the corresponding fusion multiplicity space is

$V_{ab}^c:={\mathrm{Hom}}_{\mathcal{C}}(a\otimes b,c)\mathrm{.}$

Since $c$ is absolutely simple (Absolutely simple objects), each $f,g\in V_{ab}^c$ has a unique scalar $⟨f,g⟩$ satisfying $f\circ g^{†}=⟨f,g⟩{\mathrm{id}}_c$. This fixes the inner-product convention to be linear in the first variable. The next local theorem proves that this pairing is positive definite. No multiplicity-free assumption is made: $N_{ab}^c$ may be any nonnegative integer.

A unitary ribbon fusion category is a unitary fusion category with a unitary braiding and a ribbon structure in the sense of Twist and ribbon structure. Thus the twist $\theta$ is a natural automorphism of the identity satisfying the balancing identity and the dual-compatibility condition of that root definition, and each component is unitary: ${\theta }_X^{†}={\theta }_X^{-1}$. A ribbon structure is additional data; it is not supplied by fusion rules or by a braid-group representation alone.

The twist is unitary under these hypotheses. Let $s$ be simple, choose its left dual $s^{\vee }$, and let $\eta :\mathbf{1}\to s\otimes s^{\vee }$ be the coevaluation. A nonzero endomorphism of $s$ has zero kernel and image $s$, so it is an isomorphism and $\mathrm{End}(s)$ is a division algebra. Local finiteness makes it finite-dimensional over $\mathbb{C}$. For any $a\in \mathrm{End}(s)$, finite-dimensionality gives a nonzero polynomial $p$ with $p(a)=0$. Factor $p$ over $\mathbb{C}$; since a division algebra has no zero divisors, one linear factor must vanish, so $\mathrm{End}(s)=\mathbb{C}{\mathrm{id}}_s$. Write ${\theta }_s=\lambda {\mathrm{id}}_s$. The ribbon duality condition and linearity of dualization give ${\theta }_{s^{\vee }}=\lambda {\mathrm{id}}_{s^{\vee }}$. Naturality, ${\theta }_{\mathbf{1}}={\mathrm{id}}_{\mathbf{1}}$, and balancing give $\eta ={\theta }_{s\otimes s^{\vee }}\circ \eta ={\lambda }^2{c}_{s^{\vee },s}\circ c_{s,s^{\vee }}\circ \eta \mathrm{.}$ The double braiding is unitary, so it preserves the norm of $\eta$. The zig-zag identity gives $\eta \mathrm{\ne }0$; hence the equality implies $\mathrm{\parallel }\eta \mathrm{\parallel }=\mathrm{\mid }\lambda {\mathrm{\mid }}^2\mathrm{\parallel }\eta \mathrm{\parallel }$ and $\mathrm{\mid }\lambda \mathrm{\mid }=1$. Thus ${\theta }_s$ is unitary for every simple $s$. For any object, decompose it as a finite orthogonal sum of simples; naturality makes its twist block diagonal with these unitary components, so ${\theta }_X^{†}{\theta }_X={\mathrm{id}}_X$.

Let $\mathcal{C}$ be a unitary ribbon fusion category. An object $x$ is transparent if for every simple object $y$ the double braiding

$c_{y,x}\circ c_{x,y}:x\otimes y⟶x\otimes y$

is the identity. By naturality and additivity, the double braiding on a finite direct sum is block diagonal with the double braidings on its simple summands; therefore an object is transparent exactly when each simple summand is transparent. Finite direct sums include the empty sum, so the zero object is transparent. The category is modular, or nondegenerate, when its transparent simple objects consist only of the tensor unit, equivalently when its transparent objects are exactly the finite direct sums of the tensor unit. This condition is additional to the existence of fusion spaces and a unitary braid action.

For a simple charge $a$ in a unitary ribbon fusion category, the twist component is an endomorphism ${\theta }_a\in \mathrm{End}(a)$. Since $\mathrm{End}(a)=\mathbb{C}{\mathrm{id}}_a$, there is a unique scalar ${\vartheta }_a\in \mathbb{C}$ such that

${\theta }_a={\vartheta }_a{\mathrm{id}}_a\mathrm{.}$

The scalar ${\vartheta }_a$ is the topological spin of $a$. Unitarity of the twist gives $\mathrm{\mid }{\vartheta }_a{\mathrm{\mid }}^2{\mathrm{id}}_a={\theta }_a^{†}{\theta }_a={\mathrm{id}}_a$; since ${\mathrm{id}}_a\mathrm{\ne }0$, $\mathrm{\mid }{\vartheta }_a\mathrm{\mid }=1$.

Let $a,b,c$ be simple objects in a unitary fusion category. The pairing on $V_{ab}^c=\mathrm{Hom}(a\otimes b,c)$ defined by $f\circ g^{†}=⟨f,g⟩{\mathrm{id}}_c$ is a finite-dimensional positive-definite Hermitian inner product, linear in the first variable and conjugate-linear in the second.

Assume the Axiom of Countable Choice ${\mathrm{A}\mathrm{C}}_{\omega }$ (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)). Let $\mathcal{C}$ be a unitary ribbon fusion category, let $X$ be a simple color, and let $\beta \in B_n$ and ${\beta }^{\mathrm{\prime }}\in B_m$ with $n,m\ge 1$. If their blackboard-framed, $X$-colored closures are isotopic as framed oriented links, then their root ribbon evaluations satisfy $t_n(\beta )=t_m({\beta }^{\mathrm{\prime }})$. No modularity assumption is needed. Ordinary Markov stabilization is not a framed isotopy because it changes the framing.

Let $\mathcal{C}$ be a unitary fusion category and fix a set $\mathrm{Irr}(\mathcal{C})$ of representatives for the simple isomorphism classes, with the tensor unit $\mathbf{1}$ itself representing its class. Throughout this definition, the ordered input charges $a_1,\dots ,a_n$, total charge $d$, and all internal edge charges are labels in $\mathrm{Irr}(\mathcal{C})$. Also fix a parenthesized binary fusion tree. At each internal vertex with incoming charges $x,y$ and outgoing charge $z$, choose one vector from an orthonormal basis of $V_{xy}^z$. An admissible labeling assigns a representative charge to each internal edge so that each chosen vertex basis is nonempty; a separate multiplicity-basis index is retained at every vertex. Compose the vertex morphisms, inserting the associators required by the chosen tree, to obtain a morphism from the parenthesized input tensor to $d$. These composite morphisms are the fusion-tree vectors. The family is called a fusion-tree basis when it is an orthonormal basis of the fixed-input, fixed-total fusion space. For the empty input list the tensor is $\mathbf{1}$ and its empty tree vector is ${\mathrm{id}}_{\mathbf{1}}$ when $d=\mathbf{1}$; for one input $a_1$, the only tree vector is ${\mathrm{id}}_{a_1}$ when $d=a_1$. The next local lemma proves the basis property for every finite input list and includes the dimension recursion (Fusion-tree basis and dimension recursion ↗).

The left-associated three-input tree has leaves $a,b,c$ in that order, internal charge $e$ after fusing $a,b$, and output $d$ after fusing $e,c$:

The drawing records only the parenthesization and edge labels. The basis vector is the composite morphism specified above; no proof depends on the diagram.

A fusion-basis gauge change chooses, for every triple of simple charges $a,b,c$, a unitary matrix $G_{ab}^c$ changing the orthonormal basis of $V_{ab}^c$. The convention is that the new basis columns are the old basis columns multiplied by $G_{ab}^c$. On a recoupling block this gives

$F^{\mathrm{\prime }}=G_R^{-1}F{G}_L,$

where $G_L$ and $G_R$ are the direct sums of tensor products of the vertex changes on the source and target trees. On an exchange block it gives

$R^{\mathrm{\prime }}=(G_{ba}^c{)}^{-1}R{G}_{ab}^c\mathrm{.}$

The matrices change with the bases; the underlying categorical maps do not.

Let $a,b,c$ be simple objects in a unitary braided fusion category, with braiding $c_{a,b}:a\otimes b\to b\otimes a$ as in Braiding. The typed channel-exchange map on $V_{ab}^c=\mathrm{Hom}(a\otimes b,c)$ is the pullback

${\mathcal{R}}_{ab}^c(f):=f\circ c_{a,b}^{-1}\in V_{ba}^c\mathrm{.}$

Choose orthonormal bases in the two multiplicity spaces. The matrix of ${\mathcal{R}}_{ab}^c$ in these bases is the R-symbol $R_{ab}^c$. This pullback convention fixes the orientation of the coordinate exchange; reversing the braiding orientation replaces it by the inverse convention. If the spaces are zero-dimensional, the braiding isomorphism identifies both with $0$ and the R-symbol is the unique empty $0\times 0$ matrix.

Let $\mathcal{C}$ be a unitary fusion category, let $\mathrm{Irr}(\mathcal{C})$ be the representative set fixed in Multiplicity-aware fusion-tree basis, and take $a_1,\dots ,a_n,d\in \mathrm{Irr}(\mathcal{C})$. Fix any parenthesized binary tree $T$ on the ordered input list. The fusion-tree vectors of Multiplicity-aware fusion-tree basis form an orthonormal basis of $\mathrm{Hom}(T,d)$. For $n\ge 2$, their number is

$m_T(d)={\sum }_{\begin{array}{c}\mathrm{\ell }\text{ admissible, with every internal label in }\mathrm{Irr}(\mathcal{C})\end{array}}{\prod }_{v\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{n}\mathrm{a}\mathrm{l}}{N}_{\mathrm{\ell }(v_L)\mathrm{\ell }(v_R)}^{\mathrm{\ell }(v)},$

where the sum runs over charge labelings only (the product counts the vertex multiplicity indices), the leaves carry the inputs, the root carries $d$, and each internal edge is labeled by a representative simple charge. For the empty tree, $m_0(d)=1$ if $d=\mathbf{1}$ and $0$ otherwise; for one input, $m_1(d)=1$ if $a_1=d$ and $0$ otherwise. The left-associated tree on $n\ge 2$ inputs satisfies

$m_n(d)={\sum }_e{m}_{n-1}(e)N_{e{a}_n}^d(n\ge 2)\mathrm{.}$

In a unitary braided fusion category, every R-symbol matrix in orthonormal multiplicity bases is unitary.

Let $a,b,c,d$ be simple charges in a unitary fusion category, and put $P_L=((a\otimes b)\otimes c)$ and $P_R=a\otimes (b\otimes c)$. The local fusion-tree basis result Fusion-tree basis and dimension recursion identifies

${\mathcal{H}}_L:={⨁}_e{V}_{ab}^e\otimes V_{ec}^d\text{with}\mathrm{Hom}(P_L,d),$

and

${\mathcal{H}}_R:={⨁}_f{V}_{bc}^f\otimes V_{af}^d\text{with}\mathrm{Hom}(P_R,d),$

by the corresponding tree compositions. These direct sums carry the orthonormal left- and right-tree bases. If $\mu :a\otimes b\to e$ and $\nu :e\otimes c\to d$ are vertex basis maps, write $L_{e,\mu ,\nu }:=\nu \circ (\mu \otimes {\mathrm{id}}_c):P_L\to d$. If $\rho :b\otimes c\to f$ and $\sigma :a\otimes f\to d$ are vertex basis maps, write $R_{f,\rho ,\sigma }:=\sigma \circ ({\mathrm{id}}_a\otimes \rho ):P_R\to d$. Let ${\alpha }_{a,b,c}:P_L\to P_R$ be the associator. Define the F-symbol matrix $F_d^{abc}$ by

$L_{e,\mu ,\nu }\circ {\alpha }_{a,b,c}^{-1}={\sum }_{f,\rho ,\sigma }{F}_d^{abc}[(f,\rho ,\sigma ),(e,\mu ,\nu )]R_{f,\rho ,\sigma }\mathrm{.}$

Rows index target-tree labels and multiplicity indices; columns index source-tree labels and multiplicity indices. The sums retain every fusion channel and every multiplicity coordinate. If the fixed-total fusion space is zero-dimensional, the F-symbol is the unique empty $0\times 0$ matrix.

The object-level associator is drawn below. The F matrix records its induced change of coordinates on the Hom spaces by the inverse-precomposition rule in the Definition; the diagram is only a convention aid.

Every F-symbol matrix in orthonormal fusion-tree bases of a unitary fusion category is unitary.

Let $\mathcal{C}$ be a unitary braided fusion category with associator $\alpha$ and braiding $c$. For all objects $X,Y,Z$, its two hexagons are

${\alpha }_{Y,Z,X}\circ c_{X,Y\otimes Z}\circ {\alpha }_{X,Y,Z}=({\mathrm{id}}_Y\otimes c_{X,Z})\circ {\alpha }_{Y,X,Z}\circ (c_{X,Y}\otimes {\mathrm{id}}_Z),$

${\alpha }_{Z,X,Y}^{-1}\circ c_{X\otimes Y,Z}\circ {\alpha }_{X,Y,Z}^{-1}=(c_{X,Z}\otimes {\mathrm{id}}_Y)\circ {\alpha }_{X,Z,Y}^{-1}\circ ({\mathrm{id}}_X\otimes c_{Y,Z})\mathrm{.}$

For simple input charges and a fixed simple total charge, the two route matrices in each equation are equal in fusion-tree coordinates. Each entry is the sum over all intermediate simple charges and multiplicity indices of the products of the corresponding F- and R-block entries.

The first diagram shows the two routes from $(X\otimes Y)\otimes Z$ to $Y\otimes (Z\otimes X)$; the second shows the routes from $X\otimes (Y\otimes Z)$ to $(Z\otimes X)\otimes Y$. The arrow labels are exactly the maps in the two displayed equations. They illustrate the equations; the proof above uses the typed composites and their matrix coefficients.

diagram failed to render:
((X\otimes Y)\otimes Z)
  \arrow[r,"\alpha_{X,Y,Z}"]
  \arrow[d,"c_{X,Y}\otimes\operatorname{id}_Z"']
& X\otimes(Y\otimes Z)
  \arrow[r,"c_{X,Y\otimes Z}"]
& (Y\otimes Z)\otimes X
  \arrow[d,"\alpha_{Y,Z,X}"] \\
(Y\otimes X)\otimes Z
  \arrow[r,"\alpha_{Y,X,Z}"']
& Y\otimes(X\otimes Z)
  \arrow[r,"\operatorname{id}_Y\otimes c_{X,Z}"']
& Y\otimes(Z\otimes X)

diagram failed to render:
X\otimes(Y\otimes Z)
  \arrow[r,"\operatorname{id}_X\otimes c_{Y,Z}"]
  \arrow[d,"\alpha_{X,Y,Z}^{-1}"']
& X\otimes(Z\otimes Y)
  \arrow[r,"\alpha_{X,Z,Y}^{-1}"]
& (X\otimes Z)\otimes Y
  \arrow[d,"c_{X,Z}\otimes\operatorname{id}_Y"] \\
(X\otimes Y)\otimes Z
  \arrow[r,"c_{X\otimes Y,Z}"']
& Z\otimes(X\otimes Y)
  \arrow[r,"\alpha_{Z,X,Y}^{-1}"']
& (Z\otimes X)\otimes Y

Let $a,b,c,d$ be four simple input charges and $t$ a simple total charge. The root associator identity

${\alpha }_{a,b,c\otimes d}\circ {\alpha }_{a\otimes b,c,d}=({\mathrm{id}}_a\otimes {\alpha }_{b,c,d})\circ {\alpha }_{a,b\otimes c,d}\circ ({\alpha }_{a,b,c}\otimes {\mathrm{id}}_d)$

has equal matrices along its two routes in the fusion-tree bases. If $A_1,A_2,A_3$ are the three successive matrices on the right and $A_4,A_5$ the two successive matrices on the left, then

$A_3{A}_2{A}_1=A_5{A}_4\mathrm{.}$

Each matrix coefficient sums over every intermediate simple charge and every multiplicity-basis index.

The vertices are the five parenthesizations of $a\otimes b\otimes c\otimes d$; each arrow is the indicated associator. The diagram encodes exactly the two composites stated and proved above; the proof uses the equations and coordinate expansion, not the drawing.

diagram failed to render:
((a\otimes b)\otimes c)\otimes d
  \arrow[r,"\alpha_{a\otimes b,c,d}"]
  \arrow[d,"\alpha_{a,b,c}\otimes\operatorname{id}_d"']
& (a\otimes b)\otimes(c\otimes d)
  \arrow[r,"\alpha_{a,b,c\otimes d}"]
& a\otimes(b\otimes(c\otimes d)) \\
(a\otimes(b\otimes c))\otimes d
  \arrow[r,"\alpha_{a,b\otimes c,d}"']
& a\otimes((b\otimes c)\otimes d)
  \arrow[ru,"\operatorname{id}_a\otimes\alpha_{b,c,d}"']
&

Let $\mathcal{C}$ be a unitary braided fusion category. A finite list of simple charges and total charge determines a fusion space in each parenthesization. For a crossing word with underlying object map $b_{\beta }:P_L\to P_{L^{\mathrm{\prime }}}$, define its action on the corresponding Hom spaces by $f\mapsto f\circ b_{\beta }^{-1}$, from $\mathrm{Hom}(P_L,d)$ to $\mathrm{Hom}(P_{L^{\mathrm{\prime }}},d)$. Let $U_{\beta }$ be its matrix in chosen fusion-tree bases. If each multiplicity basis is changed by the unitary gauge matrices of Fusion-basis gauge change, and $G_L,G_{L^{\mathrm{\prime }}}$ are the induced source and target tree-basis unitaries, then

$U_{\beta }^{\mathrm{\prime }}=G_{L^{\mathrm{\prime }}}^{-1}{U}_{\beta }{G}_L\mathrm{.}$

In particular, when $n\ge 2$ and all $n$ strand charges are the same simple object $a$, the adjacent exchanges act on the same fixed-total fusion space $V=\mathrm{Hom}(a^{\otimes n},d)$, satisfy the Artin relations, and define a unitary representation $\rho :B_n\to U(V)$, where $U(V)$ is the group of invertible linear maps preserving the inner product on $V$. If $V=0$, this is the unique action on the zero-dimensional space. In that case ${\rho }^{\mathrm{\prime }}(\beta )=G_L^{-1}\rho (\beta )G_L$ for every braid $\beta$. Thus basis-dependent matrix entries change, while the braid action up to unitary equivalence and consistently transformed expectation values do not.

Both arrows are the explicit composites $L$ and $R$ from step 1.1, with each crossing using the braiding $c$ and each reassociation using the named associator. The diagram records their common source and target; equality was proved from naturality and the hexagon identities above.

The claim refuted is that an algebraic Yang–Baxter or Hecke relation by itself guarantees that the resulting braid-group representation is unitary for some positive-definite inner product.

This example uses the repository's $v$-normalization $(T_i-v)(T_i+1)=0$. It does not rule out unitary representations at other parameters or after a positive dagger structure has been separately specified. The compact-group averaging lemma Averaging a Hermitian form unitarizes a finite-dimensional compact-group representation does not apply to the discrete infinite group $B_3$; the exponent-sum map Exponent sum and writhe of a braid maps it onto $\mathbb{Z}$.

Take the skeletal complex-linear category whose simple labels are $\mathbf{1},\tau$ and whose objects are finite direct sums of them, including the empty direct sum $0$. For objects $A={⨁}_s{s}^{\oplus m_s}$ and $B={⨁}_s{s}^{\oplus n_s}$, set $\mathrm{Hom}(A,B)={⨁}_s{\mathrm{Mat}}_{n_s\times m_s}(\mathbb{C})$; composition is block-matrix multiplication, dagger is conjugate transpose, and the norm is the maximum of the operator norms of its blocks. Thus direct sums are orthogonal. Define tensor product on simple labels by the fusion rules $\mathbf{1}\otimes a=a\otimes \mathbf{1}=a$ and $\tau \otimes \tau =\mathbf{1}\oplus \tau$, and extend it distributively to direct sums. For each allowed simple fusion channel $c\subset a\otimes b$, let $i_{ab}^c:c\to a\otimes b$ be the standard unit-vector isometric inclusion. Since the multiplicity space is $V_{ab}^c=\mathrm{Hom}(a\otimes b,c)$, its orthonormal basis vector is the adjoint projection $(i_{ab}^c{)}^{†}:a\otimes b\to c$; retain $i_{ab}^c$ itself for the coevaluation construction below. On simple-label blocks, tensor a pair of morphism matrices by their Kronecker product on multiplicity coordinates and by the identity on each simple fusion channel; extend over direct sums. This defines a bifunctor compatible with dagger. Use the strict unit gauge: every F or R coordinate involving $\mathbf{1}$ is the identity. Put