Balancing and the Hopf-link character identity

By balancing [F2], $c_{j,i}{c}_{i,j}=({\theta }_i^{-1}\otimes {\theta }_j^{-1}){\theta }_{i\otimes j}$. Taking the pivotal trace and using additivity [F4], decompose $i\otimes j\cong {⨁}_k{k}^{\oplus N_{ij}^k}$. Schur's lemma makes ${\theta }_i,{\theta }_j,{\theta }_k$ scalars on simples, so the trace is $({\theta }_i{\theta }_j{)}^{-1}{\sum }_k{N}_{ij}^k{\theta }_k{d}_k$, proving the balancing formula.

Define the partial trace of $f:X\otimes Z\to X\otimes Z$ over $Z$ by $P_Z(f)=(1_X\otimes {\mathrm{ev}}_{Z^{\vee }})(1_X\otimes j_Z\otimes 1_{Z^{\vee }})(f\otimes 1_{Z^{\vee }})(1_X\otimes {\mathrm{coev}}_Z)$. Substitution of the tensor coevaluation and evaluation into the trace formula shows ${\mathrm{tr}}_j(P_Z(f))={\mathrm{tr}}_j(f)$. For $f=c_{Z,X}{c}_{X,Z}$ and simple $X$, Schur's lemma gives $P_Z(f)=\alpha 1_X$; hence $\alpha d_X=s_{XZ}$. Since $d_X\mathrm{\ne }0$ by [F5], $P_Z(f)=(s_{XZ}\mathrm{/}{d}_X)1_X$.

Apply the hexagon expansion [F3] to $c_{Y\otimes Z,X}{c}_{X,Y\otimes Z}$ and take the partial trace over the final $Z$ factor. Step 1.2 replaces the middle double braiding by $(s_{XZ}\mathrm{/}{d}_X)1_X$; the remaining composite is that scalar times $c_{Y,X}{c}_{X,Y}$. Taking the trace over the remaining factors therefore gives $d_X^{-1}{s}_{XZ}{s}_{XY}$. On the other hand, decompose $Y\otimes Z$ into simples before taking its trace; additivity gives ${\sum }_k{N}_{YZ}^k{s}_{Xk}$. Equating these two evaluations and multiplying by $d_X$ yields $s_{XY}{s}_{XZ}=d_X{\sum }_k{N}_{YZ}^k{s}_{Xk}$.

Divide step 2.1 by $d_X^2$, which is nonzero by [F5]. The result is $h_X(Y)h_X(Z)={\sum }_k{N}_{YZ}^k{h}_X(k)$, and $h_X(0)=s_{X0}\mathrm{/}{d}_X=1$ because the double braiding with the unit is the identity. Thus $h_X$ is a unital fusion-ring character. Relabeling $Y=i,Z=j$ gives the stated identity.

The double braiding of a label with the unit is the identity, so $s_{i0}=d_i$. The endomorphisms $c_{j,i}{c}_{i,j}$ and $c_{i,j}{c}_{j,i}$ are conjugate via the isomorphism $c_{i,j}$; cyclicity in [F4] therefore gives $s_{ij}=s_{ji}$. To compare dual labels, apply step 1.1 to $(i^{\ast },j^{\ast })$ and reindex the summation by $k={\mathrm{\ell }}^{\ast }$. By [F7], $N_{i^{\ast }{j}^{\ast }}^{{\mathrm{\ell }}^{\ast }}=N_{ij}^{\mathrm{\ell }}$ and ${\theta }_{{\mathrm{\ell }}^{\ast }}={\theta }_{\mathrm{\ell }}$; by sphericality in [F1], $d_{{\mathrm{\ell }}^{\ast }}=d_{\mathrm{\ell }}$. The balancing sum and its scalar denominator are consequently unchanged, proving $s_{i^{\ast }{j}^{\ast }}=s_{ij}$ and $d_{i^{\ast }}=d_i$.