RT torus operators and the state-space dimension formula

Apply [F2] to the actual state space from [F3]. Its separating inverse resolves every simple summand and multiplicity, and its nonseparating inverse resolves $H$; hence it identifies the geometrically factored space with exactly the Hom space used in [F2]'s dimension calculation. The displayed formula holds for all $g,n$, including the exceptional spheres by [F2]. At $g=1,n=0$ every summand is $1$, giving $\mathrm{\mid }I\mathrm{\mid }$. The summand $\mathrm{Hom}(\mathbf{1},i\otimes i^{\ast })$ has basis $b_i$ by rigidity and simplicity, so these are a basis of the unmarked torus space.

Fix the standard oriented solid torus $D^2\times S^1$, with $m$ the positive disk-boundary and $l$ the core direction. Its boundary orientation is $(m,l)$. The reference opposite-cap reflection acts on this basis by $r=\mathrm{diag}(1,-1)$. The positive Hopf gluing exchanges the meridian of one cap with the longitude of the other; its two disk-core orientations give the exchange matrix $\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$. Removing the fixed opposite reflection gives the stated $s_{+}$, since $r{s}_{+}$ is that exchange matrix. This fixes which Hopf generator is meant and its linking orientation. In particular $m\mapsto -l$, $l\mapsto m$, rather than the inverse convention. In the capped matrix coefficient use the composition-dual test $y_j=r_j\mathrm{/}{d}_j$ because $r_j{b}_j=d_j$. The positive Hopf graph with these two coupons evaluates to $s_{ji}\mathrm{/}{d}_j$. Its closed empty-surgery normalization is $D^{-1}$ and its outgoing torus correction from [F3] is $d_j$, giving exactly $S_{ji}=s_{ji}\mathrm{/}D$. The positive core twist is the actual solid-torus map $(z,e^{i\varphi })\mapsto (e^{i\varphi }z,e^{i\varphi })$, so $m\mapsto m$, $l\mapsto l+m$ and its core normal disk makes one positive full rotation. Its ribbon operator is therefore ${\theta }_i$ on the $i$ summand with no off-diagonal terms. The cylinder Gram inverse cancels as in [F3], leaving exactly $T$. When $\lambda =\alpha$, both terms of the correction $a(R;\lambda ,\alpha )$ vanish by repeated arguments in [F4], so these really are the weight-zero matrices, not unspecified scalar multiples.

Put $A=s_{+}t=\left[\begin{array}{cc}0& 1\\ -1& -1\end{array}\right]$. Direct multiplication gives $A^3=I$ and $s_{+}^2=-I$. Composing the weight-zero $s_{+},t$ lifts introduces no weight because $t^{-1}\alpha =\alpha$, so its Maslov triple has a repeated argument. Thus $ST$ is the weight-zero lift of $A$. For its square the middle triple is $(A\alpha ,\alpha ,A^{-1}\alpha )=(\mathbb{R}l,\mathbb{R}m,\mathbb{R}(l-m))$. A sum-zero vector is $(-cl,cm,c(l-m))$, and its Walker form is $\omega (-cl,c^{\mathrm{\prime }}m)=c{c}^{\mathrm{\prime }}$, so this Maslov index is $+1$. For its third power the first and third lines coincide: $A^2\alpha =A^{-1}\alpha$. The next correction is zero. Therefore the cubed weighted lift is $(I,1)$ and [F4]'s positive weight action gives $(ST{)}^3=\kappa I$. Similarly the correction in $s_{+}{s}_{+}$ is $\mu (l,m,l)=0$, so its operator is the weight-zero $-I$ lift, equal to $S^2=C$ by [F1]. These are geometric weighted relations, not an appeal to the desired modular matrix identity.

The weight-zero inverse lift of $s_{+}$ is $s_{-}=s_{+}^{-1}$: its inverse-composition correction has repeated first and third lines, so is zero. Its operator is consequently $S^{-1}$ by functoriality. Set $B=s_{-}t=-A$. It has exactly the same action as $A$ on Lagrangian lines, hence the same accumulated cube weight $+1$, while $B^3=-I$. Its cubed operator is therefore $\kappa Z(-I,0)=\kappa C$, proving the asserted inverse-generator alternative. All matrix entries, dualities and Maslov signs use the fixed orientation from step 2.1; an orientation or inverse change cannot be hidden in a projective-scalar phrase. For example take the actual pointed data $G=\mathbb{Z}\mathrm{/}3$, $\omega =1$, $r(a,b)={\zeta }^{ab}$, $q(a)={\zeta }^{a^2}$ and pivotal identity. Bilinearity verifies both hexagons, and $q(a+b)=q(a)q(b){\zeta }^{2ab}$ verifies balancing; $q(-a)=q(a)$ and the duality snakes hold with unit evaluation/coevaluation. The induced pivotal identity gives all dimensions1, since the Drinfeld scalar ${\zeta }^{-a^2}$ times the twist is1. Its bicharacter ${\zeta }^{2ab}$ has zero radical, so [F1] makes this an actual nonselfdual modular category. It has $S_{ab}={\zeta }^{2ab}\mathrm{/}\sqrt{3}$, $T_{aa}={\zeta }^{a^2}$ and $\kappa =(1+2\zeta )\mathrm{/}\sqrt{3}$. Finite multiplication in $\mathbb{Q}[\zeta ,\sqrt{3}]$ gives $(ST{)}^3=\kappa I$ and $(S^{-1}T{)}^3=\kappa C$, where $C$ exchanges labels1 and2 and is not $I$. This corroborates the convention-sensitive general proof; the exact arithmetic diagnostic is recorded in the owner packet. The proof uses no positivity or selfduality assumption.