Nonzero global dimension and the canonical Frobenius scalar

Left and right duals of a simple $i$ are isomorphic. Indeed, rigidity identifies the unique simple $j$ with $\mathrm{Hom}(\mathbf{1},i\otimes j)\mathrm{\ne }0$ as one dual, and the unique simple $j$ with $\mathrm{Hom}(i\otimes j,\mathbf{1})\mathrm{\ne }0$ as the other; semisimplicity makes both dimensions the multiplicity of the unit in $i\otimes j$, so these simple labels coincide. Thus $i\cong i^{\ast \ast }$. Each of the two loops defining $d_i,e_i$ is nonzero: it is a nonzero inclusion of the multiplicity-one unit summand in a product of dual simples followed by a nonzero projection to that summand. Multiplicity one follows from $\mathrm{End}(i)=k$ and rigidity. Any two $a_i$ differ by a scalar $t\mathrm{\ne }0$; this multiplies $d_i$ by $t$ and $e_i$ by $t^{-1}$. Hence $\mathrm{\mid }i{\mathrm{\mid }}^2$ is well-defined and nonzero.

Put $\mathcal{D}=\mathcal{C}⊠{\mathcal{C}}^{\mathrm{r}\mathrm{e}\mathrm{v}}$. The object $\mathbf{1}\in \mathcal{C}$ is a generator for its regular left $\mathcal{D}$-module action, since $(V⊠\mathbf{1})\cdot \mathbf{1}=V$. By F2, $H(V)=\underset{}{\mathrm{Hom}}(\mathbf{1},V)$ is an equivalence from $\mathcal{C}$ to right $K$-modules. Rigidity and representability identify both $L_V=(V⊠\mathbf{1})K$ and $R_V=(\mathbf{1}⊠V)K$ with $H(V)$: applying ${\mathrm{Hom}}_{\mathcal{D}}(X⊠Y,-)$ in each case gives ${\mathrm{Hom}}_{\mathcal{C}}(X\otimes Y,V)$, by moving $V$ across the Hom using the appropriate dual. These identifications preserve the right action because both actions are composition of the representing evaluations. Thus $J:V\mapsto L_V$ is fully faithful, and the identity on the represented Hom functor defines a right $K$-linear natural isomorphism ${\rho }_V:L_V\to R_V$.

We prove the algebraic mate calculation needed to dualize sliding. In any rigid category, let $A$ be a Frobenius algebra with $b=ϵm$, and let $\mathrm{\ell }:A\to A^{\vee }$ be its perfect right-curried pairing, so $b={\mathrm{ev}}_A(\mathrm{\ell }\otimes \mathrm{i}\mathrm{d})$. Put $F=(-{)}^{\vee \vee }$, $q=({\mathrm{\ell }}^{-1}{)}^{\vee }\mathrm{\ell }:A\to F(A)$ and $b_R={\mathrm{ev}}_{A^{\vee }}({\mathrm{i}\mathrm{d}}_{F(A)}\otimes \mathrm{\ell })$. Then ${\mathrm{\ell }}^{\vee }q=\mathrm{\ell }$, hence $b_R(q\otimes \mathrm{i}\mathrm{d})=b$. For $t:U\otimes V\to \mathbf{1}$ define its cyclic transpose, of source $F(V)\otimes U$, by $R_{U,V}(t)={\mathrm{ev}}_{V^{\vee }}({\mathrm{i}\mathrm{d}}_{F(V)}\otimes [(t\otimes {\mathrm{i}\mathrm{d}}_{V^{\vee }})({\mathrm{i}\mathrm{d}}_U\otimes {\mathrm{coev}}_V)])\mathrm{.}$ The two snake identities make this operation invertible. Direct substitution proves naturality in $U$ and the rule $R_{U,W}(t({\mathrm{i}\mathrm{d}}_U\otimes h))=R_{U,V}(t)(F(h)\otimes {\mathrm{i}\mathrm{d}}_U)$ for $h:W\to V$. It also proves that two successive last-factor rotations equal rotation of their last-two-factor block: substitute the nested coevaluation formula for $V\otimes W$ and the reverse-order evaluation formula, obtaining the same nested composite.

The sliding equation for $\rho$ says that on $(V⊠\mathbf{1})(W⊠\mathbf{1})K$, sliding $W$ and then $V$, commuting the distinct exterior factors between these operations, equals sliding $V\otimes W$. To verify it without a diagram, apply ${\mathrm{Hom}}_{\mathcal{D}}(X⊠Y,-)$ and the identifications of step 1.2. In either order the two adjunctions represent the same map $X\otimes Y\to V\otimes W$. If intermediate simple decompositions are inserted, their inclusion/projection sum is $\sum q_a{p}_a=\mathrm{i}\mathrm{d}$, so erasing these sums gives the original map in both orders. The reverse-order tensor-dual evaluation is exactly the composite of the two evaluations, and the module associator equation identifies their parentheses. The unit case is evaluation against $\mathbf{1}$. Yoneda proves the sliding and unit equations.

Define ${\tau }_n(t)=R_{A^{\otimes (n-1)},A}(t)(q\otimes \mathrm{i}\mathrm{d})$ on $\mathrm{Hom}(A^{\otimes n},\mathbf{1})$, and set $t_3=b(m\otimes \mathrm{i}\mathrm{d})=b(\mathrm{i}\mathrm{d}\otimes m)$. Using the first expression and naturality in the first argument gives ${\tau }_3(t_3)=b_R(q\otimes m)=b(\mathrm{i}\mathrm{d}\otimes m)=t_3$. Thus ${\tau }_3^2(t_3)=t_3$. On the other hand, the two-factor rotation rule gives ${\tau }_3^2(t_3)=R_{A,A\otimes A}(t_3)(q\otimes q\otimes \mathrm{i}\mathrm{d})$. Using $t_3=b(\mathrm{i}\mathrm{d}\otimes m)$ and naturality in the last argument makes this $b_R(F(m)(q\otimes q)\otimes \mathrm{i}\mathrm{d})$. The nondegenerate pairing makes $h\mapsto b_R(h\otimes \mathrm{i}\mathrm{d})$ injective on $\mathrm{Hom}(A\otimes A,F(A))$, and $t_3=b_R(qm\otimes \mathrm{i}\mathrm{d})$. Therefore $qm=F(m)(q\otimes q)$. Similarly rotating $b(\mathrm{i}\mathrm{d}\otimes \eta )=ϵ=b(\eta \otimes \mathrm{i}\mathrm{d})$ gives $b_R(F(\eta )\otimes \mathrm{i}\mathrm{d})=b_R(q\eta \otimes \mathrm{i}\mathrm{d})$, so $q\eta =F(\eta )$. Thus $q$ is an algebra isomorphism. Every expression above is typed: both inputs to $F(m)$ lie in $F(A)$, and no product of an $A$ input and an $F(A)$ input is used.

We compute the canonical Frobenius scalar directly, rather than using a trace characterization of $\delta$. Fix a simple $i$ and write $j=i^{\ast }$, transporting across a chosen representative isomorphism if necessary. With $d=d_i,e=e_i,a=a_i$, choose the following projections and sections of the unit summands: $p_{ij}={\mathrm{ev}}_i,q_{ij}=(a^{-1}\otimes {\mathrm{i}\mathrm{d}}_{i^{\ast }}){\mathrm{coev}}_{i^{\ast }}\mathrm{/}e,p_{ji}={\mathrm{ev}}_{i^{\ast }}({\mathrm{i}\mathrm{d}}_{i^{\ast }}\otimes a),q_{ji}={\mathrm{coev}}_i\mathrm{/}d\mathrm{.}$ Then $p_{ij}{q}_{ij}=p_{ji}{q}_{ji}=1$. The explicit multiplication in F1 gives ${\beta }_{ij}=p_{ij}⊠q_{ij}^{\ast }$. Write the corresponding inverse-pairing component as $c_{ij}=A_i{q}_{ij}⊠p_{ij}^{\ast }$; this is its general form because the relevant two Hom spaces from the unit are one-dimensional. On $K_i=i⊠i^{\ast }$, the first exterior factor in $(\mathrm{i}\mathrm{d}\otimes \beta )(c\otimes \mathrm{i}\mathrm{d})$ is $A_i({\mathrm{i}\mathrm{d}}_i\otimes p_{ji})(q_{ij}\otimes {\mathrm{i}\mathrm{d}}_i)$; the second is $(q_{ji}^{\ast }\otimes {\mathrm{i}\mathrm{d}}_{i^{\ast }})({\mathrm{i}\mathrm{d}}_{i^{\ast }}\otimes p_{ij}^{\ast })$, which is the left dual of $(p_{ij}\otimes {\mathrm{i}\mathrm{d}}_i)({\mathrm{i}\mathrm{d}}_i\otimes q_{ji})$. No other summand contributes.

Apply this construction to $K\in \mathcal{D}$ and its Frobenius pairing. The twice-right-dual of ${\rho }_V$ is a map $(F(V)⊠\mathbf{1})K^{\vee \vee }\to (\mathbf{1}⊠F^{-1}(V))K^{\vee \vee }$, because right duality in the reversed second factor is left duality in $\mathcal{C}$. It is right $K^{\vee \vee }$-linear by twice dualizing the action square. Transport through the algebra isomorphism $q:K\to K^{\vee \vee }$ gives a right $K$-linear map ${\tilde{\rho }}_V:(F(V)⊠\mathbf{1})K\to (\mathbf{1}⊠F^{-1}(V))K$. Compose with ${\rho }_{F^{-1}(V)}^{-1}$. Full faithfulness of $J$ gives a unique isomorphism ${\delta }_V:F(V)\to F^{-1}(V)$ whose image is this composite. The twice-dual sliding equation, transported through $q$, is the sliding equation for $\tilde{\rho }$: internal $q^{-1},q$ pairs cancel, and the multiplication and unit vertices agree by step 2.2. Combining it with the inverse sliding equation for $\rho$ proves the tensor equation for $\delta$; full faithfulness reflects it in $\mathcal{C}$. Thus $g_V={\delta }_{F(V)}^{-1}:V\to F^2(V)$ is a monoidal natural isomorphism.

Monoidal natural isomorphisms of rigid tensor functors preserve duality. Here this follows directly by applying naturality and the monoidal equation for $g$ to evaluation and coevaluation: the resulting two equations say that $g_{X^{\vee }}$ is the inverse dual of $g_X$, under the canonical dual constraints. Uniqueness of a dual mate follows from the snake identities, so $g_{X^{\vee }}=(g_X^{\vee }{)}^{-1}$. Applying this twice gives $g_{F(X)}=F(g_X)$. Define $\tilde{\mathcal{C}}$ to have objects $(V,f)$ with $f:V\to F(V)$ invertible and $F(f)f=g_V$, and arrows $h$ with $F(h)f=f^{\mathrm{\prime }}h$. Tensor using $f\otimes f^{\mathrm{\prime }}$ and the tensor constraint of $F$; monoidality of $g$ proves closure. The inverse-dual identity just proved gives dual objects and their evaluation and coevaluation. The tautological map $f:(V,f)\to (F(V),F(f))$ is an arrow, with its target an object by $F(g_V)=g_{F(V)}$, and it is natural and monoidal. Hence it is a pivotal structure.

This cover is finite semisimple. Every simple $i$ is $F$-stable by step 1.1. Choose $b_i:i\to F(i)$; there is $c_i\mathrm{\ne }0$ with $F(b_i)b_i=c_i{g}_i$. Choose $t_i$ with $t_i^2{c}_i=1$. There are exactly two simple lifts $(i,\pm t_i{b}_i)$. On any multiplicity space of $i$, an arbitrary lift has the form $t_i{b}_i\otimes T$, and its equation is $T^2=\mathrm{i}\mathrm{d}$. Characteristic zero gives the two projections $(\mathrm{i}\mathrm{d}\pm T)\mathrm{/}2$, so every lift is a direct sum of these two simple lifts. Their endomorphisms are $k$, and Hom between opposite lifts vanishes. There are exactly twice as many simple labels, the tensor unit has endomorphism ring $k$, and rigidity was proved in step 4.1. Thus $\tilde{\mathcal{C}}$ is a pivotal fusion category, and its forgetful tensor functor to $\mathcal{C}$ preserves every evaluation/coevaluation composite, hence squared norms.

Over $\mathbb{C}$, squared norms in any pivotal fusion category are strictly positive. Indeed let $D_i$ be the pivotal trace of a simple $i$; substituting tensor-product evaluations and coevaluations and the monoidal pivotal equation proves $D_i{D}_j={\sum }_r{N}_{ij}^r{D}_r$, and duality gives $\mathrm{\mid }i{\mathrm{\mid }}^2=D_i{D}_{i^{\ast }}$. Let $(M_i{)}_{j,r}=N_{ij}^r$ and $D=(D_j{)}_j\mathrm{\ne }0$, since $D_{\mathbf{1}}=1$. Then $M_{i}D=D_{i}D$. Rigidity gives $N_{ij}^r=N_{i^{\ast }r}^j$, so $M_{i^{\ast }}=M_i^t$ and $M_i{M}_i^{t}D=\mathrm{\mid }i{\mathrm{\mid }}^{2}D$. Since $M_i$ is real, $\mathrm{\mid }i{\mathrm{\mid }}^2=\frac{D^{\ast }{M}_i{M}_i^{t}D}{D^{\ast }D}=\frac{\mathrm{\parallel }{M}_i^{t}D{\mathrm{\parallel }}^2}{\mathrm{\parallel }D{\mathrm{\parallel }}^2}\ge 0.$ Step 1.1 excludes zero. A pivotal structure for right double duality also supplies one for left double duality by inversion and reindexing, so this calculation applies to the cover in step 5.1 with the statement's trace convention. Each simple of $\mathcal{C}$ lifts to a simple of the cover, and forgetting preserves its two trace loops. Thus $\mathrm{\mid }i{\mathrm{\mid }}^2>0$ in $\mathcal{C}$ and ${\sum }_i\mathrm{\mid }i{\mathrm{\mid }}^2>0$. No positivity or reality of $D_i$ was used.

For arbitrary $k$, choose a finite skeletal presentation: simple labels, bases for their finite multiplicity spaces, matrices of associators, unitors, evaluations and coevaluations, their needed inverses, and the finitely many $a_i^{\pm 1}$. Their entries generate a finitely generated characteristic-zero subfield $k_0\subset k$. It embeds into $\mathbb{C}$ by choosing an algebraically independent complex image of a finite transcendence basis and extending the remaining finite algebraic extension. The split semisimple skeletal category over $k_0$ with these matrices has the same simple labels and remains split semisimple after this embedding. Its pentagon, triangle and snake equations are finite matrix identities over $k_0$, hence still hold. The two trace loops and their sum are elements of $k_0$ given by these same matrix expressions. A zero sum in $k$ would therefore give a zero sum in the complex fusion category, contradicting step 6.1. Hence $\mathrm{Dim}(\mathcal{C})\mathrm{\ne }0$.

Substitution gives $({\mathrm{i}\mathrm{d}}_i\otimes p_{ji})(q_{ij}\otimes {\mathrm{i}\mathrm{d}}_i)=e^{-1}{\mathrm{i}\mathrm{d}}_i$: its numerator is $a^{-1}({\mathrm{i}\mathrm{d}}_{i^{\ast \ast }}\otimes {\mathrm{ev}}_{i^{\ast }})({\mathrm{coev}}_{i^{\ast }}\otimes {\mathrm{i}\mathrm{d}}_{i^{\ast \ast }})a={\mathrm{i}\mathrm{d}}_i$ by the snake identity. The other first-factor composite is $(p_{ij}\otimes {\mathrm{i}\mathrm{d}}_i)({\mathrm{i}\mathrm{d}}_i\otimes q_{ji})=d^{-1}{\mathrm{i}\mathrm{d}}_i$, by $({\mathrm{ev}}_i\otimes {\mathrm{i}\mathrm{d}}_i)({\mathrm{i}\mathrm{d}}_i\otimes {\mathrm{coev}}_i)={\mathrm{i}\mathrm{d}}_i$. The inverse-pairing equation is therefore $A_i\mathrm{/}(de)=1$, so $A_i=de=\mathrm{\mid }i{\mathrm{\mid }}^2$. Finally ${\beta }_{ij}{c}_{ij}=A_i(p_{ij}{q}_{ij})⊠(q_{ij}^{\ast }{p}_{ij}^{\ast })=\mathrm{\mid }i{\mathrm{\mid }}^2$. Summing all simple blocks and using $c=\mathrm{\Delta }\eta$ gives $\lambda \mu \mathrm{\Delta }\eta =\beta c={\sum }_i\mathrm{\mid }i{\mathrm{\mid }}^2$, which is nonzero by step 7.1.