Bimodule averaging and the restricted pivotal trace pairing

The Frobenius and unit equations give $(m_A\otimes {\mathrm{id}}_A)({\mathrm{id}}_A\otimes e_A)={\mathrm{\Delta }}_A=({\mathrm{id}}_A\otimes m_A)(e_A\otimes {\mathrm{id}}_A)$, and specialness gives $m_A{e}_A={\eta }_A$. Substitution of $e_A={\mathrm{\Delta }}_A{\eta }_A$ and ${\beta }_A={ϵ}_A{m}_A:A\otimes A\to \mathbf{1}$ also gives the two inverse-pairing identities $({\mathrm{id}}_A\otimes {\beta }_A)(e_A\otimes {\mathrm{id}}_A)={\mathrm{id}}_A=({\beta }_A\otimes {\mathrm{id}}_A)({\mathrm{id}}_A\otimes e_A)$, using Frobenius followed by the unit and counit. All these equations hold for $B$ as well.

The two displayed bending maps are inverse on ordinary Hom spaces by the right-duality snake for $U$ and the left-duality snake for $V$. They preserve the required bimodule equations. To verify this without a drawing, for any object $W$ use the hexagon and naturality with $b_U$ to obtain $(c_{W,U}\otimes {\mathrm{id}}_{U^{\vee }})({\mathrm{id}}_W\otimes b_U)=({\mathrm{id}}_U\otimes c_{W,U^{\vee }}^{-1})(b_U\otimes {\mathrm{id}}_W)\mathrm{.}$ The equation with ${\tilde{b}}_V$ and the inverse braiding is obtained in the same way. Set $W=A$, substitute these equations into $T(\beta ){\rho }_X^{\mathrm{\ell }}$ and $T(\beta ){\rho }_X^r$, respectively, and then use the left- and right-intertwining equations for $\beta$. The results are exactly ${\rho }_{\mathrm{\ell }}^{U,A,V}({\mathrm{id}}_A\otimes T(\beta ))$ and ${\rho }_r^{U,A,V}(T(\beta )\otimes {\mathrm{id}}_A)$ with the plus/minus actions in [F1]. The evaluation versions of the same two naturality equations prove that $T^{-1}$ preserves the intertwining equations as well. Thus these are inverse isomorphisms of the indicated bimodule Hom spaces, without rescaling the dualities.

Define the pivotal cyclic rotation on $t:\mathbf{1}\to A\otimes A$ by ${\mathrm{rot}}_A(t)=({\mathrm{ev}}_A\otimes {\mathrm{id}}_A\otimes {\mathrm{id}}_A)({\mathrm{id}}_{A^{\vee }}\otimes t\otimes {\mathrm{id}}_A){\tilde{b}}_A$, where ${\tilde{b}}_A:\mathbf{1}\to A^{\vee }\otimes A$ is the chosen coevaluation. The equality ${\mathrm{\Phi }}_{\mathrm{\ell }}={\mathrm{\Phi }}_r$ of the two transposes of ${\beta }_A$ gives ${\mathrm{rot}}_A(e_A)=e_A$. One way to check this directly is to transpose the two inverse-pairing identities of step 1.1: their transposes say that the two resulting maps $A^{\vee }\to A$ are inverses of ${\mathrm{\Phi }}_{\mathrm{\ell }}$ and ${\mathrm{\Phi }}_r$. These maps are respectively $({\mathrm{ev}}_A\otimes {\mathrm{id}}_A)({\mathrm{id}}_{A^{\vee }}\otimes e_A)$ and $({\mathrm{id}}_A\otimes {\stackrel{}{\mathrm{ev}}}_A)(e_A\otimes {\mathrm{id}}_{A^{\vee }})$. Their equality, bent back using ${\tilde{b}}_A$, is exactly ${\mathrm{rot}}_A(e_A)=e_A$. This uses pivotal cyclic rotation, not a braided flip of $e_A$. The identical argument applies to $e_B$.

Centrality in step 1.1 and module associativity imply that $L(f)$ is left-linear. Written in formal separability slots, the required equality is $a{e}_A^{(1)}f(e_A^{(2)}x)=e_A^{(1)}f(e_A^{(2)}ax)$; it abbreviates the two composites on $A\otimes M$, with the superscripts denoting the output slots of the single morphism $e_A$, not elements of an object. If $f$ is already left-linear, its action relation and $m_A{e}_A={\eta }_A$ give $L(f)=f$. Thus $L$ is the projector onto left-module morphisms. The same argument on the right gives the assertion for $R$. Expanding the two averages, the left and right actions on each module commute and the inserted copairings occupy disjoint left and right tensor slots. This gives $LR=RL$. The image of $E=LR$ is consequently the intersection of the two images, namely the bimodule morphisms.

For arbitrary $t:\mathbf{1}\to A\otimes A$, maps $a_M:A\otimes M\to M$, $a_N:A\otimes N\to N$, $f:M\to N$ and $g:N\to M$, the pivotal contraction identity is ${\mathrm{tr}}_j\text{\hspace{0.17em}⁣}(g{a}_N({\mathrm{id}}_A\otimes f)({\mathrm{id}}_A\otimes a_M)(t\otimes {\mathrm{id}}_M))={\mathrm{tr}}_j\text{\hspace{0.17em}⁣}(a_M({\mathrm{id}}_A\otimes g)({\mathrm{id}}_A\otimes a_N)({\mathrm{rot}}_A(t)\otimes {\mathrm{id}}_N)f)\mathrm{.}$ Indeed expand each trace as its coevaluation, endomorphism and pivotal evaluation composite. Substitute the formula for ${\mathrm{rot}}_A(t)$ on the right and cancel the consecutive $A^{\vee },A$ evaluation/coevaluation by their snake. Changing the remaining trace cut from $N$ to $M$ is the cyclicity supplied by [F3]; the remaining ordered tensor slots and maps are those of the left expression. Apply this identity to the two left actions and $t=e_A$, and then use step 2.1. It gives ${\mathrm{tr}}_j(g{L}_{M,N}(f))={\mathrm{tr}}_j(L_{N,M}(g)f)$. The right-action identity is obtained by the same expansion with the two $B$ slots on the right. Composing the two adjointness identities and using commutation from step 2.2 proves the identity for $E$.

The ordinary trace pairing between ${\mathrm{Hom}}_{\mathcal{C}}(M,N)$ and ${\mathrm{Hom}}_{\mathcal{C}}(N,M)$ is nondegenerate: decompose $M,N$ into simples, and detect a nonzero matrix entry with the opposite matrix unit, whose product has trace $d_i\mathrm{\ne }0$. If a nonzero $f$ is a bimodule morphism, choose an ordinary $g$ detecting it. Since $E(f)=f$, step 3.1 gives ${\mathrm{tr}}_j(E(g)f)={\mathrm{tr}}_j(gf)\mathrm{\ne }0$, with $E(g)$ a bimodule morphism. The same reasoning on the other side proves nondegeneracy of the restricted pairing. Finite-dimensional dual-basis linear algebra gives the ordinary trace-dual bases. The zero-space case is vacuous, and no total object dimension has been divided by.

Apply step 4.1 to the two bimodules $U{\otimes }^{+}A{\otimes }^{-}V$ and $X$. Its nondegenerate pairing is ${\mathrm{tr}}_j(\alpha h)$ with $h:X\to U{\otimes }^{+}A{\otimes }^{-}V$. Transporting its second argument through the isomorphism $T$ of step 1.2 proves nondegeneracy of ${\mathrm{tr}}_j(\alpha T(\beta ))$. This establishes the stated algebraic defect-field pairing, including zero spaces, without an RT coefficient identification or division by a total dimension.