Coquasi-Bialgebras with Preantipode and Rigid Monoidal Categories

Abstract

By a theorem of Majid, every monoidal category with a neutral quasi-monoidal functor to finitely generated and projective $\Bbbk$-modules gives rise to a coquasi-bialgebra. We prove that if the category is also rigid, then the associated coquasi-bialgebra admits a preantipode, providing in this way an analogue for coquasi-bialgebras of Ulbrich’s reconstruction theorem for Hopf algebras. When $\Bbbk$ is a field, this allows us to characterize coquasi-Hopf algebras as well in terms of rigidity of finite-dimensional corepresentations.

Appendix A: A relation for the preantipode of a quasi-bialgebra

Recall from [18] that a preantipode for a quasi-bialgebra (A, Δ, ε, m, u, Φ) is a $\Bbbk$-linear map S : A → A that satisfies

$$ \sum a_{1}S(ba_{2})=\varepsilon(a)S(b)= \sum S(a_{1}b)a_{2}, \qquad \sum {\Phi}^{1}S({\Phi}^{2}){\Phi}^{3}=1, $$

(1)

for all a,b ∈ A, where $\sum {\Phi }^{1}\otimes {\Phi }^{2}\otimes {\Phi }^{3}={\Phi }$. Let us introduce also the following extended notation for the reassociator and its inverse:

$$ \begin{array}{@{}rcl@{}} {\Phi} =\sum {\Phi}^{1}\otimes {\Phi}^{2}\otimes {\Phi}^{3}=\sum {\Psi}^{1}\otimes {\Psi}^{2}\otimes {\Psi}^{3}={\ldots} \\ {\Phi}^{-1} =\sum \varphi^{1}\otimes \varphi^{2}\otimes \varphi^{3}=\sum \psi^{1}\otimes \psi^{2}\otimes \psi^{3}=\ldots \end{array} $$

Let (A,m,u,Δ,ε,Φ,S) be a quasi-bialgebra with preantipode and consider the A-actions on End(A) = Hom(A,A) defined by $\left (f\leftharpoonup a\right )(b) = f(ab)$ and $\left (a\rightharpoonup f\right )(b) = f(ba)$ for all a,b ∈ A and for all f ∈EndA. Define the elements

$$ \begin{array}{@{}rcl@{}} p :=\sum \varphi^{1}\otimes \varphi^{2}\left( \varphi^{3}\rightharpoonup S\right) & \in & A\otimes \mathsf{End}{A} , \end{array} $$

(2)

$$ \begin{array}{@{}rcl@{}} q :=\sum \left( S \leftharpoonup \varphi^{1}\right) \varphi^{2}\otimes \varphi^{3} & \in & \mathsf{End}{A}\otimes A, \end{array} $$

where $\left (x\left (y\rightharpoonup f\right )\right )(a) = xf(ay)$ and $\left (\left (f\leftharpoonup x\right )y\right )(a) = f(ax)y$ for all a,x,y ∈ A and for all f ∈EndA. Let us introduce the following notation for shortness:

$$ p:=\sum p^{1}\otimes p^{2} \qquad \textrm{ and } \qquad q:=\sum q^{1}\otimes q^{2}. $$

Lemma.

In the foregoing notation we have that for every a ∈ A

$$ \begin{array}{ll} \displaystyle{\sum p^{1}\otimes p^{2}(a) =\sum {\varphi_{1}^{1}}\psi^{1}\otimes {\varphi_{2}^{1}}\psi^{2}{\Phi}^{1}S\left( a\varphi^{2}{\psi_{1}^{3}}{\Phi}^{2}\right) \varphi^{3}\psi_{2}^{3}{\Phi}^{3},}\\ \displaystyle{\sum q^{1}(a)\otimes q^{2} =\sum {\Phi}^{1}{\varphi_{1}^{1}}\psi^{1}S\left( {\Phi}^{2}\varphi_{2}^{1}\psi^{2} a\right) {\Phi}^{3}\varphi^{2}{\psi_{1}^{3}}\otimes \varphi^{3}{\psi_{2}^{3}}.}\end{array} $$

(33)

Moreover, the following relations hold for every a,b ∈ A

$$ \begin{array}{@{}rcl@{}} \sum p^{1}a\otimes p^{2}(b) & = \sum a_{11}p^{1}\otimes a_{12}p^{2}(ba_{2}), \end{array} $$

(34)

$$ \begin{array}{@{}rcl@{}} \sum q^{1}(a)\otimes bq^{2} & = \sum q^{1}(b_{1}a)b_{21}\otimes q^{2}b_{22}. \end{array} $$

(35)

Proof

The reassociator Φ satisfies the dual relation to Eq. 3, i.e.

$$ (1_{A}\otimes {\Phi} ) \cdot (A\otimes {\Delta} \otimes A)\left( {\Phi} \right) \cdot \left( {\Phi} \otimes 1_{A}\right)= (A\otimes A\otimes {\Delta} )\left( {\Phi} \right) \cdot ({\Delta} \otimes A\otimes A)\left( {\Phi} \right). $$

In particular, it satisfies

$$ \sum {\varphi_{1}^{1}}\psi^{1}\otimes {\varphi_{2}^{1}}\psi^{2}{\Phi}^{1}\otimes \varphi^{2}{\psi_{1}^{3}}{\Phi}^{2}\otimes \varphi^{3}\psi_{2}^{3}{\Phi}^{3}=\sum \varphi^{1}\psi^{1}\otimes \varphi^{2}\psi_{1}^{2}\otimes \varphi^{3}{\psi_{2}^{2}}\otimes \psi^{3}. $$

Applying $\left (A\otimes m\right ) \left (A\otimes A\otimes m\right ) \left (A\otimes A\otimes (S\leftharpoonup a)\otimes A\right )$ to both sides we get

$$ \begin{array}{@{}rcl@{}} && \sum {\varphi_{1}^{1}}\psi^{1}\otimes {\varphi_{2}^{1}}\psi^{2}{\Phi}^{1}S\left( a \varphi^{2}{\psi_{1}^{3}}{\Phi}^{2}\right) \varphi^{3}\psi_{2}^{3}{\Phi}^{3} \\ && = \sum \varphi^{1}\psi^{1}\otimes \varphi^{2}\psi_{1}^{2}S\left( a \varphi^{3}{\psi_{2}^{2}}\right) \psi^{3} \stackrel{(31)}{=} \sum \varphi^{1}\otimes \varphi^{2}S\left( a \varphi^{3}\right) = \sum p^{1}\otimes p^{2}(a), \end{array} $$

which is the first identity in Eq. 33. The second one is proved analogously. Let us check that Eq. 34 holds as well ((35) is proved similarly). We compute

$$ \begin{array}{@{}rcl@{}} \sum p^{1}a\otimes p^{2}(b) & \stackrel{(32)}{=}& \sum \varphi^{1}a\otimes \varphi^{2}S(b\varphi^{3}) \stackrel{(31)}{=} \sum \varphi^{1}a_{1}\otimes \varphi^{2}a_{21}S(b\varphi^{3}a_{22}) \\ & \stackrel{(*)}{=}& \sum a_{11}\varphi^{1}\otimes a_{12}\varphi^{2}S(ba_{2}\varphi^{3}) = \sum a_{11}p^{1}\otimes a_{12}p^{2}(ba_{2}), \end{array} $$

where in (∗) we used the quasi-coassociativity Φ ⋅ (Δ ⊗ A)Δ = (A ⊗Δ)Δ ⋅Φ. □

Lemma.

Let (A,m,u,Δ,ε,Φ,S) be a quasi-bialgebra with preantipode and let p,q be defined as above. For all a ∈ A we have that

$$ S(a)=\sum q^{1}\left( 1_{A}\right)S\left( p^{1}aq^{2}\right) p^{2}\left( 1_{A}\right)=\sum S\left( \varphi^{1}\right) \varphi^{2}S\left( \psi^{1}a\varphi^{3}\right) \psi^{2}S\left( \psi^{3}\right) . $$

Proof

Keeping in mind that Φ^− 1 is counital, i.e. that it satisfies

$$ \left( \varepsilon\otimes A\otimes A\right)\left( {\Phi}^{-1}\right)=1_{A}\otimes 1_{A}=\left( A\otimes \varepsilon\otimes A\right)\left( {\Phi}^{-1}\right)=1_{A}\otimes 1_{A}=\left( A\otimes A\otimes \varepsilon\right)\left( {\Phi}^{-1}\right), $$

we may compute directly

$$ \begin{array}{@{}rcl@{}} && \sum S\left( \varphi^{1}\right) \varphi^{2}S\left( \psi^{1}a\varphi^{3}\right) \psi^{2}S\left( \psi^{3}\right) =\sum q^{1}\left( 1_{A}\right)S\left( p^{1}aq^{2}\right) p^{2}\left( 1_{A}\right) \\ && \stackrel{(33)}{=} \sum {\Phi}^{1}{\varphi_{1}^{1}}\psi^{1}S\left( {\Phi}^{2}{\varphi_{2}^{1}}\psi^{2}\right) {\Phi}^{3}\varphi^{2} {\psi_{1}^{3}}S\left( {\gamma_{1}^{1}}\phi^{1}a\varphi^{3}{\psi_{2}^{3}}\right) {\gamma_{2}^{1}} \phi^{2}{\Psi}^{1}S\left( \gamma^{2} {\phi_{1}^{3}}{\Psi}^{2}\right) \gamma^{3}{\phi_{2}^{3}}{\Psi}^{3} \\ && \stackrel{(31)}{=} \sum {\Phi}^{1}{\varphi_{1}^{1}}S\left( {\Phi}^{2}\varphi_{2}^{1}\right) {\Phi}^{3}\varphi^{2}S\left( \phi^{1}a\varphi^{3}\right) \phi^{2}{\Psi}^{1}S\left( {\phi_{1}^{3}}{\Psi}^{2}\right) \phi_{2}^{3}{\Psi}^{3} \\ && \stackrel{(31)}{=} \sum {\Phi}^{1}S\left( {\Phi}^{2}\right) {\Phi}^{3}S\left( a\right) {\Psi}^{1}S\left( {\Psi}^{2}\right) {\Psi}^{3}=S(a). \end{array} $$

□

Proposition.

Let (A,m,u,Δ,ε,Φ,S) be a quasi-bialgebra with preantipode. For all a,b ∈ A we have

$$ S\left( ab\right) =\sum S\left( \varphi^{1}b\right) \varphi^{2}S\left( \psi^{1}\varphi^{3}\right) \psi^{2}S\left( a\psi^{3}\right) . $$

(36)

Proof

We know from Lemma A.2 that $S(a)=\sum q^{1}\left (1_{A}\right )S\left (p^{1}aq^{2}\right ) p^{2}\left (1_{A}\right )$. Relation (36) is proved directly by applying it to $S\left (ab\right )$:

$$ \begin{array}{@{}rcl@{}} S\left( ab\right) &\stackrel{\phantom{(52)}}{=}&\sum q^{1}\left( 1_{A}\right)S\left( p^{1}abq^{2}\right) p^{2}\left( 1_{A}\right) \stackrel{(34)}{=} \sum q^{1}\left( 1_{A}\right)S\left( a_{11}p^{1}bq^{2}\right)a_{12} p^{2}(a_{2}) \\ & \stackrel{(31)}{=}& \sum q^{1}\left( 1_{A}\right)S\left( p^{1}bq^{2}\right) p^{2}(a) \stackrel{(35)}{=} \sum q^{1}(b_{1})b_{21}S\left( p^{1}q^{2}b_{22}\right) p^{2}(a) \\ & \stackrel{(31)}{=}& \sum q^{1}(b)S\left( p^{1}q^{2}\right) p^{2}(a) = \sum S\left( \varphi^{1}b\right) \varphi^{2}S\left( \psi^{1}\varphi^{3}\right) \psi^{2}S\left( a\psi^{3}\right). \end{array} $$

□

Formula (36) can be viewed as an anti-multiplicativity of the preantipode.