Internal Homs in Modules over a Hopf Algebra

Question

Given a Hopf algebra $H$, I wonder when the monoidal category of $H$-modules is left-closed, right-closed, and finally, under what circumstances right and left internal hom are isomorphic.

If I'm not mistaken, it seems that for $H$-modules $M,N$ one always has a natural $H$-action on $\text{Hom}(M,N)$ given by $(h.f)(m) := h^{\prime} f(S(h^{\prime\prime})m)$, and that provided with this $H$-action, evaluation and coevaluation $\text{Hom}(M,N)\otimes M\to N$ and $N\to\text{Hom}(M,N\otimes M)$ are $H$-linear. In particular $\text{Hom}(M,-)$ provides a right adjoint to $-\otimes M$ in $H\text{-Mod}$.

If the antipode $S$ of $H$ is invertible, one can put another $H$-action on $\text{Hom}(M,N)$, namely $(h.f)(m) := h^{\prime\prime} f(S^{-1}(h^{\prime})m)$, and this yields a right adjoint to $M\otimes -$ in $H\text{-Mod}$.

1) Where can I read about these things? I'm sure it is written down somewhere, but wasn't able to find a reference.

2) Under what circumstances are the two $H$-actions on $\text{Hom}(M,N)$ equal, or at least isomorphic? For example, they are equal if $H$ is cocommutative (in which case in particular $S=S^{-1}$). Is commutativity also sufficient?

Thank you, Hanno

Answer 1

Let me try to answer to your question. As mentioned in this answer, if $(B,m,u,\Delta,\varepsilon)$ is a bialgebra over a field $\Bbbk$, then the category of left $B$-modules is always left and right closed.

Lemma: For every left $B$-module $V$ the functor ${_B\mathsf{Hom}}(B\otimes V,-)$ is right adjoint to the functor $-\otimes V:{_{B}\mathsf{Mod}}\to {_{B}\mathsf{Mod}}$. The left $B$-action on $B\otimes V$ is the diagonal one and the left $B$-action on ${_B\mathsf{Hom}}(B\otimes V,W)$ is given by $$(b\cdot f)(a\otimes v) = f(ab\otimes v).$$ Moreover, the functor ${_B\mathsf{Hom}}(V\otimes B,-)$ is right adjoint to the functor $V\otimes -$. The left $B$-action on $V\otimes B$ is the diagonal one and the left $B$-action on ${_B\mathsf{Hom}}(V\otimes B,W)$ is given by $$(b\cdot f)(v\otimes a) = f(v\otimes ab).$$

Proof: Define the assignments $$\varphi:{_B\mathsf{Hom}}(U\otimes V,W)\to {_B\mathsf{Hom}}(U,{_B\mathsf{Hom}}(B\otimes V,W)), \\ \psi:{_B\mathsf{Hom}}(U,{_B\mathsf{Hom}}(B\otimes V,W))\to{_B\mathsf{Hom}}(U\otimes V,W) $$ given by $$\varphi(f)(u):a\otimes v\mapsto f(a\cdot u\otimes v),\\ \psi(g):u\otimes v\mapsto g(u)(1\otimes v),$$ respectively, and the ones $$\varphi':{_B\mathsf{Hom}}(V\otimes U,W)\to {_B\mathsf{Hom}}(U,{_B\mathsf{Hom}}(V\otimes B,W)), \\ \psi':{_B\mathsf{Hom}}(U,{_B\mathsf{Hom}}(V\otimes B,W))\to{_B\mathsf{Hom}}(V\otimes U,W) $$ given by $$\varphi'(f)(u):v\otimes a\mapsto f(v\otimes a\cdot u),\\ \psi'(g):v\otimes u\mapsto g(u)(v\otimes 1),$$ respectively, for all $u\in U,v\in V,f\in {_B\mathsf{Hom}}(U\otimes V,W)$ and $g\in {_B\mathsf{Hom}}(U,{_B\mathsf{Hom}}(B\otimes V,W))$. Then check that these are inverses each other. $\square$

Proposition: For a Hopf algebra $H$ with antipode $S$ the functor ${\mathsf{Hom}}(V,-)$ is right adjoint to the functor $-\otimes V:{_{H}\mathsf{Mod}}\to {_{H}\mathsf{Mod}}$. The left $H$-action on ${\mathsf{Hom}}(V,W)$ is given by $$(b\cdot f)(v) = \sum b_1f(S(b_2)v).$$ If, in addition, $S$ is invertible then ${\mathsf{Hom}}(V,-)$ is also right adjoint to the functor $V\otimes -$. The left $H$-action on ${\mathsf{Hom}}(V,W)$ is given by $$(b\cdot f)(v) = \sum b_2f(S^{-1}(b_1)v).$$

Proof: Denote by a bullet $\bullet$ the given action on a module. For example, ${_\bullet V}\otimes {_\bullet W}$ denotes the diagonal left action on $V\otimes W$. Since $H$ is a Hopf algebra, the canonical map $$\mathfrak{can}: {_\bullet H}\otimes V \to {_\bullet H}\otimes {_\bullet V}, \quad h\otimes v\mapsto \sum h_1\otimes h_2\cdot v$$ is an isomorphism of left $H$-modules with inverse $$\mathfrak{can}^{-1}: {_\bullet H}\otimes {_\bullet V} \to {_\bullet H}\otimes V, \quad h\otimes v\mapsto \sum h_1\otimes S(h_2)\cdot v.$$ If you plug it into ${_H\mathsf{Hom}}(H\otimes V,W)$ you realise that $${_H\mathsf{Hom}}({_\bullet H}\otimes {_\bullet V},{_\bullet W})\cong {_H\mathsf{Hom}}({_\bullet H}\otimes V,{_\bullet W})\cong {\mathsf{Hom}}(V,W)$$ (recall that for any algebra $A$, $A\otimes -:\mathsf{Vec}_{\Bbbk}\to {_A\mathsf{Mod}}$ is left adjoint of the forgetful functor). Now, if you induce via this isomorphism the left $H$-action on ${\mathsf{Hom}}(V,W)$ that comes from the left $H$-action on ${_H\mathsf{Hom}}({_\bullet H}\otimes {_\bullet V},{_\bullet W})$ of the previous lemma, you will find the one in the statement. Now, if in addition $S$ is invertible, then one can prove that also $$\mathfrak{can}': V\otimes {_\bullet H} \to {_\bullet V}\otimes {_\bullet H}, \quad v\otimes h\mapsto \sum h_1\cdot v\otimes h_2$$ is an isomorphism of left $H$-modules with inverse $$\left(\mathfrak{can}'\right)^{-1}: {_\bullet V}\otimes {_\bullet H} \to V \otimes {_\bullet H}, \quad v\otimes h\mapsto \sum S^{-1}\left(h_1\right)\cdot v\otimes h_2.$$ If you plug it into ${_H\mathsf{Hom}}(V\otimes H,W)$ you realize that $${_H\mathsf{Hom}}({_\bullet V}\otimes {_\bullet H},{_\bullet W})\cong {_H\mathsf{Hom}}(V\otimes {_\bullet H},{_\bullet W})\cong {\mathsf{Hom}}(V,W)$$ and the action is the claimed one. $\square$

Concerning your last questions.

1. I don't know where you may exactly read about this topic in this form. The argument above comes essentially from my PhD thesis. Maybe, if it is written somewhere, I would guess in B. Pareigis' or P. Schauenburg's works.
2. Instead, concerning when the two are isomorphic, assume that they are. By uniqueness of the adjoint, this implies that the functors $-\otimes V$ and $V\otimes -$ are isomorphic for every $V$ and naturally in $V$ (this should be the counterpart of Schauenburg, Tannaka duality for arbitrary Hopf algebras, Remark 1.2.11). Therefore, modulo silly mistakes, the category of left modules becomes a braided monoidal category and so $H$ inherits a quasi-triangular structure (Kassel, Quantum Groups, Prop XIII.1.4). I would said (but I didn't check it right now) that the converse is true as well: if $H$ is quasi-triangular, then the category of modules should be left and right closed with forgetful functor to vector spaces preserving the internal homs (I mean, it should be true in general that a one-sided closed braided monoidal category is closed on both sides). Notice that, in view of Radford, On the antipode of a quasitriangular Hopf algebra, if $H$ is quasitriangular then the antipode is invertible.

I hope this answer your questions.