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Suppose $G$ is a group, we can define the tensor representation of $V \otimes W$ by $g(x \otimes y) = gx \otimes gy$. And we can define internal Hom by $iHom(V,W)=Hom_k(V,W)$. The action of $g$ is then $(gf)(x)=g(f(g^{-1}x))$. These two are adjoint functors to Abelian groups. Similarly, we can define internal Hom for Lie representations by $gf(x)=g(f(x))-f(g(x))$. My question is, does this generalize to representations of Hopf algebras?

As I type this questions, I saw Internal Homs in Modules over a Hopf Algebra, but no proof was provided, so I think it's ok for me to ask for a reference for this result.

davik
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  • My guess is that the internal Hom for groups is obtained by identifying $Hom_k(V,W)$ with $V^*\otimes W$ and using the co-product. Presumably this would carry over to any Hopf algebra. – David Hill Feb 05 '19 at 18:05
  • I think I can define the action, but I can't show it's adjoint... – davik Feb 05 '19 at 23:50
  • How does the proofs for groups go? – David Hill Feb 06 '19 at 03:05
  • well to summarize the part I'm stuck on it's like this. for groups, if $f : A \otimes B \to C$ is a $G$ module morphism $f(gx \otimes y)= f(gx \otimes g g^{-1} y)=g(f(x \otimes g^{-1}y))$. I'm not sure how to do this for hopf algebras. – davik Feb 06 '19 at 11:11

1 Answers1

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In fact, it does and not only for Hopf algebras. Let $(B,m,u,\Delta,\varepsilon)$ be a bialgebra over a field $\Bbbk$ (for the sake of simplicity, I'm not sure it is really needed).

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 $$(b\cdot f)(a\otimes v) = f(ab\otimes v).$$

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, 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 $-\otimes V:{_{B}\mathsf{Mod}}\to {_{B}\mathsf{Mod}}$. The left $B$-action on ${\mathsf{Hom}}(V,W)$ is given by $$(b\cdot f)(v) = \sum b_1f(S(b_2)v).$$

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. $\square$

Observe that for the group Hopf algebra, $\Delta(g)=g\otimes g$, $S(g)=g^{-1}$ and you recover the situation of your first example. For the universal enveloping algebra $U(\mathfrak{g})$ of a Lie algebra $\mathfrak{g}$ you have that for every $x\in\mathfrak{g}$, $\Delta(x)=x\otimes 1+1\otimes x$, $S(x)=-x$ and you recover the second example.

Remark: It deserve to be noticed that the lemma holds true also for quasi-bialgebras and bialgebroids (for bialgebroids, see for example Proposition 3.3 in P. Schauenburg, Duals and doubles of quantum groupoids ($\times_R$-Hopf algebras), New trends in Hopf algebra theory (La Falda, 1999), Contemp. Math., vol. 267, Amer. Math. Soc., Providence, RI, 2000, 273–299.).

Remark: I would like to point out finally that indeed this provides a characterization of Hopf algebras: an algebra $H$ is a Hopf algebra if and only if its category of modules is closed monoidal with monoidal underlying functor $\omega:{_A\mathsf{Mod}} \to \mathsf{Vec}_{\Bbbk}$ preserving internal homs.

  • Just to make sure I understand the notation and claims: The subscript $B$ in front of Hom means Hom's of $B$-modules? And is the claim at first that these are the internal Hom for a bialgebra? (Which seems odd since then adding the antipode would reduce the dimension of this space). Anyway, +1 from me even before clarifying, because this is great work. – Tobias Kildetoft Feb 14 '19 at 09:58
  • Thanks @TobiasKildetoft. Yes, the subscript $B$ on the left of the hom stands for left $B$-linear and yes, the lemma is exactly giving the (right) internal homs for the category of modules over a bialgebra. In fact, you may figure out that there also exist left internal homs: $V\otimes- \dashv {_B\mathsf{Hom}}(V\otimes B,-)$, whence ${_B\mathsf{Mod}}$ is both a left and a right closed monoidal category. However, I am not sure about your claim on dimensions: on the one hand you have $B$-linear maps, on the other $\Bbbk$-linear ones. And for Hopf algebras they have the same dimension. – Ender Wiggins Feb 14 '19 at 10:23