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Let $\mathcal{H}$ be a Hopf algebra over $\mathbb{C}$. Let $\textrm{mod}_\mathcal{H}$ be the monoidal abelian category of finite-dimensional modules over $\mathcal{H}$. Fix $X\in\textrm{Obj}(\textrm{mod}_\mathcal{H})$. We know that the functor $$(-\otimes X):\textrm{mod}_\mathcal{H}\rightarrow\textrm{mod}_\mathcal{H}$$ is left adjoint to functor $$\textrm{Hom}_{\mathbb{C}}(X,-):\textrm{mod}_\mathcal{H}\rightarrow\textrm{mod}_\mathcal{H}.$$

Moreover, $\textrm{Hom}_{\mathcal{H}}(X,Y)$ is a submodule of $\textrm{Hom}_{\mathbb{C}}(X,Y)$, for all $Y\in\textrm{Obj}(\textrm{mod}_\mathcal{H})$. It seems that $\textrm{Hom}_{\mathcal{H}}(X,-)$ defines a subfunctor of $\textrm{Hom}_{\mathbb{C}}(X,-)$, since composition of morphisms of modules is a morphism of modules, as well. Am I right about that? If this is the case, does $\textrm{Hom}_{\mathcal{H}}(X,-)$ have any (left or right) adjoints? Does anyone know of any references that deal with this? Thanks in advance for answers.

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    $\begingroup$ I don't a precise reference apart elementary module theory, but I think $-\otimes_{\cal H} X \dashv \hom_{\cal H}(X,-)$... (where the left adjoint is a suitable coequalizer obtained from $-\otimes_{\mathbb C}X$). $\endgroup$
    – fosco
    Nov 25, 2019 at 8:38
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    $\begingroup$ @Fosco Doc, you cannot tensor two right $H$-modules unless you turn left at Albuquerque, first... $\endgroup$
    – Bugs Bunny
    Dec 17, 2019 at 6:57

2 Answers 2

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I am not an expert, but I think that this result is an instance of a more general phaenomenon. If this does not apply to your situation, feel free to ignore my answer or to downvote it.

It looks to me that your framework is a special case of the one described and developed in the following paper.

Hopf monads on monoidal categories, Bruguières, Lack, Virelizier. Advances in Mathematics, 227(2):745-800, 2011.

The motto of the paper is: If T is a good monad on a closed monoidal category, then its category of algebras is also closed and the monadic forgetful functor preserves internal-homs.

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Doc, let us discern what catechism you are coddling. We know that $$hom_H(X,Y)=hom_{\mathbb C}(X,Y)^H.$$ Thus, your basic functor does not take values in the category you write. Instead, $$hom_H(X,-)=mod_H \rightarrow mod_{\mathbb C}.$$ Its left adjoint functor you know and cherish already: $$-\otimes X= mod_{\mathbb C} \rightarrow mod_{H}.$$ If you impertinently persist with a functor to $mod_H$, you are taking a composition $$mod_H \xrightarrow{hom_H(X,-)} mod_{\mathbb C}\xrightarrow{T} mod_{H}$$ where the functor $T$ treats a vector space as a trivial $H$-module. The left adjoint to $T$ is the coinvariant functor $$(-)_H: mod_H \rightarrow mod_{\mathbb C}.$$ Blending these two judicious remarks together, we conclude that the left adjoint to $$hom_H(X,-)=mod_H \rightarrow mod_{H}$$ is $$(-)_H\otimes X: mod_H \rightarrow mod_{H}.$$ Gee, ain’t I a stinker?

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