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(Alternate title: Find the Adjoint: Hopf Algebra edition)

I was chatting with Jonah about his question Hopf algebra structure on $\prod_n A^{\otimes n}$ for an algebra $A$. It's very closely related to the following question:

For a $k$-algebra $A$, is there a Hopf Algebra $H(A)$ such that for any Hopf algebra $B$, we have
$\mathop{Hom}_{k\text{-alg}}(B,A)\cong \mathop{Hom}_{k\text{-Hopf-alg}}(B,H(A))$?

In other words, does the forgetful functor from $k$-Hopf-algebras to $k$-algebras have a right adjoint? There are a few related questions, any of which I'd be interested in knowing the answer to:

  • Does the forgetful functor from Hopf algebras to augmented algebras (sending the counit to the augmentation) have a right adjoint?
  • Does the forgetful functor from Hopf algebras to algebras with distinguished anti-automorphism (sending the antipode to the anti-automorphism) have a right adjoint?
  • Does the forgetful functor from Hopf algebras to algebras with augmentation and distinguished anti-automorphism have a right adjoint?

Unfortunately, I don't feel like I can motivate this question very well. My motivation is that the better I know which forgetful functors have adjoints, the better I sleep at night.

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Does this help? arxiv4.library.cornell.edu/pdf/0905.2613 –  Glen M Wilson Apr 27 '10 at 3:39
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@Glen: Yes! Could you please post that as an answer? –  Anton Geraschenko Apr 27 '10 at 5:25
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4 Answers

up vote 13 down vote accepted

The article here proves that the forgetful functor from $k$-Hopf algebras to $k$-algebras has a right adjoint. The main tool they use is the Special Adjoint Functor Theorem.

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The problem is famous and you will find in the book Sweedler. Ana Agore (one of my student in master) prove the result in june 2009 -- and her paper will apper in Comm. Algebra. Is this one http://arxiv.org/PS_cache/arxiv/pdf/0905/0905.2613v3.pdf

However, related to the problem is also another paper of her more recently: http://front.math.ucdavis.edu/1003.0318

On the other hand, two months latter (july 2009) a colleque of her (Alex Chirvasitu) give the explicit construction of it in this paper http://arxiv.org/PS_cache/arxiv/pdf/0907/0907.2881v2.pdf that allready appered in J. Algebra at the begining of 2010.

Cheers! and a nice site. Gigel Militaru

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Why dualizing the result of Takeuchi is not automatic and straightforward ? –  Zoran Skoda Jul 9 '10 at 19:47
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Composing that adjoint from the one giving the free algebra on a vector space you'd end up with a free Hopf algebra on a vector space. Those don't exist, cf. http://ncatlab.org/nlab/show/free+Hopf+algebra

Later: Ah! Anton wants an adjoint on the other side, so this is not an answer. I'll just leave it here, because anyways it is a cute piece of information.

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I'm not convinced yet. The free algebra on a vector space is a left adjoint to the forgetful functor {Algebras}→{Vector Spaces}. Composing it with a right adjoint to {Hopf Algebras}→{Algebras} wouldn't give you an adjoint to the forgetful functor {Hopf Algebras}→{Vector Spaces}. –  Anton Geraschenko Apr 27 '10 at 1:03
    
I agree that this is a cute piece of information ... I was vaguely wondering if there's an adjoint on the other side too, and now I know! –  Anton Geraschenko Apr 27 '10 at 1:39
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Here's that answer in a slightly more high-falutin' language. Let's have a category, $\mathcal{D}$. Then for any Lawvere theory, say $\mathcal{V}$, we can consider $\mathcal{V}$-algebra objects in $\mathcal{D}$; let's call this $\mathcal{D}\mathcal{V}$. If $\mathcal{D}$ has some nice structure, then the forgetful functor $\mathcal{D}\mathcal{V} \to \mathcal{D}$ has a left adjoint, the free $\mathcal{V}$-algebra functor.

But we can play this game in the opposite category as well. Only instead of simply going through the mirror and staying there, we go through, do our construction, and then come back again. So we form the category of co-$\mathcal{V}$-algebra objects in $\mathcal{D}$ by taking the opposite category of the category of $\mathcal{V}$-algebra objects in the opposite category of $\mathcal{D}$. Confused? Here it is in notation:

$$ \mathcal{D}\mathcal{V}^c = \left(\mathcal{D}^{op}\mathcal{V}\right)^{op} $$

The double-opping means that the forgetful functor from $\mathcal{D}\mathcal{V}^c$ goes to $\mathcal{D}$ and not $\mathcal{D}^{op}$. However, as we've just gone through a mirror and back again, if our category $\mathcal{D}$ has some co-nice structure, $\mathcal{D}\mathcal{V}^c \to \mathcal{D}$ has a right adjoint, as requested.

In the cases you ask about, the category $\mathcal{D}$ is the opposite category of the category of models in either Set or $\operatorname{Set}_*$ (pointed sets) of some Lawvere theory. In each case, the Lawvere theory is based on algebras and so we tend to write its coproduct as the tensor product (though, of course, it's a slightly different tensor product in each case). The category of Hopf-algebras is then formed by taking an appropriate Lawvere theory, $\mathcal{V}$: possibly monoids, possibly groups, and forming $\mathcal{D}\mathcal{V}$ and then taking the opposite category again.

So the question boils down to whether or not the category of models of a Lawvere theory in either $Set$ of $\operatorname{Set}_*$ has enough co-nice structure and whether or not what we traditionally think of as Hopf algebras truly is the category of co-whatsits in whatevers (or whether, as in the case of Banach spaces, there are a few extra things floating around that we didn't think of). Fortunately, these are both true.

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Are Hopf algebras really the category of co-whatsits in whatevers? Can you fill in the details a little more? –  Reid Barton Apr 27 '10 at 15:06
    
Does ncatlab.org/nlab/show/Hopf+algebra answer your question? –  Andrew Stacey Apr 27 '10 at 17:45
    
Ouch! Just got a downvote for this, but no explanation. Anyone care to explain? –  Andrew Stacey Apr 28 '10 at 14:33
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