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According to some form of Tannakian reconstruction, given a finite tensor category with a fiber functor to the category of vector spaces, one determines a Hopf algebra by considering tensor endomorphisms of the fiber functor. As far as I know, a similar procedure is used to reconstruct a group from its symmetric tensor category of representations.

I am curious about what happens if one is given a finite tensor category $\mathcal{C}$ and a tensor functor $\mathcal{C} \to Rep(G)$ for $G$ a finite group. It follows that there should exist a Hopf algebra $H$ (by the previous reconstruction business applied to the composition of this tensor functor with the forgetful functor $Rep(G) \to \mathrm{Vect}$) and homomorphism $\mathbb{C}[G] \to H$.

Under what conditions will $H$ be a semidirect product of $G$ with some Hopf algebra?

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Does it follow that $\mathcal{C}G$ is a Hopf subalgebra of H? I guess one should impose additional properties on the tensor functor $\mathcal{C} \to Rep(G)$ to get an inclusion of Hopf algebras. – Sebastian Burciu Mar 13 '10 at 11:20
Let $F: Rep(G') \to Rep(G)$ be an exact tensor functor. It corresponds to a morphism of affine group schemes $\phi: G\to G'$. Then $\phi$ is a closed immersion iff every object of $Rep(G)$ is a subquotient of an objet in the image of $F$. And $\phi$ is surjective and flat iff $F$ is fully faithfull and its image is stable by taking subobjects. See Deligne, Milne - Tannakian categories. – AFK Mar 13 '10 at 13:14
@Sebastian: I agree with you (I only said a homomorphism). – Akhil Mathew Mar 13 '10 at 13:24
@YBL: Thanks! <present text added to reach 15 characters> – Akhil Mathew Mar 13 '10 at 13:27
In the case of Hopf algebras, probably the same condition mentioned by YBL is necessary and sufficient to get an inclusion, although I didn't check that. – Sebastian Burciu Mar 13 '10 at 14:31

1 Answer 1

up vote 4 down vote accepted


Let $\mathcal{C}$ be a tensor category, and let $(A,\mu) \in \mathcal{C}-Alg$ be an algebra in $\mathcal{C}$. So $\mu:A\otimes A\to A$ is a morphism in $\mathcal{C}$ and $\otimes$ here means the $\otimes$ in $\mathcal{C}$ (of course there isn't another one around at this point, but I mean to emphasize it's not just the $\otimes$ of Vect).

In this context, it makes sense to talk about $A$-modules in $\mathcal{C}$, whose definition you can guess. These form a k-linear abelian category $D$ with a forgetful functor to $\mathcal{C}$ which forgets the $A$ action.

Now if $\mathcal{C}$ has a fiber functor F, then $\mathcal{C}$ is realized as the Hopf algebra $End(F)$, as you said (well $End(F)^{op}$ I think , but nevermind). The algebra $A$ can be pushed forward by $F$ to an ordinary algebra $F(A)$ in Vect. However, $D$ is not the category of $A$-modules, but a well-known proposition tells that $D$ is the category of $A\rtimes H$-modules, the semi-direct product you asked about.

Notice that no part of the discussion so far asked for any symmetric structure on $\mathcal{C}$, and also $A$ is only an algebra. To define a bialgebra in $\mathcal{C}$, however, one needs $\mathcal{C}$ to be braided, because the compatibility between $\Delta$ and $\mu$ will use the braiding. In your case braiding just means symmetry. I never worked this out in detail, but I imagine that if $A$ is actually a bi-algebra in $\mathcal{C}$, then $D$ gets endowed with a monoidal structure, and that $D$ is the category of $A\rtimes H$-modules, where $A\rtimes H$ is a bi-algebra. Likewise if $A$ is Hopf in $\mathcal{C}$, then $D$ is tensor and $A\rtimes H$ is a Hopf algebra, and $D\cong A \rtimes H$-mod.

Thus ends the part where I'm pretty sure I'm not saying anything too incorrect. Below I will try to answer your actual question. I would not trust it though until somebody smarter agrees with it.

So, now your question becomes (let's revert to considering algebras at the top and not Hopf algebras, since it should be clear how to extend): Given a functor $F:D\to \mathcal{C}$, when is $F$ the forgetful functor corresponding to some algebra $A \in \mathcal{C}$? I think for this, it will be enough to assume (in addition, of course, to assuming that $F$ is faithful and exact) that $D$ has a projective generator $M$ (although maybe this is guaranteed by a lesser assumption?). This is definitely necessary to be able to realize $D$ as some category of modules of a ring, as you desire, and I imagine that you then let $A=\underline{Hom}(M,M)$ (meaning $\mathcal{C}$-internal homs, which are distinct from $Hom_\mathcal{C}(M,M)$!), which will be an algebra in $\mathcal{C}$, and you can plug into the above.

You should definitely read Ostrik's and other papers by Etingof, Nikshych, and Ostrik about fusion and finite tensor categories.

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David, thanks for a great answer! I looked at the paper by Ostrik in part (I will need a bit more time since I know very little about tensor categories in general), but do you mean $H \rtimes A$ instead of $A \rtimes H$? I don't see how to get the action of $A$ on $H$. – Akhil Mathew Mar 13 '10 at 21:12
Akhil, You're right that the Hopf algebra H acts on A, not vice versa. Unfortunately, there are two conventions for the symbol $\rtimes$. The one I have used is the one which generalizes a common usage for groups, so that the $\triangleleft$ points towards the normal subgroup. This is also what Majid uses in his textbooks about Hopf algebras, I think. But you're right that many authors use the opposite convention. Oh well... =] – David Jordan Mar 13 '10 at 22:14
Ok, thanks! . – Akhil Mathew Mar 16 '10 at 2:01
Another good source about this is Majid: Foundations of quantum group theory, Chapter 9.4.2 braided reconstruction. The basic assumption is that the functor $V\mapsto Nat(V\otimes F,F)$ for a fibre functor $F\colon \mathcal{C}\to \mathcal{B}$ is representable. The representing object will serve as an algebra, and given more structure on $\mathcal{C}$ preserved by $F$, it will have more structure. Namely, if $\mathcal{C}$ is monoidal, it is a bialgebra; if $\mathcal{C}$ has duals, a Hopf algebra object in the category $\mathcal{V}$. This is in some sense an application of the Barr-Beck Theorem. – Zahlendreher Feb 11 '14 at 9:41
There is in general a factorization of $F$ through $B$-Mod$(\mathcal{V})\to \mathcal{V}$ but this doesn't need to be an equivalence. – Zahlendreher Feb 11 '14 at 9:42

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