# Role of fiber functor monoidal structure in Tannakian bialgebra reconstruction

Shahn Majid presents in section 9.4 of "Foundatations of Quantum Group Theory" a Tannaka-type reconstruction theorem for producing a $k$-bialgebra out of its $k$-linear monoidal category of modules $\mathcal{M}$. The theorem depends on the presense of a fiber (fogetful) functor $F: \mathcal{M} \rightarrow Vect_k$ and monoidal structure on $F$. The question is:

Is there a stronger reconstruction theorem in which $F$ carries no monoidal structure?

If not, it means there should be an example with different monoidal structures on $F$ yielding different coproducts in the bialgebra. Is there such an example?

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My guess is the answer is no. Specifically, I'd guess that there exist really different fibre functors that become isomorphic when you forget their monoidal structures. For example, two fibre functors send an object to vector spaces of the same dimension, so they become equal on objects when you replace the category of vector spaces with its skeleton. Perhaps this can be pushed further to show that sometimes (often? always?) two fibre functors will become isomorphic when you forget their monoidal structures. – anon Feb 15 '12 at 8:04

Etingof and Gelaki (and others) constructed examples of isocategorical groups, that is non-isomorphic finite groups $G$ and $H$ such that the representations categories $Rep(G)$ and $Rep(H)$ (over the complex numbers) are equivalent as monoidal categories. Now if you use the usual fiber functor $Rep(G)\to Vect$ you will reconstruct the group algebra of $G$ (or its dual). And if you use the composition $Rep(G)\simeq Rep(H)\to Vect$ you will reconstruct the group algebra of $H$.

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This doesn't quite answer my question since I have only 1 fiber functor but two monoidal structures on it – Squark Feb 15 '12 at 6:27
As underlying functors these two functors agree, as they're both just forgetting the group action. So I think it does answer your question. – Noah Snyder Feb 18 '12 at 16:11
I don't see why the composed functor is "just forgetting the group action". The equivalence between Rep(G) and Rep(H) might be some completely nontrivial functor – Squark Feb 18 '12 at 16:49
Sorry for not being clear. The functors in question are indeed isomorphic and the difference is in monoidal structure. Namely, both functors send a simple object $X$ to vector space of dimension equal to its Frobenius-Perron dimension (i.e. Frobenius-Perron eigenvalue of multiplication by class of $X$ in the Grothendieck ring of the category). The functors between semisimple categories are determined up to isomorphism by their values, which implies the claim. – Victor Ostrik Feb 18 '12 at 17:08
Thx. Can you provide a reference please? – Squark Feb 19 '12 at 20:06

The most basic form of Tannaka duality goes as follows:

Given a $k$-coalgebra $C$, we can construct its category of finite dimensional comodules ("representations"), which carries a (faithful, exact) functor to the category of finite dimensional vector spaces. On the other hand, given a $k$-linear abelian category equipped with a functor (the "fiber functor") to the category of finite dimensional vector spaces, we can construct a (not necessarily finite dimensional) $k$-coalgebra using a certain coend formula. Tannaka duality says that these two functors give an adjunction between the category of $k$-coalgebras and the 2-category of $k$-linear abelian categories equipped with a fiber functor. The counit of this adjunction is an isomorphism (so that we can reconstruct the coalgebra from its category of representations), and the unit of this adjunction is an equivalence precisely when we restrict to categories for which the fiber functor is faithful and exact (so that we can recognize such categories and fiber functors as being the ones that are categories of representations).

Note that the above result is entirely trivial (i.e., follows from the Yoneda lemma) if we take the category of not-necessarily finite dimensional comodules or require that our coalgebras be finite dimensional. On the other hand, the above result is entirely false if we replace coalgebras and comodules with algebras and modules. (Exercise: Find a nontrivial algebra that has no nontrivial finite dimensional modules. It's pretty easy!)

The adjunction I described can be "souped up" to a monoidal adjunction to handle representations of bialgebras; the algebra structure corresponds to the monoidal structure on the category. But the same caveats apply, so the theorem you state about monoidal categories is either trivial or wrong for the above reasons. The "correct" statement of monoidal Tannaka duality (and I believe more or less the one stated by Deligne) involves on the one side commutative Hopf algebras and on the other side symmetric monoidal $k$-linear abelian categories equipped with a faithful, exact, symmetric monoidal functor to finite-dimensional vector spaces. This is just a symmetric monoidal-ification of the basic form I give above.

A good reference is this paper by Daniel Schäppi (particularly the opening section), and its successor.

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Indeed, there is a symmetric monoidal 2-category whose objects are coassociative coalgebras in Vect, morphisms are homomorphisms, and natural transformations are coconjugations (njugations?). If you think of a coalgebra as an (Ab^op)-enriched category with one object, then the homomorphisms are (opposite to?) (Ab^op)-enriched functors, and the 2-morphisms are the natural transformations from this perspective. Anyway, the adjunction and equivalence that you describe are actually a symmetric monoidal adjunction/equivalence between this symmetric monoidal 2-category and the symmetric monoidal ... – Theo Johnson-Freyd Feb 15 '12 at 1:34
... 2-category whose objects are linear abelian categories with a (faithful exact) left-exact functor, where the tensor structure on this side is Deligne's tensor product of abelian categories (the representing abelian category for functors of two variables that are left-exact in each variable; also (!) the representing object for functors of two variables that are exact in each variable). So, monoidal=E_1 categories go to E_1-objects in Coalgebras, and so on. @Evan, I think you know all this (I might have learned some of it from you), but I figured I'd mention it. +1 for the answer. – Theo Johnson-Freyd Feb 15 '12 at 1:37
Thx Evan but I still don't know the answer to my simple yes/no + example question. Note that I don't restrict the modules to be finite dimensional – Squark Feb 15 '12 at 6:36

To give a different answer to your meta-question, you can certainly have a non-monoidal fiber functor from C -> Vec where C is not the category of representations of any bialgebra. Just take the category of twisted G-graded vector space Vec(G,w) where w is a nontrivial 3-cocycle. This is not the category of representations of any bialgebra, but it has an obvious non-monoidal fiber functor to vector spaces. (It is, however, the category of representations of a quasi-bialgebra.)

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This is interesting. What if I allow for quasibialgebras too? – Squark Feb 18 '12 at 17:01

The answer to your question is no. Take a finite set S with two non-isomorphic monoid structures (for example, the four element set with its two group structures). Let $k^{(S)}$ be the algebra of $k$-valued functions on $S$ with pointwise multiplication. Both monoid structures on $S$ induce a comultiplication on $k^{(S)}$. Hence they define two monoidal structures on the category of $k^{(S)}$-modules, together with strong monoidal structures on the forgetful functor $F$ to vector spaces. Thus, if you forget about the monoidal structure of $F$ you can reconstruct $k^{(S)}$ as an algebra, but you won't be able to reconstruct the comultiplication.

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@Daniel, you need to prove there is a monoidal equivalence of the resulting categories commuting with the forgetful functor – Squark Feb 18 '12 at 11:28
Ah, I misread your question. I didn't realize that you want the monoidal structure on the domain category to be fixed, and to only change the strong monoidal structure of the forgetful functor. You might want to clarify that in the original question. – Daniel Schäppi Feb 18 '12 at 12:07