# Does any tensor category correspond to a bialgebra?

I wonder how strong the power of Tannaka philosophy is, and if we accept that a tensor category is a generalized bialgebra, what difficulties we will come up against ?

Edit: Whether most tensor categories are representable, or whether for every "good enough" tensor category there exist a bialgebra with its module category isomorphic to the given category?

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Can someone with enough rep edit the typo in the title please? –  Jacques Carette Feb 28 '10 at 14:26

I'd like to explain Bruce's answer a bit more. The fusion categories Bruce mentioned have non-integer Frobenius-Perron dimensions, so it is very easy to see that they are not categories of finite dimensional modules over a bialgebra. E.g. one of the simplest of them, the so called Yang-Lie category, has simple objects $1,X$ with $X^2=X+1$. So if $X$ were a finite dimensional representation of a bialgebra, then the dimension of $X$ would be the golden ratio, which is absurd.

This, however, can be fixed if we allow weak bialgebras and weak Hopf algebras. In fact, any fusion category is the category of modules over a finite dimensional weak Hopf algebra, see arXiv.math/0203060.

As to Akhil's example (Deligne's categories), it is also true that they cannot be realized as categories of finite dimensional representations of a bialgebra (or even a weak bialgebra), but for a different reason. Namely, if X is a finite dimensional representation of a bialgebra, then the length of the object $X^{\otimes n}$ is at most ${\rm dim}(X)^n$, where ${\rm dim}$ means the vector space dimension. But in Deligne's categories, the length of $X^{\otimes n}$ grows faster as $n\to \infty$. Actually, in another paper, Deligne shows that if in a symmetric rigid tensor category over an algebraically closed field of characteristic zero, the length of $X^{\otimes n}$ grows at most exponentially, then this is the category of representations of a proalgebraic supergroup, where some fixed central order 2 element acts by parity (so essentially this IS the category of (co)modules over a bialgebra). This is, however, violently false in characteristic $p$, since if the root of unity $q$ is of order $p$, where $p$ is a prime, the the fusion categories for $U_q({\mathfrak g})$ mentioned by Bruce admit good reduction to characteristic $p$, which are semisimple symmetric rigid tensor categories with finitely many simple objects and non-integer Frobenius-Perron dimensions.

A third very simple example of a tensor category not coming from a bialgebra is the category of vector spaces graded by a finite group $G$ with associator defined by a nontrivial $3$-cocycle. This category, however, is the category of representatins of a quasibialgebra (and also of a weak bialgebra, as mentioned above).

So the conclusion is as in the previous two answers: tensor categories are more general than bialgebras. More precisely, the existence of a bialgebra for a tensor category is equivalent to the existence of a fiber functor to vector spaces, which is an additional structure that does not always exist. And if it exists, it is often not unique, so you may have many different bialgebras giving rise to the same tensor category.

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So a tensor category with a fiber functor can give us a bialgebra defined to be the endomorphisms of the fiber functor, and various fiber functors give different bialgebra, and they are morita equivalent . but if we consider the endomorphisms of identical functor which still a algebra /bialgbra,there may be some realation between the two algebra. –  Xuexing Lu Mar 1 '10 at 4:46
The bialgebras coming from different fiber functors are twist equivalent, i.e., isomorphic as algebras, and the coproducts are related by a twist (Morita equivalence is not quite the right term here). I did not understand the last sentence of your comment, though. –  Pavel Etingof Mar 1 '10 at 12:30
Twist equivalent sounds interesting! The endomorphisms of identity functor or the natural transformations from identity functor to itself form a algebra.Is it isomorphic to the algebra of endomorphisms of the bialgebra ? Is the isomorphism canonical in some sense ? –  Xuexing Lu Mar 1 '10 at 15:49
No, if ${\mathcal C}$ is the category of modules over an algebra $A$ then the endomorphism algebra of the identity functor of ${\mathcal C}$ is the center of $A$. In particular, this is the case for tensor categories and bialgebras. –  Pavel Etingof Mar 1 '10 at 18:41

The basic example is to take finite dimensional representations of $U_q(sl(2))$ and then put q a root of unity. Then quotient by the ideal of "invisible morphisms" (where f is invisible if Tr(fg)=0 for all g). This is semisimple, abelian, braided, spherical and has a finitely many isomorphism classes of simple modules.

Of course you can then repeat for $U_q(g)$ for g a semisimple Lie algebra.

Even more generally you can take the fusion category of a rational conformal field theory.

These are the examples that have been most studied and which led people to regard tensor categories as more general than bialgebras.

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Can you give any references for this construction? Google searches for "invisible morphisms" and "invisible morphism" bring up nothing but your own comments on MathOverflow! :) Is this quotienting process described in any of the standard quantum group texts? I've had a look but I can't find anything. –  Jamie Vicary Apr 8 '10 at 15:27
Jamie you want to google for "negligible tilting modules" or something like this. –  Marcel Bischoff Apr 21 at 14:40

Edit The example given here of Deligne's categories is a valid example of tensor categories that do not arise from a generalized bialgebra. However, for the correct reason, see Pavel Etingof's answer above.

An example is Deligne's Rep(S_t) for t not an integer; this is a semisimple symmetric tensor category, but it contains an object (the "regular representation" in non-integral rank) of dimension $t$. There is a blog post by David Speyer on the subject.

There is actually a more natural way of getting a tensor category with objects of non-integral dimension (that was also known earlier); this is $Rep(GL_t)$ for $t$ not an integer. It is also covered in Deligne's paper (and cf. this post by Noah Snyder). This category is something like "the free pseudo-abelian tensor category containing an object of dimension $t$", in some 2-categorical sense: any such tensor category admits a tensor functor from $Rep(GL_t)$.

(Edited in response to comments below)

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The categorical dimension is not defined in an arbitrary tensor category. You probably want to take a spherical category to define the categorical dimension. –  Bruce Westbury Feb 28 '10 at 14:07
sorry, the generalized bialgebra is not a formal terminology ,I am trying to express such an idea that whether most tensor categoryes are representable ,or whether for every "good enough" tensor category there exist a bialgebra with its module category isomorphic to the given category. –  Xuexing Lu Feb 28 '10 at 14:34
@Bruce: Ack, you're right, symmetricity is necessary. –  Akhil Mathew Feb 28 '10 at 14:41
@Akhil. For me, a tensor category is a monoidal category. There is a tensor product (functor) and associativity (natural transformation). The categorical dimension is the trace of the identity and there is a choice of left trace and right trace. You need duals to define these. I don't think a tensor functor necessarily preserves this as for quantum groups representations are vector spaces but the quantum dimension is not the dimension. –  Bruce Westbury Feb 28 '10 at 16:30
I'm sure that the OP means "monoidal k-linear category" when writing "tensor category", not some fancier notion, because the representation theory of a (biassociative, or associative and quasicoassociative) bialgebra gives one of these. But there are bialgebras whose representation theory is not symmetric or even braided, and does not have duals. –  Theo Johnson-Freyd Feb 28 '10 at 17:50