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Hopf algebras and bialgebras are sometimes introduced by saying that you've got an associative algebra $A$ and want to introduce the structure of an $A$-module on $V \otimes W$ where $V,W$ are $A$-modules (say, finite-dimensional) and $\otimes$ is the tensor product of the underlying vector spaces. My question is essentially about the converse of this idea.

Question 1. Let $A$ be an associative algebra over a field. Suppose that the tensor product of vector spaces gives a symmetric monoidal structure on $Rep(A)$. Is there necessarily a bialgebra structure on $A$ such that the comultiplication induces the $A$-module structure on $V \otimes W$ in the usual way?

I have in mind finite-dimensional representations, but more general answers would also be interesting. I'd also be interested in more specific answers - say when $A$ is semisimple, or is finite-dimensional, or Artinian, etc.

(I suppose it would also be interesting if this forced a bialgebra structure on $A$ even if the comultiplication didn't induced the same module structure on $V \otimes W$, but I have a hard time imagining that.)

I don't really know about other symmetric monoidal structures on $Rep(A)$ (even for specific $A$), but I suppose the following would also be interesting:

Question 2: Suppoes $Rep(A)$ has any symmetric monoidal structure (not necessarily the "usual" tensor product). Is there necessarily a bialgebra structure on $A$?

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    $\begingroup$ Take $V=W=A$, so you get the map $A \to A\otimes A$, $a \mapsto a(1\otimes 1)$. I'm guessing the symmetric monoidal axioms are going to tell you that's a ring homomorphism (but haven't checked at all). $\endgroup$ – Allen Knutson Oct 6 '13 at 16:19
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    $\begingroup$ You need to be careful what you mean by "the tensor product of vector spaces gives a symmetric monoidal structure." What you want to say is that there exists a symmetric monoidal structure which, after being hit with the forgetful functor, is the tensor product of vector spaces. $\endgroup$ – Qiaochu Yuan Oct 6 '13 at 16:29
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    $\begingroup$ @Allen: I suspect the symmetric monoidal axioms are only going to give you weak compatibility; see en.wikipedia.org/wiki/Weak_Hopf_algebra . $\endgroup$ – Qiaochu Yuan Oct 6 '13 at 16:58
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    $\begingroup$ @AllenKnutson: I thought of that, but couldn't quite get it to work. I think I agree with Peter Samuelson's guess, that all this leads to is an algebra $A$ such that $A \otimes A$ is an $A - A \otimes A$-bimodule that satisfies an associativity condition... $\endgroup$ – Joshua Grochow Oct 6 '13 at 18:41
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The answer to Question 2 is no. An obstruction to this being true is that $A$ can't be recovered from $\text{Rep}(A)$; we can only recover the Morita equivalence class of $A$. In particular, a monoidal product on $\text{Rep}(A)$ can be induced from a comultiplication on any algebra Morita equivalent to $A$.

For example, if $A$ is semisimple over $\mathbb{C}$ (for simplicity), then $\text{Rep}(A)$ is completely determined by $\dim Z(A)$, and in particular $A$ is Morita equivalent to $\mathbb{C}[G]$ where $G$ is any finite group with $\dim Z(A)$ conjugacy classes. Any choice of such a group gives a symmetric monoidal structure on $\text{Rep}(A)$, and these won't arise from comultiplications on $A$ in general. To be very explicit, let $A = \mathbb{C}^n$. Then any comultiplication $A \to A \otimes A$ is necessarily induced by a map of sets $[n] \times [n] \to [n]$, from which it follows that the induced tensor product has the property that a tensor product of simple representations is simple. But this isn't true for tensor products induced on representation categories of groups in general.

Monoidal products on $\text{Rep}(A)$ can be induced by even more exotic data, e.g. we could use a "comultiplication" of the form $A' \to A'' \otimes A'''$ where $A', A'', A'''$ are three different algebras Morita equivalent to $A$, or we could use an $(A, A \otimes A)$-bimodule $M$. (This exotic data still needs to satisfy suitable compatibility relations.) See also sesquialgebra.

Edit: Oh, and of course $A$ could be commutative and the monoidal product could be the tensor product over $A$!

For a positive result along these lines, Etingof, Nikshych, and Ostrik showed that any fusion category is the category of representations of a weak Hopf algebra (but here we don't require a symmetry).


Edit #2: Some thoughts on Question 1. If $V, W$ are $A$-modules then the tensor product $V \otimes W$ (the underlying field is suppressed) naturally has the structure of an $A \otimes A$-module. This gives a functor $\text{Rep}(A) \times \text{Rep}(A) \to \text{Rep}(A \otimes A)$. Let me slightly change my interpretation of "tensor product of vector spaces gives a symmetric monoidal structure" to "the monoidal structure factors as

$$\text{Rep}(A) \times \text{Rep}(A) \to \text{Rep}(A \otimes A) \xrightarrow{F} \text{Rep}(A)$$

and $F$ preserves the forgetful functor to $\text{Vect}$."

The forgetful functor is faithful and preserves colimits, so any such $F$ should also preserve colimits. Then by the Eilenberg-Watts theorem $F$ must be tensor with an $(A, A \otimes A)$-bimodule $M$. Compatibility with the forgetful functor implies that, as a right $A \otimes A$-module, $M \cong A \otimes A$, so our functor must be induced by a left $A$-module structure on $A \otimes A$. I don't think we can conclude that this comes from an algebra homomorphism $A \to A \otimes A$, but I don't know any counterexamples.

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I think the answer to (2) is "no." If $A = \mathbb C \langle x,y\rangle / ([x,y] = 1) $ is the Weyl algebra, then the category of $A$-modules has a tensor product given by $M \odot N := M \otimes_{\mathbb C[x]} N$ as a $\mathbb C[x]$-module, with the action of $y$ given by the Leibniz rule: $y\cdot m \odot n = ym \odot n + m\odot yn$. But $A$ is simple and so can't have an algebra map to $\mathbb C$, so it can't have a counit. And I'm pretty sure this monoidal structure doesn't come from any algebra map $A \to A\otimes A$ (but I don't have a proof).

For (1), Allen's suggestion sounds like the way to go, but I think there's a chance that the statement you get is "$A$ is a Hopfish algebra with $\Delta = A\otimes A$" instead of "$A$ is a bialgebra." (A Hopfish algebra is an $A-(A\otimes A)$-bimodule $\Delta$ that satisfies an associativity isomorphism, and such a bimodule gives the category of $A$-modules a monoidal product. See this paper.)

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I think that the answer to (1) is likely no in the general case where we consider finite-dimensional $A$-modules with an infinite-dimensional $A$. In this context, Allen's suggestion can't work simply because $A$ doesn't live in $\operatorname{Rep}(A)$!

The reason is that we typically can't reconstruct an infinite-dimensional algebra from its finite-dimensional representations. For instance, the algebra $A = k[x_1, x_2, \ldots] / (x_1^2 = 0, x_{i + 1}^2 = x_i)$ has no nontrivial finite-dimensional representations whatsoever! Of course, there is the usual symmetric monoidal structure on $\operatorname{Rep} A = \operatorname{Vec}$, so any bialgebra structure on $A$ will work in this case (but are there any?).

Something like Allen's suggestion should work, for purely formal reasons, for either finite-dimensional $A$ or if we consider the larger category of all $A$-modules. The dual of your question (where we ask for an algebra structure on a coalgebra inducing a given symmetric monoidal structure on the category of coalgebras) is more or less the classical theory of Tannaka duality, which gives an affirmative answer.

EDIT: I fixed the algebra $A$ to make it actually have no finite-dimensional representations. Unfortunately, this algebra does admit a bialgebra structure. I'll try to think of an example of an $A$ that has $\operatorname{Vec}$ as its category of finite-dimensional representations but provably admits no bialgebra structure.

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  • $\begingroup$ Your general point is well-taken, but I don't see that your example has no finite-dimensional representations. Can't I take any invertible finite-dimensional matrix (take $k=\mathbb{C}$ here), then take its square root, then it's square root, etc. to get a representation of your example algebra? $\endgroup$ – Joshua Grochow Oct 6 '13 at 20:44
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    $\begingroup$ There are also examples, like the Weyl algebra, that have no finite-dimensional representations. And as Quiochu pointed out, if $A$ is noncommutative, you can only recover $A$ up to Morita equivalence from its category of representations. (If $A$ is commutative then it is determined by its category of representations.) $\endgroup$ – Peter Samuelson Oct 6 '13 at 21:28
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If $A$ is an associative algebra and the category of $A$-modules is equipped with a monoidal structure such that 1) the forgetful functor to vector spaces is monoidal, and 2) the tensor product bifunctor is representable by an $A$-$A \otimes A$-bimodule, then $A$ is a bialgebra. As, condition 1) forces the trimodule $M$ which represents the tensor product to be isomorphic to $A \otimes A$ as a right $A \otimes A$-module. The left action of $A$ on this module must commute with the right action of $A \otimes A$, and so must be given by left multiplication by elements of $A \otimes A$. In this way we find an algebra homomorphism $A \rightarrow A \otimes A$. It is straightforward to check coassociativity.

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