Fix a field $k$, and let $A$ be a (commutative, coassociative, counital) $k$-bialgebra. Write $\otimes = \otimes_k$. The category $\mathrm{Mod}_A$ of $A$-modules admits the structure of a monoidal category, symmetric if $A$ is cocommutative, with underlying tensor product $\otimes$: for $M,N\in\mathrm{Mod}_A$, we have an action on $M\otimes N$ given by

$$ a\cdot (m\otimes n) = \sum_a a_{(1)} m\otimes a_{(2)} n.$$

Here, I use Sweedler's notation $\Delta(a) = \sum_a a_{(1)}\otimes a_{(2)}$ for the coproduct on $A$. From this monoidal structure, we can recover the bialgebra structure on $A$ by

$$\Delta(a) = a\cdot (1\otimes 1),$$

where $1\otimes 1\in A\otimes A$. My question is:

  • Suppose given a monoidal structure on $\mathrm{Mod}_A$ with underlying tensor product $\otimes$. When does this arise from the structure of a bialgebra on $A$?

There is also the converse question:

  • What coproducts $A\rightarrow A\otimes A$ arise from such monoidal structures on $\mathrm{Mod}_A$?

If there is a good answer to the first question, then I would also be interested in detecting when such a monoidal structure arises from a bialgebra with the property of being a Hopf algebra.

The question boils down to: given a monoidal structure on $\mathrm{Mod}_A$ with underlying monoidal product $\otimes$, what natural conditions ensure that $\Delta(a) = a \cdot (1\otimes 1)$ gives $A$ the structure of a bialgebra? For example, if $A\otimes A$ is an $A-A\otimes A$-bimodule, then $\Delta(ab) = \Delta(a)\Delta(b)$, but it is not clear to me what natural conditions force this to hold and if anything can be said if it does not.

  • $\begingroup$ Why do you require that $A$ be commutative? $\endgroup$ – Qiaochu Yuan Jul 18 '18 at 20:06
  • $\begingroup$ Also, for starters see ncatlab.org/nlab/show/sesquialgebra. A necessary condition is that $\otimes$ is $k$-linear and cocontinuous in both variables, which gets you a sesquialgebra. $\endgroup$ – Qiaochu Yuan Jul 18 '18 at 20:13
  • $\begingroup$ @QiaochuYuan because my familiarity with bialgebras stems entirely from the commutative case. I would be happy with an answer not assuming that $A$ is commutative, but I can only trust myself to write correct things when it is. (For instance, I don't recall offhand if being a Hopf algebra is a property of a noncommutative bialgebra, but I do know that it is and why it is in the commutative case.) $\endgroup$ – ne- Jul 18 '18 at 20:14
  • $\begingroup$ That page certainly looks relevant; I'll look closer at it. $\endgroup$ – ne- Jul 18 '18 at 20:15
  • 1
    $\begingroup$ Yes, the antipode remains unique, but the proof is more complicated. It's just that in the commutative case it's more typical to look at comodules (for Hopf algebras this corresponds to representations of affine schemes), and on the other hand the motivating examples of the tensor product construction you describe are noncommutative, namely group algebras and universal enveloping algebras of Lie algebras. $\endgroup$ – Qiaochu Yuan Jul 18 '18 at 20:41

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