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Let's suppose that we have an algebra $\mathcal{A}$ (I don't really care whether it's a unital one or associative). From studying Lie algebras and some of their generalizations, I am used to be able to consider tensor representations. However, their existence is tied to the fact that such algebras naturally have a comultiplication.

Are there any cases where we cannot form tensor products from representations of algebra and what is the obstruction for existence of comultiplication for them?

The only thing that I understand so far is that these algebras probably ought to have only infinite-dimensional representations, since we can easily multiply finite-dimensional matrices.

EDIT:

My original goal is rather humble: I want to understand why the associator (or 6j-symbols) in the universal (quantum) Lie algebras' representations is nontrivial, but pentagonator turns out to be trivial; then I want to think about how this might be generalized to n-associator conditions (in the context of representation theory). One answer (which I read in Deligne-Milne) is that the axioms for tensor categories are such that associator isomorphism exists (might be trivial, might be not) and the only relations are pentagon eqation (i.e. pentagonator is trivial) and hexagon equation. However, I don't understand the reasoning behind these axioms (for example, once we consider 2-groups, I bet these axioms will be wrong if one considers their reresentations, whatever they may be; so it seems that Deligne-Milne have a specific example in their heads, like finite/Lie groups or Lie algebras).

Now, there is another story, this time about quasi-Hopf algebras (which I read in Drinfeld, but didn't understand much as well), where one basically relaxes some conditions for Hopf algebra, for example, $(\mathrm{id} \otimes \Delta) \Delta = (\Delta \otimes \mathrm{id}) \Delta$ now holds up to an isomorphism, and still gets a rich situation to study from representation theory point of view. This way of thinking seems connected to the part of my original goal, concerning the n-associator. However, first, Drinfeld still has only pentagon to satisfy and, second, I still don't understand what forces associator to be nontrivial. In order to understand associator, seemingly simpler thing, first, I tried to think what might force it to become trivial or forbid its existence at all. Hence the question in the title.

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  • $\begingroup$ When you typed nital, did you mean unital? $\endgroup$ Commented Mar 5, 2021 at 21:29
  • $\begingroup$ @J.W.Tannerm yeah, I am sorry for the mistake $\endgroup$
    – user108687
    Commented Mar 5, 2021 at 21:52
  • $\begingroup$ Could you be a bit more precise? Note that if $A$ is any $\Bbbk$-algebra and $M$, $N$ are left $A$-modules then $M \otimes_{\Bbbk} N$ is an $A$-module with action $a\cdot (m \otimes n) = (a\cdot m) \otimes n$. However, what underlies the existence of a comultiplication and a counit on $A$ that make of it a bialgebra is that the category of left $A$-modules is monoidal and the forgetful functor to $\Bbbk$-vector spaces is strict monoidal. The second condition is important. For instance, since there cannot be any bialgebra structure on $Mat_2(\mathbb{C})$, this already provides an example. $\endgroup$ Commented Mar 13, 2021 at 16:02
  • $\begingroup$ @EnderWiggins I edited the question to add some more details. Unfortunately, I am not good at category theory at all and don't understand your resoaning. So could I ask to clarify the examples? When you talk about $a\cdot(m\otimes n) = (a\cdot m)\otimes n$ you mean that this doesn't work well, for example, when, say, $A$ is $\mathfrak{sl}(2)$, $m$ is a trivial representation and $n$ is a whatever one or something else (and in your language is the module not-monoidal or functor not strict monoidal)? And when you talk about $Mat_2(C)$, what algebra structure do you have in mind? $\endgroup$
    – user108687
    Commented Mar 14, 2021 at 2:47
  • $\begingroup$ @user108687 concerning your question about $\mathfrak{sl}(2)$ (which I suppose is the Lie algebra of traceless $2 \times 2$ complex matrices), the point is exactly that the "naive" representation $\rho(x)(m \otimes n) = \rho(x)(m) \otimes n$ makes of $M \otimes N$ an $\mathfrak{sl}(2)$-representation, but the obvious isomorphism $\mathbb{C} \otimes N \cong N$ (for $M = \mathbb{C}$) is not an isomorphism of $\mathfrak{sl}(2)$-representations. $\endgroup$ Commented Mar 14, 2021 at 13:19

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