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Simon Burton (Aug 08 2022 at 03:54):I seem to be confused about module objects (over a monoid object) in a monoidal category, and the monoidal (tensor) product of these.
Working in the cartesian category of sets, "modules" are monoid actions, and tensoring these involves copying, aka the coproduct. This story seems to carry over to the linear case, which is about modules over a bimonoid (or Hopf algebra). I'm hoping this is all correct so far, and includes some other important examples, such as tensoring of algebras over a monad. (This is part of my (foolish) mission to rewrite all of mathematics using string diagrams, and all this monoid business seems to be the "killer app" of string diagrams...)
My confusion arises because there is another definition of tensor product of modules, which uses a coequalizer to force the action of the monoid object on the two tensor factors to be equal. I'm hoping there is some slick explanation of why there are these two definitions: one using the coproduct, and one using a coequalizer, and why/why don't they result in the same tensor product. Is it something to do with enriched categories?
Zhen Lin Low (Aug 08 2022 at 11:21):I don't know of any definition of tensor product that only involves the coproduct. It is a theorem that in the category of commutative rings (and more generally commutative monoids), the underlying object (i.e. abelian group in the case of rings) of the coproduct is the monoidal product of the underlying objects.
Zhen Lin Low (Aug 08 2022 at 11:34):The enriched category theory perspective is that the tensor product of (bi)modules is a special case of a weighted colimit of base-valued functors ... a.k.a. profunctors. The tensor product is profunctor composition.
Simon Burton (Aug 09 2022 at 03:13):Sorry, I should have said: 'Working in the cartesian category of sets, "modules" are monoid actions, and tensoring these involves copying, aka the comultiplication.'
Zhen Lin Low (Aug 09 2022 at 03:21):I don't see how. The tensor product of a left monoid action with a right monoid action is entirely analogous to the tensor product over a ring of a left module with a right module. In particular it cannot involve comultiplication because rings do not have comultiplication.
Simon Burton (Aug 09 2022 at 04:58):Ok, thanks for your patience while I try to describe this better. The cartesian category of sets has a comultiplication which is the diagonal map. A bimonoid (or bialgebra) in this category is a (compatible) monoid and a comonoid, which turns out to be the same as just a monoid. Modules here are just monoid actions. "Tensor" of these "modules" is just cartesian product of the sets along with an action that crucially involves the diagonal (comultiplication) map.
Matteo Capucci (he/him) (Aug 09 2022 at 07:47):Simon Burton said:
"Tensor" of these "modules" is just cartesian product of the sets along with an action that crucially involves the diagonal (comultiplication) map.
Can you make this explicit?
Morgan Rogers (he/him) (Aug 09 2022 at 08:02):requires copying .
Zhen Lin Low (Aug 09 2022 at 09:04):That's not what I understand by tensoring two monoid actions.
Morgan Rogers (he/him) (Aug 09 2022 at 09:27):Indeed, Simon's confusion is between tensoring of modules for a bimonoid and tensoring of monoid actions or modules. The latter defines a tensor product which is distinct from the tensor product in the base category of sets (the tensor product used to define the monoids). The former is a construction which only requires a single type of product, using the comultiplication map of the bimonoid in order to define the module structure on the tensored modules.
Morgan Rogers (he/him) (Aug 09 2022 at 09:29):I'm pretty sure that the answer is that these cannot be reconciled, since as you've observed @Simon Burton, the tensoring of modules for a bimonoid produces a (cartesian) product of monoid actions in this special case, which is distinct from what is usually understood as the tensor product of monoid actions (which in general just produces a set, but can be made to produce a new monoid action when the monoid is commutative).
Simon Burton (Aug 09 2022 at 09:52):What is the usual tensor product of monoid actions? The one involving a coequalizer?
Simon Burton (Aug 09 2022 at 09:57):Ok, well thanks everyone for the discussion, I think this is making more sense now. I'm going to try to think about what is the "coeqalizer tensor" product of two G-sets.. In general this is going to give something degenerate, but if the orbits (of the two actions) agree somehow they will survive the coequalizer process.
Zhen Lin Low (Aug 09 2022 at 09:57):The usual tensor product of monoid actions is the one that coequalises the input actions, yes. The result has no monoid action, btw.
Simon Burton (Aug 09 2022 at 09:58):Oh, i see.. yes there's no action left after going through a co-equalizer.. hmm.
Zhen Lin Low (Aug 09 2022 at 10:02):I'm not sure what the usual name for the other product of modules over a bimonoid is... I know it has its uses in representation theory, but I think they are only distinguished by the subscript on the operator.
Simon Burton (Aug 11 2022 at 11:32):I just found this very helpful discussion in [KV1994] which goes over what we have been discussing image.png.
Simon Burton (Aug 11 2022 at 11:36):Given a group we can form its group algebra and then the tensor product of modules over this algebra is (the co-equalizer version), and I'm guessing this is the tensor product we use to get the tensor-hom adjunction to work!