An interesting analogy between algebras over a ring and promonads on a category
is formalized using the apparatus of double theories.

This is a companion discussion topic for the original entry at https://www.epatters.org/post/algebras-are-promonads
1 Like

Very nice post, Evan! Actually I should say very nice… last two years? You’re churning out amazing work in double category theory.

A question: I was under the impression a *promonad* in \mathbb D was a (loose) monad in \bf \mathbb Mod(\mathbb D)?

I can’t really sit down and check now but it’d be cool if this was equivalent to your definition and hinted at the fact \mathbb T_{\sf promnd} = 1 \boxtimes 1 for \boxtimes a kind of Gray product. In fact we’d have: \bf \mathbb Lax(1, \mathbb Mod(\mathbb D)) = \mathbb Lax(1, \mathbb Lax(1, \mathbb D)) = \mathbb Lax(1 \boxtimes 1, \mathbb D) = \mathbb Lax(\mathbb T_{\sf promnd}, \mathbb D).

However, usually 1 is the unit for Gray product so maybe this is just wishful thinkng.

Anyway, recently I was contemplating the equivalence between promonads on \sf C and ioo functors \sf C \to P and I was reminded of the classification results for actegories, but also for algebras of a ring, where you show that an action of some kind is equivalent to a map of some kind from the actor to the actee.

Perhaps this is the setting in which one can prove the father of all classification theorems?

Specifically, for algebras and actegories, this is interesting insofar as it gives a universal property to various kind of centers.

1 Like

Thank you, Matteo, you are too kind!

I agree that a promonad in \mathbb{D} is a (loose) monad in \mathbb{M}\mathsf{od}(\mathbb{D}). So that’s another characterization, and perhaps the most succinct one. Nice!

Your other remarks are also interesting. I’ll have to think more about them.

1 Like