I have a monad on an ind-category (specifically, my ind-category has a monoidal structure and I have an algebra object, so the monad is tensoring with it). It would be very useful in my work if the Eilenberg-Moore category of the monad it itself an ind-category. Is there a known criterion for the monad for this to be true? Further, is there a handy description for a generating sub-category of modules? I'm also happy to use the dual case of a comonad on a pro-category if there's a break in symmetry that makes that an easier question.
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$\begingroup$ I would guess that the conjecture for monads on ind categories is correct (exactly) when the functor part of the monad preserves filtered colimits. Dually with comonads on pro categories when the functor preserves cofilteed limits. Maybe even comonads on ind categories if filtered colimits are preserved. $\endgroup$– Paul TaylorCommented Mar 16, 2022 at 13:28
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3$\begingroup$ It is certainly true that the Eilenberg–Moore category of a finitary (i.e. filtered-colimit preserving) monad on an accessible category (i.e. Ind-category) is itself accessible (i.e. an Ind-category). $\endgroup$– varkorCommented Mar 16, 2022 at 15:48
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$\begingroup$ @varkor you should write this as an answer. I don't think there is a better criterion. $\endgroup$– Simon HenryCommented Mar 16, 2022 at 15:56
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Let $T$ be a monad on an accessible category (i.e. an $\mathbf{Ind}$-category). If the underlying endofunctor of $T$ is finitary (i.e. preserves filtered colimits), then the Eilenberg–Moore category of $T$ is also accessible. This is proven (in slightly more generality) in Theorem 2.78 of Adámek–Rosický's Locally presentable and accessible categories, for instance.
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$\begingroup$ @SimonHenry: I realised I made a mistake after posting the comment. What I meant was that I suspect in this case the Eilenberg–Moore category is the $\mathbf{Ind}$-completion of the completion under coequalisers of T-algebra homomorphisms, of the subcategory of the Kleisli category of the monad on the finitely presentable objects. $\endgroup$– varkorCommented Mar 16, 2022 at 16:37
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$\begingroup$ Thank you very much for that (and for the helpful reference). The suggestion for the generating category in your comment is also a useful direction to pursue. $\endgroup$ Commented Mar 17, 2022 at 17:18