# When does transporting a monad along an adjunction "preserve" its category of modules?

If I have an adjunction $$F \dashv G$$ where $$F \colon C \to D$$, and $$(N, m, u)$$ is a monad on $$D$$, then I can define a monad $$\widetilde{N}$$ on $$C$$ via \begin{align} (GNF, \widetilde{m} = G m_F \circ GN \varepsilon_{NF}, \widetilde{u} = G u_F \circ \eta) \ , \end{align} where $$\eta$$ and $$\varepsilon$$ are the unit and counit of the adjunction. My question is:

1. When is the Eilenberg-Moore category $$D^N$$ of $$N$$ equivalent to $$C^{\tilde{N}}$$?

Maybe that's a bit too hard, a possibly simpler question is:

1. There is an obvious functor $$D^N \to C^{\widetilde{N}}$$, sending and $$N$$-module $$(d, \rho \colon Nd \to d)$$ to the $$\widetilde{N}$$-module $$(Gd, G\rho \circ GN\varepsilon_d)$$. Can we say when this functor is an equivalence?

(PS: please don't say "Take $$C = D$$ and $$F = G = id_C$$." haha)

• Do you have any interesting examples? Are you trying to generalise something? Jan 18, 2022 at 11:32
• @ZhenLin: I kind of forgot why I even had this question. I know that I had a very specific example in mind, relating functorially two comonads on two different categories, and I sort of had the feeling that their comodules were equivalent. Anyway, here should be a (non-)example: Take algebra $A$ in vector spaces, and vector space $X$ (everything finite dim). Then $XAX^*$ is an algebra, and the canonical functor from my point $2$ above seems to be fully faithful. But I think it's only an equivalence if $X$ is the ground field. (Maybe I made a mistake) Jan 18, 2022 at 12:07
• (And for completeness, I know now that my specific example with the comonads doesn't work, so no point in spelling it out here) Jan 18, 2022 at 12:14
• Not sure if it helps, but your question 2 is equivalent to asking "When is the composition of $G$ with the forgetful functor $D^N\to D$ monadic?", so you could try to apply Beck's theorem (or one of its variants, if you have a specific example in mind). Jan 18, 2022 at 23:14

Namely, let $$T$$ be a monad on a category $$C$$, and let $$N$$ be an endofunctor on $$C^T$$. Then the construction I performed above uses the free-forgetful adjunction of $$F_T \dashv U_T \colon C \to C^T$$, and yields what they call the pushforward $$(F_T, U_T)_* N := N T$$, which is an endofunctor on $$C$$. Now, if $$N$$ is a monad, the pushforward will (as I showed above) be a monad on $$C$$, and it is called the cross product $$N \rtimes T$$.
In the paper it is then said that a sufficient condition for the comparison functor \begin{align} K_{N \rtimes T} \colon (C^T)^N \to C^{N \rtimes T} \end{align} to be an isomorphism is that $$N$$ preserve reflexive coequalizers.
And as a side note, this cross product construction really is the usual one: consider a bialgebra $$B$$ over a commutative ring $$k$$, and let $$A$$ be a $$B$$-module algebra. There is the cross or smash product construction, yielding a new $$k$$-algebra $$A \rtimes B$$ such that $$A-(B-\text{mod}) = (A \rtimes B)-mod$$. Now, algebras are monads: $$B \otimes -$$ is a monad on $$k$$-modules - its modules are exactly the $$B$$-modules -, and $$A \otimes -$$ is a monad on $$B$$-modules. Then \begin{align} (A \rtimes B) \otimes - = (A \otimes -) \rtimes (B \rtimes -) \end{align} as monads, and of course \begin{align} (k\text{-mod}^{B \otimes -})^{A \otimes -} = k\text{-mod}^{(A \rtimes B )\otimes -} \end{align}