TITLE: Re-seating a monad
QUESTION [9 upvotes]: Let $\mathcal C$ and $\mathcal D$ be categories with suitable limits and colimits for the following discussion. Is it possible to re-interpret, or "re-seat" a monad $T : \mathcal C \to \mathcal C$ as a monad over $\mathcal D$? When $T$ is finitary, I know at least one way to do this. Compute the Kleisli category $\mathcal C_T$ and consider $(\mathcal C_T)^{op}$ as a Lawvere theory. Then models of $(\mathcal C_T)^{op}$ on $\mathcal C$ are the same as $T$-algebras. If we then consider models of $(\mathcal C_T)^{op}$ on $\mathcal D$, we have a canonical, monadic forgetful functor $[(\mathcal C_T)^{op},\mathcal D] \to \mathcal D$ from which we can build a new finitary monad $T' : \mathcal D \to \mathcal D$. My question is as follows.

Is there an direct way to obtain $T'$ from $T$ (i.e. that doesn't go via the Lawvere theory)? If so, can it be extended to work for arbitrary monads?

REPLY [10 votes]: I like the question, Aleks - and I like the term "re-seating".
Yes, there is a direct way, at least in suitable circumstances.  First suppose that we're beginning with a finitary monad $T$ on $\text{Set}$.  For each set $X$, we have
$$
T(X) = \int^n T(n) \times X^n
$$
where the coend is over the category of finite sets.  (This is essentially the definition of finitariness, in disguise.  If you haven't seen this before, think of $T(n)$ as the set of $n$-ary operations in the theory $T$.)  The symbol $\times$ is of course product, but it could also be interpreted as copower: $T(n) \times X^n$ is a copower of the object $X^n$ of $\text{Set}$.  
Now take a category $\mathcal{D}$ with suitable colimits.  For $Y \in \mathcal{D}$, put
$$
T'(Y) = \int^n T(n) \times Y^n
$$
where again the coend is over the category of finite sets, and $\times$ is copower.  (We can no longer interpret $\times$ as product, because $T(n)$ and $Y^n$ belong to different categories.)  You can give $T'$ the structure of a monad, at least under mild assumptions on the behaviour of colimits in $\mathcal{D}$.  That's the same monad $T'$ you'd get if you followed the kind of procedure you describe.
Incidentally, I don't think your description of the procedure is quite right.  If, for example, $\mathcal{C} = \text{Set}$, then $(\mathcal{C}_T)^\text{op}$ is not the Lawvere theory of $T$.  The Lawvere theory is the full subcategory of $(\mathcal{C}_T)^\text{op}$ consisting of just the (free algebras on) finite sets.  
I think the coend formula for re-seating should work when $\mathcal{C}$ is an arbitrary locally finitely presentable category, not necessarily $\text{Set}$.  You'd have to replace the category of finite sets by the category $\mathcal{C}_{\text{fp}}$ of finitely presentable objects of $\mathcal{C}$.  And you'd need $\mathcal{C}_{\text{fp}}$ to act on $\mathcal{D}$, so that the copower defining $T'$ made sense.
Generalizing in a different direction, there's probably nothing special about finitariness: you could do it with any old rank.  But I suspect you do need the monad to have rank in order to re-seat a monad in the kind of way I'm suggesting.  The category of finite (or suitably small) sets provides a bridge between $\mathcal{C}$ and $\mathcal{D}$, and that's what enables the monad to be re-seated.