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A polynomial monad on a locally cartesian closed category $C$ is a monad whose underlying endofunctor is a polynomial functor and whose unit and multiplication are cartesian transformations. Since a polynomial functor is generated by a single morphism $f:B\to A$, and compositions and transformations between them can also be described in terms of their generating morphisms, a polynomial monad (unlike an arbitrary monad) can be described by a very small amount of data.

One way of constructing monads with prescribed categories of algebras is through algebraic colimits of algebraically free monads ("algebraic" meaning that the operations behave as expected on categories of algebras). It was shown by Gambino and Kock that the free monad on a polynomial endofunctor is a polynomial monad. However, I am pretty sure that a colimit of polynomial monads (and arbitrary, not-necessarily-cartesian, monad natural transformations between them) may no longer be polynomial.

Is there a class of monads on a sufficiently nice category $C$ (e.g. locally presentable and locally cartesian closed -- feel free to assume it is a Grothendieck topos if that helps) that:

  1. includes all polynomial monads,
  2. is closed under colimits of monads, and
  3. every monad in the class can be described by a "very small amount of data"?

Obviously the last criterion is not precise, but the intuitive idea should be clear by comparison to the case of polynomial monads.

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  • $\begingroup$ I think the category of accessible monads on a locally presentable lcc category C should work. It certainly satisfies 1 and 3. For 2 consider a colimit of accessible monads $T_{i}$. We can choose $\lambda$ such that C is locally $\lambda$-presentable and each of the $T_{i}$ is $\lambda$-accessible. But $\lambda$-accessible monads are closed under colimits amongst all monads (indeed they are coreflective in monads). This follows from the fact that the inclusion of $\lambda$-accessible endofunctors into the category of all endofunctors is a strict monoidal left adjoint. $\endgroup$
    – john
    Commented Apr 5, 2017 at 12:53
  • $\begingroup$ @john The category of accessible monads certainly satisfies 1 and 2, but not 3, not the way I mean it. A polynomial functor is determined by one morphism in the category; an accessible functor requries specifying an arbitrarily large cardinal $\lambda$ and a functor $C_\lambda \to C$. $\endgroup$ Commented Apr 6, 2017 at 15:02
  • $\begingroup$ I see. However, the category of finitary monads on Set is monadic over the category of signatures $Set^{N}$ for $N$ the discrete category of natural numbers. Moreover, the free monad on a signature is polynomial - indeed arises from a plain operad. And by monadicity each finitary monad is a colimit of such free monads; thus a colimit of polynomials. Since colimits of finitary monads are algebraic, a category sat. 1 and 2 must contain all finitary monads on Set. In fact, I reckon that you can extend this argument to show that your category must contain all accessible monads on Set. $\endgroup$
    – john
    Commented Apr 6, 2017 at 22:46
  • $\begingroup$ @john Good point. An accessible endofunctor of Set, however, can be described by a "very small" amount of data, since $Set_\lambda$ can be regarded as an internal category in Set and a functor $Set_\lambda \to Set$ as an internal diagram thereon. This suggests to me that maybe the answer I'm looking for might be something like "a monad obtained by internal left Kan extension from an internal diagram on an internal category". I wonder whether that can be made to work. $\endgroup$ Commented Apr 7, 2017 at 8:13
  • $\begingroup$ Hmm, I guess that the way I phrased the question, accessible monads do satisfy (3), at least in a Grothendieck topos, since then any small subcategory $C_\lambda$ can be identified with an internal category that's "discrete" in that its object-of-objects and object-of-morphisms are both coproducts of 1. But what I'd really like is some "elementary, internal" categorical description of a class of monads with properties (1) and (2). $\endgroup$ Commented Apr 7, 2017 at 8:28

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