Algebraically-free monadicity theorem

Question

The monadicity theorem characterises when a functor $u : \mathbf B \to \mathbf E$ is the forgetful functor from the category of algebras for some monad on $\mathbf E$ (up to an equivalence over $\mathbf E$). We can also define the category of algebras for an endofunctor on $\mathbf E$.

1. Is there a characterisation of those functors $u : \mathbf B \to \mathbf E$ that are the forgetful functors from the category of algebras for some endofunctor on $\mathbf E$ (up to an equivalence over $\mathbf E$)?

Such forgetful functors do not always admit left adjoints. However, when they do, they are in particular monadic (the conditions of the monadicity theorem being easy to verify). So we can expect such a characterisation to imply that $u \colon \mathbf B \to \mathbf E$ creates $u$-split coequalisers. Furthermore, when $u$ is monadic, the induced monad is algebraically-free (by definition). So (in the presence of a left adjoint) my question can really be seen as asking when a monadic functor induces an algebraically-free monad.

This motivates my second question.

2. Is there an intrinsic characterisation of those monads on $\mathbf E$ that are algebraically-free?

By "intrinsic", I mean in terms of the data $(T, \mu, \eta)$, without reference to the forgetful functor from the category of algebras.

These seem natural questions, and I expect there are answers in the literature somewhere, but so far I have not been successful in finding them.

Answer 1

I don't have a complete answer, but here's a couple of related observations.

If $T$ is the algebraically free monad on an endofunctor $P$ in a category with binary coproducts, then for all objects $X$ we have natural isomorphisms $T X \cong P(T(X)) + X$ such that the unit map is the coproduct inclusion $X \hookrightarrow P(T(X)) + X$. This can be used to show there are examples of monads that are not algebraically free on any endofunctor.

The second observation is a negative result, suggesting that the question is difficult. For monads we have the nice result that you can recover the monad up to isomorphism from the category of algebras by composing the left and right adjoints. This is no longer possible for endofunctors. One example is to define endofunctors $P_n : \mathsf{Set} \to \mathsf{Set}$ by $$P_n(X) := \begin{cases} n & X = \emptyset \\ 1 & X \neq \emptyset \end{cases}$$ For $f : X \to Y$ we take $P_n(f)$ to be the quotient map when $X = \emptyset$ and otherwise the identity on $1$. The algebras are the same for all $n > 0$: an algebra structure on $X$ is a point of $X$. Hence they generate the same algebraically free monad, so we cannot recover the endofunctor from the algebraically free monad.