Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This question want to be a follow up of the following question. In that thread I was interested in understanding relation between various presentation of algebraic theories. In particular in Eduardo Pareja Tobes answers are linked some papers that shows a correspondence between respectively finitary monads and Lawvere's algebraic theories and monads with arieties and theories with arieties.

So it seems that there almost every time a correspondence between theories (here by theories I mean the categorical presentation of a theory given by its syntactic category) and monads of some sorts.

I would like to know how further can be this relation been pushed. For instance

is true the for every theory (in the most general sense) there's a monad such that categories of models for the theory and the category of algebras for the monad are equivalent?

and on the other end

is it true the opposite implication: i.e. that for every monad there's a theory (i.e. some sort of category with some strucuture) such that the category of the models is equivalent to the category of the algebras for the monad?

Any reference would be incredibly appreciated.

share|improve this question
add comment

2 Answers

up vote 4 down vote accepted

As noted by Zhen Lin, monads (on $\mathbf{Set}$) should be thought of as morally equivalent to algebraic theories of fairly general type. By an algebraic theory I mean a single-sorted theory specified by operations and universally quantified equations between terms. The operations are allowed to be infinitary. Certainly all monads on $\mathbf{Set}$ and their algebras are describable by algebraic theories and their models. Not quite all algebraic theories are thus describable by monads, but all algebraic theories whose signature consists of a small set of operation symbols are "monadic", and there are some notable examples of monads where the signature of any algebraic theory giving rise to it must consist of a proper class of operation symbols.

First, here are some non-examples. The theory of fields is not algebraic in this sense, since the multiplicative inversion is not a globally defined operation. (It's not algebraic even if we allow multi-sorted theories. For the category of fields does not, for instance, have products -- whereas this would be true of any category of models of a multi-sorted theory.) Nor is the theory of categories algebraic in this sense -- while it can be presented as a single-sorted theory (where the sort gets interpreted as the set of morphisms), the composition is not given by a globally defined operation. (And again, it's not even algebraic in any multi-sorted sense -- if it were, then the category of models would be a regular category, which $\mathbf{Cat}$ is not).

To get an algebraic theory out of a monad $M$ on $\mathbf{Set}$, one may take the Kleisli category $Kl(M)$, viz. the full subcategory of the category of $M$-algebras whose objects are free algebras. Then define an infinitary theory $Th(M)$ by taking operation symbols of type $n \to 1$, for $n$ a possibly infinite cardinal, to be elements of the (underlying set of the) free algebra $M(n)$. This theory should be thought of as "saturated" (aka a "clone"), consisting of all definable operations of a theory; the point is that the monad multiplication $\mu: MM \to M$ takes a term $\alpha \in MM(n)$ (obtained by formally substituting $k$ operations $n_1 \to 1, \ldots, n_k \to 1$ into some $k$-ary operation, to arrive at a term of arity $n = n_1 + \ldots + n_k$) and returns an operation symbol $\mu(\alpha) \in M(n)$, which is equated to that definable term $\alpha$ in the theory.

If you like thinking in terms of Lawvere theories, then you could say that $Kl(M)^{op}$ plays the role of the infinitary Lawvere theory attached to $M$: the category of $M$-algebras is equivalent to the category of functors $Kl(M)^{op} \to \mathbf{Set}$ that preserve arbitrary products.

If the monad $M$ is finitary (preserves filtered colimits), then we can cut down to the usual Lawvere theory given by free algebras on finite sets. This generalizes: if $M$ preserves $k$-filtered colimits for some regular cardinal $k$, then one can cut down to a generalized Lawvere theory $Th_k(M)$ given by free algebras on sets of cardinality less than $k$. Then the category of $M$-algebras is equivalent to the category of functors $Th_k(M) \to \mathbf{Set}$ preserving $k$-small products. Such monads correspond to theories of rank $k$.

Not all monads $M$ on $\mathbf{Set}$ are $k$-accessible in this sense; a famous example is the category of compact Hausdorff spaces, whose underlying functor to $\mathbf{Set}$ is monadic, but where the corresponding algebraic theory is too large to have rank.

Quite a lot of this material is worked out in the nLab.

share|improve this answer
add comment

It is well-known that the theories whose category of models are monadic over $\mathbf{Set}$ are necessarily algebraic (possibly infinitary), and conversely every (possibly infinitary) algebraic theory that admits freely-generated models on every set has a category of models that is monadic over $\mathbf{Set}$. See, for instance, Linton's papers in the Seminar on triples and cohomology.

For an example of a non-algebraic theory, consider the theory of fields. This is not monadic over $\mathbf{Set}$ because e.g. the category of fields does not have products.

share|improve this answer
Subject to some caveats of course. Certainly all algebraic theories generated by a small signature are describable by monads. But the theory of complete Boolean algebras or of complete Heyting algebras, which technically is algebraic if one permits a proper class of operations, is not describable by a monad. –  Todd Trimble Jun 20 '13 at 17:46
Indeed. I had this in mind but I forgot when I was writing my answer! –  Zhen Lin Jun 20 '13 at 18:16
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.