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Stream: learning: questions

Topic: Filtered colimits and finitary monads

Bernd Losert (Jan 30 2024 at 19:22):

A monad is finitary if it preserves filtered colimits. I don't undestand this connection with filtered colimits. What I do understand is that a finitary monad is one whose algebraic theory consists of operations of finite arity. How do filtered colimits enter this picture?

Nathanael Arkor (Jan 30 2024 at 19:37):

The category of sets is the cocompletion of the category of finite sets under filtered colimits. Therefore, a finitary functor $\mathbf{Set} \to \mathbf{Set}$ is equivalent to an arbitrary functor $\mathbf{FinSet} \to \mathbf{Set}$. Finitary endofunctors can thus be seen as functors that are determined by their actions on finite sets (corresponding to the finite arities).

Nathanael Arkor (Jan 30 2024 at 19:39):

This correspondence extends from functors to monads: a finitary monad on the category of sets is equivalent to a [[relative monad]] $\mathbf{FinSet} \to \mathbf{Set}$. Relative monads thus more directly capture the intuition that your structure is determined entirely by its action on the finite arities.

Bernd Losert (Jan 30 2024 at 19:53):

I understand the FinSet -> Set part. But the correspondence with arities is still elusive. Suppose T is the free monoid monad. What does the finiteness of the arity of the monoid operation have anything to do with T preserving filtered colimits?

Nathanael Arkor (Jan 30 2024 at 20:17):

If you have a functor $\mathbf{FinSet} \to \mathbf{Set}$, you can always extend it to a functor $\mathbf{Set} \to \mathbf{Set}$ by taking a left Kan extension (this is the universal property of $\mathbf{Set}$ as the free cocompletion of $\mathbf{FinSet}$ under filtered colimits). So the reason filtered colimits appear is precisely because the filtered colimits are what you need to add to $\mathbf{FinSet}$ to get $\mathbf{Set}$. Since you are adding them freely, they must be preserved by the extension.

Bernd Losert (Jan 30 2024 at 20:23):

Yes, I already understand that part. But I don't understand the arity connection.

Mike Shulman (Jan 30 2024 at 20:24):

More concretely, the free monoid monad can be written as $TA = \coprod_{n\in\mathbb{N}} A^n$. Since coproducts commute with all limits, and finite powers are finite limits which commute with filtered colimits, the whole functor $T$ commutes with filtered colimits. But you replace the finite powers $A^n$ by infinite ones, this fails.

Nathanael Arkor (Jan 30 2024 at 20:35):

Bernd Losert said:

Yes, I already understand that part. But I don't understand the arity connection.

Consider the theory of monoids, which is an identity-on-objects functor from the free cartesian category on a single object $X$ (which is equivalent to $\mathbf{FinSet}^\circ$), to a specific cartesian category $\mathcal C$. Then, restricting the Yoneda embedding of $X$ along the identity-on-objects functor gives us a functor $\mathcal C(-, X) \colon \mathbf{FinSet} \to \mathbf{Set}$. We obtain such a functor for every algebraic theory.

More generally, we could consider $\kappa$-ary algebraic theories, with arity smaller than some (well-behaved) cardinal $\kappa$, from which we would obtain a functor $\mathcal C(-, X) \colon \mathbf{Set}_\kappa \to \mathbf{Set}$. For instance, if we just consider algebraic theories with unary operations, then we obtain a functor $\mathcal C(-, X) \colon 1 \to \mathbf{Set}$, and the corresponding monad will preserve all colimits, not just filtered colimits.

Bernd Losert (Jan 30 2024 at 20:46):

But you replace the finite powers $A^n$ by infinite ones, this fails.

OK, this makes more sense, but feels like a strange coincidence.

Bernd Losert (Jan 30 2024 at 20:47):

(deleted)

Bernd Losert (Jan 30 2024 at 20:48):

Consider the theory of monoids, which is an identity-on-objects functor from the free cartesian category on a single object $X$ (which is equivalent to $\mathbf{FinSet}^\circ$), to a specific cartesian category $\mathcal C$...

How would this work for the theory of monoid actions?

James Deikun (Jan 30 2024 at 20:52):

For the theory of actions of a specific monoid, it works the same way. The theory of monoid actions in general isn't a single-sorted algebraic theory.

Kevin Arlin (Jan 30 2024 at 21:18):

Bernd Losert said:

OK, this makes more sense, but feels like a strange coincidence.

What two things are strangely coinciding, here? Filtered colimits are interesting specifically because they're the ones that commute with finite limits, so those two classes of construction aren't occurring independently here.

Bernd Losert (Jan 30 2024 at 21:29):

What two things are strangely coinciding, here? Filtered colimits are interesting specifically because they're the ones that commute with finite limits, so those two classes of construction aren't occurring independently here.

Ah, so this is news to me. I didn't realize that's why we care about them.

Bernd Losert (Jan 30 2024 at 21:33):

Thank you everyone for your help.

Bernd Losert (Jan 30 2024 at 22:01):

James Deikun said:

For the theory of actions of a specific monoid, it works the same way. The theory of monoid actions in general isn't a single-sorted algebraic theory.

By the way, what does the Lawvere theory of monoid actions look like? Any reference? In particular, I would like to see how the one using a fixed monoid is different from the general one.

Kevin Arlin (Jan 30 2024 at 22:15):

The Lawvere theory for "A monoid and an action" is generated under finite products by generators $M$ and $S$, with maps $M^k\times S^\ell\to M^m\times S^n$ given by things like $(m_1,m_2,m_3,s_1,s_2)\mapsto (e,m_2,m_2m_1s_1,s_1):$ you can multiply in $M,$ multiply an $m$ by an $s,$ or do projections and duplications arising from being in a Cartesian category.

Kevin Arlin (Jan 30 2024 at 22:17):

The Lawvere theory for "An action of a specific monoid $M$" is generated under finite products by a single generator $S$ and has unary operations $m:S\to S$ for every $m\in M$, plus operations $S^k\to S^\ell$ coming from the fact that we're in a Cartesian category, and then mixtures of these. And we have conditions on the composites of these unary operations corresponding to relations in $M.$

Mike Shulman (Jan 30 2024 at 22:20):

Kevin Arlin said:

Filtered colimits are interesting specifically because they're the ones that commute with finite limits, so those two classes of construction aren't occurring independently here.

In addition, the definition of "filtered colimit" includes the word "finite" which can be replaced by something larger, and for any higher cardinality $\kappa$, the $\kappa$-ary algebraic theories coincide with the monads that preserve "$\kappa$-filtered colimits".

Bernd Losert (Jan 30 2024 at 22:28):

Kevin Arlin said:
Thanks. By the way, for the "action of a specific monoid" theory, if you consider models of this theory in Top, do you get continuous monoid actions? How does the theory encode that the action M × S → S must be a continuous function.

Reid Barton (Jan 30 2024 at 22:36):

The theory encodes that the action must be a morphism in whatever category you interpret into, and the morphisms of Top are continuous functions.

Bernd Losert (Jan 31 2024 at 06:31):

Reid Barton said:

The theory encodes that the action must be a morphism in whatever category you interpret into, and the morphisms of Top are continuous functions.

Yes, but in this case, it only encondes that each $m : S \to S$ is continuous, not that the whole $M \times S \to S$ is continuous.

Mike Shulman (Jan 31 2024 at 07:40):

If the monoid comes with a topology, then in order to incorporate information about that topology you'd need to consider a topological theory.

Bernd Losert (Jan 31 2024 at 08:43):

Does "topological theory" refer to the concept from https://www.sciencedirect.com/science/article/pii/S0001870807001466 ?

Mike Shulman (Jan 31 2024 at 15:28):

No, I just meant a topologically enriched Lawvere theory.

Reid Barton (Jan 31 2024 at 20:51):

Oh I missed the word "specific". In that case it only makes sense to have the monoid be discrete, if we are talking about an ordinary theory.

Bernd Losert (Jan 31 2024 at 22:54):

Mike Shulman said:

No, I just meant a topologically enriched Lawvere theory.

I would appreciate any reference to this, if you have any.

John Baez (Jan 31 2024 at 23:33):

Are there any references on enriched Lawvere theories, readable by people without tons of knowledge of enriched categories? I guess I'm reduced to recommending this:

• John Baez and Christian Williams, Enriched Lawvere theories for operational semantics.

It doesn't say anything about topologically enriched Lawvere theories, which is a special case, but it explains what enriched Lawvere theories are, and some of the basic theorems about them.

Mike Shulman (Feb 01 2024 at 01:03):

There are a bunch of references at [[enriched Lawvere theory]].

Bernd Losert (Feb 01 2024 at 09:37):

Thanks guys. I think there is still a problem even with Top-enriched Lawvere theories since the enriched theory of monoid actions by a fixed monoid will not produce continuous actions $M \times S \to S$ that are jointly continuous, only separately continuous.

Morgan Rogers (he/him) (Feb 01 2024 at 10:00):

(deleted)

Kevin Arlin (Feb 01 2024 at 18:30):

That problem is exactly what Mike was explaining how to fix. If the functor giving an $M$-space is a continuous functor from an enriched Lawvere theory then that says that the mapping $m\mapsto (\varphi_m)$ is a continuous function from $M$ to the function space $S^M,$ which is equivalent to joint continuity. Note that we probably need to be in a convenient, ie Cartesian closed, category of spaces for enriched category theory to work smoothly.

Kevin Arlin (Feb 01 2024 at 18:33):

You would get the separately continuous action by simply using the ordinary Lawvere theory and mapping into the category of spaces.

Evan Washington (Feb 01 2024 at 20:04):

Bernd Losert said:

Yes, I already understand that part. But I don't understand the arity connection.

Here's another way of seeing the relation between finitary functors and operations with finite arity. It turns out every finitary endofunctor on $\mathbf{Set}$ is a quotient of a polynomial endofunctor (Adámek, Rosický, and Vitale's Algebraic Theories, Ch. 12). Think of the polynomial endofunctor as encoding the signature (just the set of operations of each arity) and quotienting as imposing equations (between terms/operations). See Appendix A of Algebraic Theories for more on monads in particular.

Bernd Losert (Feb 01 2024 at 22:10):

Evan Washington said:

Bernd Losert said:

Yes, I already understand that part. But I don't understand the arity connection.

Here's another way of seeing the relation between finitary functors and operations with finite arity. It turns out every finitary endofunctor on $\mathbf{Set}$ is a quotient of a polynomial endofunctor (Adámek, Rosický, and Vitale's Algebraic Theories, Ch. 12). Think of the polynomial endofunctor as encoding the signature (just the set of operations of each arity) and quotienting as imposing equations (between terms/operations). See Appendix A of Algebraic Theories for more on monads in particular.

Ah, this is a good way of seeing it. Thanks.

Bernd Losert (Feb 01 2024 at 22:11):

Kevin Arlin said:

That problem is exactly what Mike was explaining how to fix. If the functor giving an $M$-space is a continuous functor from an enriched Lawvere theory then that says that the mapping $m\mapsto (\varphi_m)$ is a continuous function from $M$ to the function space $S^M,$ which is equivalent to joint continuity. Note that we probably need to be in a convenient, ie Cartesian closed, category of spaces for enriched category theory to work smoothly.

:smile: I was waiting for someone to mention that this only works out if the category is Cartesian closed.

Kevin Arlin (Feb 02 2024 at 00:20):

Yeah,

Bernd Losert said:

:smile: I was waiting for someone to mention that this only works out if the category is Cartesian closed.

Yeah, and in fact the literature on enriched Lawvere theories seems to largely assume the enrichment base is even locally presentable; but I'm sure that's primarily a convenience. Possibly some adjoints don't exist anymore that would in the lp case.