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Stream: learning: questions

Topic: Whence «strong» functors and «commutative» monads?

Ignat Insarov (Jun 12 2025 at 16:52):

I do not understand what is particularly strong about strong functors, and in what sense do commutative monads commute with anything. Is there some intuitive explanation for these names? What is their origin?

Kevin Carlson (Jun 12 2025 at 17:23):

I don't have anything for "strength", really, except perhaps that strong functors play nicely with the internal homs as well as the external homs, which seems like some kind of strengthening...As for commutative monads, it's much clearer: the monad of an algebraic theory is commutative if and only if all operations of the theory commute with each other (as happens for instance for abelian groups.)

Oscar Cunningham (Jun 12 2025 at 17:48):

Monads are monoids in the category of endofunctors, so you would hope that commutative monads were just commutative monoids in the category of endofunctors. Unfortunately the usual definition of commutative monoid requires a braided monoidal category, whereas the category of endofunctors doesn't have a braiding. However, there is a sense in which the equation for commutative monads involving the strength actually is analogous to the one for a commutative monoid. When I get the chance I'll draw the appropriate string diagrams if no one has done it first. Tag me if I forget.

Arturo De Faveri (Jun 12 2025 at 18:27):

Probably you are familiar with the fact that finitary monads on $\mathbf{Set}$ are equivalent to Lawvere theories. Commutative monads (at least finitary commutative monads on $\mathbf{Set}$) correspond to commutative theories, that is theories in which every two pair of operations $f,g$ commute: $f(g(x^1_1, \ldots x^1_m), \ldots, g(x^n_1, \ldots x^n_m)) = g(f(x^1_1, \ldots x^n_1), \ldots, f(x^1_m, \ldots x^n_m))$. Example of these theories: Abelian groups, commutative monoids,... so they actually capture usual "commutativity". If you want a commutative monad is a generalisation of commutative theories removing "finitary" and allowing the base category to be any symmetric monoidal category $\mathcal{C}$ other than $\mathbf{Set}$.

Mike Shulman (Jun 13 2025 at 01:26):

Kevin Carlson said:

strong functors play nicely with the internal homs as well as the external homs, which seems like some kind of strengthening

I think this is it. An ordinary monad involves only set-functions $\hom(X,Y) \to \hom(TX,TY)$, while a strong monad strengthens this to morphisms of internal-homs $[X,Y] \to [TX,TY]$.

Ignat Insarov (Jun 14 2025 at 20:48):

@Kevin Carlson and @Arturo De Faveri

There is this article by @Dan Marsden that I understand as saying that some commutative monads have corresponding theories with non-commutative operations. I am scared to try and check if what it says is true but it looks serious. And then it would seem to contradict what you are saying about all commutative monads having all operators in their corresponding theories commutative.

Nathanael Arkor (Jun 14 2025 at 20:55):

There are two meanings of "commutativity" here: "commutativity of one operation with another" (which is what happens in commutative monads) and "commutative/symmetric operations" (e.g. in a commutative monoid). These are generally different, though there is a relation: https://ncatlab.org/nlab/show/commutative+algebraic+theory#relation_to_commutativity_of_operations

Ignat Insarov (Jun 14 2025 at 20:57):

Oh, this looks more and more complicated by the minute.

Ignat Insarov (Jun 14 2025 at 21:01):

Am I looking at something deeply beautiful, or is this stuff inherently messed up?

John Baez (Jun 14 2025 at 21:05):

It's beautiful and not that complicated. If you have a monad corresponding a Lawvere theory, the operations all commute with each other. That doesn't imply that the binary operations obey

$b(x,y) = b(y,x)$

It implies that if you have two unary operations $d$ and $e$ they commute in this sense:

$d(e(x)) = e(d(x))$

And there's a similar formula saying that that an $n$-ary operation commutes with an $m$-ary operation. It's easy to understand with some pictures. I'm too lazy to draw those pictures now. But here's a special case: if you have two binary operations $b$ and $c$ this is what it means for them to commute with each other:

$b(c(u,v),c(x,y)) = c(b(u,x),b(v,y))$

This formula looks weird, but it's clearer if you draw the right picture. Basically you write

u v
x y

and you can apply c to the rows and then b to the columns, or b to the columns and then c to the rows, and you get the same thing.

In particular when you have a commutative monad each binary operation commutes with itself:

$b(b(u,v),b(x,y)) = b(b(u,x),b(v,y))$

If there's a unit $0$ for the binary operation $b$ this implies

$b(b(0,v),b(x,0)) = b(b(0,x),b(v,0))$

so

$b(v,x) = b(x,v)$

and $b$ is commutative in the usual sense.

John Baez (Jun 14 2025 at 21:11):

In particular, the monad for commutative monoids has a binary operation $b(x,y) = x+y$ which has a unit $0$, and the commutativity of the monad implies $x + y = y + x$ by the above argument.