## August 11, 2020

#### Posted by John Baez

We can think of a monad on $Set$ as describing some sort of algebraic gadget equipped with a bunch of operations obeying a bunch of equations. This works very nicely if we restrict attention to finitary monads, which correspond to Lawvere theories: then the operations I’m talking about are all ‘finitary’, taking some finite set of inputs. But we can also generalize to higher cardinalities: for any cardinal α, a monad of rank α on $Set$ describes some sort of gadget with operations of arity at most α.

There are nastier gadgets that have no upper bound on the arity of their operations, like complete semilattices, also known as suplattices. The point is that in such a thing, any subset has a least upper bound, no matter how large its cardinality. These are algebras of a ‘monad without rank’ — which makes me think of someone in the army who is not a private, not a lieutenant, not a colonel, not a general….

Anyway, this viewpoint on monads helps me get a feeling for commutative monads: these describe algebraic gadgets with a bunch of operations that all commute with each other, and perhaps obey other equations as well.

But first: what does it mean for an $n$-ary operation to commute with an $m$-ary operation?

I think an example makes it clear: if we’ve got an 2-ary operation $f$ and a 3-ary operation $g$ we say they commute if

$f(g(u,v,w),g(x,y,z))=g(f(u,x),f(v,y),f(w,z))$

for all $u,v,w,x,y,z$. Here you should think of the variables as arranged in a $2 \times 3$ array:

$\begin{array}{ccc} u & v & w \\ x & y & z \end{array}$

The equation says we can either apply $g$ to each of the rows and then apply $f$, or apply $f$ to each of the columns and then apply $g$. This becomes even more beautiful if we draw it using trees in 3 dimensions, and it’s almost criminal of me not to draw the picture, but nobody is paying me to write blog articles.

Notice that if $f$ and $g$ are both 1-ary operations, we get commutativity in a very standard sense:

$f \circ g = g \circ f$

Any monoid $M$ gives a monad whose operations are all 1-ary, corresponding to multiplication by elements of $M$. The algebras of this monad are sets acted on by $M$. And this monad is commutative iff $M$ is commtutative!

But more interestingly, any ring $R$ gives a monad whose algebras are $R$-modules, and this monad is commutative iff $R$ is commutative!

Another example: the monad whose algebras are commutative monoids is a commutative monad.

Cool, huh? If you want to intimidate your colleagues, just casually say this as if it were obvious: “the commutative monoid monad is a commutative monad”. But don’t do this if you want a raise.

One cool fact is that for any commutative monad $T$ on $Set$ the category $T\mathrm{alg}$ is a symmetric monoidal closed category! So, there’s a way to tensor $T$-algebras, and also an internal hom: that is, if $A$ and $B$ are $T$-algebras, the set of $T$-algebra maps from $A$ to $B$, say $[A,B]$, is also a $T$-algebra.

Puzzle 1. A grad student of mine assures me that if $S$ and $T$ are commutative monads on $Set$, then the composite $S T$ is a commutative monad too. I’m having trouble finding this fact. Is it really true? Is this for some distributive law or any distributive law? (It may work for some special distributive law.)

But I really want categorified analogues of these facts, if possible. When we look at 2-monad on $Cat$ there are various amounts of commutativity it could have. I think I’m interested in what Hyland and Power call a ‘symmetric pseudocommutative 2-monad’:

Here ‘pseudo’ means some commutativity condition holds only up to isomorphism, and ‘symmetric’ means this isomorphism is better than a mere braiding: it’s more like what you see in a symmetric monoidal category. The example to keep in mind is where $T$ is the 2-monad for symmetric monoidal categories (not necessarily strict). This is not commutative, but it is symmetric pseudocommutative.

These authors almost show that for any symmetric pseudocommutative 2-monad $T$ on $Cat$ the 2-category $T\mathrm{alg}$ is a symmetric monoidal closed 2-category. But they don’t say that.

For starters, they define $T\mathrm{alg}$ to be the 2-category of strict algebras, ‘pseudo maps’ between these, and some sort of 2-morphisms between those. That’s good. For example, if $T$ is the 2-monad for symmetric monoidal categories, its strict algebras are symmetric monoidal categories and the pseudo maps between these are symmetric strong monoidal functors.

They prove a number of things:

Proposition 18. Given a symmetric pseudocommutative 2-monad $T$ on $Cat$, $T\mathrm{Alg}$ is a symmetric 2-multicategory.

A symmetric multicategory is like a symmetric monoidal category where given two objects $a$ and $b$ you don’t actually have an object $a \otimes b$: you just know what the maps out of this object would be. They define a categorified version of this, a ‘symmetric 2-multicategory’.

Proposition 22. Given a pseudocommutative 2-monad $T$ on $Cat$, $T\mathrm{Alg}$ is a pseudoclosed 2-category.

A closed category is a category where for any two objects $a$ and $b$ you have a ‘hom-object’ $[a,b]$. Closed monoidal categories are the most famous kind, but you don’t need a tensor product in a closed category. Hyland and Power work with a categorified version of closed categories, ‘pseudoclosed 2-categories’.

Then the punchline:

Theorem 13. Given a symmetric pseudocommutative 2-monad $T$ on $Cat$, $T\mathrm{Alg}$ is a symmetric pseudoclosed 2-category.

Nice, but still no tensor product of objects!

We get the tensor product of objects here:

Theorem 14. Given a finitary pseudocommutative 2-monad $T$ on $Cat$, $T\mathrm{Alg}$ is a pseudomonoidal pseudoclosed 2-category.

Now they’re assuming $T$ is finitary, which is a bit sad even though it covers our go-to example of the monad for symmetric monoidal categories. ‘Pseudomonoidal 2-category’ means some sort of weak monoidal 2-category; I’m not sure which but I’m sure I’ll be happy with it when I find out.

What makes me nervous is that they never say they’re getting a symmetric monoidal closed 2-category or — more likely for them — ‘symmetric pseudomonoidal pseudoclosed 2-category’.

Maybe it’s obvious to experts that a symmetric 2-multicategory that’s also a symmetric pseudoclosed 2-category and a pseudomonoidal pseudoclosed 2-category is a symmetric pseudomonoidal pseudoclosed 2-category. But do I really have to check this myself?

Has there been more work on this topic?

By the way, what do I really want? I want things like this. Let $P$ be the pseudomonad for categories with chosen finite colimits and $S$ our friend the 2-monad for symmetric monoidal categories. Then I’d be delighted if someone had done this:

Puzzle 2. Show that $P\mathrm{Alg}$ is some sort of symmetric monoidal closed 2-category: sprinkle on adjectives like ‘pseudo’ to taste.

Puzzle 3. Show the same for $S\mathrm{Alg}$.

Puzzle 4. Show the same for $P S\mathrm{Alg}$, where $P S$ is made into a 2-monad in the hopefully ‘obvious’ way, with tensor products distributing over finite colimits (up to coherent isomorphism).

Puzzle 4 makes me wonder if a composite of symmetric commutative pseudomonads is again a thing of the same sort, but I’d be perfectly content if some god came down from the sky and solved the puzzle in any other way.

Posted at August 11, 2020 8:29 PM UTC

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Sorry to make the first comment a negative one, but I don’t think this is right:

There are nastier gadgets that have no upper bound on the arity of their operations, like complete Boolean algebras. These are described by ‘monads without rank’

There’s no monad at all for complete Boolean algebras. Indeed, the forgetful functor from complete Boolean algebras to sets has no left adjoint.

The way I learned this (and this is a true story) was that one day when I was a PhD student, Martin Hyland took me to lunch at his college so that we could discuss some mathematical matters. It being his college, I wasn’t paying. At some point in the conversation, he said:

“There’s no such thing as a free…”

and left a long pause.

I was sure he was going to say “lunch”.

In fact, he said “complete Boolean algebra”.

If you want an example of a monad without rank, you could use the powerset monad (whose algebras are complete lattices, with join-preserving maps as the homomorphisms), or the ultrafilter monad (whose algebras are compact Hausdorff spaces).

Posted by: Tom Leinster on August 12, 2020 1:03 AM | Permalink | Reply to this

Whoops! Thanks. I’ll change my counterexample. Yeah, there’s no monad for complete Boolean algebras; I guess the best you can do is $\kappa$-complete Boolean algebras where $\kappa$ is an inaccessible cardinal.

Posted by: John Baez on August 12, 2020 5:50 PM | Permalink | Reply to this

On Puzzle 1, I don’t think that the theory of commutative rings is commutative, is it? since $+$ doesn’t commute $\times$ in the sense intended here? and this despite it being a composite of commutative theories of monoids and groups.

PS. I would have written the commutativity law as $f(g(u,v,w),g(x,y,z))=g(f(u,x),f(v,y),f(w,z))$, maybe a typo? or maybe a formulation I didn’t know.

Posted by: Sam Staton on August 12, 2020 7:18 AM | Permalink | Reply to this

Thanks for pointing out that the monad for commutative rings is not commutative even though it’s a composite of the monad for abelian groups and the monad for commutative monoids, both of which are commutative. I will chastise that grad student. He should have noticed that counterexample. (Okay, I should have too.)

This also kills off the hope that $P S$, the pseudomonad for ‘finitary symmetric 2-rigs’ discussed in my blog post, is pseudocommutative in any reasonable sense. After all, $P S$ is a kind of categorification of the monad for commutative rings, so there’s no reason in the world that it would be pseudocommutative.

When $T$ and $T'$ are commutative monads on $Set$ it seems there’s some way to make $T T'$ into a commutative monad. I’ll do the case of finitary monads, so we can think about $T$ and $T'$ as Lawvere theories, so all their operations are finitary. To make $T T'$ into a monad we need to choose a distributive law. I may be deluded, but I feel can always choose a ‘straightforward’ distributive law that says we can commute operations in $T$ past operations in $T'$ in the ‘obvious way’, e.g.

$f \circ (g,g)=g \circ (f,f,f)$

when $f$ is a 2-ary operation of $T'$ and $g$ is a 3-ary operation of $T$. Right?

If we did this where $T$ is the monad for abelian groups and $T'$ is the monad for commutative monoids, $T T'$ would not be the monad for commutative rings! It would be the monad for things that are both abelian groups and commutative monoids, obeying rules like

$(a + b) \cdot (a' +\, b') = (a \cdot a') + (b \cdot b')$

By Eckmann–Hilton these are abelian groups.

(Thanks also for your other correction, whose merits become clear here.)

Posted by: John Baez on August 12, 2020 6:52 PM | Permalink | Reply to this

Thanks! Regarding your straightforward distributive law, I guess $T(T'(n))$ is the underlying set of commutative ring on $n$, and I am not sure that it can be put in natural bijection with the free Abelian group on $n$. So I don’t know that the presentation you describe is a presentation of a monad with $TT'$ as an underlying functor.

But I think a simpler counter-example comes from Dan Marsden’s proposal of trying to distribute the commutative monad for pointed sets $(+1)$ over itself. It seems there is no way to make this into a commutative monad. Since $(+1)\circ(+1)=(+2)$ as functors, we would have to find a commutative theory with two different constants, which is impossible.

I think what you are describing is the tensor of two theories. Perhaps you know it, if not, one recent reference is Hyland-Plotkin-Power’s Combining effects: sum and tensor Section 4.

Posted by: Sam Staton on August 13, 2020 11:26 PM | Permalink | Reply to this

Section 6 of John Bourke’s “Skew structures in 2-category theory and homotopy theory”, Journal of Homotopy and Related Structures 12 (2017), no. 1, 31-81, may have relevant information. The paper is also available on ArXiv:

https://arxiv.org/abs/1510.01467

Posted by: Nicola Gambino on August 12, 2020 2:50 PM | Permalink | Reply to this

Thanks! The best result for me seems to be

Theorem 6.7 For $T$ an accessible pseudocommutative 2-monad on $Cat$ the 2-category $T\,\mathrm{Alg}$ admits the structure of a monoidal bicategory.

Something similar was already proved by Hyland and Power, but they had expressed dissatisfaction with their argument:

Naturally, we are unhappy with the proof we have just outlined. Since the data we start from is in no way symmetric we expect some messy difficulties: but the calculations we do not give are very tiresome, and it would be only too easy to have made a slip. Hence we would like a more conceptual proof.

Bourke gives a different proof.

Posted by: John Baez on August 12, 2020 8:46 PM | Permalink | Reply to this

Bart Jacobs describes a notion of distributive law of commutative monads in “Semantics of Weakening and Contraction”. They do satisfy additional axioms beyond Beck’s originals. With these additional conditions, the composite is commutative. Maybe this is useful.

I think a very simple counter to the claim in puzzle one is just distributing (-) + 1 over itself.

Posted by: Dan Marsden on August 12, 2020 9:20 PM | Permalink | Reply to this

In general you do need some condition like finitaryness. I could believe your claim that

for any commutative monad $T$ on Set the category $T\mathrm{alg}$ is a symmetric monoidal closed category [emphasis added]

(although I don’t recall hearing it before), because of the fairly degenerate nature of $Set$. (Do you have a reference?) But to construct the tensor product of algebras for an arbitary commutative monad, you need the category of algebras to have sufficient colimits that are sufficiently respected by $T$, as described in the theorem on the nLab page, and in general I’m confident that that’s an extra condition beyond commutativity.

I see your questions are about $Cat$ specifically rather than an arbitrary 2-category, and I suppose it’s possible that $Cat$ is degenerate in a way like $Set$, but that sort of thing can’t be taken for granted.

Posted by: Mike Shulman on August 13, 2020 4:55 PM | Permalink | Reply to this

I omitted the condition that $T$ be ‘bounded’, which I guess means it ‘has rank’. Hyland and Power write:

In a similar fashion, the theory of commutative monads on a symmetric monoidal closed category V gives an abstract approach to algebra with commuting operations (generalised linear algebra). This theory is described in [11–13]. (Between them [11,13] show that a commutative monad is a symmetric monoidal monad: so the theory is covered by the abstract perspective of [17].) In the basic setting, V is complete and cocomplete and T is bounded, so that if T is commutative, the category T-Alg is itself symmetric monoidal closed. In this paper we describe an analogue of this theory at the 2-dimensional level.

Posted by: John Baez on August 13, 2020 5:14 PM | Permalink | Reply to this

Now I better understand why Hyland and Power start by constructing a symmetric 2-multicategory instead of a symmetric monoidal 2-category of $T$-algebras when $T$ is a symmetric pseudocommutative 2-monoid on $Cat$. This allows them to avoid constructing a tensor product of $T$-algebras, which I guess is where it’s crucial that $T$ have rank.

I’d never thought about this theoretical advantage of multicategories over monoidal categories: there may be situations where you can’t define a tensor product of algebraic gadgets, due to size problems — and a multicategory allows you to avoid this and still effectively to talk about maps out of the (nonexistent) tensor product.

If I’m right, someone should have done this: define a symmetric multicategory of $T$-algebras for any commutative monad on $Set$, without any rank assumption on $T$.

Has anyone seen this?

Posted by: John Baez on August 13, 2020 6:29 PM | Permalink | Reply to this

I’d never thought about this theoretical advantage of multicategories over monoidal categories: there may be situations where you can’t define a tensor product of algebraic gadgets, due to size problems - and a multicategory allows you to avoid this and still effectively to talk about maps out of the (nonexistent) tensor product.

Indeed! There may be all sorts of reasons why you can’t define a tensor product, and it may also be that you can define one but to do so would create unnecessary work. If you want to talk about maps out of a (possibly nonexistent) tensor product, you don’t need a monoidal category: all you need is a multicategory.

For example, one can speak of categories enriched in a multicategory $V$. Instead of composition consisting of maps like

$Hom(A, B) \otimes Hom(B, C) \to Hom(A, C)$

in a monoidal category, it consists of maps like

$Hom(A, B), Hom(B, C) \to Hom(A, C)$

in our multicategory, $V$.

Even if you’re enriching in something perfectly ordinary such as $V = Vect$, treating it as a multicategory rather than a monoidal category gets closer to most people’s intuitions about what a $V$-category is. After all, if you were walking down the street with a friend and wanted to tell them what a linear category was, you’d say “composition is bilinear” rather than the circuitous “composition defines a linear map out of a tensor product”. The tensor product is irrelevant!

(Of course, there are situations where you want a tensor product. But more often than is generally acknowledged, it’s an unnecessary distraction.)

Posted by: Tom Leinster on August 13, 2020 7:59 PM | Permalink | Reply to this

I’m somewhat familiar with the conceptual simplification afforded by multicategories, probably because you and Mike Shulman have explained them. I had never thought about situations where you couldn’t promote a multicategory to a monoidal category, for example due to size issues. Are there some general theorems about when you can’t do this?

‘Promote’ here is deliberately vague, because there are a couple of things you might want.

Sometimes a multicategory turns out to be the underlying multicategory of a monoidal category. So one could boldly ask if every multicategory arises this way, up to equivalence. But no, that’s too much to ask.

Here’s a silly counterexample. There’s a multicategory of finite sets whose cardinality is prime, where a multimorphism from

$f : S_1, \dots, S_n \to T$

is a function

$f : S_1 \times \cdots \times S_n \to T$

This is clearly just a smart-aleck way of talking about the monoidal category of nonempty finite sets and functions between these, with $\times$ as the monoidal structure. If I had to ‘promote’ this multicategory to a monoidal category, you can guess which monoidal category I’d choose.

I’m now guessing that there are some multicategories that are resistant to being embedded in the underlying multicategory of any monoidal category due to size issues.

Posted by: John Baez on August 13, 2020 11:12 PM | Permalink | Reply to this

Almost any operad, seen as a one-object multicategory, is an example of a multicategory that is not the underlying multicategory of any monoidal category in the world.

(For suppose we have an operad $P$ that is the underlying multicategory of a monoidal category $A$. Then $A$ must have only one object, and one-object monoidal categories are basically commutative monoids. This severely restricts what $P$ can be.)

I haven’t thought about size issues, but every small multicategory embeds into a strict monoidal category. This is a special case of Proposition 6.6.12 of Higher Operads, Higher Categories. But as remarked just before Definition 6.6.11, it’s not an especially useful theorem, in the same way that Cayley’s representation theorem for groups is less useful than one might at first think.

Posted by: Tom Leinster on August 14, 2020 4:23 PM | Permalink | Reply to this

For the benefit of bystanders, I’ll note explicitly that having rank (a.k.a. being bounded, a.k.a. being accessible) is a more general version of finitaryness. Being finitary is the same as having rank $\omega$.

John wrote:

someone should have done this: define a symmetric multicategory of $T$-algebras for any commutative monad on $Set$, without any rank assumption on $T$.

I think that’s actually fairly formal. A commutative monad on a symmetric monoidal category is the same as a monoidal monad, i.e. a monad in the 2-category $SMC_\ell$ of symmetric monoidal categories and lax symmetric monoidal functors. But $SMC_\ell$ is a full sub-2-category of the 2-category $SMulti$ of symmetric multicategories, and the latter 2-category has Eilenberg-Moore objects. I haven’t seen it written out quite like that in the literature, though.

And before you ask, no, $SMC_\ell$ doesn’t have EM-objects for monads. (It does have EM-objects for comonads though.) So passing from monoidal categories to multicategories is one of those cases where we get a better-behaved (2-)category by requiring less of its objects.

Tom wrote:

I haven’t thought about size issues, but every small multicategory embeds into a strict monoidal category.

And the construction you mention doesn’t change the size either. So every large multicategory also embeds into a large strict monoidal category. That construction is pretty easy to describe explicitly too: the objects of $F C$ are finite lists of objects of $C$, and the morphisms are finite lists of morphisms of $C$, with their domains concatenated (with an extra permutation, in the symmetric case). I think if we do this to John’s symmetric multicategory of prime sets we do actually get the symmetric monoidal category of nonempty finite sets.

It’s worth noting that this construction isn’t idempotent: if you start from a monoidal category, regard it as a multicategory, and embed it in a monoidal category, you don’t get back the monoidal category you started from (not even up to equivalence). But there is a fancier embedding that can be made idempotent: in fact, given a multicategory together with any collection of tensor products that exist in it (which might be all of them, if it happens to be a monoidal category), it can be fully embedded into a monoidal category in a way that preserves those tensor products. I’ve actually been thinking about this sort of thing a lot recently; a fancier version of it appears in my paper on $\ast$-autonomous envelopes.

Posted by: Mike Shulman on August 15, 2020 3:31 AM | Permalink | Reply to this

By the way, while the proof that tensor products of algebras over a commutative monad exist does require that the monad has rank (or some other condition ensuring that the necessary colimits of algebras exist), I don’t think I know an example of a commutative monad without rank for which tensor products of algebras provably don’t exist. The obvious example of a commutative monad without rank is the powerset monad, whose algebras are suplattices; but suplattices do have tensor products. Does anyone know an example where tensor products actually fail to exist?

In fact, I’m not even sure I know an example of a monad without rank (on a cocomplete category) for which colimits of algebras provably don’t exist! Do I?

Posted by: Mike Shulman on August 15, 2020 3:34 AM | Permalink | Reply to this

### Day and Street on multicats

Just want to mention that Brian Day and Ross Street provide some interesting generalizations of the concept of multicategory especially suited to the V-enriched context, together with some techniques for reasoning about (rewriting rules) and representing (surface diagrams) such things, in two papers:

“Lax monoids, pseudo-operads, and convolution” (2001-10-12) http://www.math.mq.edu.au/~street/Multicats.pdf

“Abstract substitution in enriched categories” (2002-03) http://www.math.mq.edu.au/~street/Substitudes.pdf or https://www.sciencedirect.com/science/article/pii/S0022404902002918

Posted by: Keith Harbaugh on August 15, 2020 4:30 PM | Permalink | Reply to this

### Re: Day and Street on multicats

Thanks! It looks like a “$V$-substitude” is what Geoff Cruttwell and I called an “unnormalized $T$-monoid” for $T$ the free strict monoidal $V$-category monad on $V Prof$: it has two different unary hom-objects, one from the underlying $V$-category and one from the multicategory structure. (“Normalization” makes the two coincide.) Unfortunately, while we knew about the first of those papers when writing ours (and cited it), we weren’t aware of the second, or we would have mentioned the connection.

Posted by: Mike Shulman on August 16, 2020 2:36 AM | Permalink | Reply to this

### Re: Day and Street on multicats

Another paper on the (potentially enriched) profunctorial approach to multicategories (which therefore also has to deal with normalization) is Hyland’s Elements of a theory of algebraic theories (2013). It seems the same ideas occurred to many people independently.

Posted by: Mike Shulman on August 17, 2020 5:15 AM | Permalink | Reply to this

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