### Pseudocommutative 2-Monads

#### 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.

I want to ask some questions about a categorified version of this story, involving pseudocommutative 2-monads.

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’:

- Martin Hyland and John Power, Pseudo-commutative monads and pseudo-closed 2-categories,
*Journal of Pure and Applied Algebra***175**(2002), 141–185.

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.

## Re: Pseudocommutative 2-Monads

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

There’s no monad

at allfor 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).