## October 1, 2023

### The Free 2-Rig on One Object

#### Posted by John Baez

These are notes for the talk I’m giving at the Edinburgh Category Theory Seminar this Wednesday, based on work with Joe Moeller and Todd Trimble.

(No, the talk will not be recorded.)

Schur Functors

The representation theory of the symmetric groups is clarified by thinking of all representations of all these groups as objects of a single category: the category of Schur functors. These play a universal role in representation theory, since Schur functors act on the category of representations of any group. We can understand this as an example of categorification. A ‘rig’ is a ‘ring without negatives’, and the free rig on one generator is $\mathbb{N}[x]$, the rig of polynomials with natural number coefficients. Categorifying the concept of commutative rig we obtain the concept of ‘symmetric 2-rig’, and it turns out that the category of Schur functors is the free symmetric 2-rig on one generator. Thus, in a certain sense, Schur functors are the next step after polynomials.

Many important categories act like categorified rings:

• $\mathsf{Set}$, the category of sets, with coproduct and product giving $+$ and $\times$

• $\mathsf{Vect}_k$, the category of vector spaces over a field $k$, with the usual $\oplus$ and $\otimes$

and many more. Since they tend to lack subtraction they are really categorified ‘rigs’. Sometimes the multiplication is the categorical product but often not. Today I’m only interested in examples where the addition is coproduct, though there are examples where it’s not, like:

• $core(\mathsf{Set})$, the groupoid of sets with the $+$ and $\times$ coming from coproduct and product in $\mathsf{Set}$.

I definitely want multiplication to distribute over addition (up to natural isomorphism), but often it distributes over more general colimits. Some categorified rigs have all colimits and multiplication distributes over all of them: for example, $(\mathsf{Set}, +, \times)$ and $(\mathsf{Vect}_k, \oplus, \otimes)$ work this way. But I’m also interested in examples that don’t have all colimits, like

• $\mathsf{FinVect}_k$, the category of finite-dimensional vector spaces over a field $k$, with the usual $\oplus$ and $\otimes$

And I’m also interested in examples that don’t even have all finite colimits, like

• the category of complex vector bundles over a space with the usual $\oplus$ and $\otimes$

I’ve been working with Joe Moeller and Todd Trimble on a class of categorified rigs with a strong linear algebra flavor, including these examples:

• $\mathsf{Vect}_k$, the category of vector spaces over a field $k$, with the usual $\oplus$ and $\otimes$

• $\mathsf{FinVect}_k$, the category of finite-dimensional vector spaces over a field $k$, with the usual $\oplus$ and $\otimes$

• the category of complex vector bundles over a space with the usual $\oplus$ and $\otimes$

They still have absolute colimits, and one great thing is that multiplication automatically distribute over absolute colimits. But what are absolute colimits? For that, let me back up and review a few things about enriched category theory.

Let $\mathsf{V}$ be any cosmos: a complete and cocomplete symmetric monoidal closed category. I only care about the case where $\mathsf{V}$ is $\mathsf{Vect}_k$ made monoidal using its usual tensor product, but it’s fun to talk more generally.

We can define $\mathsf{V}$-categories, which are like ordinary categories but with $hom(x,y) \in V$, and with composition

$\circ : hom(y,z) \otimes hom(x,y) \to hom(x,z)$

being a morphism in $\mathf{V}$. We can define $\mathsf{V}$-functors and $\mathsf{V}$-natural transformations by following our nose. We can also define a tensor product $\mathsf{C} \boxtimes \mathsf{D}$ of $\mathsf{V}$-categories, generalizing the usual product of categories. This lets us define monoidal $\mathsf{V}$-categories, which have a tensor product

$\phantom{|} \otimes \colon \mathsf{C} \boxtimes \mathsf{C} \to \mathsf{C}$

We can also define symmetric monoidal $\mathsf{V}$-categories, etc.

In the world of $\mathsf{V}$-categories, ‘absolute’ colimits are those automatically preserved by all $\mathsf{V}$-functors. More precisely: in enriched category theory we use ‘weighted’ colimits, which are defined using not only a diagram $\mathsf{D}$ but also a ‘weight’ $\Phi: \mathsf{D}^{op} \to \mathsf{V}$. Then, absolute colimits are those whose weights give weighted colimits that are preserved by all $\mathsf{V}$-functors between all $\mathsf{V}$-categories.

For example, if $\mathsf{V} = \mathsf{Set}$ we’re back to ordinary colimits and essentially the only absolute colimits are split idempotents. (You can get all the rest from these.) It’s a good exercise to check that split idempotents are preserved by all functors.

But if $\mathsf{V} = \mathsf{Vect}_k$ there are more absolute colimits:

• the initial object (often called $0$)
• binary coproducts (often called direct sums)
• splittings of idempotents (explained below)

You can get all the rest from these.

Definition. A $\mathsf{V}$-category is Cauchy complete if it has all absolute colimits.

For example, the category of complex vector bundles over a space is a $\mathsf{Vect}_k$-enriched category that has all absolute colimits! It doesn’t have coequalizers of all parallel pairs of morphisms. But given $p \colon x \to x$ that’s idempotent ($p^2 = p$) you can form the coequalizer of $p$ and $0 \colon x \to x$ — or in other words, the cokernel $coker p$. This is called splitting the idempotent $p$ because $1-p$ is also idempotent, so you can also form $coker (1-p)$ and show

$x \cong coker p \oplus coker (1-p)$

Definition. An absolute 2-rig is a monoidal $\mathsf{V}$-category $\mathsf{R}$ that is Cauchy complete.

Note that for any object $x \in \mathsf{R}$, the functors

$x \otimes - : \mathsf{R} \to \mathsf{R}$

and

$- \otimes x: \mathsf{R} \to \mathsf{R}$

automatically preserve all absolute colimits, so we say tensor products distribute over absolute colimits. For example, we have a natural isomorphism

$x \otimes (y \oplus z) \cong (x \otimes y) \, \oplus \, (x \otimes z)$

as you’d hope for in a categorified ring.

From now on I’m going to say ‘2-rig’ when I mean symmetric 2-rig: that is, one where the tensor product is symmetric monoidal. It’s just like how algebraic geometers say ‘ring’ when they mean commutative ring. I used to be annoyed by how algebraic geometers do that, but now I see why: I’m interested in the symmetric case, and it gets really boring saying ‘symmetric’ all the time.

And from now on let’s take $\mathsf{V} = \mathsf{Vect}_k$. In this case the free 2-rig on one object turns out to be an important structure in mathematics, often called the category of Schur functors! But we’ll just work it out.

There is a 2-functor

$F : \mathbf{Cat} \to \mathbf{2Rig}$

that forms the free (symmetric) 2-rig on any category. We can get it by composing three other 2-functors

$\mathbf{Cat} \xrightarrow{S} \mathbf{SymMonCat} \xrightarrow{k[\cdot]} \mathbf{SymMon}\mathsf{V}\mathbf{Cat} \xrightarrow{Cauchy \; completion} \mathbf{2Rig}$

In fact all of these are left adjoints, or technically left pseudoadjoints since we’re working with 2-categories.

Here:

1) $S$ gives the free symmetric monoidal category on a category.

2) For starters, $k[\cdot]$ gives the free $\mathsf{Vect}_k$-category on a category, by replacing each homset $hom(x,y)$ by the free vector space on that set, which we call $k[hom(x,y)]$. Since $\mathsf{V} = \mathsf{Vect}_k$, this gives a 2-functor

$k[\cdot] \colon \mathbf{Cat} \to \mathsf{V}\mathbf{Cat}$

which we can then use to get

$k[\cdot] \colon \mathbf{SymMonCat} \to \mathbf{SymMon}\mathsf{V}\mathbf{Cat}$

3) If $\mathsf{C}$ is some $\mathsf{V}$-category, $\overline{\mathsf{C}}$ is a Cauchy complete category called its Cauchy completion. Cauchy completing a symmetric monoidal $\mathsf{V}$-category we get a 2-rig.

Let’s look at two examples!

As a warmup, let’s start with the empty category $\emptyset$. The free symmetric monoidal category on the empty category, $S\emptyset$, is just the terminal category $1$. The free $Vect_k$-enriched category on this, $k[1]$, still has one object $\ast$ but now it has a one-dimensional space of endomorphisms. So

$hom(\ast, \ast) \cong k$

and composition of morphisms is multiplication in the field $k$. This is actually a symmetric monoidal $\mathsf{V}$-category. What happens when we Cauchy complete it? All idempotents already split in $k[1]$, but it doesn’t have an initial object or binary coproducts. When we throw those in we get the category of finite-dimensional vector spaces! So

$F(\emptyset) \simeq \mathsf{FinVect}_k$

Yes: the free 2-rig on the empty category is $\mathsf{FinVect}_k$. And since left pseudoadjonts preserve initial objects, this means the initial 2-rig is $\mathsf{FinVect}_k$.

That was fun. But now let’s figure out the free 2-rig on one object. More precisely, let’s work out $F 1$ where $1$ is the terminal category.

$S 1$, the free symmetric monoidal category on one object, is equivalent to the groupoid of finite sets! If we use a skeleton it has objects $n \in \mathbb{N}$, and

$hom(m,n) \cong \left\{\begin{array}{cc} S_n & if m = n \\ \emptyset & if m \ne n \end{array} \right.$

So, $S 1$ is a groupoid combining all the symmetric groups.

Next we get $k[S 1]$, the free symmetric monoidal $Vect_k$-category on one object, by linearizing the homsets of $S 1$. So, it has the same objects but now

$hom(m,n) \cong \left\{\begin{array}{cc} k[S_n] & if m = n \\ 0 & if m \ne n \end{array} \right.$

Here $k[S_n]$ is the free vector space on the symmetric group $S_n$. Its usually called the ‘group algebra’ of $S_n$ because it gets a multiplication from multiplication in the group — and this multiplication is how we compose morphisms in $k[S 1]$. At least that’s the interesting part: there are also ‘zero morphisms’ from $m$ to $n$ when $m \ne n$, and composing with these is like multiplying by zero. I’ll summarize all this by writing

$k[S 1] \simeq \bigoplus_{n \ge 0} k[S_n]$

where the direct sum is a way of glomming together $\mathsf{Vect}_k$-categories (their coproduct).

Next we get $\overline{k[S 1]}$, the free 2-rig on one object, by taking the Cauchy completion of $k[S 1]$. To understand this, I’ll now assume $k$ has characteristic zero. Then for any finite group $G$ we have

$\overline{k[G]} \simeq FinRep_k(G)$

where $FinRep_k(G)$ is the category of representations of $G$ on finite-dimensional vector spaces over $k$. So, we get

$F 1 \simeq \overline{k[S 1]} \simeq \bigoplus_{n \ge 0} FinRep_k(S_n)$

Nice! We’re seeing the free 2-rig on one object contains the representation categories of all the symmetric groups, put together in one neat package!

Now I want to prove a fun theorem about the free 2-rig on one object which explains why it’s called the category of ‘Schur functors’. This theorem actually describes, not the free 2-rig $F 1$, but its underlying category, which I’ll call $U(F 1)$.

Theorem. Let $U \colon \mathbf{2Rig} \to \mathbf{Cat}$ be the forgetful 2-functor from 2-rigs to categories, and let $[U,U]$ be the category with

• pseudonatural transformations $\alpha \colon U \Rightarrow U$ as objects
• modifications between these as morphisms.

Then

$[U, U] \simeq U(F 1)$

Proof sketch. I’ll prove a simpler theorem, but the proof of the full-fledged one works just the same way. Let’s decategorify and look at the forgetful functor from rings to sets, $U \colon \mathsf{Ring} \to \mathsf{Set}$. In this case $[U,U]$ is just the set of natural transformations $\alpha \colon U \Rightarrow U$. I’ll show you that

$[U, U] \simeq U(F 1)$

where $F \colon \mathsf{Set} \to \mathsf{Ring}$ is the ‘free ring on a set’ functor. In this decategorified case $F 1$ is just $\mathbb{Z}[x]$, the algebra of polynomials in one variable.

The proof is quick. We use a little formula for the functor $U$:

$U \cong \mathsf{Ring}(F 1, -) \qquad \qquad (\star)$

In other words, for any ring $R$, the underlying set $U R$ is naturally isomorphic to $\mathsf{Ring}(F 1, R)$, which is the set of ring homomorphisms from $\mathbb{Z}[x]$ to $R$.

This is obvious, because such a ring homomorphism can send $x$ to any element of $R$, and that determines it. But let’s prove this fact in a way that generalizes! Note that since $F$ is left adjoint to $U$ we have

$\mathsf{Ring}(F 1, R) \cong \mathsf{Set}(1, U R)$

but for any set $X$ we have $\mathsf{Set}(1,X) \cong X$, so

$\mathsf{Set}(1, U R) \cong U R$

So, we’ve shown $(\star)$.

Next we calculate:

$\begin{array}{ccll} [U,U] &\cong& \left[\mathsf{Ring}(F 1, -), \mathsf{Ring}(F 1, -) \right] & by \; (\star) \\ & \cong & \mathsf{Ring}(F 1, F 1) & by \; the \; Yoneda \; Lemma \\ & \cong & U(F 1) & by \; (\star) \; again \end{array}$

We’re done!    █

Now, when you carefully look at this proof you’ll see it has nothing to do with rings, or 2-rigs. It’s extremely general! In the decategorified case all we needed was a right adjoint functor from any category to $\mathsf{Set}$. Similarly, in the full-fledged case all we need is a right pseudoadjoint 2-functor $U$ from any 2-category to $\mathbf{Cat}$. Then we recover $U(F 1)$ from the category of pseudonatural transformations of $U$. Experts would call this result a kind of ‘Tannaka reconstruction’ theorem.

But what does this result actually mean in the cases I’m talking about?

In the case of rings, $\mathbb{Z}[x]$ acts naturally on the underlying set of any ring. Say we have a polynomial

$P = \sum_{n \ge 0} a_n x^n \in \mathbb{Z}[x]$

Then for any ring $R$, we get a map

$r \mapsto \sum_{n \ge 0} a_n r^n$

It’s not a ring homomorphism, just a map from the underlying set $U R$ to itself. And its natural in $R$.

Simple enough. But the theorem says something deeper: every natural map $U(R) \to U(R)$ comes from a polynomial in $\mathbb{Z}[x]$.

In the case of 2-rigs the story is similar. $\overline{k[S 1]}$ acts pseudonaturally on the underlying category of any 2-rig. Say we have an object

$P = \bigoplus_{n \ge 0} \rho_n \in \overline{k[S 1]}$

Remember that now $\rho_n$ is a finite-dimensional representation of $S_n$. Then for any 2-rig $\mathsf{R}$, we get a map

$x \mapsto \bigoplus_{n \ge 0} \rho_n \otimes_{k[S_n]} x^{\otimes n}$

It’s not a 2-rig map, just a functor from the underlying category $U \mathsf{R}$ to itself. It’s called a Schur functor. And it’s pseudonatural in $\mathsf{R}$.

#### Postlude

Everything I explained is in here:

so you can look there for more details and references to previous work. But for the $n$-Category Café crowd, I’ll add that I guessed a primitive version of this theorem here on the n-Café back in 2007, in response to this question of Allen Knutson:

For as long as I’ve understood Schur functors, I’ve thought about them as functors $Vect_{\mathbb{C}} \to Vect_{\mathbb{C}}$. But now that we’re going through them in a reading course on Fulton’s Young Tableaux, I discover that the input isn’t really a complex vector space, but an arbitrary module over a commutative ring. (And maybe, just maybe, a bimodule over a noncommutative one, but I doubt it.)

In particular, the Schur functor commutes with base change, aka extension of scalars.

What is the right way to describe this object, categorically? (Or should I say, 2- or 3-categorically?)

You can see me guessing the theorem back then… but it took Todd Trimble to prove it. And in the process, he whittled down my assumptions and came up with the definition of 2-rig I’m using here. It doesn’t need to have lots of colimits, just absolute colimits.

When Joe, Todd and I wrote our paper, which includes a lot more material than this, we acknowledged Allen as “prime instigator”, and blogged about it here. But I never wrote a simple explanation of how we proved the theorem above! So here it is.

Posted at October 1, 2023 3:00 PM UTC

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### Re: The Free 2-Rig on One Object

In my father’s book “lambda-rings and the representation theory of the symmetric group” he proves that the ring of symmetric functions is the free lambda-ring on one generator. Any idea how parallel that is to your one-object story?

Posted by: Allen Knutson on October 1, 2023 4:41 PM | Permalink | Reply to this

### Re: The Free 2-Rig on One Object

At nLab: Schur functor we have

Mathematicians often work with a decategorified version of $Schur$: its Grothendieck group, also known as the ring of symmetric functions. The various structures that $Schur$ possesses endow this ring with corresponding structures. Among other things, it is the free lambda-ring on one generator. As we shall see, this corresponds to the fact that $Schur$ is the free symmetric monoidal Cauchy complete linear category on one object.

Posted by: David Corfield on October 1, 2023 6:16 PM | Permalink | Reply to this

### Re: The Free 2-Rig on One Object

In the last theorem of our paper we show the Grothendieck ring of any 2-rig is a $\lambda$-ring. We show the Grothendieck ring of the free 2-rig on one object is $\Lambda$, the ring of symmetric functions. So it certainly seems believable that $\Lambda$ is the free $\lambda$-ring on one generator. But that’s not a proof.

In fact this result is a consequence of our theory. But sadly I don’t think we spelled it out. Let me do that here! In what follows ‘ring’ always means ‘commutative ring’.

The biggest result in our paper is that the free 2-rig with one generator is a 2-plethory. Whatever that means, we used it to show that the ring of symmetric functions, $\Lambda$, is a plethory.

But what the hell does that mean? It means first that the representable functor

$\mathsf{Ring}( \Lambda, -) \colon \mathsf{Ring} \to \mathsf{Set}$

lifts to a functor

$\Phi \colon \mathsf{Ring} \to \mathsf{Ring}$

but then that $\Phi$ has the structure of a comonad with a left adjoint. By some general abstract nonsense, its left adjoint

$\Psi \colon \mathsf{Ring} \to \mathsf{Ring}$

is then automatically a monad. And we show it’s the monad for $\lambda$-rings!

So: $\Psi$ sends any ring to the free $\lambda$-ring on that ring. Thus, $\Psi(\mathbb{Z}[x])$ is the free $\lambda$-ring on the free ring on one generator. In short, it’s the free $\lambda$-ring on one generator.

But can we see that it’s $\Lambda$, as your father showed? Well, look: for any ring $R$ we have

$\mathsf{Ring}(\Psi(\mathbb{Z}[x]), R) \cong \mathsf{Ring}(\mathbb{Z}[x], \Phi(R)) \cong U(\Phi(R))$

where

$U \colon \mathsf{Ring} \to \mathsf{Set}$

is the underlying set functor. But I said $\Phi$ lifts the representable functor $\mathsf{Ring}(\Lambda, -)$ to $\mathsf{Ring}$. This says exactly that

$U(\Phi(R)) \cong \mathsf{Ring}(\Lambda, R)$

So we have

$\mathsf{Ring}(\Psi(\mathbb{Z}[x]), R) \cong \mathsf{Ring}(\Lambda, R)$

But this isomorphism is natural in $R$, so by Yoneda we see that yes,

$\Psi(\mathbb{Z}[x]) \cong \Lambda$

So: $\Lambda$ is the free $\lambda$-ring on one generator!

Now I think we should have acknowledged your father as “prime instigator”.

Posted by: John Baez on October 1, 2023 10:40 PM | Permalink | Reply to this

### Re: The Free 2-Rig on One Object

We keep saying “ring of symmetric functions”, more or less to sound sort of homey and familiar to people who already know the storied background, but in this paper we never actually talk about symmetric functions as such. So in some sense “ring of symmetric functions” is a distraction from what we actually do.

What we actually do is define $\lambda$-rings more from the coalgebraic point of view: as coalgebras of a “Witt ring comonad”

$W = CRing(\Lambda, -): CRing \to CRing$

emanating from the structure of the free 2-rig. For me, the details of getting from the free 2-rig to the comonad was the most challenging thing to set up. Somewhat to my surprise, this took some careful rethinking of what decategorification is all about – surprising because I usually think of decategorification, i.e., taking isomorphism classes of objects, as something fairly simple and obvious!

But once this is done, it follows pretty readily from the comonad $W$’s being a right adjoint that $\lambda$-rings are both algebraic and coalgebraic (monadic and comonadic) over commutative rings – this categorical fact goes back to the original Eilenberg and Kelly paper, where they introduce their category of algebras construction for a monad! This implies in turn that $\lambda$-rings are indeed algebraic over sets, so that we can talk about things like free $\lambda$-rings on sets.

[Of course this point of view is closely connected with what Tall and Wraith do as well; what is novel with us is starting with the free 2-rig and applying almost purely categorical maneuvers until we get to the plethory given by the comonad $W$. There is one exception to “purely categorical” – we need to invoke “splitting fields” to get us over one critical point.]

So proving that $\Lambda$, which we take to be the Grothendieck ring of the free 2-rig, is the free $\lambda$-ring on one generator, depends greatly on how one defines $\lambda$-rings in the first place. Allen of course knows how Grothendieck defined $\lambda$-rings, or what he calls “special $\lambda$-rings”, and that it’s a bit involved and roundabout and hard to write down directly in terms of operations and equational axioms on the operations – it has a lot to do with consequences of the “splitting principle”. (Which incidentally we’ve also been thinking about recently in connection with 2-rigs.) Anyway, as I said, we define $\lambda$-rings as coalgebras over a comonad, largely because – speaking for myself here – I found it a heck of a lot easier to think about covariant hom-functors like $CRing(\Lambda, -)$ than I do the plethystic monoidal product $\odot$ on birings!

I think John already did a good job explaining why $\Lambda$ is the free $\lambda$-ring. Before I had even read the post and these comments, I got a heads-up in email from John about Allen’s question and wrote some details down before taking a peek here, and my account winds up looking similar to John’s, but FWIW here is mine.

By the Eilenberg-Kelly paper, or by string-diagrammatic arguments, the left adjoint of a right adjoint comonad, like

$CRing(\Lambda, -): CRing \to CRing,$

carries a monad structure, and the coalgebras of the comonad are naturally equivalent to algebras of its left adjoint monad $M$. (This $M$ by the way is usually denoted $\Lambda \odot -: \mathsf{CRing} \to \mathsf{CRing}$ by people who work with plethories, but that’s not necessary to know here.) In other words, the forgetful functor

$\lambda\text{-}\mathsf{Ring} \to \mathsf{CRing}$

is both monadic and comonadic, as we were saying, and in particular has a left adjoint. And so, the composite of forgetful functors

$\lambda\text{-}\mathsf{Ring} \to \mathsf{CRing} \to \mathsf{Set}$

has a left adjoint. By John’s “little formula” marked by a star, lemma 5.2 in our paper, the composite forgetful functor is represented by the free $\lambda$-ring on one generator. That by the way, as a ring, is $M(\mathbb{Z}[x])$, by composition of left adjoints and the fact that $\mathbb{Z}[x]$ is the free commutative ring on one generator.

Anyway, it’s the value at a one-element set of a composite $F$ of left adjoints

$\mathsf{Set} \overset{\mathrm{free}}{\to} \mathsf{CRing} \overset{M}{\to} \mathsf{CRing}$

whose right adjoint is, by how we set things up, the representable functor $\mathsf{CRing}(\Lambda, -): \mathsf{CRing} \to \mathsf{Set}$. But now $\Lambda$ is $F(1)$, again using John’s little formula. So the free $\lambda$-ring on one generator is $\Lambda$.

There are lots of other little by-products of the categorical story. One is that coproducts of $\lambda$-rings are given by coproducts of their underlying commutative rings, so for example, the free $\lambda$-ring on two generators is $\Lambda \otimes \Lambda$, and similarly for any finite number of generators. The free $\lambda$-ring on a general set is, as expected, a filtered or directed colimit of free $\lambda$-rings on finite subsets (and this colimit is computed at the set-theoretic level).

Posted by: Todd Trimble on October 2, 2023 4:51 PM | Permalink | Reply to this

### Re: The Free 2-Rig on One Object

By the way, my proof here that the ring of symmetric functions $\Lambda$ is the free $\lambda$-ring on one generator is a bit ‘tricky’, since I was wanting to deploy things we’d already shown and not think too much.

It might be nicer to try something else. We’ve shown that taking the Grothendieck ring turns any 2-rig into a $\lambda$-ring. We should show it’s functorial. Since 2-rigs form a 2-category but $\lambda$-rings form a mere category, we should try to prove there’s a functor

$K \colon \mathsf{2Rig}_h \to \lambda\mathsf{Ring}$

where $\mathsf{2Rig}_h$ is the category of 2-rigs and natural isomorphism classes of maps between 2-rigs. (The $h$ stands for ‘homotopy category’, and we discuss this sort of homotopy category in our paper.)

Then, if we can show $K$ is a left adjoint, it should follow that $K$ maps the free 2-rig on one object to the free $\lambda$-ring on generator!

(As Todd hinted, I am saying “the ring of symmetric functions” for “the Grothendieck ring of the category of Schur functors”, since I know an isomorphism between them.)

Posted by: John Baez on October 2, 2023 9:54 AM | Permalink | Reply to this

### Re: The Free 2-Rig on One Object

Part of what you say is very close to what we did: we sketch on page 51 how the values of the functor $K: 2\mathsf{Rig}_h \to \mathsf{CRing}$ that we defined earlier are actually $\lambda$-rings. The functoriality of the map to $\lambda$-rings is fairly immediate.

I remember spending a little time thinking about the nature of $2\mathsf{Rig}_h$ and how it might differ from $\lambda\, \mathsf{Ring}$, but I didn’t settle on a firm conclusion about this. I just remember thinking “$2\mathsf{Rig}_h$ looks very interesting”. I think I got the idea that the two categories are probably not equivalent, although it seems to be empirically true that almost all $\lambda$-rings in practice come from 2-rigs.

Posted by: Todd Trimble on October 2, 2023 5:16 PM | Permalink | Reply to this

### Re: The Free 2-Rig on One Object

we acknowledged Allen as “prime instigator”, and blogged about it here.

Posted by: RodMcGuire on October 1, 2023 6:42 PM | Permalink | Reply to this

### Re: The Free 2-Rig on One Object

Whoops! I meant here. Thanks for catching that.

Posted by: John Baez on October 1, 2023 10:31 PM | Permalink | Reply to this

### Re: The Free 2-Rig on One Object

For example, if $\mathsf{V}=Set$ we’re back to ordinary colimits and essentially the only absolute colimits are split coequalizers. (You can get all the rest from these.)

Is that really right? E.g. when you say that you can get all absolute colimits from split coequalizers, how would you get a split idempotent? Here we’re talking about ordinary categories (since $\mathsf{V} = Set$), not $Vect$-enriched categories, so you can’t add and subtract morphisms.

I would agree that essentially the only absolute colimits (over $\mathsf{V} = Set$) are split idempotents, and that you can get all absolute colimits from them. For instance, it’s a theorem that a category has all absolute colimits iff all idempotents in it split. Is that what you meant to write?

Split coequalizers are just the first in an infinite sequence of more and more complicated absolute coequalizers, a result I think is due to Bob Paré. The full sequence is described in Exercise 2.17.7 in Vol 1 of Borceux’s Handbook of Categorical Algebra.

Posted by: Tom Leinster on October 2, 2023 8:41 PM | Permalink | Reply to this

### Re: The Free 2-Rig on One Object

I might have an inkling where that came from. If you can split idempotents $e: A \to A$, say $e = s p$ where $s: X \to A$ is a section of $p: A \to X$, then the sequence

$A \underset{1}{\overset{e}{\rightrightarrows}} A \overset{p}{\to} X$

is a split coequalizer, where the splitting is given by $s: X \to A$ and $1: A \to A$. We apply that kind of doodad in lemma 2.3 (at least in the arXiv version). Perhaps that explains the conflation?

Posted by: Todd Trimble on October 2, 2023 11:34 PM | Permalink | Reply to this

### Re: The Free 2-Rig on One Object

Tom wrote:

Is that what you meant to write?

Let me take advantage of the opportunity you’re offering here and say “yes”. I certainly don’t have a good understanding of absolute colimits. I can see that the ones I mentioned are absolute, but I have no idea how one shows a certain class of them is enough to generate all of them.

(The reason is that I’ve never been interested, but I guess you’re not allowed to say that if someone asks a question about something in your talk! I guess it’s time for me to get interested.)

Posted by: John Baez on October 3, 2023 12:09 PM | Permalink | Reply to this

### Re: The Free 2-Rig on One Object

The thing is, I don’t know how you’d even make precise the statement that split coequalizers generate all absolute colimits.

For instance, suppose we wanted to make precise the following statement: a category has split coequalizers iff it has all absolute colimits. What does it even mean for a category to “have split coequalizers”? That every parallel pair has a coequalizer, and it’s split? That can’t be right, as it’s not even true in Set (which has all absolute colimits): consider a parallel pair with empty domain and nonempty codomain.

The general definition of “has absolute colimits” is a little tricky. The definition I know is this. Take a category $A$ and write $y: A \to [A^{op}, Set]$ for the Yoneda embedding. A category $A$ has absolute colimits if whenever $D$ is a diagram in $A$ such that the colimit of $y \circ D$ is absolute, then $D$ itself has a colimit in $A$ and that colimit is absolute (or equivalently, is preserved by $y$). This is a definition for ordinary categories ($V = Set$), but you can easily generalize it to arbitrary $V$.

A crucial thought behind this definition is that although absoluteness of a colimit is often defined as “preserved by any functor into any category” (a pretty wild statement: think how widely that quantification is ranging!), it’s equivalent to ask that it’s preserved by the Yoneda embedding, at least under some mild size assumptions.

Differently put, a colimit $colim a_i$ in a category $A$ is absolute iff the natural map

$colim A(a_i, b) \to A(colim a_i, b)$

is an isomorphism for all $b \in A$. For example, direct sums in an additive category are absolute because

$A(a_1, b) \oplus A(a_2, b) \cong A(a_1 \oplus a_2, b).$

Also, I gave one definition of what it means for a $V$-category to “have absolute colimits”, but I guess there’s an equivalent one in terms of existence of colimits for the absolute $V$-weights. That would be closer to what you want to say, I think, John.

Posted by: Tom Leinster on October 3, 2023 2:41 PM | Permalink | Reply to this

### Re: The Free 2-Rig on One Object

Great stuff!

Have any of you thought about whether the requirement that $k$ is a field of characteristic zero can be removed, perhaps with some tweaks to the set up? It would be very nice if it could be any ring (or better, any rig!).

I suspect however that you’d have to allow $k$ to be variable rather than keep it fixed. What I have in mind is something like the following: instead of working with the category of finite-dimensional vector spaces over a fixed $k$, you might use the category of finitely generated projective (or free?) modules, viewed as a (co)fibered category over the category of rings.

Why might this be necessary? When some people talk about polynomial functors on vector spaces over a given field $k$, they like to assume $k$ is infinite. If I recall, this is because over finite fields, polynomials don’t map injectively to functions. However, if you let the field vary over all extensions of the field you start with, then this problem disappears. If you like, this is because all algebraically closed fields are infinite. A more concrete statement that basically amounts to the same thing is that the polynomials $x$ and $x^{p^n}$ agree on the field with $p^m$ elements if and only if $m$ divides $n$.

Put this way, it all looks a bit technical. The real reason why I like the fibered category point of view is that it is usually the right one in algebraic geometry. Geometric properties are supposed to be independent of the ring of scalars. Many people would say that that’s the defining property of algebraic, as opposed to arithmetic, geometry. And since the story in your paper is mostly independent of $k$, it would be nice to make it really independent of $k$.

Posted by: James Borger on October 3, 2023 11:46 AM | Permalink | Reply to this

### Re: The Free 2-Rig on One Object

James wrote:

Have any of you thought about whether the requirement that $k$ is a field of characteristic zero can be removed, perhaps with some tweaks to the set up? It would be very nice if it could be any ring (or better, any rig!).

Indeed. We’ve proved a lot of stuff, and large portions of it don’t depend at all on working over a field of characteristic zero… or even a field… or even a ring… or even a rig.

For example most of the abstract stuff in the blog article here is incredibly general. The abstract construction of the free 2-rig on a category

$F : \mathbf{Cat} \to \mathbf{2Rig}$

as a composite of three processes

$\mathbf{Cat} \xrightarrow{S} \mathbf{SymMonCat} \xrightarrow{k[\cdot]} \mathbf{SymMon}\mathsf{V}\mathbf{Cat} \xrightarrow{Cauchy \; completion} \mathbf{2Rig}$

doesn’t require that $\mathsf{V}$ to be $\mathsf{Vect}_k$ for a field with characteristic zero. I’m pretty sure $k$ can be any commutative ring, at least.

In fact it might work when $\mathsf{V}$ is any cosmos — that is, any complete and cocomplete symmetric monoidal closed category. Remember, I defined 2-rigs at this level of generality: they are symmetric monoidal Cauchy complete $\mathsf{V}$-categories.

So, while I have the attention of all the worthies here I should ask a question!

Whenever $\mathsf{V}$ is a cosmos I believe there is a god-given symmetric monoidal cocontinuous functor

$\mathsf{Set} \to \mathsf{V}$

I believe it’s unique up to natural isomorphism too. (It must send any coproduct of copies of $1 \in \mathsf{Set}$ to the corresponding coproduct of copies of the tensor unit $I \in \mathsf{V}$.)

We can thus do base change along this functor to define a 2-functor

$\mathbf{Cat} \to \mathsf{V}\mathbf{Cat}$

and I believe we can enhance this and get a 2-functor

$v \colon \mathbf{SymMonCat} \to \mathbf{SymMon}\mathsf{V}\mathbf{Cat}$

Does this sound right so far?

If so, we have a nice way to turn any category into a 2-rig:

$\mathbf{Cat} \xrightarrow{S} \mathbf{SymMonCat} \xrightarrow{v} \mathbf{SymMon}\mathsf{V}\mathbf{Cat} \xrightarrow{Cauchy \; completion} \mathbf{2Rig}$

Then comes the really interesting part, which I’m less confident about: is the 2-functor $v$ a left pseudoadoint? If it is, all three of the above 2-functors are left pseudoadjoints, and so is their composite

$F : \mathbf{Cat} \to \mathbf{2Rig}$

We thus get a right pseudoadjoint

$U : \mathbf{2Rig} \to \mathbf{Cat}$

and we can prove

$[U,U] \simeq U(F 1)$

using the argument I described! That would be cool.

The concrete description of the free 2-rig on one object in terms of representations of symmetric groups used the fact that every finite-dimensional representation of $S_n$ is a finite direct sum of summands of the regular representation. And that’s not true in characteristic $p$. Right?

This fact is what makes the Cauchy completion of the group algebra $k[S_n]$ (viewed as 1-object $\mathsf{V}$-category) be the category of all finite-dimensional representations of $k[S_n]$.

What if we let $k$ be a general commutative ring and let $\mathsf{V} = k\mathsf{Mod}$? Then the Cauchy completion of $k[S_n]$ is the category of finite-dimensional projective modules of $k[S_n]$.

That still seems pretty nice.

Instead of working with the category of finite-dimensional vector spaces over a fixed $k$, you might use the category of finitely generated projective (or free?) modules, viewed as a (co)fibered category over the category of rings.

I believe the abstract nonsense in my talk would eat what you’re saying for breakfast and pound the table demanding lunch!

Posted by: John Baez on October 3, 2023 2:22 PM | Permalink | Reply to this

### Re: The Free 2-Rig on One Object

@ John B:

    Yummers

Posted by: jack morava on October 4, 2023 1:23 PM | Permalink | Reply to this

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