### 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 FunctorsThe 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:

- John Baez, Joe Moeller and Todd Trimble, Schur functors and categorified plethysm, to appear in
*Higher Structures*.

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.

## 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?