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Stream: theory: mathematics

Topic: Big Witt ring

John Baez (Oct 06 2024 at 17:09):

I'm trying to understand the "big Witt ring" of a commutative ring; there are various perspectives on it.

1) One slick abstract perspective is that the forgetful functor from [[lambda-rings]] to commutative rings has both the expected left adjoint but also a right adjoint

$W : \mathsf{CommRing} \to \lambda\mathsf{Ring}$

and $W(R)$ is the big Witt ring. This perspective is helpful in proportion to how well you understand lambda-rings. There are several perspectives on lambda-rings, each of which should give a different outlook on the big Witt ring. My favorite is that when you decategorify a 2-rig by taking its Grothendieck group, you get not only a ring but a lambda-ring. But there is another important perspective based on number theory, where a lambda-ring is a commutative ring equipped with commuting [[Frobenius lifts]], one for each prime $p$.

John Baez (Oct 06 2024 at 17:14):

2) The underlying set of $W(R)$ is

$\mathsf{CommRing}(\Lambda, R)$

where $\Lambda$ is the underlying ring of the free lambda-ring on one generator. I will explain why later.

$\Lambda$ is a biring, i.e. a ring object in $\mathsf{CommRing}^{\text{op}}$, and the multiplication in $W(R)$ comes from the comultiplication in $\Lambda$.

John Baez (Oct 06 2024 at 17:23):

3) Since one can show the underlying commutative ring of the free lambda-ring on one generator is

$\Lambda = \mathbb{Z}[\lambda_1, \lambda_2, \dots ]]$

we have an isomorphism of sets

$W(R) \cong 1 + t R[[t]] \subset R[[t]]$

where

$f \in \mathbb{Z}[\lambda_1, \lambda_2, \dots ]]$

is mapped to

$1 + \sum_{n = 1}^\infty f(\lambda_i) t^n$

Then the challenge is to describe the addition and multiplication on $W(R)$ in these terms. People often do this with explicit formulas, which I tend to find cryptic. The addition in $W(R)$ corresponds to the multiplication in $1 + R[[t]]$, which is why we use this description. The multiplication in $W(R)$ is more tricky. For any $a \in R$ we get an element of $W(R)$ called $(1 - a t)^{-1}$, defined by

$(1 - a t)^{-1} = 1 + a t + a^2 t^2 + \cdots$

and the multiplication $\cdot_W$ on $W(R)$ turns out to be determined once we know

$(1 - a t)^{-1} \cdot_W (1 - b t)^{-1} = (1 - a b t)^{-1}$

This formula turns out to be very useful but I don't have a good understanding of how it comes from 2).

John Baez (Oct 06 2024 at 17:31):

4) There's a ring isomorphism

$W(R) \cong R \times R \times R \times \cdots$

where $a \in W(R)$ is sent to $(g_1(a), g_2(a), \dots)$, and

$g_i : W(R) \to R$

sends $a \in R$ to something called its ith ghost component. Here's an explicit formula for the ghost components that one often sees. Start by using the isomorphism

$W(R) \cong 1 + t R[[t]]$

from 3) to write $a \in W(R)$ as a power series in $t$. Then write

$\displaystyle{ t \frac{d}{d t} \ln a = \sum_{i = 1} g_i t^i }$

This is quite cryptic at first sight, but there has got to be some conceptual interpretation, probably involving [[Adams operations]].

John Baez (Oct 06 2024 at 17:44):

5) Back to conceptual interpretations, Cartier noticed that the big Witt ring $W(R)$ can be seen as the ring of all formal curves starting at $1$ in the multiplicative group of $R$... or something like that. This is explained here, where that ring of formal curves starting at $1$ is called $C(\mathbb{G}_m, R)$.

This is probably just another way of thinking about 3), but it connects the big Witt ring to formal group laws, and in particular the "multiplicative formal group law", and I believe this should ultimately clarify the following trio of facts: a) the $K$-theory of any space is a $\lambda$-ring, b) $K$-theory is a complex oriented cohomology theory, 3) such cohomology theories are classified by formal group laws, 4) $K$-theory corresponds to the multiplicative formal group law.

(Cartier generalized the big Witt ring to an arbitrary formal group law.)

John Baez (Oct 06 2024 at 19:19):

It should be possible to clearly see how all these facts follow from the definition 1), but I'm not there yet!

Here's how 1) implies 2). We use the fact that the forgetful functor

$U : \lambda\mathsf{Ring} \to \mathsf{CommRing}$

has not only a left adjoint

$F: \mathsf{CommRing} \to \lambda\mathsf{Ring}$

but also a right adjoint

$W: \mathsf{CommRing} \to \lambda\mathsf{Ring}$

Let's use this to compute the underlying set of $W(R)$ for some commutative ring $R$.

The underlying set of any commutative ring $A$ is $\mathsf{CommRing}(\mathbb{Z}[x], A)$ since $\mathbb{Z}[x]$ is the free commutative ring on one generator.

Thus the underlying set of (the underlying commutative ring of) $W(R)$ is

$\mathsf{CommRing}(\mathbb{Z}[x], U(W(R))$

but since $U$ has a left adjoint $L$ this is

$\lambda\mathsf{Ring}(F(\mathbb{Z}[x]), W(R))$

and since $W$ has a left adjoint $U$ this is

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

By general nonsense $F(\mathbb{Z}[x])$ is the free lambda-ring on one generator. The underlying commutative ring of this, $U(F(\mathbb{Z}[x]))$, is denoted $\Lambda$. So, from this we get

The underlying set of $W(R)$ is isomorphic to the set of ring homomorphisms from $\Lambda$ to $R$.

It happens that $\Lambda$ is the ring of [[symmetric functions]], but we didn't need this in the above argument. It also happens that the ring of symmetric functions is the polynomial ring on generators $\lambda^i$ called [[elementary symmetric functions]]. This is called the fundamental theorem of symmetric functions. We used this to see why 2) implies 3).

Morgan Rogers (he/him) (Oct 07 2024 at 05:50):

Where did this come from original and why is it named after Witt? I've never heard of it, so I'm curious how it connects to other things and why you're interested in understanding it now ;)

David Corfield (Oct 07 2024 at 08:09):

There was an exchange between you and James Borger back here which touched on Big Witt vectors. Jack Morava and David Ben-Zvi chip in and get to Adams operations. Maybe a conversation worth mining.

John Baez (Oct 07 2024 at 17:18):

Morgan Rogers (he/him) said:

Where did this come from original and why is it named after Witt? I've never heard of it, so I'm curious how it connects to other things and why you're interested in understanding it now ;)

Let me dig into the history....

Okay, it turns out Witt introduced a kind of "Witt vector" associated to an algebra over a field of characteristic $p$ back in 1936, in his paper Zyklische Körper und Algebren der Characteristik $p$ vom Grad $p^n$. Struktur diskret bewerteter perfekter Körper mit vollkommenem Restklassenkörper der Charakteristik $p^n$. No, that's not the whole paper - it's just the title.

John Baez (Oct 07 2024 at 17:21):

Given a commutative ring and a prime $p$ we can make up a "$p$-typical Witt ring" containing these Witt vectors. But then someone noticed you can combine all these $p$-typical Witt rings in a single "big Witt ring", and that's what I'm talking about.

People use these for various mysterious number-theoretic tasks, which I would like to understand someday.

John Baez (Oct 07 2024 at 17:28):

But here's why I'm actually interested!

The big Witt ring is now best understood as the cofree lambda-ring on a commutative ring. Lambda-rings are important in representation theory, where they let us study things like exterior powers and symmetric powers of representations. But also, surprisingly, they're important in number theory, where they let us study things like "Frobenius lifts" - lifting the Frobenius endomorphism of a field to commutative algebras over that field.

John Baez (Oct 07 2024 at 17:30):

For a long time I've been interested, just as a hobby, in understanding why people think the Riemann Hypothesis is connected to the "field with one element". So I was very interested when lambda-rings and the big Witt ring beame important in James Borger's approach to the field with one element:

• James Borger, Lambda-rings and the field with one element.

His idea is that an algebra over the field with one element is a lambda-ring.

Digging into this, I've come to understand fairly well how lambda-rings arise from decategorifying concepts from representation theory. I've always liked representation theory, and @Joe Moeller and @Todd Trimble and I have showed that categories of representations tend to be 2-rigs, and the Grothendieck group of any 2-rig is a lambda-ring.

John Baez (Oct 07 2024 at 17:32):

But I'm less comfortable with how lambda-rings are connected to number theory and Frobenius lifts. Mainly, it seems like a miracle that lambda-rings show up in two different contexts: representation theory in characteristic zero, and number theory in characteristic $p$!

They can't really be different; they must be deeply connected, so I want to understand this.

John Baez (Oct 07 2024 at 17:36):

One way for me to start understanding this is to take the big Witt ring of a ring $R$, which doesn't seem to involve number theory or primes - it's just the cofree lambda-ring on $R$ - and see how it's built from $p$-typical lambda rings, one for each prime.

John Baez (Oct 07 2024 at 17:38):

For years I've felt I don't have the intelligence to fully understand the big Witt ring - so I wished there were some sort of "half-Witt ring" to practice on. But it's gradually starting to make sense.

Morgan Rogers (he/him) (Oct 07 2024 at 18:11):

That sounds like fun! Someone suggested to me at some point that I should try and apply some of the stuff I've done with monoids to Lambda-rings, so I'll be keeping an eye on this topic.

John Baez (Oct 07 2024 at 18:34):

What could you do with lambda-rings, speculatively speaking?

John Baez (Oct 07 2024 at 18:38):

David Corfield said:

There was an exchange between you and James Borger back here which touched on Big Witt vectors. Jack Morava and David Ben-Zvi chip in and get to Adams operations. Maybe a conversation worth mining.

Yes! One difficulty I've had is connecting the 'big picture' ideas to the nitty-gritty comutations with lambda-rings and the big Witt ring. So, I've been digging into the nitty-gritty and now maybe I'll be better able to connect it to the big picture.

John Baez (Oct 07 2024 at 18:39):

There's a nice connection between Adams operations and Frobenius operators which is starting to make sense to me.

Josselin Poiret (Oct 08 2024 at 08:27):

I haven't thought about this in a bit but I remember that the explanation in the nlab was very useful to understand why lambda rings are interesting

Morgan Rogers (he/him) (Oct 08 2024 at 16:20):

John Baez said:

What could you do with lambda-rings, speculatively speaking?

It's been just long enough that I can't remember; the general idea was to treat them as internal somethings in a topos of suitably chosen monoid actions, but that much is not exactly profound.

John Baez (Oct 08 2024 at 18:59):

Josselin Poiret said:

the explanation in the nlab was very useful to understand why lambda rings are interesting

Yes, thanks! It seems to have improved since I last read it, or maybe I just know more now. This is about what Borger calls the heterodox interpretation of a lambda-ring as a ring with a family of commuting Frobenius lifts, one for each prime.

John Baez (Oct 08 2024 at 19:22):

Equivalently it can be seen as a ring with with commuting [[p-derivations]] - a generalization of the concept of derivation. This allows us to develop a number-theoretic generalization of the concept of Taylor series, building on the known analogy between the ring of formal power series $\mathbb{Z}[[x]]$ and the ring of p-adic integers, which gives a geometrical interpretation of localizing at a prime. If we go down this fascinating road, it's good to treat a lambda-ring as a [[Joyal delta-ring]].

John Baez (Oct 08 2024 at 19:25):

But I understand a lot more about what Borger calls the orthodox interpretation of lambda-rings, which is a ring equipped with $\lambda$-operations. The idea here is that if we have a category with well-behaved "exterior power" operations, like a category of vector bundles or group representations, these will endow it's Grothendieck ring with $\lambda$-operations making it into a lambda-ring.

John Baez (Oct 08 2024 at 22:23):

I find this to be a great intro to the 'heterodox' interpretation of lambda-rings and the big Witt ring:

• Kiran Kedlaya, Notes on prismatic cohomology.

Ignore the scary title and read the nice review material!

Todd Trimble (Oct 12 2024 at 04:07):

Let me get back to you later.

John Baez (Oct 12 2024 at 05:01):

Todd and I were talking about how multiplication in the big Witt ring $W(R)$ of a commutative ring $R$ arises from comultiplication in the biring of symmetric functions $\Lambda$, via the formula

$W(R) = \mathsf{CRing}(\Lambda, R)$

Recall that as a commutative ring,

$\Lambda = \mathbb{Z}[\lambda^1, \lambda^2, \dots ]$

which we could conceivably write as

$\Lambda = \mathbb{Z}[\lambda^1(x), \lambda^2(x), \dots ]$

to emphasizing $\Lambda$ is the free $\lambda$-ring on one generator $x$. In fact Todd uses such a notation for the free $\lambda$-ring on two generators $x$ and $y$, which is $\Lambda \otimes \Lambda$.

John Baez (Oct 12 2024 at 05:06):

Since comulltiplication

$\mu: \Lambda \to \Lambda \otimes \Lambda$

is a ring homomorphism, it will be determined once we know what it does to each of the generators $\lambda^i$. I think Todd showed me how it works for $\lambda^1$ and I asked how it worked for $\lambda^2$, or something like that. Here was his reply - we've decided to talk about this here.

John Baez (Oct 12 2024 at 05:07):

You asked what does the comultiplication

$\mu: \Lambda \to \Lambda \otimes \Lambda$

applied to $\lambda^2$ look like. Short answer:

$(\mu(\lambda^2))(x, y) = (\lambda^1(x))^2 \lambda^2(y) + \lambda^2(x)(\lambda^1(y))^2 - 2\lambda^2(x)\lambda^2(y)$

where we think of $\Lambda \otimes \Lambda$ as the free lambda-ring on two generators $x, y$. This can also be written more nicely as $\sigma^2(x) \lambda^2(y) + \lambda^2(x) \sigma^2(y)$.

Longer answer: use the splitting principle, which guarantees that the 2-rig map $\overline{k\mathsf{S}} \to \mathsf{A}^{\boxtimes n}$, sending the generator $x$ to a sum $x_1 \oplus \ldots \oplus x_n$ of $n$ independent bosonic subline objects, is an extension when restricted to polynomial functors of degree $\leq n$. Since $\lambda^2$ is degree $2$, this means in effect that we can pretend the generator $x$ of $\overline{k\mathsf{S}}$ is a sum $x_1 + x_2$ of two bosonic sublines. Then the 2-rig comultiplication $\overline{k\mathsf{S}} \to \overline{k\mathsf{S}} \boxtimes \overline{k\mathsf{S}}$, taking $x$ to $x \boxtimes y$ per our first paper, induces the map of lambda-rings $\Lambda \to \Lambda \otimes \Lambda$ that takes $x = x_1 + x_2$ to $x \otimes y = (x_1 + x_2)(y_1 + y_2) = x_1y_1 + x_1y_2 + x_2y_1 + x_2y_2$. Since this lambda-ring map preserves the $\lambda^2$ operation, we calculate

$\mu(\lambda^2 x) = \lambda^2 (\mu x) = \lambda^2(x_1y_1 + x_1y_2 + x_2y_1 + x_2y_2)$

and use the exponential law for $\lambda^2$ plus vanishing of $\lambda^2(x_i y_j)$ to write this out long-hand. Sparing some gory details, this gives the answer in the short identity.

Todd Trimble (Oct 12 2024 at 16:35):

I wonder how much of this needs to be explained further?

To my taste, the symbols $\lambda^i$ and $\sigma^i$ refer primarily to certain functors, and secondarily to symmetric functions; ultimately these two points of view are united via the all-powerful splitting principle. More precisely, the ring $\Lambda$ is the Grothendieck group or ring based on isomorphism classes of Schur functors, which are essentially those endofunctors on the category of vector spaces (let's say over $\mathbb{C}$) that you can build up using $\otimes$ (together with its symmetric monoidal structure), coproducts, and splitting of idempotents. For example, $V \mapsto V \otimes V$ is a Schur functor. If $\sigma: V \otimes V \to V \otimes V$ denotes the symmetry that swaps tensor factors, then the endomorphisms

$\frac{1 + \sigma}{2}: V \otimes V \to V \otimes V, \qquad \frac{1-\sigma}{2}: V \otimes V \to V \otimes V$

are idempotent maps, and we can split these idempotents to obtain the symmetric square $S^2(V)$ and exterior square $\Lambda^2(V)$. Likewise, the symmetric power and exterior power functors, $V \mapsto S^n(V)$ and $V \mapsto \Lambda^n(V)$, are Schur functors. Tensor products and coproducts of Schur functors are again Schur functors, and Schur functors are closed under composition.

Todd Trimble (Oct 12 2024 at 16:35):

The symbols $\lambda^i$ and $\sigma^i$ can be read as the isomorphism classes $[\Lambda^i]$ and $[S^i]$, regarded as elements of the ring $\Lambda$ (whose addition is induced from taking coproducts, and whose multiplication is induced from taking tensor products). It turns out -- and this is by no means trivial -- that as a ring, $\Lambda$ is isomorphic to the polynomial algebra $\mathbb{Z}[\lambda^1, \lambda^2, \ldots]$. It is also the polynomial algebra $\mathbb{Z}[\sigma^1, \sigma^2, \ldots]$: both sets $\{\lambda^i\}$ and $\{\sigma^i\}$ serve as polynomial bases. There are other famous bases as well, which I won't mention right now, but you can read about them in famous texts such as Representation Theory by Fulton and Harris, and Symmetric Functions and Hall Polynomials by MacDonald.

Todd Trimble (Oct 12 2024 at 16:36):

But there is so much more to $\Lambda$! People go gaga over the richness of its structure and its interplay with the rest of mathematics. I'll try to indicate some main features of this structure by first pointing to similar features of a far simpler structure, namely the polynomial algebra $\mathbb{Z}[x]$, which represents the forgetful functor

$U: \mathsf{CRing} \to \mathsf{Set}$

in the sense of a natural isomorphism $U \cong \mathsf{CRing}(\mathbb{Z}[x], -)$. Now, we can remind the forgetful functor of its (tautological) ring object structure, by pointing at natural transformations $a: U \times U \to U$ (whose components are the addition functions $U(R) \times U(R) \to U(R)$) and $m: U \times U \to U$ (multiplication). At the level of the representing object $\mathbb{Z}[x]$, these transformations are induced by ring homomorphisms

$\alpha: \mathbb{Z} \to \mathbb{Z}[x] \otimes \mathbb{Z}[x], \qquad \mu: \mathbb{Z}[x] \to \mathbb{Z}[x] \otimes \mathbb{Z}[x]$

Here we should pause to note that $A \otimes B$ is the coproduct of commutative rings $A, B$, which means, by the universal property of coproducts, that $\mathsf{CRing}(A \otimes B, -) \cong \mathsf{CRing}(A, -) \times \mathsf{CRing}(B, -)$. Thus $\mathbb{Z}[x] \otimes \mathbb{Z}[x]$ is the representing object of $U \times U$.

Todd Trimble (Oct 12 2024 at 16:36):

In any category with coproducts, we can define a notion of co-ring object, dual to the notion of a ring object in a category with products. So what we can say is that according to the above, $\mathbb{Z}[x]$ is a co-ring object in the category of rings. We call it a biring. I'll leave it as a semi-advanced exercise in applying the Yoneda lemma to figure out explicit formulas for the co-ring structure on $\mathbb{Z}[x]$ that we are talking about here.

Stepping back a little: any time a representable functor $\mathsf{CRing}(A, -): \mathsf{CRing} \to \mathsf{Set}$ has ring object structure -- or in other words lifts up through the forgetful functor $U: \mathsf{CRing} \to \mathsf{Set}$ to give an endofunctor $G: \mathsf{CRing} \to \mathsf{CRing}$, the representing object $A$ becomes a biring. It can be shown that such lifts of representable functors are necessarily limit-preserving, or better yet they are right adjoints. Thus right adjoint endofunctors on $\mathsf{CRing}$ are equivalent to biring structures. The case of the biring $\mathbb{Z}[x]$ corresponds to the identity endofunctor on $\mathsf{CRing}$ (which of course is a right adjoint).

Todd Trimble (Oct 12 2024 at 16:36):

But wait, there's more! In our example, $\mathbb{Z}[x]$ carries another binary operation $\mathbb{Z}[x] \times \mathbb{Z}[x] \to \mathbb{Z}[x]$, namely polynomial composition $(p, q) \mapsto p \circ q$ (replacing the $x$ in $p(x)$ by $q(x)$). If you'll allow me to abuse language and write the identity functor as

$\mathsf{CRing}(\mathbb{Z}[x], -): \mathsf{CRing} \to \mathsf{CRing}$

for the biring $\mathbb{Z}[x]$ sitting in the contravariant slot, then the polynomial composition $\mathbb{Z}[x] \times \mathbb{Z}[x] \to \mathbb{Z}[x]$ can be read as contravariantly inducing a transformation in the other direction,

$\mathsf{CRing}(\mathbb{Z}[x], -) \to \mathsf{CRing}(\mathbb{Z}[x], \mathsf{CRing}(\mathbb{Z}[x], -))$

(these manipulations might seem a tad puzzling, and indeed it takes some fancy footwork to put it just right, but I'm going to skip over that -- you can read Schur Functors and Categorified Plethysm for details). I'll just say that the polynomial composition on $\mathbb{Z}[x]$ corresponds to the identity transformation $\mathrm{Id} \to \mathrm{Id} \circ \mathrm{Id}$ that is the comultiplication for the tautological comonad structure on $\mathrm{Id}$.

Todd Trimble (Oct 12 2024 at 16:37):

More generally, if $A$ is a biring, and if there is a comonad structure on the lifted endofunctor $\mathsf{CRing}(A, -): \mathsf{CRing} \to \mathsf{CRing}$, then the comultiplication transfers over to an operation (a function) $A \times A \to A$ that behaves similarly to polynomial composition. This operation is called plethysm, and a biring equipped with a plethysm operation is called a plethory. So in summary, giving a plethory is equivalent to giving a right adjoint comonad on $\mathsf{CRing}$; the plethory is the representing object for that comonad.

So now I can tell you that the main structural features of $\Lambda$ is that it carries a plethory structure. A rather complicated plethory structure that is still very far away from being fully understood.

Todd Trimble (Oct 12 2024 at 16:37):

Where does that plethory structure come from? Well, John mentioned at the top of this thread that the forgetful functor $U$ from the category of lambda-rings to the category of commutative rings has both a left adjoint $F$ (expected for abstract nonsense reasons) and a right adjoint $G$ (a much more specialized circumstance). So $G$ and also $U$ are right adjoints, hence the comonad $G \circ U$ is also a right adjoint. The plethory $\Lambda$ is the representing object for that right adjoint comonad.

Todd Trimble (Oct 12 2024 at 16:38):

This likely still sounds very mysterious, because for example, what are lambda-rings? Instead of saying directly what they are, I'll say in a different way where the comonad $GU$ comes from. In our paper Schur Functors and Categorified Plethysm, we explain that it is the result of decategorifying a 2-comonad that is much easier (conceptually) to understand. Namely, what we do is categorify the story I was telling, about how $\mathbb{Z}[x]$ with its plethory structure represents the identity comonad on commutative rings. In this particular categorification, commutative rings are replaced by 2-rigs, which are symmetric monoidal $\mathsf{Vect}$-enriched categories which have coproducts and idempotent splittings. The forgetful functor $\mathsf{CRing} \to \mathsf{Set}$ is replaced by the forgetful (2-)functor $\textbf{2Rig} \to \mathsf{Cat}$. This forgetful functor is representable, by a category (rather, 2-rig) of abstract Schur functors, so this category $\mathrm{Schur}$ is the replacement of $\mathbb{Z}[x]$ at the categorified level, and it is a 2-birig that represents the identity functor on the 2-category of 2-rigs. It carries a 2-plethory structure, corresponding to the tautological 2-comonad structure on this identity functor.

Todd Trimble (Oct 12 2024 at 16:38):

The air up here on this categorified mountaintop is clean and clear; the view is simple and beautiful. But then we descend, from $\mathrm{Schur}$ down to its set of isomorphism classes, or really I mean its Grothendieck ring $\Lambda$. That descent process is called 'decategorification'. And it's a little tricky. It took us months of study to make sure of our footing and the right path down to the valley. After all, we start with the "easiest" 2-comonad in the world, the identity functor on 2-rigs, and somehow this decategorifies down to an extremely non-trivial comonad on commutative rings, namely this right adjoint comonad $GU$ I mentioned. Then we can (and do) define lambda-rings to be the coalgebras of this comonad.

Okay, I've just given a thumbnail sketch of our first paper.

Todd Trimble (Oct 12 2024 at 16:38):

To return to the topic, though: John quoted my reply to a question he asked about a specific calculation, about the biring $\Lambda$. As I said, $\Lambda$, even just as a biring let alone a plethory, is a pretty complicated beast, and I believe not completely grasped in terms of giving explicit formulas for the comultiplication (co-addition is rather easier to deal with). But some small calculations you can do by hand, and I was describing to him how to go about calculating what the comultiplication $\mu: \Lambda \to \Lambda \otimes \Lambda$ does to the element $\lambda^2$, by exploiting the subject of our second paper, the splitting principle [set in the context of 2-rigs].

All this is related to the second and third posts at the top of this thread, where John is again in search of ways to wrap one's head around the comultiplication, and he was quoting some stuff he saw in a paper by Niranjan Ramachandran which gives some hints. I can come around to discussing that as well, but in order for others to be able to follow along, I thought it would help to give some background, hence this string of ear-bending comments.

John Baez (Oct 12 2024 at 17:49):

Thanks, Todd! I was distracted last night and didn't quite know the best way to kick off the conversation.

John Baez (Oct 12 2024 at 17:55):

I just noticed a small notational point that might puzzle and perturb novices:

Todd Trimble said:

To my taste, the symbols $\lambda^i$ and $\sigma^i$ refer primarily to certain functors, and secondarily to symmetric functions; ultimately these two points of view are united via the all-powerful splitting principle. [...] If $\sigma: V \otimes V \to V \otimes V$ denotes the symmetry that swaps tensor factors, then the endomorphisms

$\frac{1 + \sigma}{2}: V \otimes V \to V \otimes V, \qquad \frac{1-\sigma}{2}: V \otimes V \to V \otimes V$

are idempotent maps, and we can split these idempotents to obtain the symmetric square $\Sigma^2(V)$ and exterior square $\Lambda^2(V)$. Likewise, the symmetric power and exterior power functors, $V \mapsto S^n(V)$ and $V \mapsto \Lambda^n(V)$ are Schur functors.

This point is that the symmetric functions $\sigma^i$ are not powers of the "switch" map $\sigma: V \otimes V \to V \otimes V$ - instead there is a map sending Schur functors to symmetric functions, and the $\sigma^i$ are the symmetric functions corresponding to the "ith symmetrized tensor power" functors $V \mapsto S^i(V)$.

In the study of the symmetric group, people like lots of things named $S$, or $\mathsf{S}$, or $\Sigma$, or $\sigma$, and sometimes we get carried away and the notations conflict!

Todd Trimble (Oct 12 2024 at 17:56):

Yeah, I edited to change notation in that comment from $\Sigma^2(V)$ (which is one tradition) to $S^2(V)$, which we seem to favor.

Todd Trimble (Oct 12 2024 at 18:00):

Also, I did not explain how Schur functors give rise to symmetric functions. I wouldn't think it was completely common knowledge, how that goes. But I mentioned that the connection arises through this splitting principle that keeps coming up.

John Baez (Oct 12 2024 at 18:03):

I also have another slightly less picayune comment. If we went ahead and computed $(\mu(\lambda^n))(x,y)$, this formula:

$(\mu(\lambda^2))(x, y) = (\lambda^1(x))^2 \lambda^2(y) + \lambda^2(x)(\lambda^1(y))^2 - 2\lambda^2(x)\lambda^2(y)$

would become the 2nd term in a sequence of similar expressions: polynomials in the $\lambda^i$ where the total degree of the nth polyomial is 2n, it seems, if we count $\lambda^i$ as having degree $i$.

Are these polynomials we should recognize? Are they famous?

Todd Trimble (Oct 12 2024 at 18:12):

Mm, I imagine they're quite famous, and also pretty intensively studied, but my impression is that we ("we" being the community of mathematicians, including the experts) don't know them all yet. I wish I knew what people called them by name. Atiyah and Tall give the lackluster notation $P_n$ to these polynomials (page 258), and I think that notation might be pretty common. But your posts at the top of the thread are all about these polynomials!

Todd Trimble (Oct 12 2024 at 18:32):

So for example, the Encyclopedia of Mathematics uses the same notation $P_n$.

I've been scanning Hazewinkel for notation, and I guess it's true that the formulas can be made much more explicit if you use a different polynomial basis, like the power sums basis. He gives a number of such formulas around page 46. So maybe I'll eat my words a little, but anyway I don't know if the explicit list of polynomials purely in terms of the $\lambda^i$ are completely known. Maybe someone else can say.

(I suppose I should know Hazewinkel's article a lot better than I do, because there's evidently a lot of great stuff in it. I'm a little put off by his crapola typesetting job, but I guess that'll be more on me than on him.)

Todd Trimble (Oct 12 2024 at 18:36):

And now that I'm looking at Hazewinkel further, I see that he brings up quasisymmetric functions in section 11, which are supposed to be really important in this biz. Joachim Kock gave an interesting talk about quasisymmetric functions at the CT 2024 conference. I should be looking more into these things.

John Baez (Oct 12 2024 at 18:40):

Do you have any hint as to why quasisymmetric functions are important? I don't know any conceptual explanation of them, so I sometimes have cynically wondered if it's a case of "what are you going to do when you're an expert on symmetric functions and you run out of good ideas? Invent quasisymmetric functions and generalize all the theorems to those!" It's probably not true.

(Lusztig once said "Some people like to take theorems about groups and generalize them to quantum groups. I like to find theorems about quantum groups that aren't like theorems about groups." He came up with some amazing results....)

Todd Trimble (Oct 12 2024 at 18:49):

Eh, I don't yet. But Hazewinkel says at the beginning of section 11, "When looking at various universality properties of the Witt vectors and Symm (which is the topic of the next section) one rapidly stumbles over a (maximally) non commutative version, NSymm, and a (maximally) non cocommutative version, QSymm. This section is devoted to a brief discussion of these two objects. Somehow a good many things become easier to see and to formulate in these contexts (including certain explicit calculations). As I have said before, e.g. in [200], p. 56; [199], Ch. H1, p. 1, once one has found the right non commutative version, things frequently become more transparent, easier to understand, and much more elegant.

Todd Trimble (Oct 12 2024 at 18:54):

There's a certain amount of hardcore algebraic combinatorics to all this. Another buzzphrase that seems relevant and important to me here is this so-called Cauchy identity; see around pages 457-458 of Fulton and Harris. (Perhaps I'm jotting this down as a reminder just to myself to come back to it -- it would be very boring to the casual onlooker.)

John Baez (Oct 12 2024 at 21:15):

Thanks - I'll try to find out if anyone has a name for these polynomials $P_n$. That could unlock a lot of wisdom - or at least piles of cryptic and unmotivated identities that we might find conceptual explanations for. :upside_down:

John Baez (Oct 12 2024 at 21:16):

For other people listening in, let me try to give a conceptual explanation of these polynomials based on algebraic topology. Todd already knows this, at least implicitly, but I feel like saying it.

John Baez (Oct 12 2024 at 21:19):

The operations of direct sum and tensor product can be applied to matrices, and in particular to unitary matrices, so if $U(n)$ is the group of n $\times n$ unitary matrices then we get Lie group homomorphisms

$\oplus: U(m) \times U(n) \to U(m + n)$

$\otimes: U(m) \times U(n) \to U(m \times n)$

We can work with all $m,n$ simultaneously if we use the obvious inclusions

$U(n) \to U(n+1)$

to define the colimit of topological (or even 'smooth') groups

$\displaystyle{ U = \lim_{\longrightarrow} U(n) }$

John Baez (Oct 12 2024 at 21:22):

Then we get binary operations which are group homomorphisms

$\oplus : U \times U \to U$

$\otimes: U\times U \to U$

in addition to the group operation, which is another binary operation. (By the way, I believe books on K-theory use an Eckmann-Hiltonesque argument to show that $\oplus$ is homotopy equivalent to the group operation, and even better.)

John Baez (Oct 12 2024 at 21:24):

These maps induce maps on the classifying space for stable complex vector bundles, $BU$:

$\oplus : BU \times BU \to BU$

$\otimes: BU \times BU \to BU$

John Baez (Oct 12 2024 at 21:25):

and thus we get maps on K-theory going backward, which we can call coaddition and comultiplication:

$\alpha : K(BU) \to K(BU) \otimes K(BU)$
$\mu: K(BU) \to K(BU) \otimes K(BU)$

John Baez (Oct 12 2024 at 21:32):

Since $K$ of any space also has a ring structure, these wind up making $K(BU)$ into a 'biring'. But this biring is just our friend the free $\lambda$-ring on on generator, which @Todd Trimble has been explaining. This is called $\Lambda$. As commutative rings we have

$K(BU) \cong \Lambda \cong \mathbb{Z}[\lambda^1, \lambda^2, \dots ]$

Here $\lambda^i$, thought of as a symmetric function, is the ith elementary symmetric polynomial. But

$K(BU) \cong H^{\text{even}}(BU, \mathbb{Z})$

and thought of as an element of the integral cohomology of $BU$, $\lambda^i$ is called the ith Chern class. It's a cohomology class of degree 2i.

John Baez (Oct 12 2024 at 21:49):

So now let's think about what comultiplication

$\mu: K(BU) \to K(BU) \otimes K(BU)$

does to $\lambda^i$, in these terms!

I'll use the fact I hinted at: for a paracompact Hausdorff space $X$, the set of homotopy classes $[X,BU]$ is isomorphic to the set of stable complex vector bundles over $X$: that is, equivalence classes of vector bundles over $X$, where we count two as equivalent if they become isomorphic after summing with the same complex vector bundle.

Using this and the definition of K-theory we get

$K(X) \cong [X, BU]$

After all $K(X)$ is defined to be the set of stable complex vector bundles over $X$, made into a commutative ring using $\oplus$ and $\otimes$.

John Baez (Oct 12 2024 at 21:57):

Thus, $K(BU) \cong [BU, BU]$ is a somewhat self-referential entity: it's the commutative ring of stable vector bundles on $BU$. It's a ring because of the operations $\oplus, \otimes : BU \times BU \to BU$ acting on the covariant argument in $[BU,BU]$, and a coring because of these operations acting on the contravariant argument in $[BU,BU]$

John Baez (Oct 12 2024 at 21:59):

I'm sort of meandering, but from all this we get yet another interpretation of the elements

$\lambda^i \in \mathbb{Z}[\lambda^1, \lambda^2, \dots ] \cong \Lambda \cong K(BU)$

Namely, they must come from stable vector bundles on $BU$!

John Baez (Oct 12 2024 at 22:01):

I should describe these stable vector bundles, but I won't now. Instead, I just want to say what the

$\mu: K(BU) \to K(BU) \otimes K(BU) \cong K(BU \times BU)$

does to each element $\lambda^i$.

John Baez (Oct 12 2024 at 22:09):

It works like this: we can take any stable vector bundle on $BU$ and pull it back along the tensor product map

$\otimes : BU \times BU \to BU$

and get a stable vector bundle on $BU \times BU$. This induces a map

$K(BU) \to K(BU \times BU) \cong K(BU) \otimes K(BU)$

and this is just our friend the comultiplication $\mu$.

So what does $\mu(\lambda^i)$ mean? We start with the stable vector bundle $\lambda^i$ on $BU$, and pull it back along $\otimes: BU \times BU \to BU$. I believe every stable vector bundle on $BU \times BU$ is an integral linear combination of tensor products of stable vector bundles $\lambda^j \boxtimes \lambda^k$, where $\boxtimes$ is the 'external' tensor product of stable vector bundles: if you've got one on some space $X$ and one on some space $Y$, you can tensor them and get one on $X \times Y$.

John Baez (Oct 12 2024 at 22:13):

If so, we should be able to take $\mu(\lambda^i)$, pull it back along $\otimes$, and write it in terms of the $\lambda^j \boxtimes \lambda^k$. And I believe Todd's calculation is an example of this. He wrote

$(\mu(\lambda^2))(x, y) = (\lambda^1(x))^2 \lambda^2(y) + \lambda^2(x)(\lambda^1(y))^2 - 2\lambda^2(x)\lambda^2(y)$

but in my current notation I believe this means

$\mu(\lambda^2) = (\lambda^1)^{\otimes 2} \boxtimes \lambda^2 + \lambda^2 \boxtimes (\lambda^1)^{\otimes 2} - 2\lambda^2 \boxtimes \lambda^2$

John Baez (Oct 12 2024 at 22:17):

The first term here is not manifestly a tensor product of stable vector bundles $\lambda^j \boxtimes \lambda^k$, but it actually is: it's

$( \lambda^1 \boxtimes \lambda^2) \otimes (\lambda^1 \boxtimes 1)$

where $1$ is the trivial line bundle (the identity in the ring $K(BU)$).

John Baez (Oct 12 2024 at 22:19):

So, in short, the polynomial people call $P_i$ answers this question

Given a stable vector bundle $\lambda^i$ on $BU$, and pulling it back along $\otimes : BU \times BU \to BU$, how can we express it in terms of the stable vector bundles $\lambda^j \boxtimes \lambda^k$?

John Baez (Oct 12 2024 at 22:19):

We could also frame this in terms of Chern classes.

All this 'fluffy' stuff doesn't help us compute the polynomials $P_i$. And indeed, Todd already showed one way to do that. It simply says why we should care about these polynomials.

John Baez (Oct 16 2024 at 01:24):

@Todd Trimble had written to me some more about the big Witt ring, in which he starts analyzing some formulas from here:

• Niranjan Ramachandran, Zeta functions, Grothendieck groups, and the Witt ring.

In particular, this paper discusses a formula for multiplication in the big Witt ring which I mentioned earlier:

John Baez said:

3) Since one can show the underlying commutative ring of the free lambda-ring on one generator is

$\Lambda = \mathbb{Z}[\lambda_1, \lambda_2, \dots ]]$

we have an isomorphism of sets

$W(R) \cong 1 + t R[[t]] \subset R[[t]]$

where

$f \in \mathbb{Z}[\lambda_1, \lambda_2, \dots ]]$

is mapped to

$1 + \sum_{n = 1}^\infty f(\lambda_i) t^n$

Then the challenge is to describe the addition and multiplication on $W(R)$ in these terms. People often do this with explicit formulas, which I tend to find cryptic. The addition in $W(R)$ corresponds to the multiplication in $1 + R[[t]]$, which is why we use this description. The multiplication in $W(R)$ is more tricky. For any $a \in R$ we get an element of $W(R)$ called $(1 - a t)^{-1}$, defined by

$(1 - a t)^{-1} = 1 + a t + a^2 t^2 + \cdots$

and the multiplication $\cdot_W$ on $W(R)$ turns out to be determined once we know

$(1 - a t)^{-1} \cdot_W (1 - b t)^{-1} = (1 - a b t)^{-1}$

This formula turns out to be very useful but I don't have a good understanding of how it comes from 2).

John Baez (Oct 16 2024 at 01:26):

The first question is: why should we care about these elements $(1-at)^{-1}$? And the second is: what does the above formula for a product of them mean? And the third is: does it really determine the product on all of the big Witt ring $W(R)$?

Todd was making progress on all these questions.

Todd Trimble (Oct 16 2024 at 01:29):

Yes, sorry, I was going to say something about that! But let me collect my thoughts.

Todd Trimble (Oct 16 2024 at 01:32):

The paper of Ramachandran that John linked to mentioned that there are several reasonable choices for the (big) Witt ring multiplication. This has a lot to do with how there are various reasonable choices for a nice polynomial basis of $\Lambda$.

Todd Trimble (Oct 16 2024 at 01:36):

So going back to the message at the top of the thread: one conceptual way to define $W(R)$ is that it is the hom-set $\mathsf{CRing}(\Lambda, R)$. Thanks to the rich plethory structure on $\Lambda$ that I was sketching earlier, the hom-set acquires a commutative ring structure and even a lambda-ring structure, and indeed furnishes the right adjoint to the forgetful functor from lambda-rings to commutative rings, as John mentioned earlier.

Todd Trimble (Oct 16 2024 at 01:42):

Now it seems that a lot of sources introduce $W(R)$ as consisting of formal power series with constant coefficient $1$, i.e., elements in $1 + tR[[t]]$. So there is an isomorphism $\mathsf{CRing}(\Lambda, R) \to 1 + tR[[t]]$, and John showed how this might go: using the polynomial basis $\lambda^i$, we can define this isomorphism as sending $f: \Lambda \to R$ to $1 + \sum_{i \geq 1} f(\lambda^i) t^i$.

Another possibility is to use the polynomial basis $\sigma^i$, and define the isomorphism so as to send $f$ to $1 + \sum_{i \geq 1} f(\sigma^i) t^i$.

Todd Trimble (Oct 16 2024 at 01:48):

Each of these has its uses. But before tackling what any of this has to do with those formulas John mentioned, it might not be a bad exercise to look for a moment at how addition works in the Witt ring $1 + tR[[t]]$. (No, it is not ordinary addition of power series!) It turns out that the same formula will work whether you use the $\lambda^i$ basis or the $\sigma^i$ basis, and it's based on the co-addition on $\Lambda$ that was lightly alluded to. Maybe I'll pause a moment.

John Baez (Oct 16 2024 at 02:04):

This sounds like a good plan! Please pause all night long if you want... I'm about to have dinner, and it's 3 hours later for you.

Todd Trimble (Oct 17 2024 at 18:53):

In fact, I can take this opportunity to go a smidge deeper into our first paper. First, how does co-addition work, from first principles? I plan to be very methodical about this, which might make it look heavy in places -- I'll try to ameliorate that by surrounding some of the main conclusions by extra white space, so that readers can skip ahead to get the main points.

Todd Trimble (Oct 17 2024 at 18:53):

The plan is to see how co-addition works in the toy example of $\mathbb{Z}[x]$, and then categorify that. In discussion above I said that the explicit formula for co-addition $\alpha: \mathbb{Z}[x] \to \mathbb{Z}[x] \otimes \mathbb{Z}[x]$ can be derived as a "semi-advanced exercise" in using the Yoneda lemma, so I'll start with that. I'll use the notation $[A, B]$ to denote hom-sets (usually hom-sets that acquire extra structure). The Yoneda lemma is about representable functors. Here we have $\phi: [\mathbb{Z}[x], -] \cong U$ where $U: \mathsf{CRing} \to \mathsf{Set}$ is the forgetful functor; evaluated at a ring $R$, the isomorphism takes $f: \mathbb{Z}[x] \to R$ to $f(x) \in U(R)$. Similarly we have $[\mathbb{Z}[x] \otimes \mathbb{Z}[x], -] \cong U \times U$, instantiated by

$[\mathbb{Z}[x] \otimes \mathbb{Z}[x], -] \overset{(\pi_1, \pi_2)}{\longrightarrow} [\mathbb{Z}[x], -] \times [\mathbb{Z}[x], -] \overset{\phi \times \phi}{\longrightarrow} U \times U$

where the first product projection $\pi_1$ is induced by the first coproduct coprojection $i_1: \mathbb{Z}[x] \to \mathbb{Z}[x] \otimes \mathbb{Z}[x]: x \mapsto x \otimes 1$, and $\pi_2$ is induced by the second coproduct coprojection $i_2: \mathbb{Z}[x] \to \mathbb{Z}[x] \otimes \mathbb{Z}[x]: x \mapsto 1 \otimes x$. Taking $R = \mathbb{Z}[x] \otimes \mathbb{Z}[x]$, chase the identity element $1_{\mathbb{Z}[x] \otimes \mathbb{Z}[x]}$ through the sequence

$[\mathbb{Z}[x] \otimes \mathbb{Z}[x], R] \overset{(\pi_1, \pi_2)}{\longrightarrow} [\mathbb{Z}[x], R] \times [\mathbb{Z}[x], R] \overset{\phi \times \phi}{\longrightarrow} U(R) \times U(R) \overset{+_R}{\longrightarrow} U(R) \overset{\phi^{-1}}{\longrightarrow} [\mathbb{Z}[x], R],$

a la the proof of the Yoneda lemma. We get

$1_{\mathbb{Z}[x] \otimes \mathbb{Z}[x]} \mapsto (x \overset{i_1}{\mapsto} x \otimes 1, x \overset{i_2}{\mapsto} 1 \otimes x) \mapsto (x \otimes 1, 1 \otimes x) \overset{+}{\mapsto} x \otimes 1 + 1 \otimes x \mapsto (x \mapsto x \otimes 1 + 1 \otimes x).$

Todd Trimble (Oct 17 2024 at 18:54):

$\;$

In other words, the co-addition $\alpha: \mathbb{Z}[x] \to \mathbb{Z}[x] \otimes \mathbb{Z}[x]$ is the unique ring map taking $x$ to $x \otimes 1 + 1 \otimes x$.

$\;$

Todd Trimble (Oct 17 2024 at 18:54):

The same type of calculation shows that the comultiplication $\mu: \mathbb{Z}[x] \to \mathbb{Z}[x] \otimes \mathbb{Z}[x]$ is the unique map taking $x$ to $(x \otimes 1) \cdot (1 \otimes x) = x \otimes x$.

(One could simply guess these formulas and check that they work, but I think it's nice to know how the Yoneda lemma removes any guesswork.)

Todd Trimble (Oct 17 2024 at 18:55):

If one uses $\mathbb{Z}[x] \otimes \mathbb{Z}[x] \cong \mathbb{Z}[x, y]$, in effect identifying $x \otimes 1$ with $x \in \mathbb{Z}[x, y]$ and $1 \otimes x$ with $y \in \mathbb{Z}[x, y]$, then the co-addition becomes simply the ring map $\mathbb{Z}[x] \to \mathbb{Z}[x, y]$ taking $x$ to $x + y$, which makes everything look simple and obvious. The comultiplication takes $x$ to $xy$.

Todd Trimble (Oct 17 2024 at 18:55):

Moving in the opposite direction, suppose given a (commutative, cocommutative) bi-ring $B$. The addition on $[B, R] = \mathsf{CRing}(B, R)$ is retrieved from the co-addition $\alpha: B \to B \otimes B$ as a composite

$[B, R] \times [B, R] \overset{\sim}{\longrightarrow} [B \otimes B, R] \overset{[\alpha, 1_R]}{\longrightarrow} [B, R] \qquad (1)$

where the isomorphism obtains by the universal property of $B \otimes B$ as a coproduct. To be explicit, this isomorphism takes a pair of homomorphisms $(f: B \to R, g: B \to R)$ to the composite

$B \otimes B \overset{f \otimes g}{\longrightarrow} R \otimes R \overset{\nabla}{\longrightarrow} R$

where the codiagonal $\nabla$ is precisely the multiplication $m: R \otimes R \to R$.

Todd Trimble (Oct 17 2024 at 18:56):

$\;$

Hence the composite $(1)$ takes a pair of morphisms $(f, g)$ to the composite

$B \overset{\alpha}{\to} B \otimes B \overset{f \otimes g}{\to} R \otimes R \overset{m}{\to} R \qquad (2)$

and this defines $f + g$ in the ring $[B, R]$.

$\;$

Todd Trimble (Oct 17 2024 at 18:56):

Replacing the co-addition $\alpha$ by comultiplication $\mu$, the same construction as in $(2)$ produces the product $f \cdot g$ in $[B, R]$.

Todd Trimble (Oct 17 2024 at 18:56):

Now categorify all this. Replace $\mathbb{Z}[x]$, the free commutative ring on one generator, with the free 2-rig on one generator, which is the (additive) Cauchy completion of the $k$-linearization of the free symmetric monoidal category $\mathsf{S}$ on one generator. (To get anywhere interesting in categorifying commutative rings, you should add in some limits/colimits, and Cauchy completeness is a good place to start.) We write this as $\overline{k\mathsf{S}}$.

Todd Trimble (Oct 17 2024 at 18:57):

I think I mentioned before that this $\overline{k\mathsf{S}}$ is the representing 2-rig for the forgetful functor $\mathbf{2Rig} \to \mathsf{Cat}$, and on those grounds alone, through abstract nonsense, one can copy over (or categorify) the development above for $\mathbb{Z}[x]$, to derive a 2-birig structure on $\overline{k\mathsf{S}}$, with a categorified co-addition given by the unique (up to isomorphism) 2-rig map

$\alpha: \overline{k\mathsf{S}} \to \overline{k\mathsf{S}(x, y)}$

(where the codomain is the free 2-rig on two generators $x, y$) that sends the generator $x$ of $\overline{k\mathsf{S}}$ to the formal coproduct $x \oplus y$ in $\overline{k\mathsf{S}(x, y)}$.

Todd Trimble (Oct 17 2024 at 18:58):

It is interesting to watch what $\alpha$ does to objects like $S^n$ and $\Lambda^n$ in $\overline{k\mathsf{S}}$. I'll start with $S^n$, the $n^{th}$ symmetric power. For any 2-rig $\mathcal{R}$, there is a 2-rig $\mathcal{R}[\mathbb{N}]$ of graded $\mathcal{R}$-objects (whose symmetric monoidal tensor is given by Day convolution, coming from $\mathbb{N}$), and one way we can view the symmetric power $S^n(r) = r^{\otimes n}/S_n$ for an object $r$ of $\mathcal{R}$, which I sometimes like to write as $r^{\otimes n}/n!$, is that it is the $n^{th}$ homogeneous component of a symmetric algebra construction, which I will write as

$\exp(r) = \sum_{n \geq 0} \frac{r^{\otimes n}}{n!} = \sum_{n \geq 0} S^n(r).$

Here I'm thinking of the object $r$ as sitting in grade $1$ in $\mathcal{R}[\mathbb{N}]$, so that $S^n(r)$ sits in grade $n$. This $\exp(r)$ is the free commutative monoid on $r$, and the category of commutative monoid objects has $\otimes$ as its coproduct. This free functor gives a (partially defined) left adjoint [I say "partial" because if for example $r'$ is in degree $0$, then so would be $\exp(r')$, but maybe the 2-rig $\mathcal{R}$ we started with doesn't have infinite coproducts like $\sum_{n \geq 0} S^n(r)$ -- there's a better chance of success if the coproduct is spread across grades]. Since left adjoints preserve coproducts, we deduce an isomorphism

$\exp(r \oplus s) \cong \exp(r) \otimes \exp(s).$

Todd Trimble (Oct 17 2024 at 19:00):

$\;$
Whereupon, focusing on one grade at a time, we further deduce

$S^n(r \oplus s) \cong \sum_{j + k = n} S^j(r) \otimes S^k(s)$,

an identity which holds as a natural transformation between Schur functors of type $\mathcal{R} \to \mathcal{R}$. This holds in particular when $\mathcal{R} = \overline{k\mathsf{S}(x, y)}$.

$\;$

Todd Trimble (Oct 17 2024 at 19:01):

Decategorifying this "identity" by taking isomorphism classes, i.e., by taking Grothendieck rings, this implies that the induced co-addition $\alpha: \Lambda \to \Lambda \otimes \Lambda$ satisfies

$\alpha(\sigma^n) = \sum_{j + k = n} \sigma^j(x) \otimes \sigma^k(y)$

where the right side is visibly a convolution product.

Todd Trimble (Oct 17 2024 at 19:02):

Consider now the induced addition on $[\Lambda, R]$ ($R$ a commutative ring), sending a pair $(f, g)$ of homomorphisms $\Lambda \to R$ to the homomorphism

$\Lambda \overset{\alpha}{\longrightarrow} \Lambda \otimes \Lambda \overset{f \otimes g}{ \longrightarrow} R \otimes R \overset{m}{\longrightarrow} R.$

This composite takes $\sigma^n$ to $\sum_{j + k = n} f(\sigma^j) g(\sigma^k)$. In other words, Witt ring addition is defined by

$(f +_W g)(\sigma^n) = \sum_{j + k = n} f(\sigma^j) g(\sigma^k)$

and if we set up an isomorphism $[\Lambda, R] \overset{\sim}{\longrightarrow} 1 + tR[[t]]$ by $f \mapsto \sum_{n \geq 0} f(\sigma^n) t^n$, then the induced Witt ring addition on $1 + tR[[t]]$ is given by multiplying power series.

Todd Trimble (Oct 17 2024 at 19:02):

It turns out the same is true if we use instead the isomorphism $[\Lambda, R] \overset{\sim}{\longrightarrow} 1 + tR[[t]]$ given by $f \mapsto \sum_{n \geq 0} f(\lambda^n) t^n$. At the beginning of that story, we have the identity

$\Lambda^n(r \oplus s) \cong \sum_{j + k = n} \Lambda^j(r) \otimes \Lambda^k(s)$

as a natural transformation between Schur functors on any 2-rig $\mathcal{R}$. To see this, we replace the 2-rig $\mathcal{R}[\mathbb{N}]$ of graded objects, with its "vanilla" symmetry $\sigma(u \otimes v) = v \otimes u$ for objects $u$ in grade $p$ and $v$ in grade $q$, with the more sophisticated symmetry that introduces a sign factor, $\sigma(u \otimes v) = (-1)^{pq} v \otimes u$. Otherwise, the entire story (symmetric algebra construction as free commutative monoid, etc.) remains the same, mutatis mutandis, hence we get this "exponential identity". Therefore, Witt addition on formal power series is again multiplication, even if we opt for this other identification with $1 + tR[[t]]$.

Todd Trimble (Oct 17 2024 at 19:08):

Now that I've fully explained how addition on the big Witt ring works, I can turn my attention to how multiplication works. Multiplication of $f, g: \Lambda \to R$ is defined by the composite

$\Lambda \overset{\mu}{\longrightarrow} \Lambda \otimes \Lambda \overset{f \otimes g}{\longrightarrow} R \otimes R \overset{m}{\longrightarrow} R$

where this time we have to figure out how comultiplication $\mu$ on $\Lambda$ works. It is, of course, gotten by decategorifying (taking isomorphism classes) the unique (up to isomorphism) 2-rig map

$\overline{k\mathsf{S}} \to \overline{k\mathsf{S}(x, y)}$

that takes the generator $x$ to the formal tensor product $x \otimes y$.

Working this out in detail will involve the "splitting principle" for 2-rigs, which John will be discussing at the upcoming Octoberfest meeting, but perhaps I'll pause to take a break from this longish series of comments.

John Baez (Oct 18 2024 at 18:06):

Todd Trimble said:

Therefore, Witt addition on $[\Lambda, R]$ again corresponds to multiplication of formal power series, even if we use this second identification $[\Lambda, R] \cong 1 + tR[[t]]$.

That's great! Maybe someday we should write a little exposition of the big Witt ring. (Better than
a big exposition of the little Witt ring, explaining stuff like this. :upside_down:)

It might be fun to see what Witt addition looks like if we use the basis for $\Lambda$ given by power sum symmetric functions.

Todd Trimble (Oct 18 2024 at 21:46):

Yes, all these seem like good suggestions!

Todd Trimble (Oct 18 2024 at 21:46):

I'm going to push on with this series of posts; the next topic will be this splitting principle that I keep banging on about.

Todd Trimble (Oct 18 2024 at 21:47):

For us, the splitting principle is the idea that to establish isomorphisms between Schur functors, it is permissible to pretend that they can be decomposed as coproducts of "line objects". It is analogous to the splitting principle in K-theory, where equations are verified by acting as if vector bundles split as coproducts of line bundles.

This sounds a little flaky perhaps, so I'll put it more precisely in a moment, but first I want to say that the situation reminds me of how those Italian mathematicians from the late Renaissance who developed the cubic formula -- Cardano, Tartaglia, etc. -- made the bold move to act as if $i = \sqrt{-1}$ were a thing. Even when all the roots of the cubic polynomial were real, they were arrived at by making use of imaginary elements. (True, they were fairly uncomfortable with the situation, and it took a few centuries before mathematicians felt generally at home with splitting extensions of fields.) The analogy is apt: the coefficients of a polynomial are symmetric functions of their roots, and the roots are analogous to the line objects we are about to discuss. Don't believe the analogy? That's okay. Humor me anyway by considering a linear transformation $T: V \to V$ on a vector space. By passing to an extension of the scalar field if need be, i.e., adjoining roots, we can split $V$ into a coproduct of eigenspaces (which generically are lines), by splitting the characteristic polynomial into linear factors.

(At the risk of too much self-indulgence: this also reminds me of splitting light into a spectrum, where the energy levels of photons are given by eigenvalues of a suitable operator.)

Todd Trimble (Oct 18 2024 at 21:47):

Okay, now let me state the splitting principle more precisely. For the purposes of this thread, define a line object in a 2-rig $\mathcal{R}$ to be a (nonzero) object $r$ satisfying $\Lambda^2(r) \cong 0$. (In our paper, we say instead "bosonic subline object".) Another way of saying it is that the canonical quotient $r \otimes r \to S^2(r)$ is an isomorphism, or equivalently that the symmetry $\sigma: r \otimes r \to r \otimes r$ that transposes factors equals the identity. Or equivalently still, that the tautological action of $S_n$ on $r^{\otimes n}$ is trivial for all $n$. Finally, this last condition is equivalent to saying that for the symmetric powers $S^n$, we have $S^n(r) \cong r^{\otimes n}$ for all $n$.

Todd Trimble (Oct 18 2024 at 21:48):

Just as $\overline{k\mathsf{S}}$ with its generator $x$ is initial among 2-rigs equipped with an object, so $\overline{k\mathbb{N}}$ (the linear Cauchy completion of the linearized discrete symmetric monoidal category $\mathbb{N}$) is initial among 2-rigs equipped with a line object. Here $\overline{k\mathbb{N}}$ is equivalent to the category of $\mathbb{N}$-graded vector spaces of finite total dimension, which we in our paper denote as $\mathsf{A}$, and the line object in this universal case is a 1-dimensional vector space concentrated in grade $1$. Likewise, the walking 2-rig on $n$ line objects $L_1, \ldots, L_n$, denoted as $\mathsf{A}^{\boxtimes n}$ is equivalent to the category of $\mathbb{N}^n$-graded vector spaces of finite total dimension.