Category Theory
Zulip Server
Archive

You're reading the public-facing archive of the Category Theory Zulip server.
To join the server you need an invite. Anybody can get an invite by contacting Matteo Capucci at name dot surname at gmail dot com.
For all things related to this archive refer to the same person.

Stream: theory: mathematics

Topic: Adams operations

John Baez (Nov 15 2024 at 02:45):

This thread may be connected to the thread on the big Witt ring, but it's probably better to have it be a separate thread.

The free $\lambda$-ring on one generator, $\Lambda$, is also known as the ring of symmetric functions. I will explain its real significance in a minute. It has various important elements:

• the exterior power elements $\lambda^i$
• the symmetric power elements $\sigma^i$
• the Adams operations $\psi^i$

The first two sequences of elements can be nicely understood using 2-rigs. Given any object $r$ in any 2-rig $\mathsf{R}$ we can define its ith exterior power $\Lambda^i(r)$ and ith symmetric power $S^i(r)$, using the same constructions that work in the familiar case $\mathsf{R} = \mathsf{Vect}$. In particular we can do this in the free 2-rig on one generator $x$, which happens to be called $\overline{k\mathsf{S}}$ for reasons I could explain. So, we get important objects

$\Lambda^i(x), S^i(x) \in \overline{k\mathsf{S}}$

You can think of these as the 'walking' or 'generic' ith exterior power and ith symmetric power, respectively.

Now, the really important thing about $\Lambda$ is that it's the [[Grothendieck group]] of the free 2-rig on one generator. Thus, we get elements

$\lambda^i = [\Lambda^i(x)]$
$\sigma^i = [S^i(x)]$

in $\Lambda$.

But the [[Adams operations]] $\psi^i$ cannot be defined as the equivalence classes of certain objects in $\overline{k \mathsf{S}}$. And that makes them more tantalizing.

John Baez (Nov 15 2024 at 02:52):

They are formal differences of equivalence classes of objects in $\overline{k\mathsf{S}}$. For example:

$\psi^0 = 1$
$\psi^1 = \lambda^1$
$\psi^2 = (\lambda^1)^2 - 2 \lambda^2$

and so on. Notice that we're using exponents in two ways here: $\lambda^2$ is just the second exterior power element while $(\lambda^1)^2$ is the square of the first exterior power element, which we can define because $\Lambda$ is a ring!

John Baez (Nov 15 2024 at 02:58):

Why do we care about Adams operations? There are many reasons, but one is this.

As you probably know, you can apply a polynomial in $\mathbb{Z}[x]$ to any element of any ring. Why is this? Ultimately it's because $\mathbb{Z}[x]$ is the free ring on one generator, namely $x$. So, if we have any element $r$ of any ring $R$, there's a unique ring homomorphism from $\mathbb{Z}[x]$ to $R$ sending $x$ to $r$. And then this homomorphism sends any polynomial $P$ to some element of $R$, which we call $P(r)$. And this is how we apply a polynomial to an element of a ring!

So, the free ring on one generator acts on any ring. (Not as ring homomorphisms, mind you! I'm just saying any polynomial $P \in \mathbb{Z}[x]$ acts to send elements $r \in R$ to new elements $P(r) \in R$.)

This is a completely general trick. So, the free 2-rig on one generator acts on any 2-rig. Taking Grothendieck groups, we get a baby version of this fact: the free lambda-ring on one generator, $\Lambda$, acts on any lambda-ring.

Being the free lambda-ring on one generator, $\Lambda$ acts on any lambda-ring.

John Baez (Nov 15 2024 at 02:59):

So, the Adams operations $\psi^i \in \Lambda$, whatever they are, act to give maps from any lambda-ring $R$ to itself! And these maps have a remarkable property: they are ring homomorphisms!

This is not true of the $\lambda^i$ or the $\sigma^i$.

John Baez (Nov 15 2024 at 03:05):

For example, the symmetric power operations act on vector spaces or indeed objects in any 2-rig, but they don't obey rules like

$S^i(V \oplus W) \cong S^i(V) \oplus S^i(W)$

so we don't expect that their avatars $\sigma^i$ in $\Lambda$ would act on lambda-rings in a way that obeys

$\sigma^i(r + s) = \sigma^i(r) + \sigma^i(s)$

and indeed they don't. For example we have

$S^2(V \oplus W) = S^2(V) \oplus (V \otimes W) \oplus S^2(W)$

when $V,W$ are objects in any 2-rig, and this winds up giving

$\sigma^2(r+ s) = \sigma^2(r) + r s + \sigma^2(s)$

But the Adams operations behave mysteriously much better, e.g.

$\psi^2(r + s) = \psi^2(r) + \psi^2(s)$

John Baez (Nov 15 2024 at 03:10):

SO: the definition of the Adams operations $\psi^i$ is more mysterious than that of the the $\lambda^i$ and $\sigma^i$ (which descend from familiar operations we do to vector spaces, vector bundles, group representations and objects of other 2-rigs). But their behavior is much nicer.

And this presents a puzzle: what's the best way to understand them, so that their definition is not mysterious, and their good behavior follows automatically from that?

John Baez (Nov 15 2024 at 03:12):

Todd and I have been talking about that, and we have a guess. Roughly: while Adams operations don't arise from objects in the free 2-rig on one generator, they arise from the category of chain complexes of objects in the free 2-rig on one generator.

Or something like that: I'm being sloppy now, but we can get more precise, and we want to get completely precise about this.

John Baez (Nov 15 2024 at 05:56):

I suppose I should say a preliminary word about how to define Adams operations! Here's one way. l'll say how the Adams operation $\psi^i$ acts on a lambda-ring. This description relies on some nontrivial theorems and at first seems to only cover a special case, but it has a conceptual clarity to it that some other more elementary descriptions lack (at least to me).

Say an element $r$ of a lambda-ring $R$ is at most n-dimensional if $\lambda^i(r) = 0$ for all $i \gt n$. I'll describe $\psi^i(r)$ when $r$ is at most $n$-dimensional for some $n$.

In Theorem 6.1, Atiyah and Tall prove the following version of the splitting principle:

Splitting Principle. Suppose the element $r$ of the lambda-ring $R$ is at most $n$-dimensional. Then $R$ can be embedded in a lambda-ring $R'$ where

$r = \ell_1 + \cdots + \ell_n$

and each $\ell_i$ is at most 1-dimensional.

John Baez (Nov 15 2024 at 05:58):

Then, we have

$\psi^i(r) = \ell_1^i + \cdots + \ell_n^i$

It's not obvious, but true, that:

1) $\psi^i(r)$ is not only an element of $R'$ but actually of $R$
2) $\psi^i(r)$ is independent of how we chose $R'$ and how we wrote $r$ as a sum of elements that are at most 1-dimensional.

John Baez (Nov 15 2024 at 05:58):

All this is somewhat miraculous!

John Baez (Nov 15 2024 at 05:59):

If you believe these miracles then it's obvious that Adams operations get along with addition:

$\psi^i(r + s) = \psi^i(r) + \psi^i(s)$

at least if $r$ and $s$ are both finite-dimensional (i.e., at most $n$-dimensional for some large enough $n$).

John Baez (Nov 15 2024 at 06:00):

Simply write $r$ and $s$ as sums of elements that are at most 1-dimensional; this also gives a way to write $r+s$ as a sum of elements that are at most 1-dimensional, and then the formula for $\psi^i$ does the rest.

John Baez (Nov 15 2024 at 06:12):

The same general style of argument shows

$\psi^i(rs) = \psi^i(r) \psi^i(s)$

at least when $r$ and $s$ are finite-dimensional.

For this, we use splitting principle twice and write both $r$ and $s$ as a sum of elements that are at most 1-dimensional, say:

$r = \ell_1 + \cdots + \ell_m$
$s = \ell'_1 + \cdots + \ell'_n$

We thus have

$rs = \sum_{j = 1}^m \sum_{k = 1}^n \ell_j \ell'_k$

Then we need to show that if two elements of a lambda-ring are at most 1-dimensional, then so is their product. This is an exercise using the definition of lambda-ring.

It follows that the above formula expresses $rs$ as a sum of elements that are at most 1-dimensional!

Now we use the miraculous definition of $\psi^i$ three times. Remember, this definition says

$r = \ell_1 + \cdots + \ell_n \implies \psi^i(r) = \ell_1^i + \cdots + \ell_n^i$

So, we have

$\psi^i(r) = \ell_1^i + \cdots + \ell_m^i$
$\psi^i(s) = {\ell'}_1^i + \cdots + {\ell'}_n^i$
$\psi^i(rs) = \sum_{j = 1}^m \sum_{k = 1}^n \ell_j^i {\ell'}_k^i$

It follows that

$\psi^i(rs) = \psi^i(r) \psi^i(s)$

John Baez (Nov 15 2024 at 06:20):

So, if every element of a lambda-ring $R$ is finite-dimensional, each Adams operation

$\psi^i : R \to R$

is a ring homomorphism!

John Baez (Nov 15 2024 at 06:30):

This is not quite optimal: in fact for every lambda-ring $R$, every Adams operation $\psi^i : R \to R$ is a ring homomorphism. I only know how to show use this using a somewhat different argument. But the argument here seems more conceptual.

Let me wrap up by summarizing everything I just said. To make things terse, it helps to call an element of a lambda-ring that's at most 1-dimensional a subline.

Then the idea is this: to apply $\psi^i$ to a sum of sublines, we just take the ith power of each subline and then sum them up. It's then obvious that $\psi^i$ is additive, and since the product of sublines is a subline, we can also see $\psi^i$ is multiplicative.

David Corfield (Nov 15 2024 at 09:13):

Do you know how much of this story is coming from Adams operations simply being power operations and how much from them being power operations in a certain kind of setting, such as K-theory?

Todd Trimble (Nov 15 2024 at 18:23):

So there's also a way of defining Adams operations directly in terms of lambda-ring operations. One way of saying it is that their generating function $\psi_t$ is the logarithmic derivative of the generating function $\sigma_t$ for the symmetric powers, shifted by degree $1$:

$\sum_{n \geq 1}\psi^n t^n = t\frac{d}{dt} \log \sum_{j \geq 0} \sigma^j t^j = \frac{t\frac{d}{dt} \sum_{j \geq 0} \sigma^j t^j}{\sum_{j \geq 0} \sigma^j t^j}$

If you accept the splitting principle as discussed in the Big Witt ring conversation, then as symmetric functions $\sigma^j = \sigma^j(x_1, x_2, \ldots)$ in infinitely many variables $x_1, x_2, \ldots$, we have

$\sum_{j \geq 0} \sigma^j t^j = \prod_{i=1}^\infty \frac1{1 - x_i t}$

and it is very easy to compute the logarithmic derivative of the product. It is

$\sum_{i = 1}^\infty \frac{x_i}{1 - x_i t}$

which when we multiply by $t$ and expand the geometric series for $\frac{x_i t}{1 - x_i t}$, we get

$\sum_{n \geq 1} (x_1^n + x_2^n + \cdots) t^n$

which explains in another way how power sums arise via the splitting principle.

Also to be noted is the beautiful identity

$\frac1{\sigma_t} = \lambda_{-t}$

which gives a convenient expression for the Adams operations in terms of the $\sigma^i, \lambda^i$:

$\psi_t = t \sigma_t' \lambda_{-t} = (\sum_{j \geq 0} j\sigma^j t^j) \cdot (\sum_{k \geq 0} (-1)^k \lambda^k t^k)$

or $\psi^n = \sum_{k=0}^n (-1)^k (n-k) \sigma^{n-k} \lambda^k$.

John Baez (Nov 15 2024 at 18:38):

David Corfield said:

Do you know how much of this story is coming from Adams operations simply being power operations and how much from them being power operations in a certain kind of setting, such as K-theory?

I don't know enough about power operations in general to answer this question yet. I know essentially nothing about them. So thanks: I have more to learn now!

But when you say "K-theory", don't think about K-theory of topological spaces, where we start with the 2-rig of vector bundles and take the Grothendieck group of that. That's historically all-important, but I'm saying here that Adams operations exist whenever you have a 2-rig and take the Grothendieck group of that. So they're fundamental to "categorified ring theory". And what Todd and I are doing is lifting them from the Grothendieck group to the 2-rig itself - or more precisely, another 2-rig built from that 2-rig, which includes a notion of "negative objects".

Todd Trimble (Nov 15 2024 at 20:54):

One thing to make explicit is that the additivity of the Adams operations,

$\psi_t(r + s) = \psi_t(r) + \psi_t(s)$,

is essentially equivalent to the exponential property of the $\sigma^i$,

$\sigma_t(x + y) = \sigma_t(x) \sigma_t(y)$

because the logarithmic derivative takes products back to sums. One doesn't need to know what "$\log$" means particularly; the application $\sigma_t(x) \mapsto t\frac{\sigma_t'(x)}{\sigma_t(x)}$ takes products $\sigma_t(x) \sigma_t(y)$ to sums just by the Leibniz rule.

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

Thanks for the terse review of the formulas defining Adams operations, @Todd Trimble! I think the meaning of many of these formulas will only be clear to people who are familiar with generating functions.

Starting with an easy one, there's this:

If you accept the splitting principle as discussed in the Big Witt ring conversation, then as symmetric functions $\sigma^j = \sigma^j(x_1, x_2, \ldots)$ in infinitely many variables $x_1, x_2, \ldots$, we have

$\displaystyle{ \sum_{j \geq 0} \sigma^j t^j = \prod_{i=1}^\infty \frac1{1 - x_i t} }$

If one mentally expands each geometric series

$\displaystyle{ \frac1{1 - x_i t} }$

and then multiplies them together, one sees that the coefficient of $t^j$ is the sum of all monomials in the variables $x_1, x_2, x_3, \dots$ having total degree $j$, so

$\displaystyle{ \sigma^j = \sum_{p_1 + p_2 + \cdots = j} x_1^{p_1} x_2^{p_2} \cdots }$

This is a decategorified way of stating this fact:

If an object $V$ in a 2-rig is a coproduct of sublines $L_1, L_2, L_3, \dots$, then its jth symmetric power $S^j(V)$ is the coproduct

$\displaystyle{ S^j(V) \cong \bigoplus_{p_1 + p_2 + \cdots = j} L_1^{\otimes p_1} \otimes L_2^{\otimes p_2} \otimes \cdots }$

The tensor product here makes sense because all but finitely many of the $p_i$ must be zero, and the zeroth tensor power of an object is the unit object for the tensor product, $I$.

John Baez (Nov 15 2024 at 21:23):

But what is the categorified meaning of the quantity

$\displaystyle{ \sigma_t = \sum_{j \ge 0} \sigma^j t^j }$

in the first place? For example what's the meaning of $t$? I think Todd cracked the code, but let me report the answer.

Briefly, the symmetric algebra is a graded algebra, which we can split into homogeneous pieces or 'grades', and the exponent of $t$ records the grade.

More formally:

For any 2-rig $\mathsf{R}$ there's a 2-rig of $\mathbb{N}$-graded objects in $\mathsf{R}$. I will call this $\mathsf{R}[[t]]$ because when we take its Grothendieck group we get

$K(\mathsf{R}[[t]]) \cong K(R)[[t]]$

where the ring at right consists of formal power series in $K(R)$.

Why is this true? By definition, an $\mathbb{N}$-graded $\mathsf{R}$-object $X$ is just a sequence of objects $X_j \in \mathsf{R}$ for $j = 0, 1, 2, \dots$. Thus, passing to the Grothendieck group, we can write its equivalence class $[X]$ as

$\displaystyle{ \sum_{j \ge 0} [X_n] t^j \in K(R)[[t]] }$

where the exponent of the formal variable $t$ just keeps track of the grade.

John Baez (Nov 15 2024 at 21:25):

So, it makes sense to call the 2-rig of $\mathbb{N}$-graded $\mathsf{R}$-objects $\mathsf{R}[[t]]$.

In particular there's a symmetric algebra functor

$S \colon \mathsf{R} \to \mathsf{R}[[t]]$

sending any object $V \in \mathsf{R}$ to its symmetric algebra $S(V)$, which is the $\mathbb{N}$-graded $\mathsf{R}$-object whose jth grades is the jth symmetric powers $S^j(V)$.

Using the power series notation, the equivalence class of $S(V)$ in the Grothendieck group of $\mathsf{R}[[t]]$ is

$\displaystyle{ [S(V)] = \sum_{j \ge 0} [S^j(V)] t^n }$

Todd's formula

$\displaystyle{ \sigma_t = \sum_{j \ge 0} \sigma^j t^j }$

is just a distillation of this.

Can we understand it a bit more precisely? Yes! @Todd Trimble is using the fact that we don't need to separately consider each 2-rig $\mathsf{R}$. We can work 'universally' in the free 2-rig on one generator, called the 2-rig of Schur functors or sometimes $\overline{k\mathsf{S}}$ or sometimes more charisimatically $\mathsf{Schur}$. The jth symmetric power of this generator is often simply called $S^j$, since it represents the operation of taking the jth symmetric power.

For any object $V$ of any 2-rig, its jth symmetric power is called $S^j(V)$. But we can also talk about $S^j$ all by itself, waiting to act on some object of some 2-rig - and it's an object in $\mathsf{Schur}$.

The infinite coproduct

$\displaystyle{ S \cong \bigoplus_{j \ge 0} S^j }$

is not an object of $\mathsf{Schur}$, since that 2-rig doesn't have infinite coproducts. But it's well-defined as an $\mathbb{N}$-graded $\mathsf{Schur}$-object! So, we can write

$\displaystyle{ S \cong \bigoplus_{j \ge 0} S^j t^j \in \mathsf{Schur}[[t]] }$

where the $t^j$ is just a formal notation that tells us what grade we're in.

More precisely: given any $X \in \mathsf{Schur}$ we use $X t^j$ to mean the $\mathbb{N}$-graded $\mathsf{Schur}$-object with $X$ in the jth graded and zero everywhere else. Then given any $\mathbb{N}$-graded $\mathsf{Schur}$-object $X$, we have

$\displaystyle{ X \cong \bigoplus_{j \ge 0} X_j t^j }$

Todd Trimble (Nov 15 2024 at 21:36):

sending any object $V\in R$ to its symmetric algebra $S(V)$

Yes, this was also discussed in the big Witt thread, which I was piggybacking off of. There was the additional comment that the symmetric algebra construction is left adjoint to the forgetful functor from commutative $k$-algebras to $k$-vector spaces, and being left adjoint, it preserves coproducts. The coproduct in the category of commutative $k$-algebras is given by tensor product $\otimes_k$. Hence

$S(V \oplus W) \cong S(V) \otimes S(W)$

and this is the origin of the statement of the "exponential law", which in decategorified form reads

$\sigma_t(x + y) = \sigma_t(x) \sigma_t(y)$

when interpreted as a statement about lambda-rings.

John Baez (Nov 15 2024 at 21:52):

I wonder if we can work the grading of the symmetric algebra into that statement about $S$ being left adjoint to the forgetful functor from commutative algebras to vector spaces. I.e. can we use some high-powered formal wizardry to instantly see that $S$ is a functor from vector spaces to graded commutative algebras (with the boring 'bosonic' symmetry on graded vector spaces)?

I mean it's obvious, but Freyd would want it to be obviously obvious.

John Baez (Nov 15 2024 at 21:53):

There must be a left adjoint to the forgetful functor from graded commutative algebras to graded commutative vector spaces, and restricting this adjoint to graded commutative vector spaces that vanish outside grade zero we get the version of $S$ that I'm talking about.

Todd Trimble (Nov 15 2024 at 22:00):

I think maybe you meant vanish at grade $0$, not outside grade $0$. If $V$ sits in grade $0$, then all the $S^n(V)$ also sit in grade $0$, but then their infinite coproduct might not exist as an object in the general 2-rig $R$.

The map $\mathsf{R} \to \mathsf{R}[[t]]$ sends an object $r \in \mathsf{R}$ to its doppelganger in grade $1$ (or in grade $2$ if we are using the signed symmetry) and then applies the symmetric algebra construction, so that the copies $S^n(r)$ get spread out along the grading.

Todd Trimble (Nov 15 2024 at 22:02):

But I think you're right that this could use a little tidying up.

John Baez (Nov 15 2024 at 22:02):

Todd Trimble said:

I think maybe you meant vanish at grade $0$, not outside grade $0$

Whoops, I meant vanish outside grade $1$. I wanted to think of the usual symmetric algebra of a vector space as a special case of the free commutative graded algebra on a graded vector space, so I'm including $\mathsf{Vect}$ in $\mathsf{Vect}[[t]]$, but I should include it in grade 1, like you said.

John Baez (Nov 15 2024 at 22:06):

Anyway, all this is the easy part. And similarly, categorifying the formula

$\displaystyle{ \lambda_t = \sum_{j \ge 0} \lambda^j t_j }$

is easy if we work with $\mathbb{N}$-graded Schur functors. I was mainly trying to get our audience up to speed (if anyone is actually listening).

What we really want to discuss, I think, is how to handle the Adams operations and formulas like

$\displaystyle{ \psi_t = \sum_{j \ge 0} \psi^j t^j }$

which are not merely decategorifications of isomorphisms between $\mathbb{N}$-graded objects in 2-rigs.

The idea seems to be that each individual term $\psi^j$ comes, not from a Schur functor like $sigma^j$ or $\lambda^j$ did, but from some chain complex of Schur functors.

Todd Trimble (Nov 15 2024 at 22:21):

Yes, indeed. We've been dancing around the fact some of these explicit formulas for the $\psi^j$ involve subtractions and formal differences, which are trickier to "categorify". But there is an old idea of double-ledger accounting, a kind of forerunner to the idea of negative quantities, where credits go into a positive column $C_0$ and debits go into a negative column $C_1$. Ultimately, at a decategorified level, this is pointing to a rig map

$\mathbb{N}[\mathbb{Z}_2] \to \mathbb{Z}$

where the domain is a group rig (like a group ring, but no negatives), with elements $c_0 + c_1 x$ where $x$ is the generator of $\mathbb{Z}_2$, and the rig map takes $c_0 + c_1 x$ to $c_0 - c_1$ (i.e., takes $x$ to $-1 \in \mathbb{Z}$, which is indeed an order two element).

John Baez (Nov 15 2024 at 22:25):

(Digressing, is that bit about double-entry bookkeeping what Walters was actually getting at with his paper On partita doppia? As you know I complained about the obscurity of this title - and then regretted it when he wrote a blog article expressing mild hurt.)

Todd Trimble (Nov 15 2024 at 22:26):

We can't fully categorify this to the 2-rig level, but there are interesting proxies. Given a 2-rig $\mathsf{R}$, we can discuss the 2-rig $\mathsf{R}[\mathbb{Z}_2]$ which, to say it precisely, is the 2-rig of $\mathbb{Z}_2$-graded $R$-objects.

Todd Trimble (Nov 15 2024 at 22:27):

is that bit about double-entry bookkeeping what Walters was actually getting at with his paper On partita doppia?

Yes! It's safe to say at this remove in time that I was a referee for one of the papers that mention partita doppia. It's an idea I return to frequently.

Todd Trimble (Nov 15 2024 at 22:34):

It starts to get interesting. Given a 2-rig $\mathsf{R}$, you can form also the 2-rig $\mathsf{DG}(R)$ of $\mathbb{Z}_2$-graded chain complexes, so in other words a $\mathbb{Z}_2$-graded object $X$ equipped with differentials $X_0 \to X_1$ and $X_1 \to X_0$. Under mild conditions such as we have for the case $R$ = free 2-rig on one generator, what we called $\overline{k\mathsf{S}}$ in the big Witt thread, have an abelian and even a semisimple abelian category, where we can take quotients, and therefore take homology.

John Baez (Nov 15 2024 at 22:40):

This is extremely interesting for lots of reasons. I remember taking a course with Quillen where he said physicists working on supersymmetry were the ones who impressed on mathematicians the importance of working with $\mathbb{Z}/2$-graded objects rather than $\mathbb{N}$- or $\mathbb{Z}$-graded ones. (Not all the time, of course, but sometimes.) Complex $K$-theory was already naturally $\mathbb{Z}/2$ graded, but I'm not sure mathematicians emphasized that, as opposed to it being $\mathbb{Z}$-graded and periodic with period 2.

John Baez (Nov 15 2024 at 22:42):

I don't know if double-entry bookkeepers ever link a specific loan to a specific repayment of that loan with arrows, or something like that, but that could give a $\mathbb{Z}/2$ graded chain complex.

John Baez (Nov 15 2024 at 22:54):

By the way, we've talked forever about $K$, also known as $K_0$, but never about $K_1.$ I'm not sure how that fits into our story! It's somehow connected to the 'trace of a category' stuff you were explaining to me recently, as you can see here.

Todd Trimble (Nov 15 2024 at 22:55):

So there are two 2-rig maps $\mathsf{DG(R)} \to \mathsf{R}[\mathbb{Z}_2]$, namely the forgetful functor $U$ and the homology functor $H$. If all short exact sequences split in $\mathsf{R}$, then it's possible to show the Euler characteristic equation

"$X_0 - X_1 = H_0(X) - H_1(X)$"

which is interpreted to mean $X_0 \oplus H_1(X) \cong H_0(X) \oplus X_1$. People in K-theory often talk about things like virtual bundles which are formal differences of vector bundles, which as we were saying is problematic to categorify directly, but the Euler-characteristic equation (really roughly speaking) suggests something akin to considering a "coequalizer" of the two 2-rig maps

$H, U: \mathsf{DG(R)} \rightrightarrows \mathsf{R}[\mathbb{Z}_2]$.

So considering $(X_0, X_1)$ to be somehow equivalent to $(H_0(X), H_1(X))$, or at least think of a morphism $F$ between chain complexes to be an equivalence in this sense if it induces an isomorphism in homology $(H_0, H_1)$. This generates the notion of quasi-isomorphism between chain complexes.

Anyway, in practice it seems often to be the case that important identities in lambda-rings, including cases that ineluctably involve negatives, can be seen as decategorified shadows of quasi-isomorphisms between chain complexes.

Todd Trimble (Nov 15 2024 at 23:11):

A good example of this phenomenon (which John and Joe and I have spent time discussing) is seeing the lambda-ring equation

$\sigma_t \lambda_{-t} = 1$

as indeed a decategorified shadow of a certain quasi-isomorphism between chain complexes valued in $\overline{k\mathsf{S}}$, the free 2-rig on one generator. It's actually a sequence of quasi-isomorphisms, so in other words a quasi-isomorphism between sequences of chain complexes (and yes, this is beginning to sound a lot like what you were telling me last night about Khovanov's work, John). In other words, the equation above has a part in degree $n$ that looks like

$\sigma^n \lambda^0 - \sigma^{n-1} \lambda^2 + \sigma^{n-2}\lambda^2 - ... = 0$

for $n > 1$, and $\sigma^0\lambda^0 = 1$ in grade $n = 0$$. The collection of these for $n > 0$ is the decategorified shadow of a collection of quasi-isomorphisms expressing the fact that there is an exact $\mathbb{Z}$-graded chain complex with $n$-th part

$\ldots\to 0 \to 0 \to 0 \to S^n \otimes \Lambda^0 \to S^{n-1} \otimes \Lambda^1 \to S^{n-2} \otimes \Lambda^2 \to \ldots$

whose $\mathbb{Z}_2$-graded reduction gives a quasi-isomorphism in $\mathsf{DG(R)}$, from this chain complex to the zero chain complex. (Here $S^n \otimes \Lambda^0$ appears in grade $0$.)

Todd Trimble (Nov 15 2024 at 23:19):

As a matter of fact, this is the $n$-th entry of a sequence of quasi-isomorphisms $S^n(X) \to 0$ where $X$ is the exact chain complex

$\ldots \to 0 \to 0 \to 0 \to k \to k \to 0 \to 0 \to \ldots$

where that first $k$ appears in grade $0$. Since taking homology $H$ is a 2-rig map, it preserves symmetric power functors $S^n$, hence for $n > 0$ the calculation

$H(S^n(X)) \cong S^n(H(X)) \cong S^n(0) \cong 0$

where $H(X) \cong 0$ by exactness of $X$. Hence the trivial map $S^n(X) \to 0$ is a quasi-isomorphism.

Todd Trimble (Nov 15 2024 at 23:41):

By the way, we've talked forever about K, also known as K0​, but never about K1​. I'm not sure how that fits into our story! It's somehow connected to the 'trace of a category' stuff you were explaining to me recently, as you can see here.

Ah, thanks for mentioning this! Just like

$\pi_0 \circ \mathrm{core}: \mathsf{Cat} \to \mathsf{Set}$

has good formal properties (preserving coproducts and products, for example), important to our story of the decategorification that we use to assign the trivial 2-rig plethory to the rig-plethory $\Lambda_+$, so too does this (in some ways more interesting) "decategorification"

$\mathrm{trace}: \mathsf{Cat} \to \mathsf{Set}: C \mapsto \int^{c: C} \hom(c, c)$

and now I can better appreciate a question asked after my talk on 2-rigs at the Topos Institute, which seems to allude to this. (Unfortunately I think it might have been asked after the recording ended.)

David Corfield (Nov 16 2024 at 11:48):

John Baez said:

But when you say "K-theory"...

Hence my original "such as K-theory".

I was wondering if it had something to do with chromatic level 1 issues, and it does seem that something Adams operation-like occurs in all $K(1)$-local $E_{\infty}$ ring spectra (Lurie notes, Thm. 1.2).

But then apparently there's a form of Adams operation in so-called beta-rings (see Guillot's paper there), which arise in cohomotopy among other places.

David Corfield (Nov 16 2024 at 12:09):

John Baez said:

physicists working on supersymmetry were the ones who impressed on mathematicians the importance of working with Z/2-graded objects rather than N- or Z-graded ones

There was an argument by Kapranov that this grading is viewed differently by physicists and mathematicians (hence the first two sections of Supergeometry in mathematics and physics), and that it should properly be seen as a truncation of full 'sphere spectrum'-grading nLab super algebra.

John Baez (Nov 17 2024 at 00:35):

David Corfield said:

I was wondering if it had something to do with chromatic level 1 issues, and it does seem that something Adams operation-like occurs in all $K(1)$-local $E_{\infty}$ ring spectra (Lurie notes, Thm. 1.2).

But then apparently there's a form of Adams operation in so-called beta-rings (see Guillot's paper there), which arise in cohomotopy among other places.

Hmm, all that stuff seems too fancy for me right now - in other words, I should learn about it, but I don't understand it yet and I think Todd and I can make a lot of progress understanding Adams operations working with what I've got.

Todd and I have a nice project of categorifying Adams operations (making them something you can do with 2-rigs, instead of just the Grothendieck groups of 2-rigs), which we plan to do right here. It's about half done in some vague sense: we've done enough so I know what we're doing will work, but I want to make it better in various ways.

Maybe some later day I can try to understand how Adams operations fits into the 'chromatic hierarchy' business. But my current holy grail is understanding how the 'orthodox' approach to lambda-rings (based on $\lambda$ or $\sigma$ operations) is connected to the 'heterodox' approach (based on Adams operations or Frobenius lifts or Joyal's delta-rings).

Todd and I understand the orthodox approach quite well at a categorified level, so I want to pull the heterodox approach up to that level. And we've almost done it.

John Baez (Nov 17 2024 at 05:45):

David Corfield said:

There was an argument by Kapranov that this grading is viewed differently by physicists and mathematicians (hence the first two sections of Supergeometry in mathematics and physics), and that it should properly be seen as a truncation of full 'sphere spectrum'-grading nLab super algebra.

Thanks for reminding me of this paper. It seems more detailed than a paper with similar themes by Kapranov that I read once before - could be a trick of my memory. I really like the section "Homotopy-theoretic underpinning of supergeometry", how it describes the mathematical consequences of the fact that the first and second stable homotopy groups of spheres are $\mathbb{Z}/2$, and his concept of "supersymmetric monoidal category" where both the commutative monoid of objects and the hom-spaces are $\mathbb{Z}/2$-graded, with these two $\mathbb{Z}/2$-gradings having different meanings.

John Baez (Nov 17 2024 at 18:04):

I have an idea that keeps nagging me. It might be a mathematical will-o'-the-wisp. What's a will-o'-the-wisp? It's an

ignis fatuus (Latin for 'foolish flame'): an atmospheric ghost light seen by travellers at night, especially over bogs, swamps or marshes. This phenomenon is known in the United Kingdom by a variety of names, including jack-o'-lantern, friar's lantern and hinkypunk, and is said to mislead and/or guide travellers by resembling a flickering lamp or lantern.

John Baez (Nov 17 2024 at 18:08):

I think mathematicians often get pulled off track by will-o'-the-wisps and sink into the bog... from now on, if I think that's happening to someone I may say "That's just hinkypunk!"

John Baez (Nov 17 2024 at 18:20):

But anyway, here's my thought. I need to start with some basics before I can clearly describe it.

As Todd explained above, we can understand Adams operations at a categorified level by treating them as $\mathbb{Z}/2$-graded chain complexes of $\mathbb{N}$-graded Schur functors. This sounds complicated and ad hoc until you notice it follows a general philosophy advocated and used very successfully by Khovanov. So let me explain that.

In its simplest form, Khovanov's idea is that

To categorify formal power series with integer coefficients, use $\mathbb{Z}/2$-graded chain complexes of $\mathbb{N}$-graded vector spaces!

More precisely, he noted that an $\mathbb{Z}/2$-graded chain complex of $\mathbb{N}$-graded vector spaces that are finite-dimensional in each grade has a kind of 'Euler characteristic' which is a formal power series with integer coefficients. The $\mathbb{Z}$-graded chain complex lets us deal with minus signs, while the $\mathbb{N}$-grading lets us deal with formal power series.

We can work up to this in stages:

1) A finite-dimensional vector space $V$ has a dimension which is a natural number

$\mathrm{dim}(V) \in \mathbb{N}$

2) a $\mathbb{Z}/2$-graded chain complex

$d_0: V_0 \to V_1$, $d_1 : V_1 \to V_0$

has an Euler characteristic which is an integer

$\chi(V) \in \mathbb{Z}$

given by

$\chi(V) = \mathrm{dim}(V_0) - \mathrm{dim}(V_1) = \mathrm{dim}(H_0) - \mathrm{dim}(H_1)$

where $H_i$ is the homology $\text{ker}(d_i)/\text{im}(d_{i-1})$ where $i \in \mathbb{Z}/2$.

John Baez (Nov 17 2024 at 18:22):

3) An $\mathbb{N}$-graded vector space $V = (V_i)_{i \in \mathbb{N}}$ where each $V_i$ is finite-dimensional has a dimension or Hilbert series which is a formal power series with natural number coefficients

$\text{dim}(V) \in \mathbb{N}[[t]]$

given by

$\text{dim}(V) = \sum_{i \ge 0} \text{dim}(V_i) t^i$

John Baez (Nov 17 2024 at 18:32):

Now let's put it all together:

4) By commutativity of internalization, a

• $\mathbb{Z}/2$-graded chain complex of $\mathbb{N}$-graded finite-dimensional vector spaces

is the same as an

• $\mathbb{N}$-graded $\mathbb{Z}/2$-graded chain complex of finite-dimensional vector spaces!

The last phrase is hard to read, but we're referring to a thing with a 'bigrading': two gradings, one in $\mathbb{N}$ and the other in $\mathbb{Z}/2$.

Let's call such a thing $V = (V_i)_{i \in \mathbb{N}}$ where for each $V_i$ is a $\mathbb{Z}/2$-graded chain complex of finite-dimensional vector spaces.

Each $V_i$ has an Euler characteristic $\chi(V_i)$. Thus $V$ has an Euler characteristic which is a formal power series with integer coefficients

$\chi(V) \in \mathbb{Z}[[t]]$

given by

$\chi(V) = \sum_{i \ge 0} \chi(V_i) t^i$

John Baez (Nov 17 2024 at 18:44):

I believe we can go a bit further and show that $\mathbb{Z}/2$-graded chain complexes of $\mathbb{N}$-graded finite-dimensional vector spaces are classified up to [[quasi-isomorphism]] by their Euler characteristic $\chi(V) \in \mathbb{Z}[[t]]$. I could explain this is anyone cares.

John Baez (Nov 17 2024 at 18:49):

Now, Todd and I have been treating categorified Adams operations as something a bit fancier: $\mathbb{Z}/2$-graded chain complexes of $\mathbb{N}$-graded Schur functors.

But the possible will-o'-the-wisp that keeps beckoning me points out that this may be more elaborate than necessary, because Schur functors are already inherently $\mathbb{N}$-graded.

John Baez (Nov 17 2024 at 18:49):

So, I'm thinking that maybe $\mathbb{Z}/2$-graded chain complexes of Schur functors will be enough to categorify Adams operations!

John Baez (Nov 17 2024 at 18:51):

I think I can spell out in detail how this should work... or at least I want to try. But right now it's time to get up and have breakfast, and the above is probably already more than enough for most people to read in one sitting.

John Baez (Nov 17 2024 at 20:30):

Okay, back to explaining my current "will-o'-the-wisp". I'm trying to see if we can understand various classical equations involving things such as lambda-, sigma- and Adams operations as decategorified versions of quasi-isomorphisms between $\mathbb{Z}/2$-graded chain complexes of Schur functors.

The basic plan here is to assign an element of $\mathbb{Z}[[t]]$ to any $\mathbb{Z}/2$-graded chain complex of Schur functors, though I'm lying slightly here.

Let me eliminate that lie since it will weigh heavily on my conscience until I do! When I said "Schur functor", what I really meant is something slightly different, namely a FinVect-valued species.

A FinVect-valued species is a functor

$F : \mathsf{S} \to \mathsf{FinVect}$

where $\mathsf{S}$ is the groupoid of finite sets and $\mathsf{FinVect}$ is the category of finite-dimensional vector spaces.

John Baez (Nov 17 2024 at 20:34):

We can and will think of a FinVect-valued species as a list of finite-dimensional representations of the symmetric groups $S_n$, say

$F_n: S_n \to \text{End}(V_n)$

for some finite-dimensional (complex) vector spaces $V_n$.

Then a FinVect-valued species corresponds to a Schur functor iff the total dimension $\sum_n \text{dim}(V_n)$ is finite.

John Baez (Nov 17 2024 at 20:35):

But I don't want to require that the total dimension be finite here! So instead of saying 'Schur functor' I should say 'FinVect-valued species' - that's what I'll actually be working with.

John Baez (Nov 17 2024 at 20:37):

I will call a FinVect-valued species something like $F$, but this is short for a list

$F_n : S_n \to \text{End}(S_n) \qquad i = 0, 1, 2, 3, \dots$

Note any FinVect-valued species $F$ has a dimension

$\chi(F) = \sum_{i \ge 0} \text{dim}(V_i) t^i$

which is an element of the rig $\mathbb{N}[[t]]$ - the rig of formal power series with natural number coefficients in one variable $t$.

John Baez (Nov 17 2024 at 20:39):

Later I will talk about $\mathbb{Z}/2$-graded chain complexes of FinVect-valued species, and each of these will have an 'Euler characteristic' which is an element of $\mathbb{Z}[[t]]$. But let me start off easy.

John Baez (Nov 17 2024 at 20:58):

Whoops! I realized my plan isn't working out as expected. I feel I'm sinking into some mud now, like a guy who thinks he sees the warm and attractive lights of a house in the distance, and walks off the trail, only to sink into a bog, his last words being "that damned will-o'-the-wisp!"

John Baez (Nov 17 2024 at 21:02):

But let me see if I can fix my idea a bit. They say you should never struggle when you sink into quicksand, and they even have helpful videos on what to do when you sink into quicksand but I'm going to struggle now. I may just sink in deeper....

John Baez (Nov 17 2024 at 21:35):

Let me start by pondering this formula from earlier:

$\displaystyle{ \sigma_t = \sum_{j \ge 0} \sigma^j t^j }$

Here $\sigma^j$ is somehow a decategorified residue of the Schur functor I've been calling $S^j$, the jth symmetric power functor. Like all Schur functors, that Schur functor can be seen as FinVect-valued species

$F \colon \mathsf{S} \to \mathsf{FinVect}$

or equivalently a list of finite-dimensional representations of symmetric groups

$F_n \colon S_n \to \text{End}(V_n)$

But it's a very simple one, where $F_n = 0$ except for $n = j$. That is the key fact that makes me think we can simplify our set up a bit. The representation

$F_j \colon S_j \to \text{End}(V_j)$

happens to be the one-dimensional trivial representation but this fact is somewhat secondary.

John Baez (Nov 17 2024 at 21:41):

So what I want to do is this - a change in plan as I struggle in the bog of mathematics. For any FinVect-valued species

$F: \mathsf{S} \to \mathsf{FinVect}$

I get a list of finite-dimensional representations

$F_n \colon S_n \to \text{End}(V_n)$

and thus a list of elements

$[F_n]\in R(S_n)$

where $R(S_n) = K(\mathsf{Rep}(S_n))$ is the Grothendieck group of the category of finite-dimensional representations of $S_n$. This group has a ring structure and it's often called the representation ring of $S_n$, but I don't really need the ring structure right now.

John Baez (Nov 17 2024 at 21:47):

We can summarize this entire list in strange sort of formal power series

$\sum_{n \ge 0} [F_n] t^n$

It's strange because the coefficients don't all lie in the same abelian group: rather the nth coefficient lies in $R(S_n)$. Nonetheless there's a perfectly fine way to multiply these formal power series and get a ring of them using the "induction" operations

$R(S_n) \otimes R(S_m) \to R(S_{n+m})$

John Baez (Nov 17 2024 at 21:51):

I will write

$d(F) = \sum_{n \ge 0} [F_n] t^n$

to indicate that this power series has some properties expected of 'dimension', namely

$d(F \oplus G) = d(F) + d(G)$

$d(F \otimes G) = d(F) d(G)$

where the tensor product of FinVect-valued species here is the Cauchy product, i.e. Day convolution.

John Baez (Nov 17 2024 at 21:52):

Furthermore $d$ obeys the wonderful property

$d(F) = d(G) \iff F \cong G$

John Baez (Nov 17 2024 at 22:04):

To be continued....

John Baez (Nov 17 2024 at 23:32):

Let me quickly recap and polish things up a bit, since I took a distracting detour earlier. The 2-rig of FinVect-valued species $\mathsf{FinVect}^\mathsf{S}$ is naturally $\mathbb{N}$-graded in the sense that it's a product of linear categories

$\mathsf{FinVect}^\mathsf{S} \simeq \prod_{n \ge 0} \mathsf{FinRep}(S_n)$

where $\mathsf{FinRep}(S_n)$ is the category of finite-dimensional representations of $S_n$, and the relevant tensor product in $\mathsf{FinVect}^\mathsf{S}$, the Cauchy product, respects this grading.

John Baez (Nov 17 2024 at 23:35):

Taking Grothendieck groups, we thus have

$K(\mathsf{FinVect}^\mathsf{S}) \simeq \prod_{n \ge 0} K(\mathsf{FinRep}(S_n)) \simeq \prod_{n \ge 0} R(S_n)$

where $R(S_n) := K(\mathsf{FinRep}(S_n))$ is called the representation ring of $S_n$.

John Baez (Nov 17 2024 at 23:47):

Thus, a FinVect-valued species $F$ is the same as a list of finite-dimensional representations $F_n$ of all the symmetric groups, and we can write its equivalence class $[F] \in K(\mathsf{FinVect}^\mathsf{S})$ in the following funny way:

$[F] = \sum_{n \ge 0} [F_n] t^n$

Here the $t^n$ has no real meaning except to remind us that $[F_n] \in R(S_n)$. But it's also nice for calculations, because we can show

$[F \oplus G] = [F] + [G]$

and

$[F \otimes G] = [F] \cdot [G]$

where we multiply the coefficients of the formal power series $F$ and $G$ by inducing group representations along the inclusion $S_m \times S_n \hookrightarrow S_{m+n}$

$\cdot : R(S_m) \otimes R(S_n) \to R(S_{m + n})$

so

$[F] \cdot [G] = \sum_{i \ge 0 }\sum_{m + n = i} [F_m] \cdot [G_n] t^i$

John Baez (Nov 17 2024 at 23:58):

Two examples: each symmetric group $S_n$ has exactly two irreducible 1-dimensional representations, the trivial representation which I'll call $\Sigma^n$, and the sign representation which I'll call $\Lambda^n$. Let's call their equivalences classes in $R(S_n)$

$\sigma^n := [\Sigma^n]$

and

$\lambda^n := [\Lambda^n ]$

The list of all the trivial representations $\Sigma^n$ gives a FinVect-valued species which I'll call $\Sigma$, and it's easy to see that

$[\Sigma] = \sum_{n \ge 0} \sigma^n t^n$

Similarly the list of all the sign representations $\Lambda^n$ gives a FinVect-valued species which I'll call $\Lambda$, and it's easy to see that

$[\Lambda] = \sum_{n \ge 0} \lambda^n t^n$

These are our friends - Todd has been discussing them a lot, often denoting $[\Sigma]$ by $\sigma_t$ and $[\Lambda]$ by $\lambda_t$.

John Baez (Nov 18 2024 at 00:01):

Next I want to explain why there's some sort of 'Adams object' $\Psi$ that also gives an element in

$K(\mathsf{FinVect}^\mathsf{S}) \simeq \prod_{n \ge 0} K(\mathsf{FinRep}(S_n)) \simeq \prod_{n \ge 0} R(S_n)$

This object $\Psi$ is not a FinVect-valued species, but I want to describe it as a $\mathbb{Z}/2$-graded chain complex of FinVect-valued species, and explain how any such thing gives an element of $K(\mathsf{S}^\mathsf{FinVect})$.

But not today!

John Baez (Nov 18 2024 at 00:18):

I imagine Todd already gets the idea, and if I haven't screwed up he'll see it's just a minor tweak of the idea he already came up with and started describing here. He explained to me the idea of treating the 'Adams element' as an $\mathbb{N}$-graded chain complex of $\mathbb{N}$-graded Schur functors. It also works to use a $\mathbb{Z}/2$-graded chain complex of $\mathbb{N}$-graded Schur functors. But the main will-o'-the-wisp distracting me was whether we can use the 'inherent grading' of FinVect-valued species, described above, to treat this Adams element as a $\mathbb{Z}/2$-graded chain complex of FinVect-valued species. And I think we can.

Whether that's a real improvement is a separate question. But I wanted to strip off the 'externally imposed' $\mathbb{N}$-grading, which seems to add an extra layer of complexity, and use only the 'inherent' grading.

David Corfield (Nov 18 2024 at 10:09):

John Baez said:

Whether that's a real improvement is a separate question. But I wanted to strip off the 'externally imposed' N-grading, which seems to add an extra layer of complexity, and use only the 'inherent' grading.

I remember Urs being similarly taken with the observation that $E_{\infty}$ rings are inherently sphere-spectrum graded (here for Kapranov-ian reasons).

image.png
nLab: super algebra

David Corfield (Nov 18 2024 at 17:05):

I guess, now I come to think of it, these things aren't too far apart.

Your Schur functors are modules for $S$, the symmetric monoidal category of finite sets and bijections under disjoint union, and $E_{\infty}$ rings are modules for $\mathbb{S}$.

As Qiaochu Yuan tells us here,

$S$ is the free symmetric monoidal ∞-category on a point, while $\mathbb{S}$ is the free symmetric monoidal ∞-groupoid with inverses on a point,

and there's a natural map $S \to \mathbb{S}$, which explains a bunch of connections.

John Baez (Nov 18 2024 at 17:23):

Thanks for bringing this out.

That natural map is very appealing to me. While our friend $\mathsf{S}$ is the free symmetric monoidal category on an object, and also the free symmetric monoidal ∞-category on a point, it happens to be a groupoid (the groupoid of finite sets, for anyone who just tuned in), so without damage we can think of it as a homotopy type, and then it's the space of finite subsets of points in $\mathbb{R}^\infty$. I visualize such a subset as a swarm of gnats particles. Such a finite collection of particles in $\mathbb{R}^\infty$ is the homotopy theorist's version of a natural number.

If we include antiparticles and let particles and antiparticles be created and annihilated in pairs, we get $\mathbb{S}$. A particle-antiparticle swarm is the homotopy theorist's version of an integer.

So yes, there's a very appealing map $\mathsf{S} \to \mathbb{S}$.

John Baez (Nov 19 2024 at 00:37):

Okay, let me finish off my attempt to chase down that will-o'-the-wisp: it seems to be working - in which case it wasn't a will-o'-the-wisp after all!

John Baez (Nov 19 2024 at 00:51):

I would like to describe the Adams object as a 'derived species'. So what's that?

As explained earlier, the category of FinVect-valued species is the functor category

$\mathsf{Vect}^\mathsf{S}$

where $\mathsf{S}$ is the groupoid of finite sets. This is an abelian category so we can talk about chain complexes in it. There are at least 3 kinds of chain complexes I might be interested in:

• $\mathbb{N}$-graded chain complexes
• $\mathbb{Z}$-graded chain complexes, and
• $\mathbb{Z}/2$-graded chain complexes.

Earlier I'd been learning toward $\mathbb{Z}/2$-graded chain complexes, but by using the term 'derived species' I'm hinting at $\mathbb{N}$-graded ones. Let me work with those for now, but then I can turn these into $\mathbb{Z}$- or $\mathbb{Z}/2$-graded chain complexes by pushing along the group homomorphisms

$\mathbb{N} \to \mathbb{Z} \to \mathbb{Z}/2$

We should revisit this issue later.

John Baez (Nov 19 2024 at 01:09):

So, derived species will be my catchy nickname for an $\mathbb{N}$-graded chain complex in $\mathsf{Vect}^{\mathsf{S}}$. When I say 'chain complex', I'll mean an $\mathbb{N}$-graded chain complex for now.

However, there's more to say about this, since there various contexts in which to study $\mathbb{N}$-graded chain complexes in $\mathsf{Vect}^{\mathsf{S}}$:

• For starters, $\mathbb{N}$-graded chain complexes in $\mathsf{Vect}^{\mathsf{S}}$ naturally form a category using the usual concept of map between chain complexes. As mentioned, this is an abelian category.

• We can then take this abelian category and localize it by formally inverting quasi-isomorphisms - maps that induce isomorphisms on homology. This is called forming the derived category of $\mathsf{Vect}^{\mathsf{S}}$. I believe that when I say 'derived species', I should really mean a chain complex in $\mathsf{Vect}^{\mathsf{S}}$ viewed as an object in this derived category. And that will often be what I want to do!

• Chain complexes in $\mathsf{Vect}^{\mathsf{S}}$ also form a differential graded category if we allow ourselves to consider maps of arbitrary integer degree. Maps of degree 0 are the usual maps between chain complexes, maps of degree 1 are chain homotopies, and so on. This is probably the 'richest' context in which to work with such chain complexes. It may help you technically to convert this differential graded category as an $(\infty,1)$-category, but that's something you can do mechanically - or so I hear - and it's not really giving access to brand new information. On the other hand, there's more information in the differential graded category than in the derived category.

John Baez (Nov 19 2024 at 01:21):

@Todd Trimble has already been doing a lot of calculations up to quasi-isomorphism earlier in this conversation, and I just need to check that the most important ones can be carried out in the above derived category, which I may call the category of derived species. The difference is that he was working with

• (the derived category of) $\mathbb{N}$-graded chain complexes of $\mathbb{N}$-graded objects in $\mathsf{Schur}$

while I want to work with

• (the derived category of) $\mathbb{N}$-graded object in $\mathsf{Vect}^{\mathsf{S}}$.

John Baez (Nov 19 2024 at 01:25):

The difference between $\mathsf{Schur}$ and $\mathsf{Vect}^{\mathsf{S}}$ is not a huge deal: the former is a full subcategory of the latter, so the latter gives us a bit more wiggle room. The big difference is that Todd is using $\mathbb{N}$-graded objects in $\mathsf{Schur} \subseteq \mathsf{Vect}^{\mathsf{S}}$ while I'm trying to not use that grading, and instead use only the 'intrinsic grading' that $\mathsf{Vect}^{\mathsf{S}}$ already have.

Todd Trimble (Nov 19 2024 at 01:27):

John Baez said:

Todd Trimble has already been doing a lot of calculations up to quasi-isomorphism earlier in this conversation, and I just need to check that the most important ones can be carried out in the above derived category, which I may call the category of derived species. The difference is that he was working with

• (the derived category of) $\mathbb{N}$-graded chain complexes of $\mathbb{N}$-graded objects in $\mathsf{Schur}$

while I want to work with

• (the derived category of) $\mathbb{N}$-graded object in $\mathsf{Vect}^{\mathsf{S}}$.

No I wasn't!! I already said this privately!

John Baez (Nov 19 2024 at 01:28):

What weren't you doing, exactly?

John Baez (Nov 19 2024 at 01:28):

I said a long string of things after that first bullet.

John Baez (Nov 19 2024 at 01:29):

The main thing I'm concerned about now is that "extra grading".

Todd Trimble (Nov 19 2024 at 01:29):

I was all along working with the derived catgory of $\mathbb{N}$-graded objects in $\mathsf{Vect^S}$ (or actually, in the category of Schur objects).

John Baez (Nov 19 2024 at 01:30):

Okay, that's what I was saying right now.

Todd Trimble (Nov 19 2024 at 01:31):

Yes, and it's what I've been saying all along. So we agree that that's where we want to work.

John Baez (Nov 19 2024 at 01:33):

No, I had two bullet points. The first was my clumsy attempt at saying what you just said right now:

• the derived category of $\mathbb{N}$-graded objects in $\mathsf{Vect}^{\mathsf{S}}$

If my first bullet point seemed to be saying something different, it's because I expressed myself badly. But my long rant about the will-o'-the-wisp has all along been me trying to convince myself that we can work in a different context:

• the derived category of $\mathsf{Vect}^{\mathsf{S}}$.

John Baez (Nov 19 2024 at 01:35):

I'm trying to show how we can strip off the 'extrinsically added $\mathbb{N}$-grading and make do with the grading that already exists in $\mathsf{Vect}^{\mathsf{S}}$. This is an $\mathbb{N}$-graded 2-rig.

John Baez (Nov 19 2024 at 01:36):

I must be doing a terrible job of explaining this.

John Baez (Nov 19 2024 at 01:39):

Certainly my two bullet points were terribly written: when I wrote

• (the derived category of) $\mathbb{N}$-graded chain complexes of $\mathbb{N}$-graded objects in $\mathsf{Schur}$

versus

• (the derived category of) $\mathbb{N}$-graded objects in $\mathsf{Vect}^{\mathsf{S}}$.

what I was trying to say was

• the derived category of $\mathbb{N}$-graded objects in $\mathsf{Schur}$

versus

• the derived category of $\mathbb{N}$-graded objects in $\mathsf{Vect}^{\mathsf{S}}$

I'm just not good at this terminology.

Todd Trimble (Nov 19 2024 at 01:42):

How about you give an explicit example of what you have in mind, just to make sure I follow.

Todd Trimble (Nov 19 2024 at 01:44):

There's an example we've discussed many times, which is a certain exact chain complex

$S^n \otimes \Lambda^0 \to S^{n-1} \otimes \Lambda^1 \to \ldots$

(to which we can apply a $\mathbb{Z}_2$-reduction).

John Baez (Nov 19 2024 at 01:45):

Yes, I was trying to reach the ultimate goal of exhibiting the Adams operations $\psi^k$ as objects in the derived category of $\mathsf{Vect}^{\mathsf{S}}$. I did what I consider the bulk of the work yesterday, by showing how the $\mathbb{N}$-grading is redundant in two key cases. Now I was going to go through all your key calculations relating to Adams operations and redo them in the derived category of $\mathsf{Vect}^{\mathsf{S}}$.

John Baez (Nov 19 2024 at 01:47):

I was going to go through a bunch of your posts here and redo them in that way.

John Baez (Nov 19 2024 at 01:48):

I'll probably need to do the example you just mentioned at some point, but my mind had lined things up differently, so let's see.

Todd Trimble (Nov 19 2024 at 01:52):

John Baez said:

So, derived species will be my catchy nickname for an $\mathbb{N}$-graded chain complex in $\mathsf{Vect}^{\mathsf{S}}$. When I say 'chain complex', I'll mean an $\mathbb{N}$-graded chain complex for now.

However, there's more to say about this, since there various contexts in which to study $\mathbb{N}$-graded chain complexes in $\mathsf{Vect}^{\mathsf{S}}$:

• For starters, $\mathbb{N}$-graded chain complexes in $\mathsf{Vect}^{\mathsf{S}}$ naturally form a category using the usual concept of map between chain complexes. As mentioned, this is an abelian category.

• We can then take this abelian category and localize it by formally inverting quasi-isomorphisms - maps that induce isomorphisms on homology. This is called forming the derived category of $\mathsf{Vect}^{\mathsf{S}}$. I believe that when I say 'derived species', I should really mean a chain complex in $\mathsf{Vect}^{\mathsf{S}}$ viewed as an object in this derived category. And that will often be what I want to do!

• Chain complexes in $\mathsf{Vect}^{\mathsf{S}}$ also form a differential graded category if we allow ourselves to consider maps of arbitrary integer degree. Maps of degree 0 are the usual maps between chain complexes, maps of degree 1 are chain homotopies, and so on. This is probably the 'richest' context in which to work with such chain complexes. It may help you technically to convert this differential graded category as an $(\infty,1)$-category, but that's something you can do mechanically - or so I hear - and it's not really giving access to brand new information. On the other hand, there's more information in the differential graded category than in the derived category.

I just can't see any difference between what you call the derived category in the second bullet point, and what I was using. So maybe it's me who does a horrible job of explaining. (But whatever.)

In the first bullet point, where you say $\mathbb{N}$-graded chain complexes in $\mathsf{Vect^S}$, do you mean a chain complex whose components live in this category? Or something else?

John Baez (Nov 19 2024 at 01:54):

I mean a chain complex whose components live in $\mathsf{Vect}^{\mathsf{S}}$, with one component for each natural number.

Todd Trimble (Nov 19 2024 at 01:55):

Yes, and I claim this is what I have been using (although usually I'm using finitary $\mathsf{Vect}$-valued species).

John Baez (Nov 19 2024 at 01:57):

Really? I thought you were using a chain complex each of whose individual components was an $\mathbb{N}$-graded object in $\mathsf{Vect}^{\mathsf{S}}$, with one component for each natural number.