December 9, 2020

The Algebraic K-Theory of the Integers

Posted by John Baez

The category of groups $\mathbb{Z}^n$ and isomorphisms between these is symmetric monoidal under $\oplus$. You can build a space out of simplexes where the 0-simplexes are objects of this category, the 1-simplexes are morphisms, the 2-simplexes are commutative triangles, the 3-simplexes are commutative tetrahedra, and so on forever. This space has an operation, coming from $\oplus$, that obeys the commutative monoid axioms up to homotopy. If you ‘group complete’ this space by throwing in formal inverses, you get a space that’s an abelian group up to homotopy. It’s called the algebraic $K$-theory spectrum of the integers.

The algebraic $K$-theory spectrum of the integers has homotopy groups $\pi_0 = \mathbb{Z}$, $\pi_1 = \mathbb{Z}/2$, $\pi_3 = \mathbb{Z}/48$, and so on. These groups are called the algebraic $K$-theory groups of the integers, $K_n(\mathbb{Z})$.

The algebraic $K$-groups of the integers, and other rings, are a big deal. A lot of fairly recent progress on understanding them is due to Voevodsky, Rost, Rognes and Weibel, whose work culminated in a proof of the Bloch–Kato Conjecture connecting $K$-theory to Galois cohomology. But lots of other mathematicians deserve credit too.

Three quarters of the algebraic $K$-groups of the integers are known. That is, they’re all known except for the groups $K_{4n}(\mathbb{Z})$, which will be known if someone can prove the Kummer–Vandiver Conjecture. This is a conjecture about prime numbers.

The Kummer–Vandiver Conjecture been verified for all primes less than 163,000,000. Nonetheless, some unimpressed Wikipedia editor says

there is no particularly strong evidence either for or against the conjecture.

Brutal! But it makes sense, because there’s a probabilistic argument saying that if the conjecture is false, you’d expect just one counterexample in the first $10^{100}$ primes.

For an advanced tour of this subject, try

For more detail, starting at the very beginning, try The K-book, also by Weibel.

What I’d like to understand is the relation between stable homotopy groups of spheres and algebraic K-theory groups of the integers. For example, the third stable homotopy group of spheres is $\mathbb{Z}/24$, and this is connected to $K_3(\mathbb{Z}) = \mathbb{Z}/48$. Topologists understand this connection: there’s a homomorphism from the $n$th stable homotopy group of spheres, called $\pi_n^s$, to $K_n(\mathbb{Z})$. But I’d like to understand it.

If I had to cook up such this homomorphism, here’s what I’d guess. The stable homotopy groups of spheres are the homotopy groups of a space called the ‘sphere spectrum’. So it’s enough to get a map from the sphere spectrum to the algebraic $K$-theory spectrum of the integers. And the sphere spectrum is built from a symmetric monoidal category just like the algebraic $K$-theory spectrum is! It’s built from the category of finite sets $0,1,2,\dots$ and bijections between these, made symmetric monoidal using disjoint union. So, I’d use the obvious functor sending any finite set $n$ to the group $\mathbb{Z}^n$, and any bijection of finite sets to the corresponding isomorphism of groups. This gives a map of symmetric monoidal categories, and thus a map of spectra, and thus a bunch of homomorphisms $\pi_n^s \to K_n(\mathbb{Z})$.

Is this right? Are these the right homomorphisms?

If I knew a lot more about this stuff, I’d also understand this claim: the reason $K_3(\mathbb{Z})$ is $\mathbb{Z}/48$ is that 24 is the largest natural number $n$ such that every number relatively prime to $n$ squares to 1 mod $n$. Let’s check that:

$1^2 = 1$

$5^2 = 25 = 24 + 1$

$7^2 = 49 = 24 \times 2 + 1$

$11^2 = 121 = 24 \times 5 + 1$

$13^2 = 169 = 24 \times 7 + 1$

$17^2 = 289 = 24 \times 12 + 1$

$19^2 = 361 = 24 \times 15 + 1$

$23^2 = 529 = 24 \times 22 + 1$

Yup!

Notice anything interesting about these numbers less than and relatively prime to 24?

Posted at December 9, 2020 12:26 AM UTC

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Re: The Algebraic K-Theory of the Integers

No doubt plenty to add to nLab: algebraic K-theory.

Regarding the relation between stable homotopy groups of spheres and algebraic K-theory groups of the integers, we have at nLab: Field with one element

algebraic K-theory over $\mathbb{F}_1$ is stable cohomotopy.

Posted by: David Corfield on December 9, 2020 7:36 AM | Permalink | Reply to this

Re: The Algebraic K-Theory of the Integers

Thanks! I think the appearance of the sphere spectrum in that blog article is evidence for my guess:

Topologists understand this connection: there’s a homomorphism from the $n$th stable homotopy group of spheres, called $\pi_n^s$, to $K_n(\mathbb{Z})$. But I’d like to understand it.

If I had to cook up such this homomorphism, here’s what I’d guess. The stable homotopy groups of spheres are the homotopy groups of a space called the ‘sphere spectrum’. So it’s enough to get a map from the sphere spectrum to the algebraic $K$-theory spectrum of the integers. And the sphere spectrum is built from a symmetric monoidal category just like the algebraic $K$-theory spectrum is! It’s built from the category of finite sets $0,1,2,\dots$ and bijections between these, made symmetric monoidal using disjoint union. So, I’d use the obvious functor sending any finite set $n$ to the group $\mathbb{Z}^n$, and any bijection of finite sets to the corresponding isomorphism of groups. This gives a map of symmetric monoidal categories, and thus a map of spectra, and thus a bunch of homomorphisms $\pi_n^s \to K_n(\mathbb{Z})$.

Is this right? Are these the right homomorphisms?

I think anyone who actually understands this subject could answer my question in a millisecond. But I’ll use it as an excuse to read Weibel’s book. It’s always more fun to learn a subject when you feel like a detective on a mission to solve a mystery. Tracing back $K_3(\mathbb{Z}) = \mathbb{Z}/48$ to $\pi_3^s = \mathbb{Z}/24$ and figuring out where the extra factor of 2 comes from will make a nice hobby. And then, trying to understand how the Bernoulli numbers get involved in this business.

Posted by: John Baez on December 9, 2020 5:01 PM | Permalink | Reply to this

Re: The Algebraic K-Theory of the Integers

A definite maybe: $F_1$ keeps her secrets, and in some possibly naive sense, as $q \to 1$, $K^{alg}(F_q)$ converges to the image of the $J$-homomorphism (which is where the Bernoulli numbers, or more precisely $\zeta(1 - 2n)$, live).

BTW stable cohomotopy and stable homotopy are represented (or corepresented) by the same spectrum.

Posted by: jackjohnson on December 9, 2020 10:05 PM | Permalink | Reply to this

Re: The Algebraic K-Theory of the Integers

Tyler Lawson mentioned here

the appearance of groups whose orders are denominators of Bernoulli numbers in stable homotopy groups of spheres…

as a piece of evidence for some vague notion of a “topological Langlands correspondence”.

Posted by: David Corfield on December 10, 2020 7:15 AM | Permalink | Reply to this

Re: The Algebraic K-Theory of the Integers

For clarification (I hope): there are several interesting proposed interpretations for $F_1$, sometimes in terms of a category of modules and other times as a category of algebras. The definition of the $K$-theory of categories of modules is now pretty well-understood, and Manin observed that if $F_1$-modules are just finite sets then the $K$-theory spectrum of that category is the stable homotopy ring-spectrum; but (for example) I’m not sure anyone has investigated the $K$-theory associated to Borger’s model of $F_1$-algebras as $\lambda$-rings (which seems to have interesting connections with modules over the Bousfield localization $L_{KO}S^0$ of spectra with respect to real topological K-theory – which is closely related to the image of the $J$-homomorphism, cf eg https://www.math.ias.edu/~lurie/252xnotes/Lecture35.pdf).

Posted by: jackjohnson on December 10, 2020 3:13 PM | Permalink | Reply to this

Re: The Algebraic K-Theory of the Integers

I wrote:

Topologists understand this connection: there’s a homomorphism from the $n$th stable homotopy group of spheres, called $\pi_n^s$, to $K_n(\mathbb{Z})$. But I’d like to understand it.

My guess about how it works is right. At least, the map I guessed is Example 4.12.1 in chapter 4 of The K-book. The trick works equally well for any ring $R$: the ‘free $R$-module on a finite set’ functor is a map of symmetric monoidal categories, which gives a map of spectra, and thus maps between their homotopy groups, namely maps

$\pi_n^s \to K_n(R)$

So simple!

This book is tons of fun, by the way.

Posted by: John Baez on December 10, 2020 1:02 AM | Permalink | Reply to this

Re: The Algebraic K-Theory of the Integers

The Kummer–Vandiver Conjecture been verified for all primes less than 163,000,000. Nonetheless, some unimpressed Wikipedia editor says

there is no particularly strong evidence either for or against the conjecture.

Brutal!

I would have expected the author of this blog post to be totally on board with that editor’s assessment.

Posted by: Mark Meckes on December 9, 2020 9:14 AM | Permalink | Reply to this

Re: The Algebraic K-Theory of the Integers

Heh. Well, how much the sheer accumulation of examples boosts confidence in a conjecture depends massively on what else one knows. For something like the Collatz conjecture, the examples seem helpful. When you make up an identity that’s designed to break down when $n$ exceeds $1.54 \times 10^{43}$, then the first nonillion examples are just a joke. I don’t understand the probabilistic argument saying that if the Kummer–Vandiver conjecture is false you’d expect just one counterexample in the first $10^{100}$ cases. But if it’s plausible, we’ve got an interesting case of an important conjecture where examples aren’t very convincing. And on MathOverflow I read that from a number theorist’s point of view, this conjecture is “a little silly — its natural generalization is false and there’s no compelling reason for it to be true.”

Posted by: John Baez on December 9, 2020 4:48 PM | Permalink | Reply to this

Re: The Algebraic K-Theory of the Integers

I once sketched how some Bayesian style considerations apply to confirmation of instances increasing confidence in a mathematical hypothesis. (Chap. 5 of Towards a Philosophy of Real Mathematics). Of course, it’s all very subtle, depending on plausible analogies and so on, as Polya taught us.

Posted by: David Corfield on December 9, 2020 4:54 PM | Permalink | Reply to this

Notice anything special about these numbers?

They’re almost all prime..!

Posted by: Morgan Rogers on December 9, 2020 9:41 PM | Permalink | Reply to this

Re: Notice anything special about these numbers?

Yes! Every natural number $\le 24$ relatively prime to 24 is prime, except for 1 (which is an honorary prime).

A very round number is a natural number $n$ such that every natural number $\le n$ relatively prime to $n$ is prime, except for $1$. These are all the very round numbers:

$0,1,2,3,4,6,8,12,18,24,30$

It is easy to see that there’s no very round number bigger than $30$. First, it would have to be divisible by $2, 3$, and $5$, or else some product of two of these primes would be relatively prime to this number and smaller than it, but not prime. So the first option is $60$. $60$ is not very round because $7^2$ is relatively prime to it but not prime. In short, the problem is that there’s a prime that does not divide $60$ whose square is less than $60$. So, any round number bigger than $30$ actually needs to be divisible by $7$, too, which means it must be bigger than $2 \times 3 \times 5 \times 7 = 210$. But $11^2 < 210$ and $13^2 < 210$, so any round number bigger than $30$ must actually be bigger than $2 \times 3 \times 5 \times 7 \times 11 \times 13$. And so on: this is clearly a losing battle. At this stage, knowing there’s always a prime between $n$ and $2n$ is enough to guarantee that we can keep up this argument by induction.

There’s a spooky appearance of very round numbers in the theory of Coxeter groups (see the answer to puzzle 4 near the bottom of this page).

For example: if you get as many equal-sized balls as possible to touch a single ball of the same size in 8 dimensions, namely 240 of them, you get a pattern called the $\mathrm{E}_8$ root polytope. There’s no wiggle room.

The so-called Coxeter number of $\mathrm{E}_8$ is 240/8 = 30. Here are the numbers between 1 and 30 that are relatively prime to 30:

$1,7,11,13,17,19,23,29$

They’re all prime! And yes, there are 8 of them! If we add one to each, we get the magic numbers for $\mathrm{E}_8$:

$2,8,12,14,18,20,24,30$

There’s a lot you can do with these, but the simplest is to multiply them all:

$2 \cdot 8 \cdot 12 \cdot 14 \cdot 18 \cdot 20 \cdot 24 \cdot 30 = 696,729,600$

This gives the order of the Weyl group of $\mathrm{E}_8$: that is, the symmetry group of the pattern formed when you get as many equal-sized balls as possible to touch a single ball of the same size in 8 dimensions!

Posted by: John Baez on December 9, 2020 10:14 PM | Permalink | Reply to this

Re: The Algebraic K-Theory of the Integers

I’m intrigued by the fact 48/2 = 24 and that ‘finite sets’ are categorified natural numbers, which in turn are ‘one half’ of the integers.

Posted by: Matteo Capucci on January 6, 2021 4:33 PM | Permalink | Reply to this

Re: The Algebraic K-Theory of the Integers

Heh. I’m intrigued by that factor of 2 as well: I don’t really understand it. One way to get ahold of it is to read about the relations between $K_n(\mathbb{Z})$ and $\pi_n^s$ (or better the image of the image of the $J$-homomorphism in $\pi_n^s$, since that’s where the Bernoulli numbers show up) for higher $n$, and see what the pattern is, and why. Tackling the case $n = 3$ on its own makes it harder to figure out some patterns. For all this stuff I highly recommend The K-book by Charles Weibel.

Posted by: John Baez on January 6, 2021 6:10 PM | Permalink | Reply to this

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