The n-Category Café

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.

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.

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.

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.”

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.

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!

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.