The n-Category Café

Maybe these facts are ‘coincidences’. There are lots of obstacles to understanding them. But we can at least try to get some idea of where the numbers 24 and 240 are coming from, in the homotopy groups of spheres. They’re connected to Bernoulli numbers.

What follows is some very superficial stuff, just to get the pump primed. I just want to state some cool facts and illustrate them with a few examples.

The Bernoulli numbers are connected to a certain well-understood ‘chunk’ of the stable homotopy groups of spheres, called the image of the J-homomorphism:

$$J_k : \pi_k(\mathrm{O}(\infty)) \to \pi_k^s$$

Here

$$\pi_k^s = \lim_{n \to \infty} \pi_k(S^n)$$

is the $k$th stable homotopy group of spheres, and $\mathrm{O}(\infty)$ is the infinite-dimensional orthogonal group, generated by rotations and reflections in $\mathbb{R}^\infty$. I’ll explain the J-homomorphism later, but first let’s see what people do with it.

The homotopy groups of $\mathrm{O}(\infty)$ are well-understood thanks to Bott periodicity, and you can remember them by singing this to the tune of “Twinkle Twinkle Little Star”:

$$\mathbb{Z}_2 , \mathbb{Z}_2, 0, \mathbb{Z}, 0, 0, 0, \mathbb{Z}$$

This means that $\pi_0(\mathrm{O}(\infty)) = \mathbb{Z}_2$, and so on up to $\pi_7(\mathrm{O}(\infty)) = \mathbb{Z}$… and then it repeats, with period 8! It’s a boring song that goes on forever — the kind that annoying children sing on long road trips. But the highlights are the two $\mathbb{Z}$’s, which are connected to the quaternionic and octonionic Hopf fibrations:

$$\pi_3(\mathrm{O}(\infty)) \cong \mathbb{Z}, \qquad \pi_7(\mathrm{O}(\infty)) \cong \mathbb{Z}$$

The J-homomorphism maps these to — and in fact as it happens onto — the 3rd and 7th stable homotopy groups of spheres:

$$\mathbb{Z}_{24} \cong \mathrm{im} J_3 \cong \pi_3^s$$

$$\mathbb{Z}_{240} \cong \mathrm{im} J_7 \cong \pi_7^s$$

So in these two cases, knowing $\mathrm{im} J_k$ tells you everything. But the J-homomorphism is not onto in general, and its image is easier to understand than the rest of the stable homotopy group $\pi_k^s$, so I’ll focus on that.

Note: the cases we care about happen when $k$ is one less than a multiple of 4. Frank Adams showed something wonderful: when $k = 4n -1$, the image of the J-homomorphism is a cyclic group whose order is connected to Bernoulli numbers!

Namely, the order of $\mathrm{im}J_{4n-1}$ is the denominator of

$$B_{2n}/ 4n$$

where $B_{2n}$ is a Bernoulli number.

Here are some examples:

• $n = 1$. Here $B_{2n} = 1/6$, so the denominator of $B_{2n}/4n$ is $4 \times 6 = 24$, so $\mathbb{Z}_{24} \cong \mathrm{im}J_3 \subseteq \pi_3^s$. In this case $\mathbb{Z}_{24}$ is all of the stable homotopy group $\pi_3^s$.

• $n = 2$. Here $B_{2n} = -1/30$, so the denominator of $B_{2n}/4n$ is $8 \times 30 = 240$, so $\mathbb{Z}_{240} \cong \mathrm{im}J_7 \subseteq \pi_7^s$. In this case $\mathbb{Z}_{240}$ is all of the stable homotopy group $\pi_7^s$.

• $n = 3$. Here $B_{2n} = 1/42$, so the denominator of $B_{2n}/4n$ is $12 \times 42 = 504$, so $\mathbb{Z}_{504} \cong \mathrm{im}J_{11} \subseteq \pi_{11}^s$. In this case $\mathbb{Z}_{504}$ is all of the stable homotopy group $\pi_{11}^s$.

• $n = 4$. Here $B_{2n} = -1/30$, so the denominator of $B_{2n}/4n$ is $16 \times 30 = 480$, so $\mathbb{Z}_{480} \cong \mathrm{im} J_{15} \subseteq \pi_{15}^s$. In this case $\mathbb{Z}_{480}$ is not the whole stable homotopy group $\pi_{15}^s \cong \mathbb{Z}_480 \times \mathbb{Z}_2$. Just when you were getting complacent!

Another pattern that breaks down is this: the numerator of a Bernoulli number doesn’t need to be 1. But all we care about is their denominators. And Adams proved these were connected to the image of the J-homomorphism using Von Staudt’s theorem.

This theorem says the denominator of $B_{2n}$ is the product of all primes $p$ such that $p-1$ divides $2n$. Let’s try it out!

• $n = 1$. The primes $p$ such that $p-1$ divides $2n = 2$ are 2 and 3, so the denominator of $B_{2n}$ is $2 \times 3 = 6$, and the denominator of $B_{2n}/4n$ is $4 \times 6 = 24$.

• $n = 2$. The primes $p$ such that $p -1$ divides $2n = 4$ are 2, 3 and 5 so the denominator of $B_{2n}$ is $2 \times 3 \times 5 = 30$, and the denominator of $B_{2n}/4n$ is $8 \times 30 = 240$.

• $n = 3$. The primes $p$ such that $p -1$ divides $2n = 6$ are 2, 3 and 7 so the denominator of $B_{2n}$ is $2 \times 3 \times 7 = 42$, and the denominator of $B_{2n}/4n$ is $12 \times 42 = 504$.

• $n = 4$. The primes $p$ such that $p-1$ divides $2n = 8$ are 2, 3, and 5 so the denominator of $B_{2n}$ is $2 \times 3 \times 5 = 30$, and the denominator of $B_{2n}/4n$ is $16 \times 30 = 480$.

Yup, it’s working! Note that the order of $\mathrm{im} J_{4n-1}$ always winds up being $8n$ times a bunch of distinct odd primes.

We could ask if the weird pattern I mentioned at the start keeps on going. That is: is there some nice lattice packing of balls in dimension 12 where each ball touches 504 others? Is there one in dimension 16 where each ball touches 480 others, or perhaps twice as many?

I don’t know. The Coxeter–Todd lattice is a cool lattice in 12 dimensions, but each ball touches 756 others. Like 504, this number has 2, 3 and 7 as prime factors. Does that count for something? The Barnes–Wall lattice is a cool lattice in 16 dimensions, but each ball touches 4320 others. Like 480, this number has 2, 3 and 5 as prime factors. Does that count for something? Who knows!

There’s a lot left to understand… the stuff about lattices may be a dead end, but the connection between $\mathrm{im} J$ and Bernoulli numbers is worth a lot more study. I just wanted to lay some facts on the table.

The J-homomorphism

Oh yeah — and what’s the J-homomorphism?

You can get the orthogonal group $\mathrm{O}(n)$ to act on an $n$-sphere. It’s obvious how it acts on an $(n-1)$-sphere — use the unit sphere in $\mathbb{R}^n$. But if you take $\mathbb{R}^n$ and add a point at infinity you get an $n$-sphere, and since $\mathrm{O}(n)$ acts on $\mathbb{R}^n$ it acts on this $n$-sphere. Even better, it acts in a way that preserves a point, namely the origin — or if you prefer, the point at infinity. So we get a map

$$j: \mathrm{O}(n) \times S^n \to S^n$$

which is basepoint-preserving.

This lets us take information about the homotopy groups of $\mathrm{O}(n)$ and get information about the homotopy groups of spheres. That’s the idea behind the J-homomorphism! It will be a homomorphism

$$J_k: \pi_k(\mathrm{O}(n)) \to \pi_k(S^{n+k})$$

It goes like this. Let’s work in the world of nice topological spaces and basepoint-preserving maps; this has a smash product $X \wedge Y$ where we take the usual cartesian product $X \times Y$ and smash down everything like $(\ast, y)$ and $(x,\ast)$. Since our map above is basepoint-preserving, it gives a map

$$j: \mathrm{O}(n) \wedge S^n \to S^n$$

Now, if we have an element of $\pi_k(\mathrm{O}(n))$, we can represent it with a map $\alpha: S^k \to \mathrm{O}(n)$ and then take the smash product

$$\alpha \wedge 1 : \; S^k \wedge S^n \; \to \; \mathrm{O}(n) \wedge S^n$$

But $S^k \wedge S^n \cong S^{n+k}$, so we get a map

$$S^{n+k} \to \mathrm{O}(n) \wedge S^n$$

Now we can compose this with $j$ and get a map

$$S^{n+k} \to S^n$$

This represents an element of $\pi_{n+k}(S^n)$. And because that element does not depend on our original choice of a representative $\alpha$, we’ve gotten a map

$$J_k : \pi_k(\mathrm{O}(n)) \to \pi_{n+k}(S^n)$$

It’s actually a homomorphism: the J-homomorphism. Then we can take the limit as $n \to \infty$ and get

$$J_k : \pi_k(\mathrm{O}(\infty)) \to \pi_k^s$$

This is great — it was discovered by George W. Whitehead, who taught me homotopy theory at MIT. But I’ve always found it a bit twisty, so I’m glad to see some slicker viewpoints on the nLab.

For example, we can take the action

$$j: \mathrm{O}(n) \times S^n \to S^n$$

and express it as a homomorphism

$$\mathrm{O}(n) \to \mathrm{Aut}(S^n) \hookrightarrow \Omega^n S^n$$

where $\Omega^n S^n$ is the topological monoid of basepoint-preserving maps from the sphere to itself. And this induces a homomorphism

$$\pi_k(\mathrm{O}(n)) \to \pi_k (\Omega^n S^n) \cong \pi_{n+k}(S^n)$$

which is the J-homomorphism.

Or, if we want to go straight to the stable case, we can work with the spectrum $\mathrm{O}$, which is a nice way of thinking about $\mathrm{O}(\infty)$, and the sphere spectrum $\mathrm{S}$, which is a way of thinking about $\Omega^n S^n$ in the limit as $n \to \infty$. Using these we can think of the J-homomorphism as a map of spectra

$$J: \mathrm{O} \to \mathrm{GL}_1(\mathrm{S})$$

Here $\mathrm{GL}_1(R)$ makes sense for any ring spectrum $R$: it’s roughly the invertible $1 \times 1$ matrices with entries in $R$. The sphere spectrum is a ring spectrum — the most fundamental ring spectrum of all — and its invertible elements come from guys in $\mathrm{Aut}(S^n)$ for any sphere $S^n$. So this is essentially just a slick packaging of what I’d said before.

Anyway, I’ve drifted away from my original goal, which is to understand how the Bernoulli numbers get into the game… but with luck I’ll get back to this someday, maybe after I retire.