## December 10, 2020

### Bernoulli Numbers and the J-homomorphism

#### Posted by John Baez

I’m planning to stop teaching at U. C. Riverside in June 2021. I’ll only be 60, but what’s the use of quitting work when you’re too old to have fun?

I want to spend more time doing research and writing expository papers and books, and I’ve saved up enough money to do this. I’ll still do serious work, like trying to save the planet with applied category theory. But I’ll also delve into all sorts of puzzles that I haven’t had enough time for yet.

Here’s one. You may have heard about the funny way the number 24 shows up in the homotopy groups of spheres:

$\pi_{n+3} (S^n) \cong \mathbb{Z}_{24}$

whenever $n$ is big enough, namely $n \ge 5$. If you try to figure out where this comes from, you’re led back to a map

$S^7 \to S^4$

called the quaternionic Hopf fibration. This by itself doesn’t make clear where the 24 is coming from — but you can’t help but notice that when you pack equal-sized balls as densely as is known to be possible in the quaternions, each one touches 24 others.

Coincidence? Maybe! But it’s also true that

$\pi_{n+7} (S^n) \cong \mathbb{Z}_{240}$

when $n$ is big enough. And if you try to figure out where this comes from, you’re led back to a map

$S^{15} \to S^8$

called the octonionic Hopf fibration. And you can’t help but notice that when you pack equal-sized balls as densely as possible in the octonions, each one touches 240 others!

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.

Posted at December 10, 2020 5:46 PM UTC

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### Re: Bernoulli Numbers and the J-homomorphism

Congrats on your retirement! Are you thinking to still have grad students (possibly under the table)? When I was at Princeton there were lots of such students, who for some reason all had Bill Browder as their claimed advisor.

As for the 24 and 240 at the beginning, consider the automorphism groups of those lattices (that’s $D_4$ and $E_8$, right?), fixing the sphere around the origin. Do those groups contain cyclic groups that act simply transitively on the spheres touching the central sphere? If so, what’s the Coxeter-shortest possible generator of such a subgroup?

Also, what happened with the complex Hopf fibration? $\pi_3(S^2) \neq {\mathbb{Z}} {}_6$. Is there a homotopy vs. stable homotopy issue here?

Posted by: Allen Knutson on December 11, 2020 2:08 AM | Permalink | Reply to this

### Re: Bernoulli Numbers and the J-homomorphism

I’ll be allowed to have grad students after I retire. But I quit taking new ones a few years ago, when I had seven. After my two remaining students — Christian William and Jade Master — are done, I want to see what life without students is like. I have three papers with students to finish and I’m trying hard to finish two by the end of this year. I’m looking forward to some freedom, where I feel like I can wake up and do whatever I want.

I plan to travel a bit and talk to good mathematicians when the plague ends. Brendan Fong and David Spivak are trying to set up an institute for applied category theory in the Bay area — the Topos Institute — and I hope to work there. But it may turn out that I need grad students to stay happy. It’ll be interesting to see.

Posted by: John Baez on December 11, 2020 6:40 AM | Permalink | Reply to this

### Re: Bernoulli Numbers and the J-homomorphism

Sounds like a fabulous plan to me 👍

Posted by: bertie on December 11, 2020 10:34 AM | Permalink | Reply to this

### Re: Bernoulli Numbers and the J-homomorphism

Allen wrote:

As for the 24 and 240 at the beginning, consider the automorphism groups of those lattices (that’s $D_4$ and $E_8$, right?), fixing the sphere around the origin.

Right. I think in the first case you have a choice between two Weyl groups: $D_4$ and the bigger $F_4$. They give isomorphic lattices, right, both with 24 nonzero vectors of shortest possible length forming the vertices of a 24-cell. Right?

Do those groups contain cyclic groups that act simply transitively on the spheres touching the central sphere?

I’ve thought about things sort of like this, but not quite this. Could this really be possible?

Posted by: John Baez on December 11, 2020 6:58 AM | Permalink | Reply to this

### Re: Bernoulli Numbers and the J-homomorphism

Allen wrote:

Also, what happened with the complex Hopf fibration?

I didn’t talk about that one because it doesn’t work.

$\pi_3(S^2) \ncong {\mathbb{Z}} {}_6$.

Yeah. Of course we should note that also

$\pi_7(S^4) \cong \mathbb{Z} \times \mathbb{Z}_{12} \ncong \mathbb{Z}_{24}$

and

$\pi_{15}(S^8) \cong \mathbb{Z} \times \mathbb{Z}_{120} \ncong \mathbb{Z}_{240}$

It’s just that the quaternionic and octonionic Hopf fibrations give elements of $\pi_7(S^4)$ and $\pi_{15}(S^8)$ which when suspended enough times generate the stable homotopy groups

$\pi_{3}^s \cong \mathbb{Z}_{24}$

and

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

The complex Hopf fibration gives an element of $\pi_3(S^2) \cong \mathbb{Z}$ which when suspended enough gives a generator of the stable homotopy group

$\pi_1^s \cong \mathbb{Z}_2$

But we don’t get $\mathbb{Z}_6$ showing up! My best excuse is that only the stable homotopy groups $\pi^s_{4n-1}$ have a really interesting $\mathrm{im} J$. But that’s just a wild guess.

Posted by: John Baez on December 11, 2020 7:12 AM | Permalink | Reply to this

### Re: Bernoulli Numbers and the J-homomorphism

“Note that the order of im J_{4n-2} always winds up being a power of two times a bunch of distinct odd primes.”

I am confused about the n=3 case. This is 12 x 42 = 504. But 12 is not a power of two (3 appears twice in 504).

Two small typos: A right parenthesis is missing on \pi_7(O(\infty)) (just after “Twinkle Twinkle Little Star”), and 2m should be 2n in the statement of Von Staudt’s theorem.

Posted by: anon on December 11, 2020 3:17 AM | Permalink | Reply to this

### Re: Bernoulli Numbers and the J-homomorphism

Thanks for catching those mistakes!

What I should have said is that the order of $im J_{4n-1}$ is always $8n$ times a product of distinct odd primes.

By Von Staudt’s theorem the denominator of $B_{2n}$ is 2 times a product of distinct odd primes. The order of $im J_{4n-1}$ is $4n$ times that.

And by the way, the distinct odd primes always include 3.

Posted by: John Baez on December 11, 2020 6:00 AM | Permalink | Reply to this

### Re: Bernoulli Numbers and the J-homomorphism

Congratulations! I must say I’m rather envious of your future freedom, though perhaps there is something in being governed by a teaching schedule. I’ve had a lot of fun exploring the philosophy of history this term, and it seems I’ll be teaching philosophy of language next year. But then of course you don’t need classes to induce you study something.

Curious to see all this here about the homotopy groups of the spheres, the J-homomorphism, and the quaternionic Hopf fibration, when over at the nForum we’re following Urs’s program for M-theory, where the quaternionic Hopf fibration governs cohomotopy in degrees 4 and 7 twisted by the J-homomorphism.

Posted by: David Corfield on December 11, 2020 8:16 AM | Permalink | Reply to this

### Re: Bernoulli Numbers and the J-homomorphism

I’ll see if I need some external structure to keep from drifting into lassitude. It’ll be interesting to see. If I do, I hope I have the sense to find such structure.

I have a bunch of papers and books I want to write. I’ve spent the last few years dreaming about them. Writing books can be tiring, so I may do collections of essays which I put on the arXiv one at a time, or blog about, to give myself some immediate gratification.

Tim Hosgood has LaTeXed up 2610 pages of This Week’s Finds, and right now I’m starting to polish these up a bit. So, it’s not like I’ll be short of work. If I’m not careful I can get tired just thinking about it.

Posted by: John Baez on December 11, 2020 6:16 PM | Permalink | Reply to this

### Re: Bernoulli Numbers and the J-homomorphism

There is a strengthening of the Smith–Minkowski–Siegel mass formula called the Siegel–Weil formula, which helps explain the appearance of $240$, $24$ and $6=3\cdot2$ in the densest lattice packings in $8$, $4$ and $2$ dimensions respectively, and their relationship to the octonions, quaternions and complex numbers. In particular, the first value turns out to be “the same” as the $240$ arising from the theory of Bernoulli numbers and the zeta function, and thus should be “the same” as the one appearing in the image of the $J$ homomorphism: the others are also related to $24$ and $2$, but in a more indirect way.

The Siegel–Weil formula implies that a certain linear combination with rational coefficients of the theta series of even lattices with the same dimension $n$ and discriminant $D$ (let’s restrict to even $n$ for simplicity) is a certain Eisenstein series. More explicitly, at least in the cases we care about, the Eisenstein series has weight $k=n/2$ and character $\chi = \left(\frac{(-1)^k D}{\cdot}\right)$ (a Kronecker symbol).

Two key facts are:

• The coefficient of order 1 of the theta series of an even lattice, let’s call it $m$, equals the number of vectors of norm 2.

• The coefficient of order 1 of the Eisenstein series in question (normalized so that its constant term is $1$) equals $2/L(1-k,\chi)$, and all higher-order coefficients are integral multiples of this one.

The latter fact is not too difficult to prove once we recall that these Eisenstein have the form $E_{k,\chi}(e^{2\pi i \tau}) \propto \sum_{(m,n)\neq(0,0)} \chi(n) (m+n\tau)^{-k}$.

In particular, for even unimodular lattices $\chi$ is the trivial character, and $L(1-k,\chi)$ is simply $\zeta(1-k) = -B_k /k$ (hence only in the unimodular case we will have a clean relationship with ordinary Bernoulli numbers). In any other case, we have $L(1-k,\chi) = -B_{k,\chi} /k$, where $B_{k,\chi}$ is a generalized Bernoulli number.

Now, in dimensions $2, 4, 8$ (and also in dimension $1$, which we’re excluding) something very special happens: the densest lattice packings come from lattices that are maximal orders in rational division algebras of class number 1. The latter property (which one can think of as a sort of “unique factorization into primes”) implies that the linear combination of theta series in the Siegel–Weil formula has only one term, so the theta series equals the Eisenstein series, and we get the identity $m = 2/L(1-k,\chi)$. This is a very rare property for lattices, I asked a question about it on MathOverflow and it turns out that there is only two other sporadic examples in dimension $16$.

-For the lattice $E_8$ of octonionic integers, by the above discussion, the coefficient is $m=2/\zeta(1-(8/2)) =2/\zeta(-3) = 240$. Thus the fact that the $E_8$ lattice is the unique even unimodular eight-dimensional lattice already forces it to have $240$ minimal-norm vectors.

-In four dimensions there are no even unimodular lattices; the next best thing is an even $2$-modular lattice, namely the $D_4$ lattice of Hurwitz integers. Since $N$-modular lattices have discriminant $N^{n/2}$, the corresponding Dirichlet character $\left( \frac{(-2)^{4/2}}{\cdot}\right) = \chi_2$ is the principal character of order $2$. For the principal character of order $p$, where $p$ is any prime, we have the identity $B_{k,\chi_p} = B_k (1-p^{k-1})$, so in this case

$m = \frac{2}{\zeta(1-2) (1-2^{2/1})} = -\frac{24}{1-2} = 24.$

Note that the fact that $m$ turns out to be exactly $24$ here is sort of a mathematical coincidence due to the fact that $2-1 = 1$. In fact, there exist only four other $4$-dimensional $p$-modular lattices with $p=3, 5, 7, 13$ with the same property of being quaternion orders of class number 1; for those we have $m=12, 6, 4, 2$ respectively. However, in all cases we obtain a divisor of $24$, so in some sense the behavior of quaternion orders of class number $1$ is still controlled by the homotopy-theoretic $24$.

-Finally, in two dimensions the densest packing (i.e. the hexagonal lattice of Eisenstein integers) is $3$-modular. The corresponding character is $\left( \frac{(-3)^{2/2}}{\cdot}\right) = \left( \frac{-3}{\cdot}\right)$, and the coefficient is (by a known formula)

$m = \frac{2}{\sum_{n=1}^{3} \frac{n}{-3}\left( \frac{-3}{n}\right)} = \frac{2}{-\frac{1}{3} + \frac{2}{3} + 0} = 6.$

Again there is a similar formula for the other rings of integers of quadratic imaginary fields with class number $1$ (Gaussian integers, Kleinian integers,… up to $D=163$). In an interesting reversal of the quaternionic situation, now all the $m$’s we obtain are multiples of $2$.

A brief comment about other dimensions: in dimension $16$ (the sporadic example that I mentioned above), the two unique even unimodular lattices $E_8\times E_8$ and $D_{16}^+$ (famous from heterotic string theory) have the same number of smallest-norm vectors, equal to $m=2/\zeta(1-(16/2)) =2/\zeta(-7) = 480$. On the other hand, in dimension $24$, the coefficient of order $1$ in the Eisenstein series is $65520/691$ (and all higher-order coefficients are integral multiples of this one). In this case, the Siegel–Weil formula involves more than one theta series, so that alone is not enough to extract any immediate relationship between $65520$ and the number of smallest-norm vectors in the Niemeier lattices, but perhaps some variant of it can be used to prove the fact you quote for the specific case of extremal lattices.

Now the question would be, is there any object that relates to generalized Bernoulli numbers in the same way that the usual sphere spectrum/$J$-homomorphism relate to ordinary Bernoulli numbers? I found a paper by N. Zhang that tries to answer exactly this question (see in particular Table 1), though unfortunately I’m too dumb to understand it yet.

(Sorry for the long post, and for any possible mistakes. I’m not an expert on this topic myself, but I had wondered the same question not so long ago; hopefully it could help you clarify some part of the mystery).

Posted by: pregunton on December 13, 2020 1:02 PM | Permalink | Reply to this

### Re: Bernoulli Numbers and the J-homomorphism

The $n$Lab article on Bernoulli numbers points to formal group laws as the reason the former show up in algebraic topology:

In algebraic topology/cohomology Bernoulli numbers appear as the coefficients of the characteristic series of the A-hat genus (see there), and they (or equivalently, their generating functions) also appear in the expression for the Todd class.

The Bernoulli numbers are also proportional to the constant terms of the Eisenstein series and as such appear in the exponential form of the characteristic series of the Witten genus.

Finally they appear as the order of some groups in the image of the J-homomorphism (cf. Adams 65, section 2).

Of course, all of these cases are related to formal group laws. Formal groups bear also some other connections to Bernoulli numbers and generalizations like Bernoulli polynomials.

There’s a nice retirement project for you. Make sense for mere mortals of what they’re talking about on this old thread - David Ben-Zvi here and Jack Morava and James Borger from here - and developments over the past 11 years.

Posted by: David Corfield on December 11, 2020 9:11 AM | Permalink | Reply to this

### Re: Bernoulli Numbers and the J-homomorphism

I looked at how the size of $\mathrm{im} J$ in the stable homotopy groups $\pi_{11}^s$ and $\pi_{15}^s$ were related to the number of nearest neighbors of each point in the Coxeter–Todd and Barnes–Wall lattices in dimensions 12 and 16, respectively. But I should have tried the really interesting example of a $4n$-dimensional lattice: the Leech lattice!

In the Leech lattice in dimension 24, each point has

$196560 = 2^4 \cdot 3^3 \cdot 5 \cdot 7 \cdot 13$

nearest neighbors. Let’s compare the order of $\mathrm{im} J_{23} \subseteq \pi_{23}^s$.

We know it’s $4n$ times the denominator of $B_{2n}$ where now $n = 6$. And by Von Staudt’s theorem we know that denominator is the product of primes $p$ such that $p-1$ divides $2n = 12$. These primes are $2,3,5,7$ and $13$. So the order of $\mathrm{im} J_{23}$ is

$24 \cdot 2 \cdot 3 \cdot 5 \cdot 7 \cdot 13$

which equals

$65520 = 2^4 \cdot 3^2 \cdot 5 \cdot 7 \cdot 13$

So it’s not $196560$, but it’s exactly $1/3$ times as big (unless I made a mistake somewhere).

Note that for the Coxeter–Todd lattice in 12 dimensions, each point has $2^2 \cdot 3^3 \cdot 7$ nearest neighbors, while $\mathrm{im} J_{11}$ has order $2^3 \cdot 3^2 \cdot 7$, exactly $2/3$ as big.

For the Barnes–Wall lattice in 16 dimensions, each point has $2^5 \cdot 3^3 \cdot 5$ nearest neighbors, while $\mathrm{im} J_{15}$ has order $2^5 \cdot 3 \cdot 5$, exactly $1/3^2$ as big.

So like Mr. Jones in the Dylan song, I feel like something is happening here but I don’t know what it is. I should see what people have written about the number theory of lattices. There might be some simple reason why, when you have balls densely packed in a lattice in $\mathbb{R}^{4n}$, the number of balls touching each ball has prime factors $p$ such that $p-1$ divides $2n$.

Posted by: John Baez on December 11, 2020 3:41 PM | Permalink | Reply to this

### Michael

John, the explanation might have to do with looking at extremal lattices and their theta functions. Here I’m using the term extremal as being best possible with respect to theorems (and a general theory) developed by Conway & Sloane, Nebe, Venkov and several others. It turns out that in this sort of situation, the extremal lattice (if it exists) has kissing number that is naturally divisible by the denominator of the relevant Bernoulli number. This follows from the theory of modular forms modulo p considered by Serre, Swinnerton-Dyer and others. They created this to explain and understand what was behind the extraordinary congruences for Ramanujan’s tau function. Thus we loop back to the Leech lattice, whose theta function involves the tau function.

If people are interested, a lot of the above is embedded in papers of Gabriele Nebe and Neil Sloane, who both have online publication lists. Of course I have to mention the bible here, Conway & Sloane’s Sphere Packings, Lattices and Groups or SPLAG. The specific results about divisibility for extremal lattices were first published by Bannai and others about 15-20 years ago.

Posted by: Mike on December 13, 2020 12:31 AM | Permalink | Reply to this

### Re: Bernoulli Numbers and the J-homomorphism

Oh, great — I’d like to learn about this! Earlier today I noticed that the Smith—Minkowski–Siegel mass formula gives results like this:

In dimensions that are multiples of 8, when we sum over isomorphism classes of even unimodular lattices we get

$\sum {1\over|\operatorname{Aut}(\Lambda)|} = {|B_{n/2}|\over n}\prod_{1\le j < n/2}{|B_{2j}|\over 4j}$

This is an interesting example of groupoid cardinality.

So for example in dimension 8, where there’s just one isomorphism class, namely the $\mathrm{E}_8$ lattice, we get

$\begin{array}{ccl} \displaystyle{\frac{1}{|Aut(\mathrm{E}_8)|} } &=& \displaystyle{ {|B_4|\over 8}{|B_2|\over 4}{|B_4|\over 8}{|B_6|\over 12} } \\ &=& \displaystyle{ {1/30\over 8}\;{1/6\over 4}\;{1/30\over 8}\;{1/42\over 12} } \\ &=& \displaystyle{ {1\over 696729600} } \end{array}$

where $Aut(\mathrm{E}_8)$ is usually called the $\mathrm{E}_8$ Weyl group.

Now this is not directly about kissing numbers, but it’s connected because the 240 shortest vectors in the $\mathrm{E}_8$ lattice form a homogeneous space for $Aut(\mathrm{E}_8)$. And more excitingly, the factors

$\frac{|B_{2j}|}{4j}$

are also showing up in the image of the J-homomorphism! More precisely, the denominator of this fraction (in lowest terms) is order of

$im J_{4n-1} \subseteq \pi_{4n-1}^s$

So, one thing I want to do is better understand why the Bernoulli numbers are showing up in the Minkowski–Siegel mass formula, and also why their denominators are showing up in the formula for the order of the image of the J-homomorphism. My hope is that by digging in this area I’ll learn some interesting things. Ideally there’d be some more direct connection between the J-homomorphism (or something connected to that) and certain nice lattices.

But now I want to understand this too:

It turns out that in this sort of situation, the extremal lattice (if it exists) has kissing number that is naturally divisible by the denominator of the relevant Bernoulli number.

This is great — this must explain what I was seeing!

Posted by: John Baez on December 13, 2020 1:49 AM | Permalink | Reply to this

### Re: Bernoulli Numbers and the J-homomorphism

Thanks, John. Regarding why Bernoulli numbers show up here in mass formulas and related contexts, it has to do with zeta values. Pregunton, in his post earlier today on the Siegel-Weil theorem, goes into this in more detail, as well as the connections with interesting lattices.

About congruences and divisibility for modular functions, I can’t think of anything better than Serre’s and especially Swinnerton-Dyer’s original articles, which are all well written.

About Im J being exactly as large as it is, I haven’t thought about this in a long time, but J. Frank Adams and his papers and books in the 60s take this, and the Adams Conjecture as a major theme. Someone more knowledgable than I in homotopy theory can, no doubt, say more.

Posted by: Mike on December 13, 2020 6:24 PM | Permalink | Reply to this

### Re: Bernoulli Numbers and the J-homomorphism

I’ve been reading Adam’s four-part paper on the J-homomorphism, and it’s a lot of fun: it pushes the limits of my abilities, but I could eventually understand it all if I work hard, since I’ve already spent ages studying about K-theory, $\lambda$-rings, and other such prerequisites.

One thing that bugs me so far is that I don’t see a really good reason, in this paper, for the order of $im J_{4n-1}$ to be the denominator of $B_{2n}/4n$. Instead he bounds the order from above — in a certain sense — and also bounds it below, using two separate very clever arguments, which I can’t imagine anyone coming up with if they didn’t already know the answer ahead of time… perhaps using a line of thought that was too hard to make rigorous.

A nice quick overview of these two arguments is here:

So, I think I need to read some papers by Atiyah, Milnor, Quillen and others that Adams is building on. In particular, the Todd class is instantly, visibly connected to Bernoulli numbers, so I think I need to better understand how the Todd class are related to the J-homomorphism.

Posted by: John Baez on December 13, 2020 6:46 PM | Permalink | Reply to this

### Re: Bernoulli Numbers and the J-homomorphism

IIRC Adams (J IV (1966) §7.1b) proves a sharper result: the $J$-homomorphism maps a copy of $\mathbb{Z}$ to a cyclic group in degree $4k-1$, and the image of a generator can be identified with $\frac{1}{2} \zeta(1-2k) \in \mathbb{Q}/\mathbb{Z}$. Deligne, I think, talks about this in his long paper about three points on the projective line; see for ex § 1.1 of

for further references.

In other news, congratulations on your coming elevation! May I suggest

Posted by: jackjohnson on December 11, 2020 6:39 PM | Permalink | Reply to this

### Re: Bernoulli Numbers and the J-homomorphism

Thanks! But why will it be helpful to read about “a maze which malignant, insect-like aliens are seeking to traverse in order to subjugate Earth”?

Posted by: John Baez on December 13, 2020 1:52 AM | Permalink | Reply to this

### Re: Bernoulli Numbers and the J-homomorphism

Call me paranoid but it looks like they (lizard people? vampires? spiders and snakes?) are already here….

More seriously, the metaphor of a maze reminds me in some ways of mathematics, in which topics/areas are linked by surprising unexpected transitions. It takes a lot of courage? balance? good sense? to explore—which seems to me exemplified by this blog.

Posted by: jackjohnson on December 13, 2020 1:22 PM | Permalink | Reply to this

### Re: Bernoulli Numbers and the J-homomorphism

The reference to “the third homotopy groups of spheres” is followed by an isomorphism $\pi_{n + 3}(S^n) \cong \mathbb{Z}/24\mathbb{Z}$. Should it be $\pi_3(S^n)$? (The link “quaternionic Hopf fibration” says that $\pi_7(S^4) \cong \mathbb{Z} \times \mathbb{Z}/12\mathbb{Z}$, so the isomorphism as written doesn’t seem to work for $n = 3$.)

Posted by: L Spice on December 11, 2020 7:33 PM | Permalink | Reply to this

### Re: Bernoulli Numbers and the J-homomorphism

I said $\pi_{n+3}(S^n)$ is $\mathbb{Z}_24$ when $n$ is big enough, and that’s true. We just need $n \ge 5$. On the other hand $\pi_3(S^n)$ is trivial for $n \ge 4$.

Remember, $\pi_j(S^n)$ is trivial whenever $j \lt n$: every map from a sphere into a higher-dimensional sphere is null-homotopic.

On the other hand $\pi_{n+k}(S^n)$ settles down to a fixed value, independent of $n$, whenever $n \ge k+2$, thanks to the Freudenthal suspension theorem. This range $n \ge k+2$ is called the stable range.

So yes, $\pi_7(S^4)$ is $\mathbb{Z} \times \mathbb{Z}_{12}$. The quaternionic Hopf fibration gives a generator of the $\mathbb{Z}$ here. But when you apply the suspension homomorphism

$\pi_{7}(S^4) \to \pi_{8}(S^5)$

to this generator, you enter the stable range and get the generator of $\pi_8(S^5) \cong \mathbb{Z}_{24}$. Then applying the suspension homomorphism

$\pi_{8}(S^5) \to \pi_{9}(S^6)$

you get the generator of $\pi_9(S^6) \cong \mathbb{Z}_{24}$. And so on forever.

You can see the stable range colored in periwinkle here, and you can see how the homotopy groups settle down as we move across any row when we hit the stable range. Unfortunately we can’t see from this chart what the suspension homomorphisms do, exactly.

(I don’t use fancy names for shades of purple very much so I’m not sure ‘periwinkle’ is right, but it looks more like periwinkle than lavender.)

Posted by: John Baez on December 11, 2020 8:23 PM | Permalink | Reply to this

### Re: Bernoulli Numbers and the J-homomorphism

Ah, got it; thanks! I misunderstood the phrase “third homotopy group”, because (despite the context of large $n$) I didn’t make the connection to stable homotopy, and so I was expecting it to be a group $\pi_3$ rather than $\pi_{n + 3}$.

Posted by: L Spice on December 11, 2020 8:53 PM | Permalink | Reply to this

### Re: Bernoulli Numbers and the J-homomorphism

Actually I should have written “third stable homotopy group of spheres” or something like that — the phrase “third homotopy group of spheres” was incorrect for what I was actually talking about. So I’ll fix that!

Posted by: John Baez on December 11, 2020 9:06 PM | Permalink | Reply to this

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