The J-homomorphism is a well-known and classical map $\pi_n (O(k)) \to \pi_{n+k} (S^k)$, or after stabilizing with respect to $k$, a map $J_n:\pi_n (O) \to \pi_{n}^{st}$, from the stable homotopy of orthogonal groups to stable homomotopy of spheres. The main results on $J_n$ were proven by Adams in a classic series of four papers.

The results are: the image of $J_{4r-1}$ (in which case the source group is $\mathbb{Z}$) is cyclic of order $den (\frac{B_r}{4r})$ (denominator of Bernoulli numbers). For $n\equiv 0,1 \pmod 8$, $J_n$ is injective (the source is $\mathbb{Z}/2$). Another theorem by Adams is that the unit map $\pi_{n}^{st} \to \pi_n (BO)= \pi_{n-1} (O)$ hits all $\mathbb{Z}/2$-groups in the image.

These results play an important role in differential topology (for example in the classification of exotic spheres), which is why from time to time, I struggle to understand these results. But I am rather foreign to stable homotopy theory and I am scared away by this battle with homological algebra and stable homotopy theory and never manage to get the main points from Adams' papers.

However, for the case $n=4r-1$, there is a version of the proof without leaving the mathematical terrain I am used to navigate in. There are two parts: proving that the image of $J$ has \it{at least} the order $den (\frac{B_r}{4r})$ requires an invariant an a device for its computation. The invariant is the $e$-invariant $e: \pi_{4n-1}^{st} \to \mathbb{Q}/\mathbb{Z}$ and the computation of $e \circ J$ is done using characteristic classes. All this is well-explained in Hatcher's book project on $K$-theory, with a hands-on definition of the $e$-invariant.

Proving that the image of $J$ has \it{at most} the order $den (\frac{B_r}{4r})$ requires a construction of a nullhomotopy. It follows from the Adams conjecture and some Bernoulli numerology. Besides the first proofs of the Adams conjecture by Quillen and Sullivan, there exist two proofs which I understand (by Becker-Gottlieb and a simplification of it by E. Brown, which I wrote up some years ago).

Here are my questions:

1. Is there an argument for the injectivity of $J_{n}$, $n \equiv 0,1 \pmod 8$, which is similarly direct as the argument in Hatcher's book?

2. The $J$-homomorphism gives a map of spectra $\Sigma^{-1} KO \to S$. What is the composition with the unit map $S \to KO$, and what is a low level explanation for it? EDIT: this is too naive, see Neil's comment below.

3. Is there a low-level description for the image of the unit map $S \to KO$ on homotopy groups?

4. Or do have to learn it the hard way?

• The J-homomorphism is only an infinite loop map with respect to the multiplicative infinite loop structure on $Q_0S^0$, so you get a map of spectra from $\Sigma^{-1}kO$ to $gl_1(S^0)$ rather than $S^0$. Here $\pi_k(gl_1(S^0))=\pi_k(S^0)$ for $k>0$, but other properties of $gl_1(S^0)$ are not too well understood. Also, you need to use the connective spectrum $kO$ rather than the periodic $KO$ here. – Neil Strickland Jan 31 '14 at 21:14
• One simple description of the unit map $S\rightarrow KO$ is that at the level of the corresponding infinite loop spaces, it is the limit of the classifying map induced by the regular representation $\Sigma _n\rightarrow O(n)$. – user43326 Jan 31 '14 at 21:16
• For question 1, I think there is a similar sort of argument to show injectivity in these cases, though it needs a bit more input, notably Adams operations in real K-theory. I have some handwritten notes on this from 20 years ago that would take some work to decipher after this long a time. My recollection is that I extracted these from Adams' J(X) - IV paper. My plan was, and still is, to include this in that unfinished book mentioned in the question, though as the years pass the chances of this ever happening become increasingly slim. – Allen Hatcher Feb 1 '14 at 22:41
• Question 1: The case $n=0$ mod 8 can be proved using surgery theory of framed manifolds, and probably also the other case. Indeed, a map $S^{8n} \to O$ defines a stable framing of $S^{8n}$, and this framed manifold represents its image in $\pi_n^{st}$, any framed cobordism showing that it is in the kernel can by surgery be turned into a disc (dim$=8n+1$), and hence a null-homotopy of the map. The odd case takes more work (if possible - which I think it is) since the surgery does not lead to a disc but may have cells of dimension $4n+1$. Is this considered "similarly direct"? – Thomas Kragh Feb 2 '14 at 9:54
• @Thomas: I do not understand this argument. The framing of $S^{8n}$ gives an element of $\pi_{8n}^{st}$ (this is the geometric description of the $J$-homomorphism). Now one has to show that if the original element in $\pi_{8n}(O)$ is nonzero, then the stable framing is NOT nullbordant. – Johannes Ebert Feb 2 '14 at 10:17

This answer probably won't be coming from the perspective that you want, since it'll use even more stable homotopy theory than Adams did. But I think it's pretty clear in its own way.

First let me make a small correction to Neil's comment. Actually, from the J-homomorphism construction you only get a map of spectra from the connective cover of $\Sigma^{-1}ko$ to $gl_1(S^0)$. So you have to kill that bottom Z in ko. But actually the better thing to do is to "find" the same Z below $gl_1(S^0)$ instead, by changing the target to $pic(S^0)$, the classifying spectrum for invertible spectra. So the J-homomorphism should be thought of as a map of spectra $ko\rightarrow pic(S^0)$. In these terms it has an easy intuitive description too: $ko$ classifies vector bundles, $pic(S^0)$ classifies stable spheres, and the J-homomorphism sends a vector bundle to its one-point compactification. Now you see why it's the multiplicative structure on $S^0$ that's relevant, because this operation sends direct sum of vector bundles to smash product of spheres.

This isn't just me being pedantic, by the way. It's good to understand exactly what structure the J-homomorphism carries, and it's also very handy to have that extra Z around. One measure of this is the following:

Claim: Let $J: ko \rightarrow pic(S^0)$ be any map of spectra which sends the unit class in $\pi_0 ko$ to the class of the 1-sphere in $\pi_0 pic(S^0)$. Then the image of J on homotopy groups satisfies the lower bound'' estimates established by Adams for the J-homomorphism.

Unless I'm misreading your question it's these lower bound estimates that you're interested in. So what I'm saying is that for those purposes, all you need to know about the J-homomorphism, besides the natural structure it carries, is that it sends 1 to 1'' on that bottom Z we attached (or rather, never bothered to forget).

The way we'll prove the above claim is to give a more modern version of Adams' e-invariant argument. This will be based on Rezk's study of logarithmic operations coming from the Bousfield-Kuhn functor.

The motivation is that, as Neil says, the spectrum $pic(S^0)$ (or $gl_1(S^0)$) is kind of mysterious. That's too bad for us, since apparently we need to understand it to prove the above claim. However, the Bousfield-Kuhn functor tells you that for any prime p, the "part of a spectrum which mod p K-theory understands", meaning the K/p-localization of the spectrum, only depends on the n^{th} space of the spectrum, for any n you like. This is some amazing consequence of periodicity. The upshot is that the K/p-localization doesn't care that $pic(S^0)$ has this exotic infinite loop structure, and you get a natural map $$pic(S^0)\rightarrow L_{K/p}S^{1}$$ which exhibits the latter as the K/p-localization of the former.

Now, a priori this map is somewhat mysterious on the level of homotopy groups. But actually Rezk made a fantastic study of it in his paper on logarithmic cohomology operations (which the above is essentially an instance of.) One consequence of Rezk's work is a calculation of the effect of the above map on $\pi_0$. It is as follows. First, recall that $\pi_{-1} L_{K/p}S$ identifies with $Hom(Z_p^\times,Z_p)$ (noncanonically, this is just $Z_p$, the $p$-adic integers.) Then, the above map $log:pic(S^0)\rightarrow L_{K/p}S^{1}$ sends the class of the $1$-sphere to the homomorphism $Z_p^\times\rightarrow Z_p$ given by $$x\mapsto \frac{1}{2p}log(x^{p-1}).$$ (I might have a sign wrong, but that doesn't matter for our purposes.) Here $log$ stands for the $p$-adic logarithm. It's an interesting thing that the above expression makes sense and is primitive. That is, the set of all $(p-1)^{st}$ powers of $p$-adic units is exactly the domain of convergence of $log$, and, also, $2p$ is the GCD of all the values of $log$.

We deduce from this that the composition $log\circ J:ko\rightarrow L_{K/p}S^{1}$ sends the unit class in $\pi_0ko$ to that same homomorphism. Now the point is that $K/p$-local homotopy is computationally friendly. Also, $ko$ is nearly $K/p$-local: its $K/p$-localiztion is the $p$-adic $KO$, via the connective cover map $ko\rightarrow KO$. Thus it's not hard to calculate that a homotopy class of maps $ko\rightarrow L_{K/p}S^{1}$ is completely determined by its effect on $\pi_0$. So actually the above minimal information tells us exactly what $log\circ J$ is. Then if we look on higher homotopy groups, we see that the image of $log\circ J$ has size given by Adams' upper bound. This implies the claim.

By the way, I think the cleanest way to perform the above calculations is to fix the generator $g = exp(\frac{2p}{p-1})$ for the group $Z_p^\times/\pm$ of p-adic Adams operations acting on p-adic $KO$. Then you get to write down the fiber sequence $$L_{K/p}S \rightarrow KO\overset{1-\psi^g}{\rightarrow} KO,$$ and the result is that $log\circ J$ identifies with $ko\rightarrow KO\rightarrow L_{K/p}S^{1}$, where the last map is the boundary map in the above sequence. (Again my signs might be off, sorry for that.)

• Oh, by the way, I shouldn't give the wrong impression here. I breezily talk about all these "computations". But I'm terrible at computations, and besides, I did these computations a while ago, and I'm partly relying on memory. So they could be wrong! – Dustin Clausen Feb 1 '14 at 13:52
• +1, even if I have to admit that I understand even less than in Adams' work. But it sounds impressive. – Johannes Ebert Feb 1 '14 at 15:56
• Well, understandable would be better than impressive! I hope someone else can come along and answer your question as intended. – Dustin Clausen Feb 1 '14 at 16:44
• Hi @DustinClausen, I'm sorry but I don't quite get why having "direct sum of vector bundles getting sent to smash product of spheres" tells us that only the multiplicative structure is relevant. Could you elaborate on that? My (possibly wrong) understanding would be: to see that the J-map is a multiplicative map would be to check that the map on cohomology theories: $K(X) \rightarrow Sph(X)$ is a map on multiplicative but not additive structures. This is tensor product of bundles on the left but I am not quite sure what it is on the right. It would be great if you could clarify this! – Elden Elmanto Mar 20 '14 at 21:05
• Hi Elden, I mean that in your map of cohomology theories, if you take the direct sum on the left (not tensor product!) and the fiberwise smash product on the right, then the map preserves these structures. It's the smash product on the right that I was calling "multiplicative", because it's related to the product in the E-infinity ring S^0. – Dustin Clausen Mar 29 '14 at 22:33