# Which framed manifolds are in the image of J?

The stable homotopy groups of spheres are in natural correspondence with framed cobordism classes of framed manifolds.

Is there a good way to tell if the class of framed manifold is in the image of the J homomorphism?

• According to mathoverflow.net/questions/100860/…, they're the framed spheres. – Qiaochu Yuan Jun 29 '14 at 2:15
• (or framed cobordant to a framed sphere)... why don't you post that as an answer? – Tilman Jun 30 '14 at 11:02
• But every framed manifold except those few with Kervaire invariant one is framed cobordant to a framed sphere. That is the point of the Kervaire invariant, unless I have it wrong. – Mark Hovey Jun 30 '14 at 11:11
• @Mark: I could also have it wrong (I don't actually know anything about this topic which is why I'm reluctant to post an answer), but it seems like the statement is "framed cobordant to a (possibly exotic) sphere," whereas the image of J is about the ordinary non-exotic spheres. – Qiaochu Yuan Jun 30 '14 at 18:06

## 1 Answer

I hope to leave a comment but it is a bit too long.

The correspondence between framed cobordism classes and elements in stable homotopy groups come from the Thom-Pontrjagin Construction: Given a framed manifold in a large Euclidean space, we send the manifold to 0, project the framed tublar neighborhood(The frame comes from the frame of tangent bundle and trivial frame of Euclidean space with dimension high enough) via the frame and all the other point to infinity. In another direction, given an element in a stable homotopy group, you can find a continuous map between spheres as a representative, make it smooth, make it transversal to the origin, then look at the inverse image. The inverse image comes with a frame because a point is framed and you pull it back.

Now we look at the definition of J-homomorphism. Actually here we are doing basically the same thing: You look at the standard embedding of $S^n$ into $\mathbb{R}^{n+k}$, then if you want to give a frame on tublar neighborhood you are looking at $\pi_n(SO(k))$, and use the same construction stated above, you obtain an element in $\pi_{n+k}(S^n)$. Both $\pi_n(SO(k))$ and $\pi_{n+k}(S^n)$ are stable(For maybe different reasons) when $n$ is large enough and when going stable, the map commutes with J, so we can define stable J.

Now things becomes more or less obvious here. The image of J is represented by frames over the standard embedded sphere in Euclidean spaces. If my memory is correct, when your Euclidean space has very high dimension, different embeddings are isotopic so the ways of embedding make no differences.

There is one more thing: In odd dimension all framed manifolds are framed cobordant to framed spheres, but these spheres may not be standard, i.e. they may have exotic differential structure on it. So it is not true that in odd dimension the stable J is surjective: image of J is those frames on standard spheres, but the stable homotopy group contains frames on exotic spheres.

A good reference to me is Kosinski's Differential Manifolds, where I can find all details and proofs.