# 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

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.