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I got this very nice answer to one of my previous questions to which I have the following follow up question:

What I understand, or let's say, think I understand so far is: one main problem of interest in algebraic topology is to determine $\pi_n (S^m)$ for all $n$ and all $m$ $\geq 0$. We know some cases, for example we know that for $m > n$, $\pi_n (S^m) = \{0\}$ and for $n = m$, $\pi_n(S^m) = \mathbb Z$. We also know, due to the Freudenthal suspension theorem, that $\pi_{n + k}(S^n) \cong \pi_{n + k + 1}(S^{n + 1})$ for $n > k + 1$. We also know some other combinations of $n$ and $m$ but we have not managed to determine all of them.

Quite recently, in 2009, the following theorem, known as the Kervaire Invariant One problem was proved:

The Kervaire invariant $\kappa : \Omega_{4k + 2}^{fr} \to \mathbf{Z}_2$ is trivial unless $4k + 2 = 2,6,14,30,62$.

Here $\Omega_{4k + 2}^{fr}$ denotes the group of cobordism classes of framed $4k + 2$ manifolds.

Question: How does that let me compute $\pi_n(S^m)$? When I write compute, I really mean "gather information". It would be good enough to show that there are no surjective maps $S^n \to S^m$ then we'd have $\pi_n(S^m) = \{0\}$. But I don't see how to use the Kervaire invariant to gather any information about $\pi_n(S^m)$. Thanks for your help.

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    http://en.wikipedia.org/wiki/Stable_homotopy_group_of_spheres#Framed_cobordism2012-09-08
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    The Kervaire problem isn't completely solved: it might still be nontrivial in dimension 126!2012-09-08
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    Knowing the solution of the Kervaire invariant problem is a crucial step for knowing the group of smooth structures on a sphere (http://en.wikipedia.org/wiki/Exotic_sphere#The_monoid_of_smooth_structures_on_spheres_in_a_given_dimension). The last chapters of Kosinski's differential manifolds treat this problem. Once the solution for the Kervaire invariant problem is given, the problem of determining the group of smooth structures reduces to a calculation of homology groups of spheres, which is a problem in its own.2012-09-09
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    @AaronMazel-Gee I know : )2012-09-09

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