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.