Higher homotopy groups meaning

Question

I am developing intuition for higher homotopy groups but it's very hard for me to visualize what $\pi_2$ represents (and $\pi_n$ for that matter). I know that $\pi_2(S^2) \cong \mathbb{Z}$ and can kind of see "wrapping" a sphere around itself an integer number of times. But, something like $\pi_3(S^2)$ doesn't make very much intuitive sense to me at all. How am I supposed to think about these groups? When is it supposedly obvious that $\pi_i(X)$ is trivial and when is it nontrivial?

Answer 1

The question in its current form is too general. However, the case $\pi_n(S^{n-1})$ is relatively easy and was discovered first.

In Milnor's Topology from differential viewpoint, you can find a quite intuitive explanation of the fact that elements of $\pi_3(S^2)$ correspond to framed cobordism classes of $1$-dimensional smooth submanifolds of $S^3$, that is, certain natural equivalence classes of circles in $S^3$ with a framing. (Essentially, this duality is induced by taking preimage of the "north pole" $f^{-1}(n)$ for any $f: S^3\to S^2$ for $[f]\in \pi_3(S^2)$.) A framing on a circle in $S^3$ is the choice of two "normal" vectors in each point of the circle. If you have one such framing, any other framing can be obtained by a loop $S^1\to SO(2)$ that acts on the 2 framing vectors pointwise. Not surprisingly, homotopic curves gives rise to cobordant framings. Clearly, $\pi_1(SO(2))\simeq \mathbb{Z}$.

Using the same approach, the case $\pi_n(S^{n-1})\simeq \mathbb{Z}_2$ for $n>3$ can be reduced to $\pi_1(SO(n-1))\simeq \mathbb{Z}_2$ for $n>3$.