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I know $\pi_4(S^2)$ is $\mathbb{Z}_2$. However, I don't know how to visualize it. For example, it is well known that $\pi_3(S^2)=\mathbb{Z}$ can be understood by Hopf Fibration. Elements in $\pi_3(S^2)=\mathbb{Z}$ can be understood as describing the number of links of the $U(1)$ fibers in $S^2$.

So, do we have similar picture for $\pi_4(S^2)$? And, do we have similar topological invariant as links in previous case?

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If you compute the generator of $\pi_4(S^3)$ with the Hopf map, does this generate $\pi_4(S^2)$? – Jason DeVito Apr 26 '13 at 2:29
If you compose... – Jason DeVito Apr 26 '13 at 2:37
(Ignore my previous two comments. The answer is "yes" because the Hopf map induces isos on all $\pi_n$ except $n=2$ from the LES of homotpy groups.) – Jason DeVito Apr 26 '13 at 18:46
Jason's suggestion does indeed work -- $\eta^2$ is a generator of $\pi_4(S^2)$. More generally, there's a framework due to Sinha and Walter that describes elements of homotopy in terms of linking numbers. I've never tried to understand it, but the relevant papers are here:, – Aaron Mazel-Gee Apr 29 '13 at 7:07

One way I see is the following.

To start with, $\pi_{4}(S^{3}) \simeq \pi_{4}(S^{2})$ because of the existence of the Hopf fibre sequence $S^{1} \rightarrow S^{3} \rightarrow S^{2}$ and moreover the isomorphism is given by composition with the Hopf map. Hence, as was observed in the comments, you might as well try to visualize the nontrivial element of $\pi_{4}(S^{3})$.

Thom-Pontryagin construction tells you that $\pi_{4}(S^{3})$ is the same as framed 1-submanifolds of $S^{4}$ up to framed bordism (in $S^{4} \times [0, 1]$). This is a geometric answer to the problem, but let me just give you an idea how one might prove that this group is indeed $\mathbb{Z} _{2}$.

Dimension 1 submanifolds of $S^{4}$ are easy, as they are just disjoint sum of circles, but we have to remember about the framings. If the normal bundle of such $S^{1} \subseteq S^{4}$ is frameable (ie. trivial), then the space of all framings inducing a given orientation on it is homeomorphic to $Map(S^{1}, GL_{+}(3))$, as these are all "change of frames" of a trivial bundle.

It follows that the isotopy classes of framings of a trivial rank $3$ bundle on a circle are given by

$\pi_0(S^1, GL_{+}(3)) \simeq \pi_1(GL_{+}(3)) \simeq \pi_1(SO(3)) \simeq \pi_1(\mathbb{RP}^{3}) \simeq \mathbb{Z} _{2}$.

It is the circle with this "twisted" framing that corresponds to the non-trivial element of $\pi_{4}(S^3)$ under the Thom-Pontryagin construction.

Of course, this is a non-trivial result, although not very difficult either. Essentially, one needs to show that the "twisted" and "untwisted" circle are not cobordant and also that any disjoint sum of circles is cobordant to just one. I couldn't find any easily accessible source for this fact, but this is covered in Appendix B of "Instantons and Four-Manifolds" of Freed and Uhlenbeck, together with nice pictures (including a drawing of a dinosaur).

Added: Another, perhaps easier way is to note that by Freudhental suspension theorem the map $\pi_{3}(S^{2}) \rightarrow \pi_{4}(S^{2})$ is an epimorphism. Hence, the suspension of the Hopf map $\eta: S^{3} \rightarrow S^{2}$, say $\Sigma \eta$, is the sole generator of $\pi_{4}(S^{3})$. However, the easiest way to show that $\Sigma \eta$ is of order $2$ is perhaps less geometric and more in the realm of homotopy theory (one uses Steenrod squares, see Schwede's Untitled Book Project about Symmetric Spectra, p. 15).

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