This is really just a handwavy but perhaps more "visual" description of Pontryagin's result as cited by solbap in the comments above. Though I've written a huge block of text, there are some reasonably concrete three-dimensional pictures that you can build up in your head in this case, but it does take quite a bit of practice.

First, I assume that you are familiar with Pontryagin's construction relating the homotopy classes of maps to the k-sphere with framed (co-)bordism classes of codimension k submanifolds.

Check out Milnor's book *Topology from the Differentiable Viewpoint* if you're not familiar with this. Because your user profile says that you are interested in condensed matter physics, I'll add that this idea is used in the case of $k=2$ to draw some nice pictures of "homotopies around defects" in this paper of Teo and Kane.

## Warmup, $\pi_3(S^2)$

As a warmup, let's try to visualize homotopy classes of maps from $S^3$ to $S^2$, i.e. the situation of the Hopf fibration. Pontryagin's construction says that we should be looking at bordism classes of framed codimension 2 submanifolds in $S^3$. 3-2=1, so we should be looking at 1-dimensional submanifolds, i.e. links in $S^3$. Here we have framed links in $S^3$ which can be visualized by drawing each component of the link with another parallel copy that winds around it, much like a ribbon.

You should convince yourself that all components in these framed links can be merged together into a single unknot with some integer framing. Thus what matters ultimately is the classification of possible framings. Imagine taking a 2D slice of $S^3$ transverse to a point $p$ of the framed link and placing the point $p$ at the origin of that plane. Then the framing at that point is just a choice of the $x$- and $y$- axes (i.e. a 2-dimensional frame). As we carry this plane along the original unknot, this choice of axes can rotate in that plane and so the classification of framings is naturally an integer. You may check that the inverse image of the North pole of the Hopf fibration is an unknot, and the inverse image of any other point on the sphere is an unknot which is linked once with it. Finally, you should see how you can build up any other homotopy class from "adding" Hopf fibrations together by putting multiple copies of this framed unknot together (possibly with opposite orientations), which gives a visualization of the group structure on the set of homotopy classes.

In this way you get a visualization of $\pi_3(S^2)$ by means of some pictures of framed circles. I can't resist here adding a link to this paper of DeTurck et al which gives some beautiful illustrations and description of the homotopy classes of maps from $T^3$ to $S^2$ with this tool.

## $\pi_4(S^3)$

Now, you are interested in the case of homotopy classes of maps from $S^4$ to $S^3$. In this case you are now looking at framed links in $S^4$. You can still arrange for the link to become a single framed unknot by a sequence of bordisms. However, the framing can no longer be drawn with simply just a single parallel knot. Consider taking a 3-dimensional slice transverse to a point $p$ on the link in $S^4$ and let us place $p$ at the origin of our $R^3$ that we sliced with. In $S^4$, the framing of the link yields a choice of a 3-dimensional frame in this $R^3$ slice. And just as the relevant topological invariant of the framing in $S^3$ was how this frame rotates as we travel along the $S^1$ corresponding to our link component, leading to an element of $\pi_1(S^1)$ (the winding number), in $S^4$, we must now track how this 3-d frame rotates as we follow the $S^1$ of the link component. But now we are considering a continuous loop of choices of 3-dimensional orientations, i.e. an element of $\pi_1(SO(3))$, which is well known to be $\mathbb{Z}/2\mathbb{Z}$.

With this key ingredient of the 3-dimensional framing, hopefully you can see that $\pi_4(S^3)=\pi_1(SO(3))=\mathbb{Z}/2\mathbb{Z}$.

This post imported from StackExchange at 2014-03-22 17:01 (UCT), posted by SE-user j.c.