# Homotopy $\pi_4(SU(2))=Z_2$

I am a physics student, recently I read a paper using Homotopy $\pi_4(SU(2))=Z_2$, I guess mathematicians have some visualization or explanation of this result. So I come here ask for help.

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<google.com/search?q=Π4(S3)> – Francois Ziegler Dec 9 '12 at 1:58
$SU(2)$ is homeomorphic to $S^3$ so you are asking why is $\pi_4(S^3) = \mathbb{Z}/2$? Pontryagin came up with a method to do such computations using bordisms of framed submanifolds; see thm 21 on page 99 of math.rochester.edu/people/faculty/doug/otherpapers/pont4.pdf. – solbap Dec 9 '12 at 2:28
Actually use $\pi_4$, not $\Pi_4$, which is a different construction. Also, $\pi_4(S^2) = \pi_4(S^3)$. – David Roberts Dec 9 '12 at 2:28
This question is somewhat terse -- it's not clear whether you want to know a generating class (as in the Hopf map comment), or why this is the only class (a nontrivial calculation), or how to detect whether a map is nullhomotopic (along the lines checking the framing of a preimage of a point, as in solbap's comment). Some clarification of what you'd really like to know would help us understand how to help. – Tyler Lawson Dec 9 '12 at 3:45
@Yingfei please add link to original question when cross-posting in order people do not repeat already done work - this is usual practice. – Alexander Chervov Dec 9 '12 at 5:21

This calculation of $\pi_4(S^3)$ is also obtained in the paper R. Brown and J.-L. Loday, Topology, 26 (1987) 311-334, and also available here. In that paper, $S^3$ is regarded as the double suspension $SS$ of the circle $S^1$, which is itself seen as an Eilenberg-Mac Lane space $K(\mathbb Z,1)$. We obtain in Proposition 4.10 a determination of $\pi_4$ of the double suspension $SS$ of a $K(G,1)$ for any group $G$ as the kernel of a morphism $G \tilde{\wedge} G \to G$ defined by the commutator map, where the group $G \tilde{\wedge} G$ is the quotient of the free group on the set $G \times G$ by a set of relations satisfied by commutators. Hence the result for $\pi_4 SK(G,1)$ is easy to calculate if $G$ is abelian: in fact in that case it is $G \otimes G$, the tensor product of abelian greoups, factored by the relations $g \otimes h + h \otimes g$ for all $g,h \in G$.

Part of the intuition behind this is that the suspension $SX$ of a space $X$ is regarded as the union $C^+X \cup C^- X$ of two cones with intersection $X$, and this union is one to which our van Kampen type theorem for squares of spaces can apply. In fact we are dealing with the triad $(SX; C^+X, C^-X)$, which is the union of two triads $(C^+X;C^+X,X)$, $(C^- X; C^-X,X)$, and the union of these two "trivial" triads creates something in dimension $3$. If $\pi_2 X=0$, this gives a complete determination of $\pi_3 SX$, and further work gives a result on the double suspension in the given case.

So the intuition is that in homotopy theory, identifications in low dimensions have high dimensional homotopy implications, and to cope with this for gluing purposes we need algebraic structures with structures in a range of dimensions. The hope is that someday such structures will have applications in physics!

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I fixed the link to your paper. – Akhil Mathew Dec 10 '12 at 2:40
@Akhil: thanks! – Ronnie Brown Dec 10 '12 at 9:41

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 $\pi_4(S^3)=\pi_1(SO(3))=\mathbb{Z}/2\mathbb{Z}$.

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Thank you! This is a very very nice answer~ – Yingfei Gu Dec 11 '12 at 10:09
Thanks, feel free to ask for clarification; it can be hard to describe the pictures without a chalkboard. – j.c. Dec 13 '12 at 11:59
Thanks for answering as well as pointing me to the reference of Teo and Kane. I am reviewing my old post in MO and stack exchanges, and again found your answer useful for my current thoughts. Recently i learned about the relation between stable homotopy groups of sphere and framed cobordism groups and understand more about this issue. And realized your answer actually indicates the mapping named as J homomorphism. And reading that helps me understand more of Algebraic topology. – Yingfei Gu Mar 29 '15 at 16:21