# Homotopy $\pi_4(SU(2))=\mathbb{Z}_2$

Recently I read a paper using $$\pi_4(SU(2))=\mathbb{Z}_2.$$ Do you have any visualization or explanation of this result?

More generally, how do physicists understand or calculate high dimension homotopy group?

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–  student Dec 10 '12 at 22:21

It is very hard to visualize these homotopy classes, since they correspond to maps $S^4\rightarrow SU(2)\approx S^3$. The homotopy groups of spheres (and any other space) are typically very difficult to calculate in generality and physicists typically ask mathematicians. But there exist simple results in the so-called "stable range" where there is a regular structure (Bott periodicity, $\pi_k(U(N)) = \pi_{k+2}(U(N))$ for large enough $N$), and there exist tools to calculate homotopy groups of certain spaces, such as the long exact sequence of a fibration.

For the case of spheres see the table on wikipedia, where the chaotic behavior is clear and $\pi_4(S^3)$ is listed. There is a very good review by Mermin (1) where you can learn how to visualize and calculate simple homotopy groups.

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+1 too. At least we agree it's hard! –  twistor59 Dec 8 '12 at 14:21
Thanks! I've read Mermin's RMP paper before and did learn a lot, but for this problem, the fibration method seems not work, at least not in a apparent way. –  Yingfei Gu Dec 8 '12 at 14:27
@Yingfei Gu, I naively thought one might use some sort of generalization of Hopf fibration (which let you calculate $\pi_3(S^2))$. But it seems that it is not possible to do this in an obvious way. –  Heidar Dec 8 '12 at 14:33
@Heidar recently i am revisiting my posts on MO and stack-exchange. And realized this issue is essential about relation between stable homotopy groups of sphere and framed cobordism groups. And the answer in MO also pointed out the case $[S^n,S^{n-1}]$ for large n is in fact in the image of J-homomorphism(However, $[S^{n},S^{n-2}]$ doesn't, need a little more trick(I guess it was called pontryagin pairing) to prove it is $\mathbb{Z}_2$ as well.). Now I agree the best way to visualize the answer is via Pontryagin-Thom's construction. Thanks for discussion about this issue 2 years ago. –  Yingfei Gu Mar 29 at 16:35

Since $SU(2)$ is topologically a three-sphere $S^3$, you can begin by investigating the homotopy groups of spheres. Unfortunately, although there are some regular results, such as $\Pi_n(S^n)=\mathbb{Z}$, and $\Pi_m(S^n)=0$ for $m<n$, I don't think there is a single method to calculate $\Pi_m(S^n)$ for $m>n$. Individual results for $m>n$ are chaotic. So, I think the answer to your last question is "they would ask a mathematician!", because this (algebraic topology) is a very large topic.

For your specific case, there is a reference given on math overflow, but I don't have the book unfortunately.

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Yes, section 4 of this ref describes the process. I don't fully understand but you start with the Hopf fibration $S^3 \rightarrow S^1$ ($S^3$ is an $S1$ bundle over $S^2$), and then apply suspension operation to increase the dimensionalities of the spheres. –  twistor59 Dec 8 '12 at 14:42