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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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Cross post: mathoverflow.net/questions/115866/homotopy-pi-4su2z-2 –  student Dec 10 '12 at 22:21

3 Answers 3

This question has been posted also at http://mathoverflow.net/questions/115866/homotopy-pi-4su2z-2 with both geometric and algebraic (that was mine!) type of answers. The geometric answers tell of Pontjyagin's method of constructing explicit representations of maps to spheres. The algebraic methods tells of the answer from a general theorem which gives some results on homotopy groups of complexes, and when the conditions under which it works are satisfies, gives detailed and calculable algebraic answers.

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Thank you for pointing out the answers from mathoverflow. They are very interesting! –  Heidar Dec 10 '12 at 21:33

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.

(1): Rev. Mod. Phys. 51, 591–648 (1979)

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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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Sorry, I had not seen your answer. +1 –  Heidar Dec 8 '12 at 14:19
Thanks for nice reference. There seems having explanation by suspension of SU(2), I am trying to imagine it. –  Yingfei Gu Dec 8 '12 at 14:24
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

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