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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 my memory is not mistaken, Mike Hopkins discussed $\pi_4(S^2)$ geometrically in his talk at the Atiyah birthday conference in 2009. The talk can be found here: maths.ed.ac.uk/~aar/atiyah80.htm –  Christian Nassau Apr 26 '13 at 12:44
The way you are understanding $\pi_3(S^2)$ is via the Pontryagin-Thom construction, which yields a correspondence between bordism classes of framed submanifolds (framed links in $S^3$) and homotopy classes of maps (elements of $\pi_3(S^2)$) via looking at inverse images. In the case of $\pi_4(S^2)$, the inverse images are (generically) framed surfaces in $S^4$, which are at least to me much harder to visualize. I did describe the situation for $\pi_4(S^3)$ here, which might help: mathoverflow.net/questions/115866/homotopy-pi-4su2z-2/… –  j.c. Apr 26 '13 at 13:04

3 Answers 3

Let me expand upon jc's comment above. For convenience, I'll pass to the stable homotopy group for a little while. The first stable stem is $\pi_1^s = {\mathbb Z}/(2)$. A representative class is the figure-8 in the plane. The number of double points, modulo-$2$ is the only invariant. There is a construction due to Koschorke or Koshorke and Sanderson. Lift the figure-8 into 3-space, and put a figure-8 in the normal plane. You'll get a twisted torus. It is constructed as a figure-8 times interval with one full-twist. This immersed torus represents a generator of $\pi_2^s$. It lifts to an embedding in $4$-space. That is your generator of $\pi_4(S^2)$. Another representative is the standard: $(e^{i\theta}, e^{i\phi})$ torus in ${\mathbb C}^2$. However, a rigid projection of the latter always has branch points. I don't know how to perturb this to project to the twisted figure-8 torus.

The twisting along the double curve of the twisted torus is a mod-2 invariant. Two full twists gets you to an immersion from which the double curve can be removed.

The homotopy theorists proof that these constructions work is to observe that the framing on the figure-8 (tngt+normal) is induced by the Lie framing on the circle. And the Lie framing represents a generator. Ditto for the twisted torus.

The papers that explain all this are from the era 1978-1988. They include works by Koshorke, Koshorke and Sanderson, Peter Eccles, and a couple of papers of mine.

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For me, the most satisfactory explanation of the isomorphism $\pi_4(S^2)\cong \mathbb{Z}/2$ is in Pontryagin's beautiful survey

Smooth manifolds and their applications in homotopy theory. 1959 American Mathematical Society Translations, Ser. 2, Vol. 11 pp. 1–114 American Mathematical Society, Providence, R.I.

He derives the isomorphism by using the Pontryagin-Thom construction. Pontryagin has a soft touch and patience with the details. In Section 15 he deals with $\pi_{n+2}(S^n)$. I had a very enjoyable intellectual experience reading this paper.

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The most explicit and geometric generator for $\pi_4(S^2)$ is probably the achiral genus-1 Lefschetz fibration $f : S^4 \to S^2$ with two Lefschetz singularies (one positive and one negative, with parallel vanishing cycles). In my knowledge, this Lefschetz fibration was first considered by Matsumoto ("On 4-manifolds fibered by tori", Proc. Japan Acad. 58 (1982), 298-301), and it is also described in Example 8.4.7 in the Gompf-Stipsicz book "4-Manifolds and Kirby Calculus".

I don't have now a proof of the homotopic non-triviality of this map, but I give this as a challenge!

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Why do you prefer that over the suspension of the Hopf fibration? –  Ryan Budney May 9 '13 at 19:08
Because the Suspension of the Hopf fibration is a Map from S^4 to S^3. –  ThiKu May 10 '13 at 5:25
OK, you can compose the suspension with the Hopf fibration, to obtain a map $S^4 \to S^2$. In fact, this is strictly related with the map I considered, but I think that defined in the way I did is nicer as a smooth map. –  Daniele Zuddas May 10 '13 at 7:07

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