# How to see the quaternionic hopf map generates the stable 3-stem?

I am looking for a direct proof that the quaternionic hopf map generates (after suspension) the 3rd stable homotopy group of spheres. There are some related MO questions, for example:

But the focus in these seems more to be on how to calculate that the image of the hopf map is 24-torsion, and I want to see why the hopf map generates this group.

## background

The quaternionic Hopf map is a fibration $$S^3 \to S^7 \stackrel{\nu}{\to} S^4$$ which induces an isomorphism $$\pi_k(S^4) \cong \pi_k(S^7) \oplus \pi_{k-1}(S^3).$$ The Freudenthal suspension theorem implies that the suspension map $$\pi_7S^4 \to \pi_8 S^5 \cong \pi^{st}_3$$ is surjective. Thus there is a surjective map: $$\mathbb{Z}\langle \nu \rangle \oplus \pi_6 S^3 \to \pi_3^{st}$$

In the analogous setting of the complex Hopf fibration (the classical case) this is enough to conclude that the image of the hopf map generates $\pi_1^{st}$ (the extra factor $\pi_2S^1 = 0$ vanishes). However in this case the extra factor is non-trivial. We can look up that $\pi_6 S^3 \cong \mathbb{Z}/12$.

## The Question

Now if we know, by some argument, both facts that $\pi_6 S^3 \cong \mathbb{Z}/12$ and that $\pi_3^{st} \cong \mathbb{Z}/24$, then the above subjectivity implies that the image of $\nu$ must be a generator. However we can see by the previous MO questions that these calculations can be quite involved and non-obvious.

I will be teaching a topics course this summer and I would like to include a proof that the image Hopf map generates $\pi_3^{st}$. I want to use this in some later calculations. I would like to do this without having to explicitly calculate $\pi_6 S^3$ and $\pi_3^{st}$ since both those calculations seem like they will be quite involved and then the proof that the Hopf map generates is somewhat unsatisfying. I am hoping that there is a more direct argument, which hopefully doesn't use too much elaborate machinery. (Though what one considers elaborate machinery depends on ones perspective).

Some of the possible tools that will be available are: a direct geometric argument, Thom spectra, the Pontryagin-Thom correspondence between bordism groups and homotopy groups of these spectra, and the AHSS. I am considering later in the coure presenting the Galatius-Madsen-Tillmann-Weiss theorem, so it could be possible to use an h-principle argument in combination with those techniques if that helps, though I don't see that it would. Another possible tool is Whitehead products as they will show up in a limited form at the very end of the course (I seem to remember that there is a connection between Whitehead products and the kernel of the suspension homomorphism, so maybe these help).

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The first thing to comes to mind is derive it from $\pi_3 MSpin=0$. Is there a geometric proof that 3-dimensional spin manifolds bound? – Charles Rezk Feb 20 '13 at 2:28
Charles, yes there are direct proofs via Kirby calculus. Usually these algorithms are used to compute things like the Rochlin invariant (an obstruction to bounding a homology ball), but the algorithms are fairly general. – Ryan Budney Feb 20 '13 at 9:18
Not an answer, but some related material: Koschorke and Sanderson used the self-intersection (SI) maps of immersed 3-manifolds in 4-space to compute this group. Other results of Koschorke are relevant. Mike Freedman's paper on self-interestions of immersions also contains information. Peter Eccles wrote about the (SI) maps. Finally, the standard Froisart Morin sphere eversion can be shown to represent a generator. All of the results I mentioned are in published in the era (1978-1986). – Scott Carter Feb 20 '13 at 13:05
Chris: Ryan's argument isn't circular. Starting with a Heegaard splitting of M (from a triangulation or Morse function), you get M as surgery on a link. This already shows $\Omega_3$ is trivial. An algorithm of S. Kaplan encodes a spin structure on M in terms of a "characteristic sublink" of this link, and shows how to find a bounding spin manifold by eliminating the characteristic sublink. This route seems easier to me than the AHSS, but as you say this depends on one's perspective. – Danny Ruberman Feb 20 '13 at 18:52
Perhaps this paper on the vanishing of the third spin cobordism group is relevant: maths.ed.ac.uk/~aar/papers/stipsicz.pdf The author has tried to make the proof "as elementary as possible". – Sam Gunningham Feb 20 '13 at 19:28