TITLE: third stable homotopy group of spheres via geometry?
QUESTION [68 upvotes]: It is ''well-known'' that the third stable homotopy group of spheres is cyclic of order $24$. It is also ''well-known'' that the quaternionic Hopf map $\nu:S^7 \to S^4$, an $S^3$-bundle, suspends to a generator of $\pi_8 (S^5)=\pi_{3}^{st}$. It is even better known that the complex Hopf map $\eta:S^3 \to S^2$ suspends to a generator of $\pi_4 (S^3) = \pi_{1}^{st} = Z/2$. For this, there is a reasonably elementary argument, see e.g. Bredon, Topology and Geometry, page 465 f:

By the long exact sequence, $\pi_3 (S^2)=Z$, generated by $\eta$.
By Freudenthal, $\pi_3 (S^2) \to \pi_4 (S^3) = \pi_{1}^{st}$ is surjective.
Because $Sq^2: H^2(CP^2;F_2) \to H^4(CP^2;F_2)$ is nonzero, the order of $\eta$ in $\pi_{1}^{st}$ is at least $2$ (the relation between these things is that $\eta$ is the attaching map for the $4$-cell of $CP^2$).
By a direct construction, $2\eta$ is stably nullhomotopic. Essentially, $\eta g = r \eta$, where $r,g$ are the complex conjugations on $S^2=CP^1$ and $S^3 \subset C^2$. $g$ is homotopic to the identity, $\eta=r\eta$. The degree of $r$ is $-1$, so after suspension (but not before), composition with $r$ becomes taking the additive inverse. Therefore $\eta=-\eta$ in the stable stem.

My question is whether one can mimick substantial parts of this argument for $\nu$. Here is what I already know and what not:

There is a short exact sequence $0 \to Z \to \pi_7 (S^4) \to \pi_6 (S^3) \to 0$ that can be split by the Hopf invariant. Thus $\nu$ generates a free summand.
is the same argument as for $\eta$. 
using the Steenrod operations mod $2$ and mod $3$ on $HP^2$, I can see that the order of $\nu$ in $\pi_{3}^{st}$ is at least $6$.
this is a complete mystery to me and certainly to others-:)). How can I bring $24$ in via geometry? How do I relate the quaternions and $24$? What one sees immediately is that one has to be careful when talking about conjugations in the quaternionic setting, in order to avoid proving the false result ''$2 \nu=0 \in \pi_{3}^{st}$''.

I know that this result goes back to Serre, but I cannot find a detailed computation in his papers and it seems that the calculation using the Postnikov-tower and the  Serre spectral sequence is a bit lengthy. There are three other approaches I know but they are much less elementary:
Adams spectral sequence, J-homomorphism (enough to show that the order of $\nu$ is $24$), framed bordism (supported by things like Rochlin's theorem and Hirzebruch's signature formula).
Any idea? P.S.: if there is a similar argument for the octonionic Hopf fibration $S^{15} \to S^8$ (the stable order is 240), that would be really great.

REPLY [14 votes]: There is a "half-geometric" observation that since
$$
\pi_3^s\cong H_3(\tilde A_N),\ N\gg0
$$
where $\tilde A_N$ is the universal central extension of the $N$th alternating group, elements of $\pi_3^s$ might be described by certain 3-cycles (hopefully of geometric origin).
Seemingly this geometry is easier to discern after the embedding $\tilde A_N\hookrightarrow\mathrm{St}_N(\mathbb Z)$ into the Steinberg group of the integers.
(For a ring $R$, the group $\mathrm{St}_N(R)$ is the universal central extension of $\mathrm E_N(R)$ - the group which for most decent rings is the same as $\mathrm{SL}_N(R)$; one has $K_3(R)\cong H_3(\mathrm{St}(R))$ which in good cases stabilizes after some $\mathrm{St}_N(R)$.)
In "The generalized Grassmann invariant" (late 70ies) K. Igusa introduced technique of pictures to construct a homomorphism
$$
\chi:K_3(\mathbb Z[\pi])\to H_0(\pi;\mathbb F_2[\pi]);
$$
in his own words, "The definition of $\chi$ comes from very intuitive geometric considerations, but unfortunately the algebraic analogue is rather clumsy." 
Anyway geometry is still there as this map is strongly related to pseudoisotopy of compact manifolds (with $\pi_1=\pi$ and $\pi_2=0$). The image of $\pi_3^s(B\pi_+)\to K_3(\mathbb Z[\pi])$ is in the kernel of $\chi$, and for trivial $\pi$ all this enabled him to detect an element of order 48 in $K_3(\mathbb Z)$; $\pi_3^s$ is a subgroup of index 2 there. His picture representing a generator of $K_3(\mathbb Z)$

has been haunting me for decades. It shows something like "$1/2$ of the generator of $\pi_3^s$ living outside $\tilde A_N\hookrightarrow\mathrm{St}_N(\mathbb Z)$" and there surely must be something simpler which represents the generator of $\pi_3^s$ inside $\tilde A_N$ itself. This should not be that difficult as $\tilde A_N$ comes equipped with a very explicit and nice embedding into $\mathrm{Spin}(N)$...