[Math] third stable homotopy group of spheres via geometry

at.algebraic-topologyhomotopy-theorystable-homotopy

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:

  1. By the long exact sequence, $\pi_3 (S^2)=Z$, generated by $\eta$.
  2. By Freudenthal, $\pi_3 (S^2) \to \pi_4 (S^3) = \pi_{1}^{st}$ is surjective.
  3. 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 $CP^2$).
  4. 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:

  1. 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.
  2. is the same argument as for $\eta$.
  3. 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 $6$.
  4. 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.

Best Answer

You said you don't want to talk about framed manifolds, but that's a good way of seeing the 24. $\nu$ is represented by $SU(2)$ in its invariant framing. Take a K3 surface. It's framed, and it has Euler characteristic 24. Take a vector field that has 24 isolated zeroes of index 1. If you cut out a little disk around each of these 24 zeroes, the boundary will be an $S^3 = SU(2)$ with its invariant framing. So the K3 surface minus these 24 little disks is a null-bordism of $24\nu$. Probably not suitable for your course, as you would have to explain framed bordism and K3 surfaces, but cute nonetheless I think. By the way, the analog for $\eta$ is the two-sphere (Euler characteristic 2).

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