Question

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.

Answer 1

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).

Answer 2

This is really a comment try to make clear the point Tilman was trying to make but it is too long. A K3 surface has trivial canonical bundle (after all that and simple connectivity is the definition) and hence the bundle of self dual two forms is trivial (since on a complex surface we have $\Lambda^+= \Lambda^{2,0} \oplus R \omega$, $\omega$ being the Kahler form). In fact is follows from Yau's theorem that K3 surfaces admit hyperkahler metrics so there is a metric where the Levi-Civita connection is trivial on $\Lambda^+$.

Two forms act on vector fields on a four-manifold via contraction then duality under this actions self-dual forms act like imaginary quaternions (so quaternions do figure in the story). Thus taking a orthonormal basis of self-dual forms $\omega_1,\omega_2,\omega_3$ and your vector field $X$ you get a framing (not stable) away from the zeros of $X$, by looking at $(X,(\iota_X \omega_1)^*,(\iota_X \omega_2)^*,(\iota_X \omega_3)^*)$. Then arrange that the vector field is pointing out around each little 3-sphere surrounding a zero then you see that the induced framing of each little sphere is the Lie-group framing. I believe this observation is due to Atiyah.

Answer 3

Not an answer, but a cute thing that might be relevant for your question:

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Here's another nice way of seeing that $\eta$ is stably of order two.
We use the Thom-Pontryagin constuction to identify $\pi_3(S^2)$ with the cobordism group of framed 1-manifolds in $\mathbb R^3$ (the framing is on the 2-dimensional normal bundle). The element $\eta$ corresponds to the unknot, with a framing that ``twists once''.

We consider the following two elements of $\pi_3(S^2)$ $$ S^3\hspace{.2cm}\xrightarrow{\cdot 2}\hspace{.2cm}S^3\hspace{.2cm}\xrightarrow{\eta}\hspace{.2cm}S^2 $$ and $$ S^3\hspace{.2cm}\xrightarrow{\eta}\hspace{.2cm}S^2\hspace{.2cm}\xrightarrow{\cdot 2}\hspace{.2cm}S^2 $$ The first element corresponds to the disjoint union of two unknots, each one with a framing that "twists once". It is cobordant to the unknot with a framing that "twists twice", and represents the element $2\in \pi_3(S^2)\cong \mathbb Z$.
The second element represents the two-fold cabling of the unknot: this is the Hopf link. Eah one of the two circles has a framing that "twists once", but they are also linked to each other. With an explicit cobordism, one can show that this framed link is framed-cobordant to the unknot with a framing that "twists 4 times". It therefore represents the element $4\in \pi_3(S^2)\cong \mathbb Z$.

Now, the stable homotopy groups of spheres form a ring in which the integer multiples of the unit are central (!). So, from the equation $2\circ \eta=\eta\circ 4$ we can deduce that $2\eta=0$ stably.

Answer 4

Expanding upon Andre Henriques's answer, $\pi_3^s(P^\infty) \cong {\mathbb Z}/8$. The homotopy group is isomorphic to the cobordism group of non-orientable surfaces in 3-space. Boy's surface is a generator. See also Tony Phillips's movie.

The Froisart-Morin eversion of a $2$-sphere can be capped off at either end to give an immersed $3$-sphere in $4$-space. It has one quadruple point, a closed curve of triple points, and its double point set is a non-orientable surface (connected sum of 3 projective planes. By combining results of Koschorke from about 1979-1982, one can see that the multiple point invariants (Kahn-Priddy maps) give that this sphere is a generator of the ${\mathbb Z}/(24)$. Also, Koschorke's figure-8 construction applied to Boy's surface coincides with a map on stable homotopy that maps the ${\mathbb Z}/(8)$ injectively to the ${\mathbb Z}/(24)$.

An alternate view of the generator of $\pi^s_2$ is to take an immersed generic projection of the standard (tropical) torus from $4$-space into $3$-space. I have played with this in mathematica, linear projections yield surfaces with 4 branch points, so the torus has to be perturbed in $4$-space before projecting. If you take a figure 8, multiply it by an interval, and then put a full twist in it, you get the mod-2 generator. The belt trick undoes two full-twists.