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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.

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The stable order of $\sigma$ is 240, see math.cornell.edu/~hatcher/stemfigs/stems.html –  André Henriques Nov 4 '10 at 21:06
    
I corrected it, thanks. –  Johannes Ebert Nov 4 '10 at 21:11

6 Answers 6

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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I'm probably missing the point, but why is the K3 surface framed? Its signature is -16 so its Pontryagin number is non-zero. –  Paul Nov 5 '10 at 2:51
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Of course, the K3 surface is not framed, or its Euler characteristic would be 0. But it does admit a framing away from 24 points. –  Oscar Randal-Williams Nov 5 '10 at 7:11
    
Paul's question still holds: why is K3 framed away from 24 points? (equivalently: why is K3 framed away from one point?) Is it just by obstruction theory, or is there also a geometric way of seeing this? –  André Henriques Nov 5 '10 at 10:33
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I think it was forgotten to mention that K3 is spin. Therefore, if you delete one point from K3, you get something homotopic to a 3-dimensional CW complex and since BSpin(4) is 3-connected, TK3 is trivial away from that point. –  Johannes Ebert Nov 5 '10 at 11:12
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@ Oscar. "framed" in the context of $\pi_n^s$ is (usually) short for "stably framed", and so the Euler characteristic doesn't obstruct, e.g. $S^2$ is stably framed. @Johannes's. Apply your comment to the perpendicular bundle to the given nowhere-zero section. Note $w_2$ is zero for that bundle, so obstruction theory gives you three more sections, unique because K3-24pts is simply connected. But I don't see what makes K3 different from $S^4$ or $S^2\times S^2$ here. There has to be a further reason why for $K3$ the framing restricts to the Lie group framing on the boundary $S^3$s. –  Paul Nov 5 '10 at 13:56

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. If each zero of the vector field is a source then there are 24 zeroes (that is the Euler characteristic of a K3).

I believe this observation is due to Atiyah.

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


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.

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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.

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I am sorry to repeat a selfpromoting text I metioned at some other place of MO, but my paper below is addressed precisley to the computation of $\pi(3)$ by geometry (immersion theory).

http://link.springer.com/journal/10958/113/6/page/1

The paper: A. Szűcs: Two theorems of Rokhlin. This gives the computation of π(3) by elementary tools.

But it uses the following non-trivial fact: The 3-dimensional spin cobordism group is trivial.

You can read a proof of this latter in the same issue in the paper by A. Stipsicz.

But a more simple proof for this is in:

Melvin, P.; Kazez, W. 3-dimensional bordism. Michigan Math. J. 36 (1989), no. 2, 251–260.

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I really like your account. Thank you for linking to it! I do have a lingering question. Does your argument also use something like Smale's theorem on immersions? (which I think could be considered quite "non-elementary") or can you get by with a more elementary argument? For example I am worried about the proof of the key Lma 1.4 where you refer to "Hirsch theory". –  Chris Schommer-Pries Jul 19 '13 at 17:54
    
Thank you Chris! Lemma 1.4 by now has a simple proof in the paper (mentioned in the introduction) by Rourke - Sanderson. (This is the so called Compression theorem. Actually I think its proof can be even further simplified), The first paragraph of the paper aims to answer that type of quite understandable worry, that you also had. –  András Szűcs Jul 19 '13 at 19:56

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 $K_3(\mathbb Z)$ Igusa's picture for the generator of K_3(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 $\mathrm{Spin}(N)$...

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