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I've read in the textbooks that the non-trivial generator $\eta_n$ of $\pi_{n+1}(S^n)$ is the suspension of the Hopf map $S^3\to S^2$, and the generator $\chi$ of $\pi_5(S^3)$ is given by $\eta_3 \circ \eta_4$. Fine.

My question is, how I can visualize them? Is there a nice explicit way to describe these maps $\eta_3$ and $\eta_3\circ \eta_4$ ? How about the generator of $\pi_6(S^3)$ ?

(Other questions on MO look more serious. Hopefully this question is not out of place ...)

EDIT: anyone with rudimentary understanding of basic homotopy theory would say $\eta$ and $\eta\circ\eta$ are explicit enough, but I just can't visualize the suspension. I would be happy with a nice description of $SU(2)$ bundles over $S^n$, as my first exposure to homotopy is through quantum field theory...

Further edit: Thanks everyone for answers, I'm almost inclined to accept Per's answer, but I'm not still satisfied :p

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Should it be on instead? It's surely not a research-level question... – Yuji Tachikawa Sep 11 '10 at 4:34
Don't be so hard on your question. Visualization is an important tool for understanding, but one that is sometimes hard to get through the usual channels. I think there are many research mathematicians who could learn something from a good answer to this question; I know I look forward to reading the answers. – Tom Church Sep 11 '10 at 6:31
up vote 5 down vote accepted

(This is a bit late, but I hope you find it interesting!)

Here's smooth representation of the generator of $\pi_4(Sp(1))$ (and so the same homotopy group of $S^3$ and $SU(2)$). Consider $S^4 = \mathbb{HP}^1$, and $Sp(1)$ the unit sphere in $\mathbb{H}$. Then the following function $t\colon \mathbb{HP}^4 \to Sp(1)$ represents the nontrivial homotopy class $S^4 \to S^3$: $$ t[p;q] = \frac{2p\bar{q}i\bar{p}q - |p|^4 + |q|^4}{|p|^4 + |q|^4} $$ where [p;q] are homogeneous coordinates on $\mathbb{HP}^4$. I don't know if this has appeared previously (I would love to know!), but I presented this as part of some slides at the Australian Mathematical Society's annual conference last year (see slide 6), and originally worked it out with a pointer from Michael Murray to the Hopf fibration described using quaternions (that is, $Sp(1) \to S(Im\mathbb{H})$, the unit sphere in the pure imaginaries). That this map is the generator (i.e. is not null-homotopic) I calculated following the answer at my question Detecting homotopy nontriviality of an element in a torsion homotopy group.

Note that this function followed by the inclusion $Sp(1) \hookrightarrow Sp(2)$ (as the top left entry) is the generator of $\pi_4(Sp(2))$ (by results of Mimura and Toda). And thus we also get a representative for the generator of $\pi_4$ of $Spin(5) = Sp(2)$.

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Through the Pontrjagin-Thom construction, a framed $n-k$ manifold in $S^n$ determines a map from $S^n$ to $S^{n-k}$. $\eta$ is represented by $S^1$ in $S^3$ with framing which "twists around once". The suspension of $\eta$ is represented by $S^1$ in $S^4$ lying in the equatorial $S^3$ with framing which is the product of this "twist once" framing within $S^3$ and the trivial framing in the normal direction, etc.

The composite is represented by an $S^1 \times S^1$ with a framing which is "twist around once" on each factor.

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The $S^1$ that represents $\eta$ is unknotted, I guess? In that case, what do you get from other knots? – Mariano Suárez-Alvarez Sep 20 '10 at 16:00
@Mariano - the $S^1$ arises in the Pontryagin-Thom construction as the inverse image of a regular value of the map from $S^3$ to $S^2$. Homotopy classes of these maps are in correspondence with framed cobordism classes of links. Thus knottedness (or even number of components) of the inverse image is not a homotopy invariant, but "framing number" is. – j.c. Sep 22 '10 at 1:34
Is it possible to see that this generator when added to it is homotopic to zero? – Yuji Tachikawa Sep 27 '11 at 13:26
In the answer it should read "framed k-manifold". For the circle in $S^3$, how many normal framings are there? Well the framing can twist around any number of times, so you get $\mathbb{Z}$ many ($\pi_1 SO(2)$) and this corresponds to $\pi_3 S^2$. However when you suspend then there are only two ways to twists around, the trivial non-twisted way and the other way corresponding to $\pi_1(SO(n))$. – Chris Schommer-Pries May 17 '14 at 8:09
@Chris you can edit it, you know... – David Roberts May 17 '14 at 22:35

The main thing to visualize is the Hopf fibration of $S^2$, its suspensions, and their various compositions.

Let $f \colon S^3 \to S^2$ be the Hopf fibration.

When you suspend $f$ to get $g \colon S^4 \to S^3$, you effectively embed a 2-sphere as the equator of a 3-sphere and extend the mapping in parallel to 2-spheres of latitude. Thus away from the poles you still have circles as preimages.

You can see that $f$ and $g$ compose to give a map $h \colon S^4 \to S^2$. To get a sense of how this looks as a fibration, you can work backwards. First, the preimage of a point in $S^2$ under $f$ is a circle in $S^3$. As noted above, each pointwise preimage of this circle under the suspension $g$ is again generically a circle. When the different circles fit together cleanly, it looks like you get a torus fibration, where the tori twist and interlink within each latitudinal 3-sphere of $S^4$ analogously to the meshing of circles in $S^3$ for the Hopf fibration. If you now suspend this situation, you get a torus fibration over $S^3$ that looks like $h$ within each 2-sphere of latitude.

(I'm still not happy with this description but decided to post it in the hope it might spark some ideas.)

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Is there some simple argument that shows that this map is not null-homothopic? – Dmitri Sep 11 '10 at 18:19

You can read some John Baez

which contains exactly your answer :-)

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Thanks, but John only says it's the suspension... :p – Yuji Tachikawa Sep 11 '10 at 17:59

$S^3$ is isomorphic to $SO(3)$, which is a real Lie group and therefore is a differentiable and oriented real 3-fold, which has the double cover by $SU(2)$ which is the universal covering of $SO(3)$ or exponential of $su(2)$. Higher dimensional terms come from Bott's periodicity theorem. This is the QFT explanation.

In other words, thanks to the complex structure, we have a triangulation (approximation by CW (cell) complex attaching several n-dimensinal cell $e^n$ to a point set ${0}$, and Eilenberg-Steenrod axioms of singular homology of integral coefficient [with torsion module]). Then the textbook of Chern-Weil theory of characteristic class or some classic foliation (Postnikov tower of fibration) can tell you that there is a $E_2$ spectral sequence of double complex [see Bott-Tu GTM82, P.251-252] which is computable by exact sequence. This is the algebraic topology answer (non-simply connected space).

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I'm finding it hard to parse your answer! For instance, $SO(3)$ has $SU(2)$ as its double cover and not the other way around. What higher dimensional terms are you referring to and how does Bott periodicity come in? I can imagine referring to Bott's result about the cell structure of $\Omega G$ but that's not clear either. And nothing that you have said helps to visualize the higher homotopy elements. One, of course, knows that there are algebraic tools to compute what the referred elements are but that wasn't the question! – Somnath Basu Sep 20 '10 at 5:44
I thought Yuji wanted several other comments, but if he wanted geometric topology, I should mention lens space, which I am not working on. (Should I call it by Spin group?) I was considering whether I should write about loop group and DGA model, but it is outside of my reach today. (And I am not working on differential topology of exotic space today.) – Makoto Sep 20 '10 at 6:01
If you mean by "visualize" some graphs like string topology and higher category, you can see many pictures of Riemann surfaces, trees (operad), and braid groups. But I think it is not the ordinary mathematics based on logical algebra. Also, torsion part is not so difficult to handle and I didn't care about it. In my opinion, the "theoretical tool" is the jet space. – Makoto Sep 20 '10 at 21:25
You might want to revise the first sentence, since $S^3$ is isomorphic to $SU(2)$, not the quotient $SO(3)$. – S. Carnahan Sep 21 '10 at 5:59
It is not fair to revise my post after someone noticed it, though I found some difficulty with "complexification". – Makoto Sep 21 '10 at 6:56

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