How can I visualize the nontrivial element of $\pi_4(S^3)$ and $\pi_5(S^3)$ ?

Question

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

Answer 1

(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 $Spin(5) = Sp(2)$.

Answer 2

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.

Answer 3

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.)

Answer 4

You can read some John Baez

http://math.ucr.edu/home/baez/week102.html

which contains exactly your answer :-)

Answer 5

$S^3$ 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).