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I need to prove that this has an element of infinite order.

Do I use Frudenthal suspension theorem. Any hints on what to use to prove this?

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Why do you need to prove this? – Mariano Suárez-Alvarez Apr 7 '12 at 18:33
@Mariano, all of us here are just junkies of mathematics, in search of a fix... C'mon man, don't hold out on him like that! – Asaf Karagila Apr 8 '12 at 7:55

(not a proof, but hopefully suggests a direction for you)

For any continuous map $\phi:S^{2n-1}\rightarrow S^n$ we can define the Hopf invariant $h(\phi)$ as in . The Hopf invariant defines a homomorphism $h:\pi_{2n-1}(S^n)\rightarrow \mathbb{Z}$.

In our particular case, there is a special fibre bundle called a Hopf bundle $$S^3\rightarrow S^7\rightarrow S^4$$ whose projection map is in $\pi_7(S^4)$. In fact, the projection map from any Hopf bundle has Hopf invariant $1$, so $h:\pi_7(S^4)\rightarrow \mathbb{Z}$ is surjective. In particular this shows that the projection map $S^7\rightarrow S^4$ has infinite order in $\pi_7(S^4)$

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There are a number of ways at approaching this.

1) As 'you' mentions above, there is the Hopf fibration $$S^3 \to S^7 \to S^4$$

The following is an exercise in Hatcher's book

Theorem: For a fiber bundle $F \to E \to B$ such that the inclusion $F \to E$ is homotopic to a constant map the long exact sequence in homotopy breaks up into split short exact sequences $\pi_n(B) \simeq \pi_n(E) \oplus \pi_{n-1}(F)$. In particular the above fibration gives isomorphisms $$\pi_n(S^4) \simeq \pi_n(S^7) \oplus \pi_{n-1}(S^3)$$

and so using the well known fact that $\pi_n(S^n) = \mathbb{Z}$, the result follows.

2) Using Serre's mod $\mathcal{C}$ theory you can prove that $\pi_{2n-1}(S^n)$ is the direct sum of an infinite cyclic group and a cyclic group. A proof is given in Spanier, for example (pp. 516). A proof that avoids spectral sequences is given in Chapter 20 of Tammo Tom Dieck's Algebraic Topology book.

3) If you are feeling very brave, you can push the methods given in the Appendix to Chapter 12 of Mosher and Tangora's book to fully compute $\pi_7(S^4)$ (this is a bit of overkill for the problem!)

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Awesome answer :) – you Apr 8 '12 at 16:43

I think the following should be right:

We have the following part of a long exact sequence induced by the fibration: $S^3 \hookrightarrow S^7 \rightarrow S^4$: $$ ... \rightarrow \pi_7(S^3) \rightarrow \pi_7(S^7) \rightarrow \pi_7(S^4) \rightarrow \pi_{6}(S^3) \rightarrow \pi_6(S^7) = 0$$

We first claim that the map induced by the inclusion $i_*: \pi_7(S^3) \rightarrow \pi_7(S^7)$ is zero. Well if $\eta: S^7 \rightarrow S^3$ is a representative of a class in $\pi_7(S^3)$, then $i \circ \eta$ is nullhomotopic since the map $i: S^3 \rightarrow S^7$ can be made to miss a point for $3 < 7$ (the content of the fact that $\pi_3(S^7) = 0$). Therefore, the induced map is zero.

Next, observe that $\Sigma: \pi_{6}(S^3) \rightarrow \pi_7(S^4)$ splits the short exact sequence $$0 \rightarrow \pi_7(S^7) \cong \mathbb{Z} \rightarrow \pi_7(S^4) \rightarrow \pi_{6}(S^3) \rightarrow \pi_6(S^7) = 0$$. Therefore $\pi_7(S^4) \cong \mathbb{Z} \oplus \pi_{6}(S^3)$

This can be generalized to the $S^{15}\rightarrow S^8$ bundle too!

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