# Why is $\pi_7(\mathbb S^4)=\mathbb Z \oplus \mathbb Z_{12}$?

I'm trying to visualize this fact, not prove it.

If we consider the (quaternionic) Hopf fibration $$p:\mathbb S^7 \to \mathbb S^4$$, where $$\mathbb S^7$$ is the unit sphere in $$\mathbb H^2$$ (we denote the quaternions by $$\mathbb H$$), the sphere $$\mathbb S^4:=\mathbb H \mathbb P^1$$ is the quaternionic projective line, and $$p(z,w)=[z:w]$$, then $$p$$ is an element of $$\pi_7(\mathbb S^4)$$ of infinite order. So, this map is responsible for the $$\mathbb Z$$ part of $$\pi_7(\mathbb S^4)$$.

Who is the element of order $$12$$?

Start with the quaternion Hopf fibration $$S^3\xrightarrow{i} S^7\xrightarrow{\nu}S^4.$$ The map $$\nu$$ is an element of Hopf invariant one so generates the infinite cyclic summand in $$\pi_7S^4$$. The map $$i$$ is the fibre inclusion. Now loop the fibration to get a homotopy fibration sequence $$\dots\rightarrow\Omega S^7\xrightarrow{\Omega\nu}\Omega S^4\xrightarrow\delta S^3\xrightarrow{i} S^7\rightarrow\dots$$ where $$\delta$$ is the fibration connecting map. Since $$\pi_3S^7=0$$ the map $$i$$ is null-homotopic. This implies that $$\delta$$ has a right homotopy inverse (ie. section) $$\sigma:S^3\rightarrow \Omega S^4$$. You can see easily using the adjunction $$\pi_3\Omega S^4\cong\pi_4S^4$$ that $$\sigma$$ is the suspension map (ie. the adjoint of the homeomorphism $$\Sigma S^3\cong S^4$$).
Now let $$\mu$$ denote the loop multiplication on $$\Omega S^4$$ and consider the composite $$\varphi:S^3\times \Omega S^7\xrightarrow{\sigma\times\Omega\nu}\Omega S^4\times\Omega S^4\xrightarrow\mu\Omega S^4.$$ It's not difficult to see that this map is a weak equivalence (recall that the loop addition on $$\Omega S^4$$ induces the addition on $$\pi_*S^4$$). Since both the spaces involved have CW homotopy type, $$\varphi$$ is a homotopy equivalence by the Whitehead Theorem.
In any case we get $$\pi_7S^4\cong\pi_6\Omega S^4\cong\pi_6(S^3\times\Omega S^7)\cong\pi_6S^3\oplus\pi_6\Omega S^7\cong \pi_6S^3\oplus \pi_7S^7.$$ From the way we defined $$\varphi$$ we see that the summand $$\pi_7S^7\cong\mathbb{Z}$$ corresponds cyclic summand in $$\pi_7S^4$$ generated by $$\nu$$. What's left is the summand $$\pi_6S^3\cong \mathbb{Z}_{12}$$, which is generated by a class normally denoted $$\nu'$$. From the above remarks we see that the corresponding summand in $$\pi_7S^4$$ is explicitly generated by the suspension $$\Sigma \nu'$$. This is who the element of order 12 is.
Now the class $$\nu'\in\pi_6S^3$$ itself actually has quite a nice description. Recall that $$S^3$$ can be identified as the unit sphere in the quaternions. The quaternionic multiplication induces a product of $$S^3$$ which turns it into a (nonabelian) Lie group. Consider the commutator map $$c':S^3\times S^3\rightarrow S^3,\qquad (x,y)\mapsto xyx^{-1}y^{-1}.$$ Note that this map is trivial when restricted to the wedge $$S^3\vee S^3$$. This means that it factors over the smash $$S^3\wedge S^3=(S^3\times S^3)/(S^3\vee S^3)$$ to give a map $$c:S^6\cong S^3\wedge S^3\rightarrow S^3.$$ I. James has shown that this map generates $$\pi_6S^3\cong\mathbb{Z}_{12}$$.