How to show that higher Hopf fibrations are not nullhomotopic?

Question

Here Proving that hopf map from $S^3 \to S^2 $ is not null homotopic is an aswer showing Hopf map is not null homotopic. How would one show that the higher fibrations $S^3 \to S^7 \to S^4$ and $S^7 \to S^{15} \to S^8$ are not null homotopic?

Answer 1

There are geometric ways to see this as well as homotopical. Geometrically, a very similar analysis to the linked problem works. Namely, taking the cofiber of $S^7 \rightarrow S^4$ gives something homeomorphic to $\mathbb{H}P^2$ (quaternionic projective space) and taking the cofiber of $S^{15} \rightarrow S^8$ gives $\mathbb{O}P^2$ (octonionic projective space), but there is subtlety in this last one since it is a little tricky to actually define octonionic projective spaces, in particular they don't actually exist for dimensions higher than 2. But after you have done this analysis, you can use Poincare duality to see that the cohomology ring is not trivial, so the maps you ask for are not nullhomotopic.

The homotopical approach is much easier. If the map $S^{2n-1} \rightarrow S^n$ were nullhomotopic, the homotopy fiber is easily seen to be $\Omega S^n \times S^{2n-1}$, but the latter is not homotopy equivalent to a finite complex (seen by computing its homology with the Serre spectral sequence). Then just recall the fiber of a fibration is equivalent to the homotopy fiber.