Which spheres are fiber bundles?

Solution 1:

Odd dimensional spheres are all total spaces of fiber bundles. One has the Hopf fibration $$S^1\rightarrow S^{2n+1} \rightarrow \mathbb{C}P^n.$$

Spheres of dimensiona $4n+3$ are are total spaces in another Hopf fibration $$S^3\rightarrow S^{4n+3}\rightarrow \mathbb{H}P^n.$$

Here, $\mathbb{C}P^n$ denotes the complex projective space (of dimension $2n$) and $\mathbb{H}P^n$ denotes the quaternionic projective space (of dimension $4n$). Noting that $\mathbb{C}P^1 = S^2$ and $\mathbb{H}P^1=S^4$, these generalize the the two fibrations you mentioned.

If you allow disconnected fibers, then every sphere is a fiber bundle because a covering space is nothing but a fiber bundle with discrete fiber and spheres double cover the real projective spaces.

If you do not allow disconnected fibers, then even dimensional spheres are not fiber bundles, at least where the fiber is a smooth manifold (except in the trivial cases of the fiber being a point or the fiber being the whole sphere). This is because because if one has $$M\rightarrow S^n \rightarrow X,$$ then the foliation of $S^n$ by $M$ gives rise to a subbundle of $TS^n$: all vectors tangent to $M$.

On the other hand $TS^n$ has no subbundles when $n$ is even. For if there was as subbundle $\nu$, we'd have a splitting $TS^n = \nu\oplus \nu^\perp$. Since the Euler class is multiplicative, we have $0\neq 2 = e(TS^{n}) = e(\nu)\cup e(\nu^\perp)$. This implies $e(\nu)\neq 0$, which implies that $\nu$ is either rank $0$ or rank $n$ (since these are the only two degrees $S^n$ has nontrivial cohomology). This implies that $M$ is either dimensino $0$ (so a point), or dimension $n$ (so, $M = S^n$).

I don't know what happens if you allow $M$ to be a more wild kind of topological space.