# Which spheres are fiber bundles?

The Hopf fibration is a fiber bundle with total space $S^3$, and there are similar constructions for $S^7$ and $S^{15}$. Are there any other ways to regard a sphere as a nontrivial fiber bundle?

My guess is no. Perhaps it would help to look at the long exact sequence of homotopy groups.

I'm not requiring that the base and fiber be manifolds, smooth or otherwise. As a side note, I'm also curious to know whether the base and fiber are generally manifolds if the total space is a manifold -- I'm guessing yes.

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For your side note, the answer is no. The dogbone space is a non-manifold whose product with $\Bbb{R}$ is $\Bbb{R}^4$: en.wikipedia.org/wiki/Dogbone_space –  Micah Sep 1 '12 at 4:32
This is one reason I love math: all my guesses were wrong! –  Hew Wolff Sep 3 '12 at 18:06

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.

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Much food for thought, thank you @JasonDeVito. I'm still curious about possible non-manifold fibers and bases: does anyone have more information there? –  Hew Wolff Sep 3 '12 at 18:05
Also, do we know whether there are any bundles (say with connected fibers) for odd-dimensional spheres apart from the ones you listed? –  Hew Wolff Sep 10 '12 at 19:39
I know for a fact that there are bundles of the form $S^3\rightarrow S^7 \rightarrow S^4$ which are not bundle isomorphic to the usual Hopf bundle. I don't know of any examples where the fiber or base space are different from the ones you and I listed, but I also don't know of a theorem which somehow guarantees that this should be the case. –  Jason DeVito Sep 10 '12 at 19:44
Some quick googling just turned up ams.org/journals/bull/1962-68-03/S0002-9904-1962-10747-2/…. In it, Browder says he proves elsewhere that if $S^n\rightarrow B$ is a fiber bundle with $B$ a polyhedron, then the fiber is homotopy equivalent to $S^1$, $S^3$, or $S^7$. (Due the Poincare conjecture, we can now strengthen this to homeomorphic for $S^7$ and diffeomorphic for $S^3$ and $S^1$). –  Jason DeVito Sep 10 '12 at 19:54

I've been confusing myself into thinging the below constructions would answer the question. I'll leave them in case they might interest anyone and give ideas.

For $p,q>1$, describe points $X\in S^{pq-1}$ as $p\times q$ matrices with norm 1: i.e. $X=(x_{ij})$ where $\sum_{i,j}x_{ij}^2=1$. Let $U\in SO(p)$ act on $S^{pq-1}$ by $X\mapsto UX$. This gives us an action of $SO(p)$ on $S^{pq-1}$, and by letting $B=S^{pq-1}/SO(p)$, we get a fibration $$SO(p)\rightarrow S^{pq-1}\rightarrow S^{pq-1}/SO(p).$$

Problem: The action is not free!

If I'm not overly confusing myself here, there's always $$F\rightarrow S^{pq-1}\rightarrow S^{q-1}$$ where you describe points $X\in S^{pq-1}$ as $p\times q$ matrices with norm 1, i.e. $\sum_{i,j}x_{ij}^2=1$, which map to $Y\in S^{q-1}$ by $y_j=\sum_i x_{ij}^2$ and on which matrices in $U\in SO(p)$ act on the fibers by $X\mapsto UX$.

Problem: Fibres depend on the number of coordinates in $S^{q-1}$ are zero. The action is not free (see above) and irrelevant for the argument.

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