Is it known whether, for any fibration of a sphere with connected fibres, the fibre has to be a sphere? $S^1$, $S^3$ or $S^7$?

2$\begingroup$ No, obviously not, because every projection map $S^n \times F \to S^n$ is a fibration. $\endgroup$ – Johannes Hahn Dec 7 '14 at 12:38

7$\begingroup$ @Johannes Hahn: I think the question wants the sphere as total space: a fibration of a sphere, not over a sphere. $\endgroup$ – Oscar RandalWilliams Dec 7 '14 at 12:46
It was shown in
Browder, William. Fiberings of spheres and $H$spaces which are rational homology spheres. Bull. Amer. Math. Soc. $\bf 68$ 1962 202–203.
that for any fibration $F \to S^n \to B$, where $B$ is a nontrivial polyhedron and $F$ is connected, $F$ must have the homotopy type of $S^1,S^3$ or $S^7$. This builds on previous work of SpanierWhitehead and Borel.


$\begingroup$ I think one should also add $S^0$ to the list. $\endgroup$ – Alex Degtyarev Dec 7 '14 at 20:21

$\begingroup$ Right, $F$ must be connected  I added this. $\endgroup$ – Andreas Thom Dec 7 '14 at 22:24

$\begingroup$ May there be a link with the fact that the possible dimensions of composition algebras are only 1, 2, 4, and 8 (Jacobson 1958)? $\endgroup$ – Wolfgang Dec 10 '14 at 8:19

$\begingroup$ already found the positive answer in math.rochester.edu/people/faculty/doug/otherpapers/… $\endgroup$ – Wolfgang Dec 10 '14 at 8:27