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 $S^7$?

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    $\begingroup$ 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
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    $\begingroup$ @Johannes Hahn: I think the question wants the sphere as total space: a fibration of a sphere, not over a sphere. $\endgroup$ – Oscar Randal-Williams 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 non-trivial 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 Spanier-Whitehead and Borel.

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  • $\begingroup$ Minor quibble: $B$ cannot be a point. $\endgroup$ – John Klein Dec 7 '14 at 15:54
  • $\begingroup$ 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

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