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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2$\begingroup$ No, obviously not, because every projection map $S^n \times F \to S^n$ is a fibration. $\endgroup$– Johannes HahnCommented Dec 7, 2014 at 12:38
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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 Randal-WilliamsCommented Dec 7, 2014 at 12:46
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1 Answer
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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.
answered Dec 7, 2014 at 13:29
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$\begingroup$ Minor quibble: $B$ cannot be a point. $\endgroup$ Commented Dec 7, 2014 at 15:54
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$\begingroup$ I think one should also add $S^0$ to the list. $\endgroup$ Commented Dec 7, 2014 at 20:21
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$\begingroup$ Right, $F$ must be connected - I added this. $\endgroup$ Commented Dec 7, 2014 at 22:24
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$\begingroup$ May there be a link with the fact that the possible dimensions of composition algebras are only 1, 2, 4, and 8 (Jacobson 1958)? $\endgroup$– WolfgangCommented Dec 10, 2014 at 8:19
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$\begingroup$ already found the positive answer in math.rochester.edu/people/faculty/doug/otherpapers/… $\endgroup$– WolfgangCommented Dec 10, 2014 at 8:27