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Nov 22, 2019 at 11:44 history edited Neil Strickland CC BY-SA 4.0
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Nov 5, 2019 at 22:32 comment added Ian Agol I think the Smale conjecture might imply that the foliation by $S^3$s is equivalent to a principle $SU(2)$ bundle. en.wikipedia.org/wiki/Smale_conjecture I'm not sure if this is helpful in a classification though.
Nov 5, 2019 at 19:20 comment added Kevin Casto Oh duh, of course there is, because $\mathbb{H}P^\infty = BS^3$. Same with $S^1 \to S^d \to S^d \times \mathbb{C}P^\infty$
Nov 5, 2019 at 19:11 comment added Kevin Casto @NeilStrickland Fair enough! I do think the fibration question is interesting though. e.g., is there a fibration $S^3 \to S^d \to B$, where $H^*(B) = H^*(S^d \times \mathbb{H}P^\infty)$?
Nov 5, 2019 at 18:11 comment added Neil Strickland @KevinCasto we are assuming that we have an actual fibre bundle rather than just a fibration, so $B$ will have dimension $d-k$ and so cannot have cohomology in degrees above that.
Nov 5, 2019 at 17:59 comment added Kevin Casto If $d < k$ it seems like you get $H^*(B) = \mathbb{Z}[x,y]/y^2$ where $|x| = k+1$ and $|y| = d$. It seems like there could be nontrivial examples of this? Or am I missing something?
Nov 3, 2019 at 13:47 comment added Igor Belegradek The original reference is [Browder, William, Higher torsion in H-spaces, Trans. Amer. Math. Soc. 108 (1963), 353–375] where it is shown that the fiber must be $S^1, S^7, S^3$ in which cases the respective bases are homotopy $CP^m$, $S^8$, and (I think) the rational homology $HP^m$. There are some examples in the last case which aren't homotopy $HP^m$.
Nov 3, 2019 at 13:29 history answered Neil Strickland CC BY-SA 4.0