Four Sphere Fibrations Does there exist a manifold $M$ and a compact Lie group $H$ such that we have a fibration $H \to S^4 \to M$, where $S^4$ is the four sphere?
 A: MR0148071 (26 #5580)
Browder, William
Fiberings of spheres and H-spaces which are rational homology spheres.
Bull. Amer. Math. Soc. 68 1962 202–203.
55.40 (55.50)
Let $F\to S^n \to B$ denote a fiber bundle over a polyhedron $B$ whose total space is an $n$-sphere $S^n$. Spanier and White-head have shown that the fiber $F$ is an $H$-space, and Borel has shown that $F$ has the rational homology of a sphere. The author completes these results by announcing the following. (1) $F$ has the homotopy type of $S^1$, $S^3$, or $S^7$. (2) If $X$ is any $H$-space (connected) with $H^∗(X;\mathbb{Z})$ finitely generated and $H^∗(X;\mathbb{Q})=H^∗(S^n;\mathbb{Q})$, then X has the singular homotopy type of a sphere or a projective space of dimension 1, 3, or 7. 
A: For connected $H$ the long exact homotopy sequence implies $\pi_1M=0$, which by dimension reason leaves only the possibilities $M=S^2$ or $M=S^3$. But $S^4$ is neither an $S^1$-bundle over $S^3$ (because it is simply connected) nor a $T^2$-bundle over $S^2$ (for example because it is not symplectic, or again because of simply connectedness). 
For disconnected $H$, compactness implies that $\pi_0H$ is finite, hence $\pi_1M$ is finite, thus either $S^4$ is a circle bundle over a spherical 3-manifold (which contradicts simple connectedness) or it is a torus bundle over the projective plane, which also contradicts simply connectedness).
So in any case the answer is No.
