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?



Four Sphere Fibrations
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.


