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?

$\begingroup$ I remember a paper of Browder about fibrations of spheres. I will see if I can find it. $\endgroup$ – Ben McKay Feb 10 '16 at 13:22

1$\begingroup$ You probably want to exclude $\mathbb{RP}^4$ by connectedness assumptions. $\endgroup$ – Matthias Wendt Feb 10 '16 at 13:23
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 3manifold (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.

$\begingroup$ Are you saying $\pi_1M\simeq 0$ implies that $M=S^2$ or $S^3$? For instance, consider $S^4$ as a subset of $\mathbb{R}^4\times\mathbb{R}$ consisting of $(x,t)$ with $x^2+t^2=1$. Then, let $\mathbb{Z}/2$ act on $S^4$ by sending $(x,t)$ to $(x,t)$. Am I making a mistake here? $\endgroup$ – user51223 Feb 10 '16 at 14:04

$\begingroup$ Yes, by the Solution of the Poincaré conjecture in Dimension 3, or the classification of surfaces in Dimension 2. However, if you allow H to be 0dimensional (and thus M 4dimensional), then indeed H=Z/2Z and M=RP^4 is another possibility (actually the only one). $\endgroup$ – ThiKu Feb 10 '16 at 14:41

$\begingroup$ well, I do not see any condition on the dimension of $H$ in the posted question; so that last part of your answer may need a small correction. $\endgroup$ – user51223 Feb 10 '16 at 15:09

$\begingroup$ If $H$ is not 0dimensional, then the dimension of $M$ is at most 3 and the first part of the answer applies. $\endgroup$ – ThiKu Dec 11 '18 at 9:28
MR0148071 (26 #5580) Browder, William Fiberings of spheres and Hspaces 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 Whitehead 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.

$\begingroup$ The Euler characteristic gets rid of the possibility of $H=S^1$ I think (can't foliate the 4sphere by curves, i.e. no nowhere vanishing vector field). That leaves $H=S^3$ as fiber, which is also easy I think, since then the base has to be $S^1$, and you look at cohomology. $\endgroup$ – Ben McKay Feb 10 '16 at 13:38

$\begingroup$ The review for the later paper "Higher torsion in Hspaces" has an additional connectedness assumptions. These exclude $\mathbb{RP}^4$. And $\mathbb{Z}/2$ is the only finite group acting freely on $S^4$. $\endgroup$ – Matthias Wendt Feb 10 '16 at 13:39