Base and fiber Riemannian homogeneous implies total space smooth homogeneous

Question

Let $ F,B $ be Riemannian homogeneous manifolds. Suppose that $ E $ is the total space of a fiber bundle $$ F \to E \to B $$ Is $ E $ always smooth homogeneous (admits a smooth transitive action by a (possibly noncompact) finite dimensional Lie group)?

I think this is true for dimension of $ E \leq 3 $. I'll come back later and list out all the gory details for low dimensions. But just curious if anyone has an immediate counterexample.

Edit:

In response to xsnl comment:

My answer here covers everything

https://mathoverflow.net/questions/6142/circle-bundles-over-rp2/416631#416631

but here's basically what it says for the sake of completeness:

The circle bundles over $ \mathbb{R}P^2 $ are described in wikipedia page for Seifert fiber space https://en.m.wikipedia.org/wiki/Seifert_fiber_space and they are all homogeneous.

This is the wikipedia description of the bundles with orientable total space

"{b; (n2, 1);} (b integral.) This is the prism manifold with fundamental group of order 4|b| and first homology group of order 4, except for b=0 when it is a sum of two copies of real projective space, and |b|=1 when it is the lens space with fundamental group of order 4."

They are all the Riemannian homogeneous prism manifolds ( $ b \geq 2 $ ) and the Riemannian homogeneous lens space with four element cyclic fundamental group (b=1) (which coincidences with unit tangent bundle of $ \mathbb{R}P^2 $) and finally $ b=0 $ which is $ \mathbb{R}P^3 \# \mathbb{R}P^3 $ and is only smooth homogeneous not Riemannian homogeneous. In particular it is smooth homogeneous for the euclidean group $ E_3 $ of three space. See Connected sum of two copies of $ RP^3 $

This is the wikipedia description of the two bundles with non orientable total space

"{b; (n1, 1);} (b is 0 or 1.) These are the non-orientable 3-manifolds with S2×R geometry. If b is even this is homeomorphic to the projective plane times the circle, otherwise it is homeomorphic to a surface bundle associated to an orientation reversing automorphism of the 2-sphere."

They are both Riemannian homogeneous see Mapping torus of the antipodal map of $ S^2 $

Answer 1

This is not true in general. In fact, it begins to fail in dimension $4$. I'd bet it fails in all higher dimensions as well, but don't immediately see a nice proof.

For example, $S^2$ is Riemannian homogeneous. A linear $S^k$ bundle over $S^2$ is classified by $[S^2, BSO(k+1)] = [S^1, SO(k+1)] = \pi_1(SO(k+1))$.

For $k \geq 2$, $\pi_1(SO(k+1)) \cong \mathbb{Z}/2\mathbb{Z}$, so there are precisely two $S^k$-bundles over $S^2$.

In the case $k=2$, the total space of the non-trivial $S^2$-bundle over $S^2$ is another well known manifold: $\mathbb{C}P^2\sharp-\mathbb{C}P^2$. The way I think about this bundle structure is as follows: Starting with $\mathbb{C}P^2$ and deleting a small ball, what's left is diffeomorphic to the disk bundle in the tautological bundle over $S^2 = \mathbb{C}P^1$. In particular, $\mathbb{C}P^2\setminus{\text{ball}}$ is a bundle over $S^2$ with fiber a disk.

To form $\mathbb{C}P^2 \sharp -\mathbb{C}P^2$, we join two copies of $\mathbb{C}P^2\setminus{\text{ball}}$ by gluing the boundaries by the identity. The identity map obviously respects the projection map, so we get a well defined projection $\mathbb{C}P^2\sharp -\mathbb{C}P^2$ with fiber two copies of $D^2$ glued along their boundary by the identity - that is, with fiber $S^2$.

(More generally, $\mathbb{C}P^n\sharp -\mathbb{C}P^n$ is the total space of an $S^2$-bundle over $\mathbb{C}P^{n-1}$.)

The point of all this: the total space of the unique non-trivial $S^2$-bundle over $S^2$ is diffeomorphic to $\mathbb{C}P^2 \sharp -\mathbb{C}P^2$. And note that $S^2\times S^2$ is not homotopy equivalent to $\mathbb{C}P^2 \sharp -\mathbb{C}P^2$ because their cohomology rings are different.

Lastly, I claim that $\mathbb{C}P^2\sharp -\mathbb{C}P^2$ is not homogeneous. To see this, first note that since it is simply connected and compact, it is homogeneous if and only if it is Riemannian homogeneous. Thus, if $G/H=\mathbb{C}P^2 \sharp -\mathbb{C}P^2$ for some $G$ and $H$, then we may assume that $G$ itself is compact and connected. The long exact sequence in homotopy groups associated to the fibration $H\rightarrow G\rightarrow G/H$ now shows that $\pi_0(H) = \ast$, that is, that $H$ is connected. Moreover, since $\pi_2(G) = \pi_2(H) = 0$, while Hurewicz implies $\pi_2(\mathbb{C}P^2 \sharp -\mathbb{C}P^2)\cong H_2(\mathbb{C}P^2 \sharp -\mathbb{C}P^2)\cong \mathbb{Z}^2$, it follows that $\pi_1(H)$ contains a $\mathbb{Z}^2$. In particular, $H$ has a cover of the form $T^2\times H'$ for some connected compact Lie group $H'$.

The isotropy representation now shows that $H\subseteq SO(4))$. Since $SO(4)$ has rank $2$, $H$ has rank at most $2$, which now implies that $H' = \{e\}$ so $H = T^2$. As $4 = \dim G - \dim H = \dim G - 2$, $\dim G = 6$. Moreover, since $\chi(\mathbb{C}P^2 \sharp -\mathbb{C}P^2) = 4\neq 0$, $G$ and $H$ have the same rank. So, $G$ is a $6$-dimensional compact Lie group with rank $2$. It follows that $G$ is covered by $SU(2)\times SU(2)$.

If $\pi:SU(2)\times G$ is the universal cover, then $(SU(2)\times SU(2))/\pi^{-1}(H)$ is canonically diffeomorphic to $G/H$, so we can work with $SU(2)\times SU(2)$. The long exact sequence in homotopy groups coming from the bundle $\pi^{-1}(H)\rightarrow SU(2)\times SU(2)\rightarrow \mathbb{C}P^2 \sharp -\mathbb{C}P^2$ shows that $\pi^{-1}(H)$ is connected. Since $\pi^{-1}(H)$ is a compact connected Lie group covering $H = T^2$, $\pi^{-1}(H) = T^2$ as well. But $T^2\subseteq SU(2)\times SU(2)$ is a maximal torus, so is uniquely embedded up to conjugacy. Thus, we may as well assume $T^2\subseteq SU(2)\times SU(2)$ is the "usual" one. But then, $SU(2)\times SU(2)/T^2 = S^2\times S^2\neq \mathbb{C}P^2 \sharp -\mathbb{C}P^2$.