Let $ U_1 \to E \to B $ be a $ U_1 $ principal bundle. Suppose that $ B $ is homogenous (admits a transitive action by a Lie group) and compact. Then must it be the case that $ E $, the total space of the bundle, also admits a transitive action by some Lie group?
If $ B $ is dimension 1 then this is true. Because then $ B=S^1 $ and $ E= T^2 $.
If $ B $ is dimension 2 this is also true. Because then $ B= S^2,T^2,\mathbb{R}P^2,K^2 $.
If $ B=S^2 $ then $ E $ is a homogeneous lens space $ L_{n,1}=SU(2)/C_n $. (they are classified by Euler class $ \mathbb{Z} $ see https://math.stackexchange.com/questions/2204744/classification-of-principal-g-bundles-over-closed-oriented-surfaces-with-g-c/2204804#2204804 since $ S^2 $ is closed and orientable)
If $ B=T^2 $ then $ E= H(3,\mathbb{R})/\Gamma_r $ (see https://math.stackexchange.com/questions/4367670/is-every-nil-manifold-a-nilmanifold). (again they are classified by Euler class $ \mathbb{Z} $ since $ T^2 $ is closed and orientable )
For non-orientable closed surface the bundles are classified by $ \mathbb{Z}/2\mathbb{Z} $ rather than $ \mathbb{Z} $.
If $ B=\mathbb{R}P^2 $ then $ E $ is either $ \mathbb{R}P^2 \times S^1 $ or the mapping torus of the antipodal map on the sphere. Both are homogeneous indeed Riemannian homogeneous see https://math.stackexchange.com/questions/4362827/mapping-torus-of-the-antipodal-map-of-s2/4366823#4366823 For related questions see https://math.stackexchange.com/questions/3183562/classification-of-circle-bundles-over-mathbbrp2 or Circle bundles over $RP^2$
If $ B=K^2 $ then $ E $ is either $ K^2 \times S^1 $ or the unique nontrivial $ U_1 $ principle bundle over $ K^2 $. The total space of the trivial bundle is a solvmanifold with a transitive action by $ SE(2) \times U_1 $. The nontrivial $ U_1 $ principal bundle over $ K^2 $ happens to have total space homeomorphic to a nonprincipal circle bundle over $ T^2 $. But this question Torus bundles and compact solvmanifolds shows that every circle bundle over a torus is a solvmanifold so the nontrivial $ U_1 $ principal bundle over $ K^2 $ must also be homogeneous for some solvable group.
Note that this conjecture does not hold for general circle bundles over homogeneous manifolds. For example there are two non principal circle bundles over $ K^2 $ whose total spaces are not homogeneous. Indeed these are the other two of the four non orientable compact flat three manifolds, they can also be viewed as the other two of the four mapping tori of $ K^2 $. They are non principal bundles $ S^1 \to S^1 \rtimes_b K^2 \to K^2 $ both with non orientable total space. For $ b=0 $ this is the mapping torus of the Y homoeomorphism of $ K^2 $, it has first homology $ \mathbb{Z}^2 \times C_2 \times C_2 $. For $ b=1 $ this is the mapping torus of $ K^2 $ for the mapping class corresponding to the combination of a Dehn twist and a Y homoemorphism. It has first homology $ \mathbb{Z}^2 \times C_4 $. Both have total space which are not homogeneous see the argument given here https://math.stackexchange.com/questions/4423862/is-a-mapping-torus-of-a-solvmanifold-always-a-solvmanifold?noredirect=1&lq=1 and the arguments given here https://math.stackexchange.com/a/4374850/758507
I'm not sure if this is still true for $ U_1 $ principal bundles over compact homogeneous 3-manifolds.
Idea for a counterexample: $ U(1) $ principal bundles are classified by $ H^2(M,\mathbb{Z}) $. For a 3-manifold we have $ H^2(M,\mathbb{Z})=H_1(M,\mathbb{Z})=\pi_1(M)/[\pi_1(M),\pi_1(M)] $.
The compact homogeneous 3-manifolds with $ S^3 $ geometry together with their $ H^2(M,\mathbb{Z}) $ are: \begin{array} \\ &SU(2)/2I & 1 \\ &SU(2)/2O & 2 \\ &SU(2)/2T & 3 \\ &SU(2)/C_n & 1 \\ &SU(2)/Dic_{n} & 2^2 \\ \end{array}
The compact homogeneous 3-manifolds with $ S^2 \times E^1 $ geometry together with their $ H^2(M,\mathbb{Z}) $ are: \begin{array} \\ &S^2 \times S^1 & 1 \\ & \mathbb{R}P^2 \times S^1 & 1 \\ &S^2 \rtimes S_1 & 1 \\ & \mathbb{R}P^3 \# \mathbb{R}P^3 & 2^2 \\ \end{array}
Perhaps some of the three non-trivial principal $ U_1 $ bundles over $ \mathbb{R}P^3 \# \mathbb{R}P^3 $ have a total space which is not homogeneous? This is already kind of a weird manifold so that seems like the best bet to me. Another option that might be easier to work with is $ SU(2)/Q_8=SU(2)/Dic_2 $ which is the manifold of complete flags in $ \mathbb{R}^3 $, it can also be described as the unit tangent bundle of $ \mathbb{R}P^2 $. This manifold has 3 nontrivial $ U(1) $ principal bundles as well. Maybe some of them have total space which is not homogeneous?
Cross-posted from MSE https://math.stackexchange.com/questions/4759062/is-the-total-space-of-a-u-1-principal-bundle-over-a-compact-homogeneous-spac
