Does anybody know if orientable, closed $3$-manifolds that are circle bundles over $RP^2$ have been classified? One can determine the isomorphism classes of bundles using obstruction theory, but I am interested in what total spaces can appear. I am not assuming the bundle is principal.
Thank you.
Such manifolds are examples of Seifert fibered spaces, which have, indeed, been classified. A good reference is Montesinos "Classical Tessellations and Three-Manifolds". Basically, such manifolds (over any nonorientable surface base) are classified by their Euler class, which measures the obstruction to the existence of a section.
According to Scott's paper "The geometries of 3-manifolds", the only closed manifolds that admit an $S^2 \times \mathbb{R}$ geometry are the two $S^2$ bundles over $S^1$, $P^2 \times S^1$ and $P^3 \# P^3$. It seems like the last one is the only candidate (unless I'm missing a way in which $S^2 \times S^1$ can be viewed as such a bundle).
Edit. As pointed out below, I spoke too soon. Of course such bundles could have spherical geometry too. Scott's paper is a nice reference for that too!
I think that they have Seifert fiber space presentation as: $(On,1|(1,b))$.
Or $(On,1|(1,b),(a_1,b_1),...,(a_r,b_r))$, if you allow an orbifold with cone points in $RP^2$.
You can look at the cases by decomposing $RP^2=Mo\cup_{\partial}D$, so the orientable 3-manifold will be the
1) orientable $Q=Mo\tilde{\times}S^1$, the twisted circle bundle over the mobius band, very well known being equivalent to the orientable I-bundle over the Klein bottle, with boundary a torus $T$, 2) and a Dehn-filling in the remaining disk $D$, with a whichever fibered solid torus or tori.We could say that $(On,1\mid (1,b))=Q\cup_T W(1,b)$, for a fibered $(1,b)$ solid torus $W$
There are two principal circle bundles over $ \mathbb{R}P^2 $ and an infinite family of non principal circle bundles over $ \mathbb{R}P^2 $.
Applying LES homotopy to the bundle $$ S^1 \to M \to \mathbb{R}P^2 $$
We have that $$ \mathbb{Z} \to \pi_1(M) \to C_2 \to 0 $$
If the first map is injective we have
$$ 0 \to \mathbb{Z} \to \pi_1(M) \to C_2 \to 0 $$
If the first map is not injective then we have $$ 0 \to C_m \to \pi_1(M) \to C_2 \to 0 $$
Recall that a circle bundle is orientable as a bundle iff it is a principal circle bundles. So using classifying spaces one can show there are only two principal circle bundles. See for example
https://math.stackexchange.com/questions/3183562/classification-of-circle-bundles-over-mathbbrp2
These two bundles have Seifert invariants {b; ($n_1$, 1);} (b is 0 or 1). They are exactly the two non orientable compact manifolds with $ S^2 \times E^1 $ geometry. They correspond to the case $$ 0 \to \mathbb{Z} \to \pi_1(M) \to C_2 \to 0 $$ above. The first one (b=0) is the trivial bundle $ S^1 \times \mathbb{R}P^2 $ with fundamental group $ \mathbb{Z} \times C_2 $. The second one (b=1) has fundamental group $ \pi_1(M) \cong \mathbb{Z} $ and can be described as the mapping torus of the antipodal map on $ S^2 $. Both of these manifolds are Riemannian homogeneous with respect to the natural metric from $ S^2 \times E^1 $ and they both have isometry group $ SO_3(\mathbb{R}) \times O_2(\mathbb{R}) $. For details see
Mapping torus of orientation reversing isometry of the sphere
https://math.stackexchange.com/questions/4322584/s2-times-r-geometry
https://math.stackexchange.com/questions/4360976/natural-group-action-on-mapping-torus
Next up is the infinite family of non principal circle bundles. These have Seifert invariants {b; ($n_2$, 1);} ($b=0,1,2,3,4,\dots$), denote the manifolds by $ M_b $. They correspond to the case $$ 0 \to C_m \to \pi_1(M) \to C_2 \to 0 $$ above (with the exception of $ b=0 $). For $ b \neq 0 $ we have $ m=2b $ and the fundamental group is a finite group of order $ 4b $. The $ M_b $, $ b \neq 0 $, admit $ S^3 $ geometry, in other words a round metric: a metric with constant positive curvature. $ M_1 $ is exceptional and happens to be the lens space $ L_{4,1}\cong SO_3(\mathbb{R})/C_2 \cong SU_2/C_4 $ with $ \pi_1(M_1) \cong C_4 $. It is the only Lens space that occurs here. $ M_1 $ is isomorphic as a bundle to the unit tangent bundle of $ \mathbb{R}P^2 $. Since $ M_1 $ is a lens space it has large isometry group with respect to the round metric, in particular $ SO_3(\mathbb{R}) \times O_2(\mathbb{R}) $.
The $ M_b $ for $ b\geq 2 $ are exactly the standard prism manifolds. $ \pi_1(M_b) $ is the binary dihedral group of order $ 4b $. In other words $$ M_b \cong SO_3(\mathbb{R})/D_{2b} \cong SU_2/\tilde{D_{2b}} $$ where $ D_{2b} $ is the dihedral group of order $ 2b $ and $ \tilde{D_{2b}} $ is the binary dihedral group of order $ 4b $. All the prism manifolds $ M_b $, $ b\geq 3 $ have isometry group $ O_3(\mathbb{R}) $ with respect to the round metric.
The prism manifold $ M_2 $ is somewhat exceptional. The dihedral group of order 4 is degenerate: it is abelian and isomorphic to the Klein four group $ V_4 \cong D_4 \cong C_2 \times C_2 $. And its extension, the binary dihedral group of order $ 8 $, is actually isomorphic to the 8 element quaternion group $ Q_8 $. These groups have larger normalizer ($ V_4 \trianglelefteq S_4 $, with quotient $ S_4/V_4 \cong S_3$). So the manifold $ M_2 $ has especially large isometry group with respect to the round metric $$ Iso(M_2) \cong SO_3(\mathbb{R}) \times S_3 $$ There are several interesting perspectives on $ M_2 $. As mentioned above $ M_2 \cong SO_3(\mathbb{R})/D_4 \cong SU_2/Q_8 $. Also $ M_2 $ is the manifold of complete flags in $ \mathbb{R}^3 $. And, as a bundle, $ M_2 $ is the projectivized tangent bundle of $ \mathbb{R}P^2 $.
Finally we turn our attention to $ M_0 $, certainly the most exceptional of the bunch. Note that all the manifolds above were Riemannian homogeneous and $ S^3 $ geometry. $ M_0 $ is neither. In fact $ M_0 $ is a third bundle where we have $$ 0 \to \mathbb{Z} \to \pi_1(M) \to C_2 \to 0 $$ and like the two principal bundles with infinite fundamental group it admits $ S^2 \times E^1 $ geometry. $ M_0 $ is actually the connected sum $$ M_0 \cong \mathbb{R}P^3 \# \mathbb{R}P^3 $$ So its fundamental group is a free product $$ \pi_1(M_0)\cong C_2 * C_2 $$ Another interesting thing to note is that although it admits a geometry ( $S^2 \times E^1$) unlike every other circle bundle over $ \mathbb{R}P^3 $ it is not Riemannian homogeneous for any metric. That said it is smooth homogeneous, admitting a transitive action (not by isometries) by the Euclidean group $ E_3 $. For more details see
https://math.stackexchange.com/questions/4365938/connected-sum-of-two-copies-of-rp3
