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Let $ M $ be a 3-manifold which is the total space of a fiber bundle (with connected nontrivial fiber and base) with either the fiber or base a hyperbolic surface. Then does $ M $ always admit a Thurston geometry?

The fundamental group of a hyperbolic surface is not virtually solvable. Therefore any bundle with a hyperbolic surface as base or fiber cannot have a virtually solvable fundamental group. So it can only admit $ \tilde{SL_2}, H^2 \times E^1 $ or $ H^3 $ geometry.

Bearing this in mind here are some examples of bundles with these three geometries:

$ \tilde{SL_2} $: The unit tangent bundle of any hyperbolic surface.

$ H^2 \times E^1 $: The mapping torus of any periodic (finite order) mapping class of a hyperbolic surface. Including the trivial product of a circle with any hyperbolic surface.

$ H^3 $: The mapping torus of any pseudo-Anosov mapping class of a hyperbolic surface.

Another thing to note: The fundamental group of a circle bundle always contains a normal copy of $ \mathbb{Z} $ by LES homotopy. Thus a circle bundle over a hyperbolic surface $$ S^1 \to M \to \Sigma_g $$ has a fundamental group which contains a normal $ \mathbb{Z} $ subgroup. Therefore, a circle bundle over a hyperbolic surface can only admit $ \tilde{SL_2} $ or $ H^2 \times E^1 $ geometry (such examples of circle bundles are given above with the unit tangent bundle admitting $ \tilde{SL_2} $ geometry and the trivial circle bundle admitting $ H^2 \times E^1 $ geometry).

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No, $M$ does not always admit a Thurston geometry. The thing to keep in mind is that (outside of a tiny number of exceptions that you probably know) the Thurston geometries occur only when the manifold is irreducible, and in that case only when the manifold is either atoroidal or is toroidal and has one of the geometries other than $\mathbb H^3$ that allow many many toruses, namely product geometries, circle fibered geometries, NIL geometry, SOLV geometry.

One can avoid those situations by constructing a 3-manifold $M$ that fibers over $S^1$ with a fiber $F$ that is a closed, oriented surface of genus $g \ge 2$, that is not atoroidal, and that does not have any circle fibered geometry. For that purpose, take a homeomorphism $h : F \to F$ for which there is a simple closed curve $C \subset F$ with connected complement, such that $h(C)=C$ and such that $h \mid F-C$ is a pseudo-Anosov homeomorphism. The mapping torus construction $$M_h = F \times [0,1] / (x,1) \sim (f(x),0) $$ produces the desired counterexample; it has, up to isotopy, just a single incompressible torus, namely $T = C \times [0,1] / (x,1) \sim (f(x),0)$; and it has hyperbolic geometry on the complement $M_h-T$.

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  • $\begingroup$ I see. Do all these counter examples involve hyperbolic surfaces? Do you know of any counterexamples where the fiber and base are not hyperbolic? For example some circle bundle over $ \mathbb{R}P^2 $ that doesn't admit any Thurston geometry? $\endgroup$ Feb 18, 2022 at 19:43
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    $\begingroup$ That's another question, but I believe circle bundles over $\mathbb RP^2$ all have either $S^2 \times S^1$ or $S^3$ geometry. $\endgroup$
    – Lee Mosher
    Feb 18, 2022 at 20:26
  • $\begingroup$ Fair enough. I've separated the two questions: math.stackexchange.com/questions/4385644/… $\endgroup$ Feb 18, 2022 at 20:40
  • $\begingroup$ I see, so what you've done here is explicitly constructed a homeomorphism of a hyperbolic surface whose mapping class is neither periodic nor pseudo-Anosov. Is it true that a mapping torus of a hyperbolic surface admits a Thurston geometry if and only if the mapping class is either periodic or pseudo-Anosov? $\endgroup$ Feb 19, 2022 at 11:44

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