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We know by Thurston's Geometrization Conjecture, that every closed 3-manifold admits a prime decomposition: it must be the connected sum of prime 3-manifolds.

My question is: if $\mathbb{X}$ is one of Thurston's 8 geometrical models, and $\Gamma < \text{Diffeo}(\mathbb{X})$ is a cocompact subgroup (not necessarily a subgroup of isometries) then the quotient $M=\mathbb{X}/\Gamma$ is a prime 3-manifold? Or it could admit a non-trivial decomposition (in the sense that it could admit more than one prime manifold glued together)?

Thanks in advance for any help!

Edit: To be clear, what I truly want to know is if it is possible for a quotient $M=\mathbb{X}/\Gamma$ by a cocompact subgroup $\Gamma < \text{Diffeo}(\mathbb{X})$, that acts freely and discrete on $\mathbb{X}$, to have distinct regions locally isometric to distinct model geometries.

Odylo Abdalla Costa
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    Let $X=H^2\times \Bbb R$, there is a quotient $\Sigma_g\times S^1$ of this where $\Sigma_g$ is a surface of genus $g$, $g\geq 2$. This is not prime, being a sum of tori crossed with $S^1$. – s.harp Jul 10 '23 at 21:55
  • Thanks, @s.harp. That's true. Probably I didn't pose well my question: in your case, $\Sigma_{g}\times S^{1}$ is modeled by one geometry, namely $\mathbb{H}^{2}\times \mathbb{R}$ since it is obtained via a quotient by a cocompact subgroup of $Isom(\mathbb{H}^{2}\times \mathbb{R})$.

    But is there an example where a quotient of the type I wrote ($\mathbb{X}/\Gamma$ for $\Gamma < Aut(\mathbb{X})$ cocompact) where it can have distinct regions locally isometric to distinct model geometries?

     – Odylo Abdalla Costa Jul 10 '23 at 22:08
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    @s.harp: You misunderstand the definition of a prime 3-manifold. Your example is prime. – Moishe Kohan Jul 10 '23 at 22:19
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    You forgot a number of assumptions on $\Gamma$ such as discreetness and freeness of the action. – Moishe Kohan Jul 10 '23 at 22:20
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    Just a remark: the fact that every manifold admits a prime decomposition is way more elementary than the Geometrization conjecture, and one only needs orientable and compact (boundary is allowed) – Henrique Augusto Souza Jul 10 '23 at 23:11
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    Also, by $\operatorname{Aut}(\mathbb{X})$ do you mean the group of self homeomorphisms of $\mathbb{X}$? – Henrique Augusto Souza Jul 10 '23 at 23:12
  • @HenriqueAugustoSouza for example. But it can even be $\text{Diffeo}(\mathbb{X})$. – Odylo Abdalla Costa Jul 10 '23 at 23:14
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    One usually does not regard the full group of homeomorphisms/diffeomorphisms as the automorphism group of a Thurston's geometry, I am not sure where did you get this idea from. In any case, once you add the orientability assumption, then there is only one counter-example to primeness, namely, $RP^3 # RP^3$, as a quotient of the $S^2\times R$ geometry. This is a somewhat nontrivial result. – Moishe Kohan Jul 10 '23 at 23:18
  • @MoisheKohanonstrike hmm right... yes, I regarded the group of automorphisms (diffeomorphism more precisely) because my original question was the other way around in some sense: I had a compact manifold $\mathbb{X}/\Gamma$ with $\Gamma< \text{Aut}(\mathbb{X})$ and wanted to know what's its geometry, in the spirit of Thurston's 8 geometries... just from the information above I can find conclude it is only one of the eight? Or it could be more than one? – Odylo Abdalla Costa Jul 10 '23 at 23:24
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    This is the only one geometry. The key is that this is the only geometry with nontrivial $\pi_2(X)$. – Moishe Kohan Jul 10 '23 at 23:49
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    @OdyloAbdallaCosta the manifold in the answer shows that your $M$ may fail to admit any geometry at all: https://math.stackexchange.com/questions/3578183/homogeneous-universal-cover-implies-existence-of-a-geometric-structure – Henrique Augusto Souza Jul 11 '23 at 00:01
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    And the question in your edit is very different from the rest of what you wrote. It seems that you do not really understand the statement of the geometrization theorem. You can start by reading the relevant Wikipedia article. – Moishe Kohan Jul 12 '23 at 17:24

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