I suppose the solid torus in $\mathbb{R}^3$ is not a geometric manifold. Since I am not an expert in this area, I would like to ask whether there is some easy way to see this.



Existence of a geometric structure on a solid torus
It all depends on your definition of a "geometric manifold."


1. One definition would require the existence of a complete finite volume locally homogeneous Riemannian metric (from Thurston's list of eight 3D geometries). This definition is used, for instance by Walter Neumann in
his [Notes on Geometry and 3-Manifolds](https://www.math.columbia.edu/%7Eneumann/preprints/budfinal.pdf):


*Neumann, W. D.*, Notes on geometry and 3-manifolds, Böröczky, Károly jun. (ed.) et al., Low dimensional topology. Proceedings of the summer school, Budapest, Hungary, August 3-14, 1998. Budapest: János Bolyai Mathematical Society. Bolyai Soc. Math. Stud. 8, 191-267 (1999). [ZBL0944.57012](https://zbmath.org/?q=an:0944.57012).


Then the answer is negative: Such a metric does not exist. I do not know of any illuminating proof of this, except based on a case-by-case analysis, but it helps to know that every lattice in a solvable Lie group is uniform, see [here](https://en.wikipedia.org/wiki/Lattice_(discrete_subgroup))): This reduces the number of cases one has to handle.


2. Or, maybe, you do not want to impose the finite volume assumption. This is the viewpoint adopted by Peter Scott in


*Scott, Peter*, [**The geometries of 3-manifolds**](http://dx.doi.org/10.1112/blms/15.5.401), Bull. Lond. Math. Soc. 15, 401-487 (1983). [ZBL0561.57001](https://zbmath.org/?q=an:0561.57001).


and Bill Thurston in


*Thurston, William P.*, [**Three dimensional manifolds, Kleinian groups and hyperbolic geometry**](http://dx.doi.org/10.1090/S0273-0979-1982-15003-0), Bull. Am. Math. Soc., New Ser. 6, 357-379 (1982). [ZBL0496.57005](https://zbmath.org/?q=an:0496.57005).


Then the solid torus is geometric for all but two of the eight geometries. Namely, take a cyclic group $\Gamma$ of isometries of a homogeneous space $X$ (one of the 8 geometries, excluding $S^3$ and $S^2\times {\mathbb R}$) not fixing any point in $X$. Then $X/\Gamma$ is homeomorphic to the (open) solid torus.


3. Or, "geometric" is in the sense of admitting an $(G,X)$-structure in Thurston's sense (or maybe Ehresmann). This is the viewpoint adopted by Thurston in


*Thurston, William P.*, Three-dimensional geometry and topology. Vol. 1. Ed. by Silvio Levy, Princeton Mathematical Series. 35. Princeton, NJ: Princeton University Press. x, 311 p. (1997). [ZBL0873.57001](https://zbmath.org/?q=an:0873.57001).


Then solid torus $M$ does admit **any** such structure as long as $X$ is 3-dimensional: This is because $M$ admits an embedding in $X$. Thurston discusses three kinds of geometric structures: Modeled on pseudogroups, groups and "model geometries." In the book, he never defines a "geometric manifold" (in general).


Or, maybe you mean something completely different...


Edit. Per Henry's comment: Here, geometric structures are always taken on manifolds without boundary. It is common to apply the adjective "geometric" to manifolds with boundary. Then the geometric structure (whatever it is) is understood to be on the interior of the manifold (the manifold minus its boundary).


