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.

3$\begingroup$ It is $SL_2\mathbb{R}$. I think that is a geometric manifold, with a homogeneous Einstein Lorentz metric, but perhaps not a geometric manifold in some other sense. $\endgroup$ – Ben McKay Feb 8 at 14:05

4$\begingroup$ I suggest giving a precise definition of "geometric manifold" then people can answer your question. If you go for a fairly traditional Thurstonstyle definition, your question should be answered in the Peter Scott paper, as this is a Seifert manifold. $\endgroup$ – Ryan Budney Feb 8 at 14:35
It all depends on your definition of a "geometric manifold."
 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 3Manifolds:
Neumann, W. D., Notes on geometry and 3manifolds, Böröczky, Károly jun. (ed.) et al., Low dimensional topology. Proceedings of the summer school, Budapest, Hungary, August 314, 1998. Budapest: János Bolyai Mathematical Society. Bolyai Soc. Math. Stud. 8, 191267 (1999). ZBL0944.57012.
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 casebycase analysis, but it helps to know that every lattice in a solvable Lie group is uniform, see here): This reduces the number of cases one has to handle.
 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 3manifolds, Bull. Lond. Math. Soc. 15, 401487 (1983). ZBL0561.57001.
and Bill Thurston in
Thurston, William P., Three dimensional manifolds, Kleinian groups and hyperbolic geometry, Bull. Am. Math. Soc., New Ser. 6, 357379 (1982). ZBL0496.57005.
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.
 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., Threedimensional geometry and topology. Vol. 1. Ed. by Silvio Levy, Princeton Mathematical Series. 35. Princeton, NJ: Princeton University Press. x, 311 p. (1997). ZBL0873.57001.
Then solid torus $M$ does admit any such structure as long as $X$ is 3dimensional: 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).

1$\begingroup$ Perhaps it's worth clarifying that the "complete finite volume locally homogeneous Riemannian metric" should be on the interior. The OP admits to being nonexpert, and I remember finding this point confusing when I was learning this stuff. $\endgroup$ – HJRW Feb 8 at 16:40
