Two points of confusion about Thurston and Perelman's Geometrization Theorem

Question

I am a physicist and having trouble parsing through the Thurston and Perelman's Geometrization Theorem.

I have two main confusions:

1. The simplest version of geometrization deals with closed three-manifolds, namely compact without boundary. In that case, the interior of the manifold will have only one of the eight geometries. But, one of the model geometries is $\mathbb{R}^{3}$ which is not compact. Does that mean that the geometries appearing in the classification are compact quotients, $\mathbb{R}^{3}/\Gamma$ where $\Gamma$ is the isometry group of the quotient?

2. The classification is not of the entire three-manifold itself, but rather it is a two-step classification. A geometrization of $\Sigma$ is a decomposition $H \cup G$ where the union is over tori. If the tori are incompressible, then there is a further decomposition into domains modeled on the Thurston geometries. Does this mean that the metric on $\Sigma$ is not one induced from the Thurston geometries, but rather the metric on the domain parts of $\Sigma$ are?

Answer 1

I will gradually edit this answer as I learn more about 2 and 3 manifolds.

Attempt 2 (?):

Any closed three-manifold $\Sigma$ admits a unique (up to deletion/insertion of spheres) connected-sum decomposition into prime manifolds $M_{i}$. A prime manifold $M_{i}$ has a (isotopically) unique minimal decomposition by incompressible tori, which are embedded tori $T^{2} \to M_{i}$ that are also $\pi_{1}$ injective.

Thus, understanding $\Sigma$ amounts to understanding prime manifolds. Accordingly, there are two main types of prime manifolds: atoroidal (there is no incompressible torus and the decomposition of $M_{i}$ trivializes) and Seifert-fibered spaces. The latter spaces are disjoint unions of circles such that all but finitely many circles are cores of (identified) solid tori in $M_{i}$. If the torus is not identified, then all but finitely many circles will be cores of $S^{1} \times B^{2}$, otherwise they will be cores of $[0,1] \times B^{2}/(x,1) \simeq (x e^{\frac{ik}{n} 2 \pi},1)$, where $k \in Z_{n}$. (I believe this is a twisted bundle over $B^{2}$ as opposed to the solid torus which is a trivial circle bundle over the ball. If that is true, then every Seifert-fibered space is just a bundle over a ball.)

The geometrization is then a statement about the interiors of each piece in the torus decomposition of a prime manifold, which translates to $\Sigma$ as it is decomposable into connected sums of such prime manifolds. There are eight model geometries, and each such piece (in the torus decomposition) will be modeled on one of the eight.

---

Attempt 1 (wrong):

The geometrization program is really a three step procedure. Take $\Sigma$ a closed three-manifold. This admits a unique decomposition into prime manifolds, which are manifolds without essential separating two-spheres. After throwing away the spheres, you may try to cut away the next simplest surface, the two-torus. This quotients out the tori in each prime manifold appearing in the decomposition, leaving you with irreducible manifolds. To make this a two step process, one may simple choose to decompose $\Sigma$ into irreducibles.

This may have a boundary, so the geometrization theorem states that the interior of each irreducible is modeled on one of the eight Thurston geometries. As the different geometries are not compatible, one cannot glue metrics on each interior smoothly to provide get the metric on $\Sigma$.