Trying to understand the geometrization conjecture The Thurston's geometrization conjecture says:
Every closed orientable 3-manifold decomposes canonically into pieces whose interior has a locally homogeneous complete metric.
I'm trying to understand what a canonical decomposition means in this context. As far as I understand, that "canonical pieces" can be obtained as follows:
Let $M$ be a closed orientable 3-manifold. For the Knesser-milnor theorem, it can be decomposed as a conected sum
$$M\cong M_1 \sharp \cdots \sharp M_k$$
(unique up to permutation and homeomorphism) and each $M_i$ is prime. If $M_i$ is prime, then is either irreducible or $S^1 \times S^2$, so the KM decomposition can be rewrited as
$$M\cong N_1 \sharp \cdots \sharp N_s \sharp S^1 \times S^2 \sharp \cdots \sharp S^1 \times S^2$$
where the $N_i$ are irreducible. Finally, we aply the Jaco-Shalen decomposition, that says that if $N$ is a closed orientable irreducible 3-manifold, then there exist a family $\mathcal T$ of incompressible tori such that each component of $M\setminus \mathcal{T}$ is either a Seifert manifold or atoroidal. 
So, I undertstand that these small pieces which admit a geometry are the atoroidal or the Seifert manifolds. Am I right? 
And what happens whith $S^1 \times S^2$?Is a Seifert manifold?
Thanks!
 A: The thing is that the JSJ decomposition was introduced few years before Thurston came up with his conjecture. Accordingly, the JSJ decomposition does not completely match Thurston's (but is very close). 
Suppose that $M$ is a closed connected irreducible 3-manifold (I do not assume orientability, but all my examples will be orientable). From the JSJ viewpoint, we should be finding a minimal collection of pairwise disjoint Seifert submanifolds $Z_1,...,Z_n$ of $M$ with incompressible boundary, such that if $Z$ is a Seifert submanifold with incompressible boundary in $M$ then $Z$ is isotopic to a submanifold in one of the $Z_i$'s. In particular, the boundary of each $Z_i$ is a collection of 2-sided tori and Klein bottles. 
The fact that such a system of Seifert submanifolds exists is not at all obvious and is due to Jaco-Shalen (JS) and Johannson (J) who proved this independently in mid 1970s. 
Warning: Some of these Seifert submanifolds could be of the form $T^2\times [0,1]$ or $K^2\times [0,1]$, where $K^2$ is the Klein bottle. 
Remark. 1. Jaco, Shalen and Johannson were not really interested in decomposing $M$, but finding the characteristic submanifold of $M$, which is the union of Seifert submanifolds $Z_i$.  


*They worked in greater generality where $M$ is allowed to have (possibly higher genus) boundary. Then characteristic submanifolds are unions of Seifert submanifolds and interval bundles. The latter contain (up to isotopy) all essential annuli and Moebius bands in $M$. 


Define a collection of subsurfaces in $M$ which are boundary components of the pieces $Z_i$, where in case we get two isotopic tori (or Klein bottles), we discard one of these. (This can easily happen if one of $Z_i$ is, say, $T^2\times [0,1]$.)  
Cutting $M$ open along the these surfaces results in a collection  of manifolds $M_k$ each of which is either atoroidal or Seifert (or both!). 
Remark. One of the fastest ways to define a Seifert manifold is to say that it admits a foliation by circles. Then it is obvious that, say, $S^2\times S^1$ and $S^3$ are Seifert. It is less obvious but true that, with exception of hyperbolic and Sol-manifolds, all geometric 3-manifolds are Seifert.  
In contrast, Thurston would find a minimal decomposition of $M$ along incompressible  tori and Klein bottles such that removing these subsurfaces results in a (typically disconnected) open 3-manifold which admits a geometric structure (i.e. a complete locally homogeneous Riemannian metric). --- Different components typically will have different geometry. One may (but does not have to) insist on the locally homogeneous metric of $M$ having finite volume --- This is (mostly) automatic. Another thing which is a bit ambiguous is if we should insist that the splitting tori and Klein bottles are 2-sided.  From my viewpoint, it is better to allow 1-sided subsurfaces, otherwise, there could be  a finite covering $M'\to M$ such that the lift of such Thurston's geometric decomposition from $M$ to $M'$ does not result in a geometric decomposition (we may loose minimality). 
As an example, consider the mapping torus of an Anosov diffeomorphism of $T^2$, i.e. a diffeomorphism whose action on $H_1(T^2)$ has two eigenvalues $\lambda^{\pm 1}$ with $|\lambda|\ne 1$. Such a manifold admits the Sol geometry and, hence, should not be decomposed (from Thurston's viewpoint). However, $M$ is not Seifert, so, from JSJ viewpoint, we should cut it along a torus resulting in i$T^2\times [0,1]$. Thus, in this example, the JSJ decomposition is not the same as the geometric decomposition. 
Here is another example. Let $M_1$ be a compact oriented hyperbolic manifold whose boundary is a single torus $T^2$. Let $M_2$ be an orientable manifold which is the total space of a non-orientable interval bundle over the Klein bottle $K^2$. Thus, $M_2$ is a Seifert manifold whose boundary is a single torus. Glue $M_1$ and $M_2$ along the  boundary tori any way you want and obtain a closed orientable irreducible 3-manifold $M$. The JSJ decomposition of $M$ would amount to undoing this gluing process and the components of the decomposition will be $M_1$ (an atoroidal component) and $M_2$ which is both atoroidal and Seifert! However, from Thurston's viewpoint, what you should be doing is cutting $M$ open along a 1-sided Klein bottle $K^2$ (contained in $M_2$). The result is a 3-manifold $M_1$. To obtain $M$ from $M_1$ then we are gluing its only boundary torus to itself. Which of these two decompositions is the "right one" is open to a debate, but for many purposes, is irrelevant. If you prefer the first decomposition, then taking a 2-fold covering $M'\to M$ and lifting the above decomposition to $M'$ we obtain a decomposition 
into 3 pieces, two of which are homeomorphic to $M_1$ and one is $T^2\times [0,1]$. Thus, this decomposition is neither JSJ nor Thurston's. In contrast, if we lift the 1-sided Klein bottle from $M$ to $M'$ we obtain a 2-sided incompressible torus which yields both JSJ and Thurston's decompositions of $M'$. Personally, I prefer the latter as more canonical and "more minimal". 
Apart from these two issues (Sol-manifolds and manifolds containing 1-sided Klein bottles), Thurston's decomposition is the same as JSJ.   
