# JSJ-Decomposition - proof of finiteness in Hatcher's Notes

I'm currently trying to understand the proof of the (elementary) JSJ-Decomposition in Allen Hatcher's notes on 3-Manifolds.

Specifically, I have trouble understanding the application of proposition 1.7 to corollary 1.8.

Proposition 1.7 states:

For a compact irreducible 3-Manifold M there is a bound on the number of components in a system $S = S_1 \cup \dots \cup S_n$ of disjoint closed incompressible surfaces $S_i \subset M$ such that no component of $M | S$ ("M split along S") is a product $T \times I$ with $T$ a closed surface.

Corollary 1.8 states:

In a compact connected irreducible 3-Manifold M there exists a (possibly empty) collection T of disjoint incompressible tori, such that each component of $M|T$ is atoroidal.

In the proof, one starts with $T=\emptyset$, and then applies the following steps repeatedly:

1. If $M|T$ is atoroidal, we are done.
2. Otherwise, there exists an incompressible torus in $M|T$ which is not boundary-parallel (by definition of atoroidal). Add this torus to T. Go to step 1.

To show that this procedure will finish in a finite number of steps, the following argument is given:
Assume otherwise. Then there exist arbitrarily large collections T of disjoint incompressible tori, such that no component of $M|T$ is a product of $I$ and a torus, in contradiction to proposition 1.7.

I don't see why proposition 1.7 applies here - the condition in 1.7 is stronger, namely that no component of $M|T$ is a product of $I$ and a closed surface.

So my question is: why and how is prop. 1.7 applicable here?

I hope this question is suitable for MathOverflow. If not, a hint where I might get an answer would be nice.

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If a component $C$ of $M|T$ is a product of some closed surface $S$ and $I$, won't $S$ have to be just a component of the boundary of $C$ and therefore one of the tori in $T$? –  Andreas Blass Jun 9 '11 at 12:51
Look up the definition of atoroidal -- it means that if you have an incompressible surface $S$ in $M$, then $M | S$ has a component diffeomorphic to $S \times I$. –  Ryan Budney Jun 9 '11 at 17:01

## 1 Answer

Since we can only do JsJ-decomposition for toroidal manifold, as you say, we can assume that manifold M is toroidal. cutting M along the essential torus, the induced manifold or manifolds are both not the surface I-bundle and possible toroidal. We can do Jsj-decomposition again and again until that each component is atoroidal. And off course, each component is not surface I-bundle ,since we cut manifold along the essential torus.

Prop 1.7 implies that the maximal time for jsj-decomposition is finite.That implies 1.8.

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Thanks to your answer and the comments above I think I understand now - we never get a surface bundle, since the incompressible tori we use to split are not boundary-parallel. –  mogron Jun 10 '11 at 7:26