# 3-manifold theorem reference request or proof

The following is a theorem of which I have great interest in but cannot find anything about on the internet,

Every 3-manifold of finite volume comes from identifying sides of some polyhedron

I'm fairly certain that "identifying sides of some polyhedron" may be a simplification of the technical terminology. I believe it is just referring to gluing faces of polyhedron to form closed 3-manifolds. Such examples are given by the Seifert-Weber space, the Poincare homology sphere, the 3-dimensional real projective space, the $\frac{1}{2}$ twist cube space, etc. I'm assuming the proof is based off of Moise's theorem and proceeds as follows,

Let $M$ be an arbitrary closed 3-manifold. By Moise's theorem we have that $M$ can be tetrahedralized, so we let $T$ be the tetrahedralization of $M$ consisting of tetrahedrons $t_{1},...,t_{n}$. Pick an arbitrary tetrahedra $t_{1}$ of $T$ and proceed to glue $t_{2}$ to $t_{1}$, forming a new polyhedron $P_{2}$, and then glue $t_{3}$ to $P_{2}$ resulting in $P_{3}$, and so on. After all tetrahedra $t_{1},...,t_{n}$ have been glued, we have some resulting polyhedron $P_{n}$. From here, then somehow show that $P_{n}$ can be glued to $M$?

Any references to papers, expository writing, a proof of, or even the formal statement and name of this theorem would be greatly appreciated!

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You should note in your question that this has already been asked elsewhere and similarly on the other side. (math.stackexchange.com/questions/100029/…) –  Mariano Suárez-Alvarez Jan 18 '12 at 4:26
Usually no, it is not a good idea; for your particular question, I am pretty sure you'll get more useful answers here than there though. But you should note the fact that there is another copy of the question elsewhere (for all sorts of reasons: people might answer here when the answer has already been answered there, thereby wasting effort; &c) –  Mariano Suárez-Alvarez Jan 18 '12 at 5:16
The answer is yes. The idea is once you have a triangulation of a connected manifold, you can take your polyhedron to be a regular neighbourhood of a maximal tree in the dual 1-skeleton. The identifications on the boundary come from inflating the regular neighbourhood to the point where it fills the triangulation. –  Ryan Budney Jan 18 '12 at 5:34
It seems to me that you have already answered your question in the second "blocked text". The faces of $P_n$ are identified in pairs by looking at the corresponding faces of the triangulation, and by definition this gluing gives back $M$. What else is there to say? –  John Pardon Jan 18 '12 at 5:54
Is "finite volume" being used to mean "compact" here? –  Mariano Suárez-Alvarez Jan 18 '12 at 6:59