1
$\begingroup$

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!

Improve this question
$\endgroup$
9
  • 1
    $\begingroup$ You should note in your question that this has already been asked elsewhere and similarly on the other side. (math.stackexchange.com/questions/100029/…) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jan 18, 2012 at 4:26
  • 1
    $\begingroup$ 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) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jan 18, 2012 at 5:16
  • 2
    $\begingroup$ 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. $\endgroup$
    – Ryan Budney
    Commented Jan 18, 2012 at 5:34
  • 2
    $\begingroup$ 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? $\endgroup$
    – John Pardon
    Commented Jan 18, 2012 at 5:54
  • 2
    $\begingroup$ Is "finite volume" being used to mean "compact" here? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jan 18, 2012 at 6:59

1 Answer 1

Reset to default
6
$\begingroup$

I think the reference that you are looking for is this article by Cannon, Floyd, and Parry.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Scott - This works, but it is more "high powered" than what the original poster is asking for. Isn't it? $\endgroup$
    – Sam Nead
    Commented Jan 18, 2012 at 9:35
  • $\begingroup$ Sam: Probably so. I heard Jim talk about it a few years ago, and thought it was very interesting. $\endgroup$
    – Scott Carter
    Commented Jan 19, 2012 at 0:04
  • $\begingroup$ @Scott Carter: I'm attempting to read through this and it is a little dense, any additional papers that may make the reading a bit easier or should I pull one of the "descend down through the references until I reach familiar territory"? $\endgroup$
    – Samuel Reid
    Commented Jan 19, 2012 at 2:16
  • 1
    $\begingroup$ @Samuel Reid, I have not read the article myself. I heard Jim Cannon talk about it. He usually gives very lucid talks, but one finds that there are terribly dense details in the text. My understanding of his work is that it is well worth trying to slog through it. $\endgroup$
    – Scott Carter
    Commented Jan 19, 2012 at 23:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .