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I googled it and wikipidead it too : but apparently there is no definition of an ideal triangle on a punctured torus ( i.e a compact [ hyperbolic ] surface with one genus and one boundary component, or torus minus an open disk ). How do you define it ? As I know in the case on just one point removed off the torus, we take the standard tringulation in the rectangle forming the fundamental domain of torus, remove the four vertices off the rectangle and consider the quotient by a $ Z+Z $ action.

So, in this case, should we remove a quartar of a disk from each corner of the rectangle ( fundamental domain or universal cover ) and consider its quotient by a $ Z + Z $ action ?

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"punctured" means "one point removed", so it is the case you already know. –  Bruno Martelli Nov 2 '10 at 14:40
    
Yes, but I am looking for torus minus an open disk. –  Analysis Now Nov 2 '10 at 14:44
    
Given a non-compact manifold $M$, you can take the definition of an ideal triangulation of $M$ to be a triangulation of the one-point compactification of $M$ with a single $0$-cell, being the point at infinity. Some people vary this a little, having $n$ $0$-cells where $n$ is the number of ends of $M$. For most purposes this distinction makes little difference. –  Ryan Budney Nov 2 '10 at 14:46
    
I'm not clear what kind of answer you're looking for. Take a torus with an open disk removed, and put a hyperbolic metric on it. Then triangles with their vertices on the boundary are ideal triangles. –  Hugh Thomas Nov 2 '10 at 17:24
    
"Punctured" usually means "remove a point" while "holed" means "remove an open disk". But terminology can vary. –  Sam Nead Nov 2 '10 at 22:54

1 Answer 1

To define an ideal triangulation one typically starts with a hyperbolic metric. Suppose that $T^2 = S^1 \times S^1$ is the two-torus and $X = T - D$ is the torus minus an closed embedded disk. There are essentially two ways to give $X$ a complete hyperbolic metric. As everyone has remarked, and you have remarked as well, if the metric has finite area then you can triangulate $X$ with exactly two ideal triangles with all vertices at the "ideal point at infinity."

On the other hand, the other "kind" of metric on $X$ is a metric of infinite area. Since an ideal triangle has area $\pi$ it will take infinitely many ideal triangles to triangulate $X$. These triangulations can be very wild. In particular they are not typically determined by a finite amount of data. Thus the phrase "ideal triangulation" virtually always refers to the case when $X$ has finite area.

EDIT: Of course I misread the question! The question asks about ideal triangulations of a hyperbolic surface with geodesic boundary. Here the answer pretty much has to be "spun triangulations". That is: fix such a metric on $X = T - D$ when $D$ is an open disk. Fix a collection of three proper embedded geodesic arcs in $X$, perpendicular to the geodesic boundary. These cut $X$ into a pair of right-angled hexagons. Now we spin. Choose an orientation for $\partial X$. Drag all of the endpoints of all the arcs in the direction of the orientation. At each point of time during this process we have three proper geodesic arcs, but the angles they make with the boundary go to zero. Take the Hausdorff limit and you will get a spun ideal triangulation of $X$. I'll end by remarking that you can build a spun triangulation of a pair of pants ($S^2$ minus three open disks with disjoint closures).

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I mentioned torus minus an open disk, not closed. For a hyperboplic metric isnt it the case, that by Gauss-Bonnet, the metric has area $ 2\pi $, or else we join two isometric copies of these object to get a closed orinted surface of genus 2 which has area $ 4\pi $. –  Analysis Now Nov 2 '10 at 22:25
    
The area computation above is for torus minus an open disk though., otherwise we can make the open end a funnel to have infinite area, which I guess youn were taling about. –  Analysis Now Nov 2 '10 at 22:27
    
I'm sorry - I misread your question. I'll edit the answer. –  Sam Nead Nov 2 '10 at 22:41

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