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Start with the two dimensional hyperbolic plan, H2.

Then make a huge (non-ideal) equilateral polygon (with a number of sides that is large and a multiple of four). I want to identify the sides of the polygon to make a smooth space with no conical singularities.

Obviously this is only possible if the sum of the exterior angles is 2$ \pi$. (Then by the Gauss-Bonnet theorem this sets of the size of the polygon in terms of the number of sides.) Suppose the sum of the interior angles is 2$\pi$.

My question: how much freedom do I have in deciding which side to identify with which side?

I know one way to do it - label the successive sides $A, B, \bar{A}, \bar{B}, C, D, \bar{C}, \bar{D}$ going round, and then identify $A$ with $\bar{A}$ etc. My question is whether this is the only way to do it.

In my example the sides are identified with another side that is nearest-neighbor-but-one. Can I make an identification that is smooth in which they are identified with much more distant sides?

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    $\begingroup$ Have you seen the Harer & Zagier paper? (There are many followups too, the latest one contains most of the references I think; there is also a related question here on MO, mathoverflow.net/q/125840/41291) $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Aug 4, 2016 at 19:30
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    $\begingroup$ The basic premise of the question is faulty: You require the angle sum of the polygon equal to $2\pi$ only if all the vertices of the polygon belong to a single equivalence class after the identification. These equivalence classes are traditionally called "cycles". Then the correct requirement is that for each cycle of the sum of angles at its vertices equals $2\pi$. $\endgroup$
    – Misha
    Commented Feb 1, 2017 at 20:49

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