Fundamental polygons and surfaces

  • #1


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The theorem of classification of closed surfaces says that any closed surface is homeomorphic to a fundamental polygon in the plane.

I was wondering if any fundamental polygon can be made into a closed surface by adjoining an appropriate atlas to it.

The topological requirements of a closed surface (compactness and connectedness) are certainly met by the fund. polygon but can we give it an atlas, and if so, is the resulting 2-manifold boundaryless?

Sure as hell appears so to me but I want to make sure.

Side question: Say we have a polygon with sides identified, and say that some side "a" is identified with n sides. Is n a topological invariant?

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Answers and Replies

  • #2
A fundamental polygon is just a polygon. Do you mean a polygon with edges identified in a certain pattern? Are you asking if any polygon can have it's edges identified in such a way as to make it a closed surface? Only if it has an even number of sides. Specifically, if it has 2m sides, you can make the surface of genus m. Othewise somewhere there must be an odd number of edges identified, and here the structure will either be a boundary (if there is one edge) or a place where the structure fails to be a manifold (if there are 3 or more).
  • #3
Oops, I thought the addition of the word "fundamental" was to refer to the fact that its sides were (potentially) indentified. Why waste breath and ink on the word "fundamental" then??

Btw - My side question is tightly related to the main question because in the proof of the classification thm, we show that any surface is homeomorphic to a polygon with pairwise identified sides. So if the answer to my side question comes out 'yes', then the main question will be answered.

Btw² - I have edited my side question to what I really had in mind.
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  • #4
Fundamental polygons are polygons associated to a surface. I've never seen this term used the other way around, to build a surface from a polygon. Maybe it is, but I would just call that a quotient space.

As far as your side question, each (interior of an) edge belongs to an equivalence class of edges. If n of these are identified, then the result locally looks like the product of the wedge product of n half open intervals with an open interval, ie, [0,1)x(0,1) if n=1, (0,1)x(0,1) if n=2, two intersecting planes if n=4, and so on. These spaces are all non-homeomorphic, so no homeomorphism can take a point in an eq class of m edges to one with n edges unless n=m. Note this can only be a manifold if n=1 or n=2 always, and only a closed one if n=2 (which implies the total number of edges is even).
  • #5
Jesus, I know I keep saying this but how do you know so much??

I hope that someday, when I grow up, I'll be Status X.

And thanks too!

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