I'm re-learning topological constructs and came across fundamental polygons as a way to describe certain topologies. I have an intuitive understanding but any mathematical understanding was left behind in a 20 year old college class.

The examples there give a handful of resultant topologies, all based on a four-sided fundamental polygon: sphere, Möbius strip, Klein bottle, torus, real projective plane. I have three questions.

  1. Are there other topologies for a 4-gon?

  2. More generally, how many topologies can you construct in this fashion for an n-gon?

  3. What about if you disallow discontinuities? (I don't know the proper term for this, but e.g. labeling three edges "A" would cause it, as would labeling two adjacent edges A but with arrows in the same direction.)


For the 4-gon, you have exactly these four cases. The reason is that we do disallow discontinuities, i.e. we are interested in surfaces = twodimensional manyfolds: In the surface, the (identified) sides become a single curve that locally looks like the $x$-axis within $\mathbb R^2$, i.e. there must be exactly two sides (corresponding to lower and upper half plane). This means that exactly two edges must be labelled "A" etc. We might allow the two A edges to be adjacent with "wrong" directions. But such cases be better resolved by splitting the two edges in half (and have a "legal" $(n+2)-gon). Indeed, you may note that the projective plane can be considered as coming from an "illegal" 2-gon this way (while the sphere might also be considered to actually be a glued 2-gon) With these restrictions, we have only few combinatorical possibilities: If the "red" edges are adjacent, we necesarily obtain the sphere. If they are not adjacent, we have one of the three remaining situations: Both colour pairs are "parallel", or both are "antiparallel" or we have a mixed case.

We can derive a simple upper bound for the topologies possibly constructed from the $2n$-gon: There are $(2n-1)\cdot(2n-3)\cdot\ldots\cdot3\cdot1=\frac{(2n)!}{2^nn!}$ possibilities to partition the $2n$ edges into pairs (fix the first, select his partner among the rest, fix the first available, select his partner among the rest, etc.). For each such partition, we may label each pair either "parallel" or "antiparallel". Thus we end up with $\frac{(2n)!}{n!}$ as an upper bound. Note that the correct number is a lot smaller (for example we obtain $\frac{4!}{2!}=12$ instead of $4$ for the 4-gon):

  • If adjacent edges have the same label, only one orientation is allowed
  • Rotating the polygon produces the same topology (the rotation induces a homeomorphism)
  • Some of the topologies might be obtainable from a $2k$-gon with $k<n$.
  • $\begingroup$ Thanks! That's very nearly there. I wonder if we can get a tighter upper bound... $\endgroup$ Sep 29 '12 at 16:30
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    $\begingroup$ If you want to include the Mobius strip, then you must also allow some edges not to be identified with any other edge. This means you will get some surfaces with boundary. In this case you shouldn't forget the annulus (say, from identifying only the left and right sides with no twist). I'm not sure whether that would still be considered a fundamental polygon. $\endgroup$
    – Hew Wolff
    Sep 30 '12 at 19:48
  • $\begingroup$ @HewWolff: At worst, this would add a third variant to what can be done with a pair of matching edges, but I don't believe either that this is usually considered fundamentel polygon. $\endgroup$ Sep 30 '12 at 21:00

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