I was playing with some shapes recently and came across a surface $M$ with the following fundamental polygon: link

This has surface word $$a c b f_1 a^{-1} c^{-1} f_2 b^{-1} f_3,$$

where the $f_i$ are boundary edges that are not identified to any other. I'd like to know how to identify this surface. I know this might be orientable because every paired edge has an inverse. I also see that there are two boundary circles $\mathbb{S}^1$ formed by $f_1 f_2$, and $f_3$ (these connect back up to themselves). Under identification, I count 3 vertices and 5 edges ($a, b, c, (f_1 f_2),$ and $f_3$). Given there is 1 face, this should have $$\chi (M) = 3 - 5 + 1 = -1$$ I recall that we can cap the two boundary circles to form a surface $M^*$ without boundary, which has $$\chi (M^*) = \chi (M) + 2 = 1$$ This doesn't fit with the classification of compact surfaces however. If it's orientable, then $\chi (M^*)$ would be even. Further, wouldn't $\chi (M^*) = 1$ imply that $M^* \cong \mathbb{P}^2$, the projective plane?

I don't actually know what I'm doing here. I know there might be some argument using word operations, or cutting and pasting, but I don't know where to start. Moreover, I think a larger question I have is how to even deal with unpaired edges in fundamental polygons in the first place. I've searched around quite a bit, but everything I've found just deals with surfaces without boundary. The best I've found is that the classification theorem extends to compact surfaces with boundary in a way that states that these surfaces are homeomorphic to any of $\mathbb{S}^2, n\mathbb{P}^2,$ or $n\mathbb{T}^2$ with a finite number of disks removed.

Any pointers or resources on this would be greatly appreciated. If anyone is interested, the place this turns up is from a very convoluted maze level in the 1980 Atari game Adventure. I think it'd be fun to figure out what surface this lies on!


1 Answer 1


There are six edges, not five: there is not one edge $(f_1f_2)$, instead there are two edges $f_1$ and $f_2$ which meet at a vertex.

So $\chi(M)=-2$ and $\chi(M^*)=0$. The surface $M^*$ is therefore a torus, and $M$ itself is a torus with two holes punched out of it.

  • $\begingroup$ Oh awesome, that makes a lot of sense thank you! I'm guessing this works in general to deal with any polygonal representation of a surface with boundary? i.e. identify orientability, determine the (correct) Euler characteristic, then remove the appropriate number of disks? $\endgroup$
    – Evan
    Nov 24, 2018 at 20:04
  • $\begingroup$ That's right, that's pretty much what the classification theorem for compact connected surfaces says. $\endgroup$
    – Lee Mosher
    Nov 24, 2018 at 20:44

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