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I am currently learning about polygonal presentations of surfaces.

In the notation I'm using (following Lee's "Topological Manifolds"), $\langle a, b \ |\ aba^{-1}b^{-1}\rangle$ is a presentation of the torus $\mathbb{T}^2$, and $\langle a,b\ |\ abab \rangle$ is a presentation of the real projective plane $\mathbb{P}^2$. Both of these examples can be thought of as specifying labellings and orientations of the edges of a square, which in turn specify how to glue the edges together to obtain the respective surfaces.

As a fun exercise for myself, I'm trying to list all possible topological spaces (surfaces?) that can result from gluing together the edges of a triangle. I conjecture that the following five four presentations represent all possible such spaces (up to homeomorphism), and also conjecture that they fall into the given homeomorphism classes: $$\langle a \ | \ aaa\rangle \approx \text{?}$$ $$\langle a \ | \ aaa^{-1}\rangle \approx \text{?}$$ $$\langle a, b \ | \ aab\rangle \approx \text{?}$$ $$\langle a, b, c \ | \ abc\rangle \approx \mathbb{D}^2 \text{ (closed disk)}$$

Questions: Are these, in fact, all of them (up to homeomorphism), or are there some that I've missed? Are any two on this list homeomorphic (meaning I've double-counted)? And are there any common descriptions of the homeomorphism classes with question marks? (I realize that "common descriptions" is vague.)

EDIT: By "five" I of course meant "four." That is, $$\langle a, b \ | \ aa^{-1}b\rangle \approx \mathbb{D}^2,$$ which is geometrically clear upon drawing the picture.

Note: These are in fact polygonal presentations, and not group presentations. Because we are dealing with triangles (which are in some sense degenerate), we cannot always read off the fundamental group directly from the polygonal presentation as if it were a group presentation. The example in the "EDIT" illustrates this.

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With respect to your homeomorphic question, if two spaces are homeomorphic then they have the same fundamental group. So, what are the groups you have here? You have 4 groups - two are isomorphic. Can you work out which two they are? Then work out if the spaces can be homeomorphic...(Also, I'm not sure if "gluing all the edges of a triangle together" makes sense...but I'm no topologist, so it probably does...) – user1729 Jul 21 '11 at 9:11
Oh, right, the cone and the disk. (Sorry, it's 3 AM here.) Thanks for that. :-) – Jesse Madnick Jul 21 '11 at 9:16
@Jesse Madnick: on a related note, I once had a triangle with these identificactions: let's call A,B,C the vertices starting from the bottom left hand one; then the edges AC and BC towards C were identified and the whole AB edge was identified with C. Maybe this is not something you're interested in because you can't express it in your notation, but maybe it is. :) – Andy Jul 21 '11 at 10:12
Your edit isn't what I meant! I was pointing out that if two spaces are homeomorphic then their fundamental groups are isomorphic. You have now edited it so you have $\pi_1(\mathbb{D}^2)=\mathbb{Z}=F_2$, which is a contradiction. Two of your groups are isomorphic to \mathbb{Z}, the other three are pairwise non-isomorphic ($C_3$ (cyclic of order 3), $F_2$ (free on two generators) and the trivial group...). – user1729 Jul 21 '11 at 11:35
Oh. I looked at Lee's book on Google books; apparently these are "polygonal presentations", and are not explicitly connected to groups. Yikes! I thought you were just being extremely loose with your words. – Dylan Moreland Jul 21 '11 at 20:00

$aab$ is a Möbius band. $aaa$ is not a surface, nor is $aaa^{-1}$, as with three things coming together you're not locally homeomorphic to a plane or half-plane.

EDIT: In the comments, I mention one can prove that $aab$ is a Möbius band by invoking the theorem on classification of surfaces (although I accidentally wrote characterization instead of classification). Another way to do it involves surgery. First note that $aab=aabc$. Then draw a line from the $aa$ corner to the $bc$ corner, and cut along that line to get two triangles, $adc$ and $abd^{-1}$. Now put the two triangles back together, but identifying the edges labeled $a$; you get $dcdb^{-1}$, which you recognize as a Möbius band.

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Thank you for this answer, but could you perhaps elaborate a bit -- either on why $aab$ is a Mobius band, or whether $aaa$ or $aaa^{-1}$ are spaces that appear in other contexts/guises? – Jesse Madnick Jul 21 '11 at 9:45
Hm, ok, well, I usually think of the Mobius band as $acab$, which you're claiming is the same as $aab$... But we can't just add sides at will in general, can we? Otherwise, $aa^{-1}b$ (the closed disk) would be the same as $aca^{-1}b$ (the cylinder). – Jesse Madnick Jul 21 '11 at 9:54
I would prove $aab$ is a Mobius band by showing it's non-orientable, has one boundary, and has the same Euler characteristic as a Mobius band, and then I'd appeal to the characterization theorem for surfaces - but you may not have seen all of that yet. As for $aaa$, think in 3-space about the xy-plane together with the upper half of the xz-plane. Sure it's a space of some kind, but it's not a manifold, not a surface, because of what's happening in a neighborhood of any point on the x-axis. I don't know if it has a name or any uses. It isn't $aaa$, but suffers the same objection. – Gerry Myerson Jul 21 '11 at 10:00
If you make $acab$ out of paper you see the $c$ and the $b$ come together to form a single edge; this doesn't happen with $aca^{-1}b$. – Gerry Myerson Jul 21 '11 at 10:02
@Andy, yes, thanks. Jesse, $aaa$ or $aaa^{-1}$ or any attempt to identify edges in groups of more than 2 is going to result in something that's not a surface. – Gerry Myerson Jul 21 '11 at 12:51

In regards to $aaa$ this has a somewhat common description as the pseudo projective plane of order $3$. In general the pseudo projective plane of order $n$ can be defined in a similar way by taking a regular $n$-gon with the presentation $a^n$. Also my topology professor referred to $aaa^{-1}$ as the dunce cap space.

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As for the dunce cap, – ttt Jul 21 '11 at 20:40

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