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If we take a square and identify opposite sides, we get a torus. If we change the direction of one pair of sides we get the Klein bottle. If we change the direction of both sides, we get first a Möbius strip, then when I tried to glue the opposite sides I got a sphere with a singularity (ie the band is twisted into a point), is this process allowed, ie is it a topological sphere, is it possible to glue this together without a singularity if I had some extra spatial dimensions?

And given an $2n$-gon of sides $a_1$ to $a_{2n}$, which identifications of sides (pairwise and with $4$ possible directions), result in a smooth compact manifold, is there some rule to determine which manifold, and are all identifications allowed? How many diemsnions are needed for a smooth gluing process?

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The first shape you describe is actually the real projective plane. I'm not sure about your second question, though. might give you more info. – Lopsy Jan 27 '12 at 21:23
up vote 3 down vote accepted

All gluings result in a compact surface. It's obvious that every point in the interior of the polygon is locally homeomorphic to $\mathbb{R}^2$, and it's not so hard to see that this is also true of the interior of the edges, so it only remains to inspect the vertices, and this isn't so bad either: by "following the gluing" clockwise or counterclockwise you eventually produce a cycle that closes up a neighborhood of each vertex. This is easier for me to see geometrically than to spell out rigorously.

I know of two straightforward algorithms to determine the homeomorphism type of the corresponding surface. The first uses Mayer-Vietoris to compute the first homology $H_1(M, \mathbb{Z})$, which it's known completely determines the surface. The second is combinatorial and takes some diagrams to explain properly, but is totally elementary; it should be explained anywhere the classification of (compact) surfaces is proven in the usual way.

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Actually orientability and Euler characteristic are a complete set of invariants of compact surfaces. (You don't need to know homology.) – Grumpy Parsnip Jan 27 '12 at 23:34
@Jim: sure, but computing orientability and Euler characteristic is about as easy as computing the first homology. – Qiaochu Yuan Jan 27 '12 at 23:40
I know, but you can compute Euler characteristic and orientability without any knowledge of homology. I'm not claiming this is a deep comment. – Grumpy Parsnip Jan 28 '12 at 0:06

If you glue both pairs of sides in a square after switching the orientation of one side in each pair, so that the glueing instructions are

enter image description here

we see that in fact you started with a $2$-disk and identified each pair of antipodal points in the boundary. That's one of the standard constructions of the projective pane.

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Let's work by cut and paste. Suppose that the square is $[0,1]\times[0,1]$ with horizontal edges labeled $a$ and vertical edges labeled $b$. Orient the edges by the word $baba$ traversed counter-clockwise from the origin. Cut the square along $c$, the line $x=y$, oriented by an arrow pointing out from the origin. We now have two triangles with edges labeled by the words $c^{-1}ba$ and $bac$. Glue the $a$'s together. (You have to reflect one triangle.) We get a 4-gon with labels $bb^{-1}cc$. Now glue the $b$'s together to obtain a 2-gon with labels $cc$. That's $\mathbb{R}P^2$.

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