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Feb 7, 2014 at 6:21 comment added Neil Hoffman @Gian Actually, $T^2/(\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z})$ could be a compact orbifold. In fact, the torus covers itself via such deck transformations. My answer above gives all of the possible orbifolds of the form you wanted together with a realization. However, other realizations exist - $S^2(2,2,2,2)$ can cover itself via an order 2 covering map, and so it can also be realized as $T^2/(\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z})$.
Feb 5, 2014 at 20:04 history edited Neil Hoffman CC BY-SA 3.0
fixed a small typo in the quotient group of RP^2(2,2)
Feb 5, 2014 at 18:21 vote accept Gian
Feb 5, 2014 at 18:20 comment added Gian Very nice answer, thanks a lot! Just a very brief stupid question: what if I consider $T^2/Z/(3Z \times 3Z)$ or $T^2/Z/(4Z \times 4Z)$? They are not compact orbifolds, right? But in this case, do they have any mathematical meaning?
Feb 3, 2014 at 23:33 history answered Neil Hoffman CC BY-SA 3.0