# Topological invariants of toroidal orbifolds

Which are the most powerful topological invariants of toroidal orbifolds?

In particular I am looking for topological invariants of two-dimensional toroidal orbifolds such as $T^{2}/Z_{k}\times Z_{k}$ and $T^{2}/Z_{k}$.

I'm wondering if the corresponding orbifold Euler characteristics are zero or not and how I can calculate them.

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The seven orbifolds that are quotients of the torus you are interested in are: $T^2$, the Klein bottle (which can be realized as $T^2/\mathbb{Z}/2\mathbb{Z}$), $S^2(2,2,2,2)$ (which can be also realized as $T^2/\mathbb{Z}/2\mathbb{Z}$ however the group does not act freely in this case), $S^2(2,3,6)$ (which can be realized as $T^2/\mathbb{Z}/6\mathbb{Z}$), $S^2(3,3,3)$ (which can be realized as $T^2/\mathbb{Z}/3\mathbb{Z}$), $S^2(2,4,4)$ (which can be realized as $T^2/\mathbb{Z}/4\mathbb{Z}$), and $RP^2(2,2)$ (which can be realized as $T^2/(\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z})$). The other 3 compact Euclidean 2 orbifolds are the quotients of the plane by the Euclidean triangle groups. In these cases, the group acting on the torus is dihedral.
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? – Gian Feb 5 '14 at 18:20
@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 $T^2/(\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z})$. – Neil Hoffman Feb 7 '14 at 6:21
Orbifold Euler characteristic satisfies the same formula with respect to orbifold coverings that ordinary Euler characteristic satisfies for ordinary coverings: the Euler characteristic of the cover equals the covering degree times the Euler characteristic of the base. Since $T^2$ has Euler characteristic zero, its quotient orbifolds by finite group actions must also have orbifold Euler characteristic zero. Neil Hoffman's answer enumerates them all.