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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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2 Answers 2

up vote 6 down vote accepted

There are 10 orbifolds that can be covered by the torus i.e. 10 compact Euclidean 2 orbifolds. However, only 7 of them are quotients of the torus by a cyclic group or Abelian product of cyclic groups. The intuition that Euler characteristics are zero is correct. Formulas for the orbifold Euler characteristic appear throughout the literature. I like Chapter 13 of Thurston's notes in terms of a reference. Specifically, it has a table of all of the 2-orbifolds with non-negative Euler characteristic, which is helpful in this context. Genevieve Walsh's survey Orbifolds and Commensurability is also quite relevant.

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.

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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 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 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.

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