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In his thesis, Bieberbach solved Hilbert 18 problem and proved that any compact, flat Riemannian manifold is a quotient of a torus. I need a reference to an orbifold version of this result: any compact, flat Riemannian manifold $M$ is a quotient of a torus.

It should not be hard to prove: we should take the development map and it should give a local isometry from the orbifold universal cover of $M$ to ${\Bbb R}^n$. The corresponding monodromy action defines a homomorphism from the orbifold fundamental group of $M$ to the group of affine isometries. The rotational part of its image is finite by Margulis lemma.

However, I am pretty sure it's published somewhere, and it's always safer (and more ethical) to cite.

Thanks in advance.

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  • $\begingroup$ If you need a textbook reference you could use "Bieberbach Groups and Flat Manifolds" by L. S. Charlap or "Spaces of constant curvature by J. A Wolf. $\endgroup$
    – Igor Belegradek
    Commented Feb 9, 2019 at 17:59
  • $\begingroup$ does it have the result stated for orbifolds? $\endgroup$
    – Misha Verbitsky
    Commented Feb 9, 2019 at 18:05
  • $\begingroup$ They don't use the word "orbifold". Everything is stated for discrete isometry groups of $\mathbb R^n$. Which is the same thing because flat orbifolds are good. $\endgroup$
    – Igor Belegradek
    Commented Feb 9, 2019 at 18:07
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    $\begingroup$ It seems you are unaware of the fact that complete nonpositively curved orbifolds are good (i.e., developable). This is due to Gromov (I think) and proved e.g. in Bridson-Haefliger "Metric spaces of nonpositive curvature". $\endgroup$
    – Igor Belegradek
    Commented Feb 9, 2019 at 19:08
  • $\begingroup$ thanks, I would look in this book $\endgroup$
    – Misha Verbitsky
    Commented Feb 10, 2019 at 19:05

2 Answers 2

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Bieberbach‘s 1911-12 paper (part 1, part2) proves a result about groups rather than manifolds, and it does not assume the groups to be torsion-free. In today’s language it says that a discrete, cocompact group of Euclidean isometries contains its subgroup of translations (which is necessarily a free Abelian group) as a subgroup of finite index. So you can just cite Bieberbach.

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  • $\begingroup$ I also need to prove that the development map is globally defined. For example, for an orbifold CP^1 with one conical point, there is no globally defined development map, because it is simply connected. $\endgroup$
    – Misha Verbitsky
    Commented Feb 9, 2019 at 18:04
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    $\begingroup$ This example is not flat. It is not a good orbifold (i.e., does not have a manifold cover) and in particular has no universal covering, on which the developing map could be globally defined. $\endgroup$
    – ThiKu
    Commented Feb 9, 2019 at 19:13
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    $\begingroup$ However, if an orbifold has a universal covering, then the standard construction of a developing map works. $\endgroup$
    – ThiKu
    Commented Feb 9, 2019 at 19:15
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    $\begingroup$ And a geometric orbifold (e.g., a flat orbifold) always has a manifold cover and hence a universal covering. This should be in Thurston‘s lecture notes. $\endgroup$
    – ThiKu
    Commented Feb 9, 2019 at 19:23
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Besides Charlap's book (already mentioned in comments above), I think the following reference can be very helpful in situating these concepts with more modern language, despite not specifically mentioning "orbifolds":

P. Buser, A geometric proof of Bieberbach’s theorems on crystallographic groups, Enseign. Math. (2), 31 (1985), 137–145.

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