Bieberbach theorem for compact, flat Riemannian orbifolds
$begingroup$
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.
dg.differential-geometry gr.group-theory riemannian-geometry geometric-group-theory affine-geometry
asked 9 hours ago
Misha VerbitskyMisha Verbitsky
5,15111936
$endgroup$
add a comment |
$begingroup$
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.
dg.differential-geometry gr.group-theory riemannian-geometry geometric-group-theory affine-geometry
asked 9 hours ago
Misha VerbitskyMisha Verbitsky
5,15111936
$endgroup$
$begingroup$
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
8 hours ago
$begingroup$
does it have the result stated for orbifolds?
$endgroup$
– Misha Verbitsky
7 hours ago
$begingroup$
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
7 hours ago
1
$begingroup$
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
6 hours ago
add a comment |
$begingroup$
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.
dg.differential-geometry gr.group-theory riemannian-geometry geometric-group-theory affine-geometry
asked 9 hours ago
Misha VerbitskyMisha Verbitsky
5,15111936
$endgroup$
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.
dg.differential-geometry gr.group-theory riemannian-geometry geometric-group-theory affine-geometry
dg.differential-geometry gr.group-theory riemannian-geometry geometric-group-theory affine-geometry
asked 9 hours ago
Misha VerbitskyMisha Verbitsky
5,15111936
asked 9 hours ago
Misha VerbitskyMisha Verbitsky
5,15111936
edited 9 hours ago
Misha Verbitsky
asked 9 hours ago
Misha VerbitskyMisha Verbitsky
5,15111936
asked 9 hours ago
Misha VerbitskyMisha Verbitsky
5,15111936
asked 9 hours ago
Misha VerbitskyMisha Verbitsky
5,15111936
5,15111936
$begingroup$
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
8 hours ago
$begingroup$
does it have the result stated for orbifolds?
$endgroup$
– Misha Verbitsky
7 hours ago
$begingroup$
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
7 hours ago
1
$begingroup$
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
6 hours ago
add a comment |
$begingroup$
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
8 hours ago
$begingroup$
does it have the result stated for orbifolds?
$endgroup$
– Misha Verbitsky
7 hours ago
$begingroup$
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
7 hours ago
1
$begingroup$
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
6 hours ago
$begingroup$
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
8 hours ago
$begingroup$
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
8 hours ago
$begingroup$
does it have the result stated for orbifolds?
$endgroup$
– Misha Verbitsky
7 hours ago
$begingroup$
does it have the result stated for orbifolds?
$endgroup$
– Misha Verbitsky
7 hours ago
$begingroup$
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
7 hours ago
$begingroup$
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
7 hours ago
1
1
$begingroup$
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
6 hours ago
$begingroup$
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
6 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered 9 hours ago
ThiKuThiKu
6,07712036
$endgroup$
$begingroup$
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
7 hours ago
$begingroup$
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
6 hours ago
$begingroup$
However, if an orbifold has a universal covering, then the standard construction of a developing map works.
$endgroup$
– ThiKu
6 hours ago
$begingroup$
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
6 hours ago
add a comment |
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1 Answer
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votes
$begingroup$
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.
answered 9 hours ago
ThiKuThiKu
6,07712036
$endgroup$
$begingroup$
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
7 hours ago
$begingroup$
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
6 hours ago
$begingroup$
However, if an orbifold has a universal covering, then the standard construction of a developing map works.
$endgroup$
– ThiKu
6 hours ago
$begingroup$
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
6 hours ago
add a comment |
$begingroup$
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.
answered 9 hours ago
ThiKuThiKu
6,07712036
$endgroup$
$begingroup$
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
7 hours ago
$begingroup$
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
6 hours ago
$begingroup$
However, if an orbifold has a universal covering, then the standard construction of a developing map works.
$endgroup$
– ThiKu
6 hours ago
$begingroup$
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
6 hours ago
add a comment |
$begingroup$
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.
answered 9 hours ago
ThiKuThiKu
6,07712036
$endgroup$
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.
answered 9 hours ago
ThiKuThiKu
6,07712036
answered 9 hours ago
ThiKuThiKu
6,07712036
answered 9 hours ago
ThiKuThiKu
6,07712036
answered 9 hours ago
ThiKuThiKu
6,07712036
6,07712036
$begingroup$
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
7 hours ago
$begingroup$
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
6 hours ago
$begingroup$
However, if an orbifold has a universal covering, then the standard construction of a developing map works.
$endgroup$
– ThiKu
6 hours ago
$begingroup$
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
6 hours ago
add a comment |
$begingroup$
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
7 hours ago
$begingroup$
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
6 hours ago
$begingroup$
However, if an orbifold has a universal covering, then the standard construction of a developing map works.
$endgroup$
– ThiKu
6 hours ago
$begingroup$
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
6 hours ago
$begingroup$
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
7 hours ago
$begingroup$
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
7 hours ago
$begingroup$
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
6 hours ago
$begingroup$
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
6 hours ago
$begingroup$
However, if an orbifold has a universal covering, then the standard construction of a developing map works.
$endgroup$
– ThiKu
6 hours ago
$begingroup$
However, if an orbifold has a universal covering, then the standard construction of a developing map works.
$endgroup$
– ThiKu
6 hours ago
$begingroup$
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
6 hours ago
$begingroup$
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
6 hours ago
add a comment |
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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
8 hours ago
$begingroup$
does it have the result stated for orbifolds?
$endgroup$
– Misha Verbitsky
7 hours ago
$begingroup$
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
7 hours ago
1
$begingroup$
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
6 hours ago