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 Feb 9 at 16:26
Misha VerbitskyMisha Verbitsky
5,20611936
$endgroup$
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 Feb 9 at 16:26
Misha VerbitskyMisha Verbitsky
5,20611936
$endgroup$
$begingroup$
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
Feb 9 at 17:59
$begingroup$
does it have the result stated for orbifolds?
$endgroup$
– Misha Verbitsky
Feb 9 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
Feb 9 at 18:07
4
$begingroup$
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
Feb 9 at 19:08
$begingroup$
thanks, I would look in this book
$endgroup$
– Misha Verbitsky
Feb 10 at 19:05
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 Feb 9 at 16:26
Misha VerbitskyMisha Verbitsky
5,20611936
$endgroup$
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 Feb 9 at 16:26
Misha VerbitskyMisha Verbitsky
5,20611936
asked Feb 9 at 16:26
Misha VerbitskyMisha Verbitsky
5,20611936
edited Feb 9 at 16:37
Misha Verbitsky
asked Feb 9 at 16:26
Misha VerbitskyMisha Verbitsky
5,20611936
asked Feb 9 at 16:26
Misha VerbitskyMisha Verbitsky
5,20611936
asked Feb 9 at 16:26
Misha VerbitskyMisha Verbitsky
5,20611936
5,20611936
$begingroup$
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
Feb 9 at 17:59
$begingroup$
does it have the result stated for orbifolds?
$endgroup$
– Misha Verbitsky
Feb 9 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
Feb 9 at 18:07
4
$begingroup$
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
Feb 9 at 19:08
$begingroup$
thanks, I would look in this book
$endgroup$
– Misha Verbitsky
Feb 10 at 19:05
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
Feb 9 at 17:59
$begingroup$
does it have the result stated for orbifolds?
$endgroup$
– Misha Verbitsky
Feb 9 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
Feb 9 at 18:07
4
$begingroup$
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
Feb 9 at 19:08
$begingroup$
thanks, I would look in this book
$endgroup$
– Misha Verbitsky
Feb 10 at 19:05
$begingroup$
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
Feb 9 at 17:59
$begingroup$
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
Feb 9 at 17:59
$begingroup$
does it have the result stated for orbifolds?
$endgroup$
– Misha Verbitsky
Feb 9 at 18:05
$begingroup$
does it have the result stated for orbifolds?
$endgroup$
– Misha Verbitsky
Feb 9 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
Feb 9 at 18:07
$begingroup$
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
Feb 9 at 18:07
4
4
$begingroup$
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
Feb 9 at 19:08
$begingroup$
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
Feb 9 at 19:08
$begingroup$
thanks, I would look in this book
$endgroup$
– Misha Verbitsky
Feb 10 at 19:05
$begingroup$
thanks, I would look in this book
$endgroup$
– Misha Verbitsky
Feb 10 at 19:05
add a comment |
2 Answers
2
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 Feb 9 at 17:01
ThiKuThiKu
6,36012137
$endgroup$
$begingroup$
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
Feb 9 at 18:04
2
$begingroup$
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
Feb 9 at 19:13
2
$begingroup$
However, if an orbifold has a universal covering, then the standard construction of a developing map works.
$endgroup$
– ThiKu
Feb 9 at 19:15
3
$begingroup$
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
Feb 9 at 19:23
add a comment |
$begingroup$
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.
answered Feb 25 at 3:36
Renato G. BettiolRenato G. Bettiol
3,9161642
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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 Feb 9 at 17:01
ThiKuThiKu
6,36012137
$endgroup$
$begingroup$
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
Feb 9 at 18:04
2
$begingroup$
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
Feb 9 at 19:13
2
$begingroup$
However, if an orbifold has a universal covering, then the standard construction of a developing map works.
$endgroup$
– ThiKu
Feb 9 at 19:15
3
$begingroup$
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
Feb 9 at 19:23
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 Feb 9 at 17:01
ThiKuThiKu
6,36012137
$endgroup$
$begingroup$
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
Feb 9 at 18:04
2
$begingroup$
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
Feb 9 at 19:13
2
$begingroup$
However, if an orbifold has a universal covering, then the standard construction of a developing map works.
$endgroup$
– ThiKu
Feb 9 at 19:15
3
$begingroup$
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
Feb 9 at 19:23
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 Feb 9 at 17:01
ThiKuThiKu
6,36012137
$endgroup$
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 Feb 9 at 17:01
ThiKuThiKu
6,36012137
answered Feb 9 at 17:01
ThiKuThiKu
6,36012137
answered Feb 9 at 17:01
ThiKuThiKu
6,36012137
answered Feb 9 at 17:01
ThiKuThiKu
6,36012137
6,36012137
$begingroup$
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
Feb 9 at 18:04
2
$begingroup$
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
Feb 9 at 19:13
2
$begingroup$
However, if an orbifold has a universal covering, then the standard construction of a developing map works.
$endgroup$
– ThiKu
Feb 9 at 19:15
3
$begingroup$
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
Feb 9 at 19:23
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
Feb 9 at 18:04
2
$begingroup$
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
Feb 9 at 19:13
2
$begingroup$
However, if an orbifold has a universal covering, then the standard construction of a developing map works.
$endgroup$
– ThiKu
Feb 9 at 19:15
3
$begingroup$
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
Feb 9 at 19:23
$begingroup$
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
Feb 9 at 18:04
$begingroup$
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
Feb 9 at 18:04
2
2
$begingroup$
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
Feb 9 at 19:13
$begingroup$
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
Feb 9 at 19:13
2
2
$begingroup$
However, if an orbifold has a universal covering, then the standard construction of a developing map works.
$endgroup$
– ThiKu
Feb 9 at 19:15
$begingroup$
However, if an orbifold has a universal covering, then the standard construction of a developing map works.
$endgroup$
– ThiKu
Feb 9 at 19:15
3
3
$begingroup$
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
Feb 9 at 19:23
$begingroup$
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
Feb 9 at 19:23
add a comment |
$begingroup$
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.
answered Feb 25 at 3:36
Renato G. BettiolRenato G. Bettiol
3,9161642
$endgroup$
add a comment |
$begingroup$
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.
answered Feb 25 at 3:36
Renato G. BettiolRenato G. Bettiol
3,9161642
$endgroup$
add a comment |
$begingroup$
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.
answered Feb 25 at 3:36
Renato G. BettiolRenato G. Bettiol
3,9161642
$endgroup$
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.
answered Feb 25 at 3:36
Renato G. BettiolRenato G. Bettiol
3,9161642
answered Feb 25 at 3:36
Renato G. BettiolRenato G. Bettiol
3,9161642
answered Feb 25 at 3:36
Renato G. BettiolRenato G. Bettiol
3,9161642
answered Feb 25 at 3:36
Renato G. BettiolRenato G. Bettiol
3,9161642
3,9161642
add a comment |
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
Feb 9 at 17:59
$begingroup$
does it have the result stated for orbifolds?
$endgroup$
– Misha Verbitsky
Feb 9 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
Feb 9 at 18:07
4
$begingroup$
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
Feb 9 at 19:08
$begingroup$
thanks, I would look in this book
$endgroup$
– Misha Verbitsky
Feb 10 at 19:05