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
5,16111936
$endgroup$
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
5,16111936
$endgroup$
$begingroup$
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
3
$begingroup$
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
5,16111936
$endgroup$
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
5,16111936
asked Feb 9 at 16:26
Misha VerbitskyMisha Verbitsky
5,16111936
edited Feb 9 at 16:37
Misha Verbitsky
asked Feb 9 at 16:26
Misha VerbitskyMisha Verbitsky
5,16111936
asked Feb 9 at 16:26
Misha VerbitskyMisha Verbitsky
5,16111936
asked Feb 9 at 16:26
Misha VerbitskyMisha Verbitsky
5,16111936
5,16111936
$begingroup$
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
3
$begingroup$
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
3
$begingroup$
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
3
3
$begingroup$
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 |
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 Feb 9 at 17:01
ThiKuThiKu
6,17712037
$endgroup$
$begingroup$
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
2
$begingroup$
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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "504"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f322835%2fbieberbach-theorem-for-compact-flat-riemannian-orbifolds%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
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
6,17712037
$endgroup$
$begingroup$
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
2
$begingroup$
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
6,17712037
$endgroup$
$begingroup$
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
2
$begingroup$
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
6,17712037
$endgroup$
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
6,17712037
answered Feb 9 at 17:01
ThiKuThiKu
6,17712037
answered Feb 9 at 17:01
ThiKuThiKu
6,17712037
answered Feb 9 at 17:01
ThiKuThiKu
6,17712037
6,17712037
$begingroup$
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
2
$begingroup$
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
2
$begingroup$
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
2
2
$begingroup$
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 |
Thanks for contributing an answer to MathOverflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f322835%2fbieberbach-theorem-for-compact-flat-riemannian-orbifolds%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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
3
$begingroup$
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