The developing map of conformally flat manifold
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There is one sentence I don't understand in some paper.
"A simply connected and conformally flat three mainifold can be conformally immersed into $S^3$" by the means of a developing map.
Is any reference about this short argument? Maybe it is a direct consequence from definition. Could anyone explain a little bit to me?
dg.differential-geometry riemannian-geometry
asked Nov 28 at 5:04
H-H
1485
add a comment |
up vote
7
down vote
favorite
There is one sentence I don't understand in some paper.
"A simply connected and conformally flat three mainifold can be conformally immersed into $S^3$" by the means of a developing map.
Is any reference about this short argument? Maybe it is a direct consequence from definition. Could anyone explain a little bit to me?
dg.differential-geometry riemannian-geometry
asked Nov 28 at 5:04
H-H
1485
1
The general story is explained in Sharpe's book Differential Geometry: Cartan's Generalization of Klein's Erlangen Programme, but the argument is precisely Petrunin's below.
– Ben McKay
Nov 28 at 17:00
add a comment |
up vote
7
down vote
favorite
up vote
7
down vote
favorite
There is one sentence I don't understand in some paper.
"A simply connected and conformally flat three mainifold can be conformally immersed into $S^3$" by the means of a developing map.
Is any reference about this short argument? Maybe it is a direct consequence from definition. Could anyone explain a little bit to me?
dg.differential-geometry riemannian-geometry
asked Nov 28 at 5:04
H-H
1485
There is one sentence I don't understand in some paper.
"A simply connected and conformally flat three mainifold can be conformally immersed into $S^3$" by the means of a developing map.
Is any reference about this short argument? Maybe it is a direct consequence from definition. Could anyone explain a little bit to me?
dg.differential-geometry riemannian-geometry
dg.differential-geometry riemannian-geometry
asked Nov 28 at 5:04
H-H
1485
asked Nov 28 at 5:04
H-H
1485
asked Nov 28 at 5:04
H-H
1485
asked Nov 28 at 5:04
H-H
1485
asked Nov 28 at 5:04
H-H
1485
1485
1
The general story is explained in Sharpe's book Differential Geometry: Cartan's Generalization of Klein's Erlangen Programme, but the argument is precisely Petrunin's below.
– Ben McKay
Nov 28 at 17:00
add a comment |
1
The general story is explained in Sharpe's book Differential Geometry: Cartan's Generalization of Klein's Erlangen Programme, but the argument is precisely Petrunin's below.
– Ben McKay
Nov 28 at 17:00
1
1
The general story is explained in Sharpe's book Differential Geometry: Cartan's Generalization of Klein's Erlangen Programme, but the argument is precisely Petrunin's below.
– Ben McKay
Nov 28 at 17:00
The general story is explained in Sharpe's book Differential Geometry: Cartan's Generalization of Klein's Erlangen Programme, but the argument is precisely Petrunin's below.
– Ben McKay
Nov 28 at 17:00
add a comment |
1 Answer
1
active
oldest
votes
up vote
9
down vote
accepted
Since $mathbb{S}^3$ is conformally flat, we can think that any point of our manifold $M$ admits a neighborhood that is conformally equivalent to an open set in $mathbb{S}^3$.
If two such neighbohoods overlap then the corresponding gluing map between corresponding open sets in $mathbb{S}^3$ is a composition of inversions (Liouville's theorem).
So after applying a composition of inversions to one of the open sets you can assume that the gluing map is identity.
This way you can extend the parametrization of a neighborhood to an immersed neighborhood of any path.
The composition of inversions at a neighborhhod of the end point of path depends continuously on the path and therefore has to be the same for homotopic paths.
Since $M$ is simply connected it gives a well defined conformal immersion $Mhookrightarrow mathbb{S}^3$.
answered Nov 28 at 5:19
Anton Petrunin
26.4k578196
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
accepted
Since $mathbb{S}^3$ is conformally flat, we can think that any point of our manifold $M$ admits a neighborhood that is conformally equivalent to an open set in $mathbb{S}^3$.
If two such neighbohoods overlap then the corresponding gluing map between corresponding open sets in $mathbb{S}^3$ is a composition of inversions (Liouville's theorem).
So after applying a composition of inversions to one of the open sets you can assume that the gluing map is identity.
This way you can extend the parametrization of a neighborhood to an immersed neighborhood of any path.
The composition of inversions at a neighborhhod of the end point of path depends continuously on the path and therefore has to be the same for homotopic paths.
Since $M$ is simply connected it gives a well defined conformal immersion $Mhookrightarrow mathbb{S}^3$.
answered Nov 28 at 5:19
Anton Petrunin
26.4k578196
add a comment |
up vote
9
down vote
accepted
Since $mathbb{S}^3$ is conformally flat, we can think that any point of our manifold $M$ admits a neighborhood that is conformally equivalent to an open set in $mathbb{S}^3$.
If two such neighbohoods overlap then the corresponding gluing map between corresponding open sets in $mathbb{S}^3$ is a composition of inversions (Liouville's theorem).
So after applying a composition of inversions to one of the open sets you can assume that the gluing map is identity.
This way you can extend the parametrization of a neighborhood to an immersed neighborhood of any path.
The composition of inversions at a neighborhhod of the end point of path depends continuously on the path and therefore has to be the same for homotopic paths.
Since $M$ is simply connected it gives a well defined conformal immersion $Mhookrightarrow mathbb{S}^3$.
answered Nov 28 at 5:19
Anton Petrunin
26.4k578196
add a comment |
up vote
9
down vote
accepted
up vote
9
down vote
accepted
Since $mathbb{S}^3$ is conformally flat, we can think that any point of our manifold $M$ admits a neighborhood that is conformally equivalent to an open set in $mathbb{S}^3$.
If two such neighbohoods overlap then the corresponding gluing map between corresponding open sets in $mathbb{S}^3$ is a composition of inversions (Liouville's theorem).
So after applying a composition of inversions to one of the open sets you can assume that the gluing map is identity.
This way you can extend the parametrization of a neighborhood to an immersed neighborhood of any path.
The composition of inversions at a neighborhhod of the end point of path depends continuously on the path and therefore has to be the same for homotopic paths.
Since $M$ is simply connected it gives a well defined conformal immersion $Mhookrightarrow mathbb{S}^3$.
answered Nov 28 at 5:19
Anton Petrunin
26.4k578196
Since $mathbb{S}^3$ is conformally flat, we can think that any point of our manifold $M$ admits a neighborhood that is conformally equivalent to an open set in $mathbb{S}^3$.
If two such neighbohoods overlap then the corresponding gluing map between corresponding open sets in $mathbb{S}^3$ is a composition of inversions (Liouville's theorem).
So after applying a composition of inversions to one of the open sets you can assume that the gluing map is identity.
This way you can extend the parametrization of a neighborhood to an immersed neighborhood of any path.
The composition of inversions at a neighborhhod of the end point of path depends continuously on the path and therefore has to be the same for homotopic paths.
Since $M$ is simply connected it gives a well defined conformal immersion $Mhookrightarrow mathbb{S}^3$.
answered Nov 28 at 5:19
Anton Petrunin
26.4k578196
edited Nov 28 at 5:29
answered Nov 28 at 5:19
Anton Petrunin
26.4k578196
answered Nov 28 at 5:19
Anton Petrunin
26.4k578196
answered Nov 28 at 5:19
Anton Petrunin
26.4k578196
26.4k578196
add a comment |
add a comment |
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The general story is explained in Sharpe's book Differential Geometry: Cartan's Generalization of Klein's Erlangen Programme, but the argument is precisely Petrunin's below.
– Ben McKay
Nov 28 at 17:00