Are manifolds admitting a circle foliation covered by manifolds with a (non-trivial) circle action?An example of a complex manifold without a finite open coverDoes there exist smooth circle action on manifolds M^4n with exactly three fixed points such that nneq 1a question about the isotropy subgroup of circle action on manifolds with isolated fixed pointNontrivial examples of non-trivial principal circle bundlesManifolds with prescribed fundamental group and finitely many trivial homotopy groupsCompact Lie group action on non-Hausdorff (but CGWH) space with Hausdorff quotient6-manifolds admitting SO(3) action with 2 orbit typesObstruction to a general S^1-action
Are manifolds admitting a circle foliation covered by manifolds with a (non-trivial) circle action?
An example of a complex manifold without a finite open coverDoes there exist smooth circle action on manifolds M^4n with exactly three fixed points such that nneq 1a question about the isotropy subgroup of circle action on manifolds with isolated fixed pointNontrivial examples of non-trivial principal circle bundlesManifolds with prescribed fundamental group and finitely many trivial homotopy groupsCompact Lie group action on non-Hausdorff (but CGWH) space with Hausdorff quotient6-manifolds admitting SO(3) action with 2 orbit typesObstruction to a general S^1-action
$begingroup$
More precisely, is there a criterion that decides the above question?
I am particularly interested in the smooth setting: is a smooth manifold with a smooth regular foliation by circles covered by a smooth manifold with a smooth non-trivial circle action?
I am mostly interested in finite covers.
dg.differential-geometry at.algebraic-topology
asked 11 hours ago
Caterina C.Caterina C.
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add a comment |
$begingroup$
More precisely, is there a criterion that decides the above question?
I am particularly interested in the smooth setting: is a smooth manifold with a smooth regular foliation by circles covered by a smooth manifold with a smooth non-trivial circle action?
I am mostly interested in finite covers.
dg.differential-geometry at.algebraic-topology
asked 11 hours ago
Caterina C.Caterina C.
412 bronze badges
New contributor
Caterina C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
3
$begingroup$
Do you want the action to be free?
$endgroup$
– Thomas Rot
10 hours ago
1
$begingroup$
Maybe look at simply connected $4$-manifolds, by a result of Fintushel if there is circle action their topology is very restricted (just connect sums of $ mathbbCP^2$, $ barmathbbCP^2$,$S^2 times S^2$ and $S^4$). It could be that circle foliations aren't as retricted- I don't know.
$endgroup$
– Nick L
10 hours ago
1
$begingroup$
Do you want the leaves of the circle action to project to the leaves of the initial foliation by circles ?
$endgroup$
– BS.
8 hours ago
1
$begingroup$
The action does not necessarily need to be free. Also the orbits of the circle action do not need to be the lifted leaves of the original foliation. Thanks a lot for your answers till now!
$endgroup$
– Caterina C.
4 hours ago
add a comment |
$begingroup$
More precisely, is there a criterion that decides the above question?
I am particularly interested in the smooth setting: is a smooth manifold with a smooth regular foliation by circles covered by a smooth manifold with a smooth non-trivial circle action?
I am mostly interested in finite covers.
dg.differential-geometry at.algebraic-topology
asked 11 hours ago
Caterina C.Caterina C.
412 bronze badges
New contributor
Caterina C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
More precisely, is there a criterion that decides the above question?
I am particularly interested in the smooth setting: is a smooth manifold with a smooth regular foliation by circles covered by a smooth manifold with a smooth non-trivial circle action?
I am mostly interested in finite covers.
dg.differential-geometry at.algebraic-topology
dg.differential-geometry at.algebraic-topology
asked 11 hours ago
Caterina C.Caterina C.
412 bronze badges
New contributor
Caterina C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 11 hours ago
Caterina C.Caterina C.
412 bronze badges
New contributor
Caterina C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 11 hours ago
Caterina C.Caterina C.
412 bronze badges
New contributor
Caterina C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 11 hours ago
Caterina C.Caterina C.
412 bronze badges
asked 11 hours ago
Caterina C.Caterina C.
412 bronze badges
412 bronze badges
New contributor
Caterina C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Caterina C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3
$begingroup$
Do you want the action to be free?
$endgroup$
– Thomas Rot
10 hours ago
1
$begingroup$
Maybe look at simply connected $4$-manifolds, by a result of Fintushel if there is circle action their topology is very restricted (just connect sums of $ mathbbCP^2$, $ barmathbbCP^2$,$S^2 times S^2$ and $S^4$). It could be that circle foliations aren't as retricted- I don't know.
$endgroup$
– Nick L
10 hours ago
1
$begingroup$
Do you want the leaves of the circle action to project to the leaves of the initial foliation by circles ?
$endgroup$
– BS.
8 hours ago
1
$begingroup$
The action does not necessarily need to be free. Also the orbits of the circle action do not need to be the lifted leaves of the original foliation. Thanks a lot for your answers till now!
$endgroup$
– Caterina C.
4 hours ago
add a comment |
3
$begingroup$
Do you want the action to be free?
$endgroup$
– Thomas Rot
10 hours ago
1
$begingroup$
Maybe look at simply connected $4$-manifolds, by a result of Fintushel if there is circle action their topology is very restricted (just connect sums of $ mathbbCP^2$, $ barmathbbCP^2$,$S^2 times S^2$ and $S^4$). It could be that circle foliations aren't as retricted- I don't know.
$endgroup$
– Nick L
10 hours ago
1
$begingroup$
Do you want the leaves of the circle action to project to the leaves of the initial foliation by circles ?
$endgroup$
– BS.
8 hours ago
1
$begingroup$
The action does not necessarily need to be free. Also the orbits of the circle action do not need to be the lifted leaves of the original foliation. Thanks a lot for your answers till now!
$endgroup$
– Caterina C.
4 hours ago
3
3
$begingroup$
Do you want the action to be free?
$endgroup$
– Thomas Rot
10 hours ago
$begingroup$
Do you want the action to be free?
$endgroup$
– Thomas Rot
10 hours ago
1
1
$begingroup$
Maybe look at simply connected $4$-manifolds, by a result of Fintushel if there is circle action their topology is very restricted (just connect sums of $ mathbbCP^2$, $ barmathbbCP^2$,$S^2 times S^2$ and $S^4$). It could be that circle foliations aren't as retricted- I don't know.
$endgroup$
– Nick L
10 hours ago
$begingroup$
Maybe look at simply connected $4$-manifolds, by a result of Fintushel if there is circle action their topology is very restricted (just connect sums of $ mathbbCP^2$, $ barmathbbCP^2$,$S^2 times S^2$ and $S^4$). It could be that circle foliations aren't as retricted- I don't know.
$endgroup$
– Nick L
10 hours ago
1
1
$begingroup$
Do you want the leaves of the circle action to project to the leaves of the initial foliation by circles ?
$endgroup$
– BS.
8 hours ago
$begingroup$
Do you want the leaves of the circle action to project to the leaves of the initial foliation by circles ?
$endgroup$
– BS.
8 hours ago
1
1
$begingroup$
The action does not necessarily need to be free. Also the orbits of the circle action do not need to be the lifted leaves of the original foliation. Thanks a lot for your answers till now!
$endgroup$
– Caterina C.
4 hours ago
$begingroup$
The action does not necessarily need to be free. Also the orbits of the circle action do not need to be the lifted leaves of the original foliation. Thanks a lot for your answers till now!
$endgroup$
– Caterina C.
4 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Dennis Sullivan constructed a non-vanishing Lispschitz vector field on a closed $5$-manifold such that all orbits are periodic but they amazingly have unbounded lengths (!). An addendum by Nicholas Kuiper shows that the flow can be chosen smooth, and another by William Thurston gives another construction which is real analytic.
answered 9 hours ago
BS.BS.
8,1992 gold badges30 silver badges45 bronze badges
$endgroup$
2
$begingroup$
It is not clear to me how this answers the question. Why can't this manifold be covered by a manifold with a circle action?
$endgroup$
– Nick L
9 hours ago
1
$begingroup$
@Nick L : I assumed (perhaps wrongly) that the OP asked for the circle action to have orbits the leaves of the lifted foliation.
$endgroup$
– BS.
8 hours ago
add a comment |
$begingroup$
For circle foliations of compact $3$-manifolds, this is essentially answered by a theorem of Epstein: every such foliation is a Seifert fibration. Most Seifert fibrations are finitely covered by a product (surface)x(circle), and in these cases one of course has a free circle action.
One of the exceptional cases of Seifert fibrations not covered by a product is the fibration of $S^3$ obtained as follows. Take the linear foliation of slope p/q on a solid torus (completed by the core of the solid torus as another leaf), the linear foliation of slope q/p on another solid torus (again with the core as another leaf), and glue the two solid tori along their boundaries by sending one meridian to the longitude to the other and one longitude to the meridian of the other. The result is $S^3$ with a circle foliation, which does not seem to come from a free circle action if p and q are bigger than 1.
answered 10 hours ago
ThiKuThiKu
6,7931 gold badge24 silver badges42 bronze badges
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Dennis Sullivan constructed a non-vanishing Lispschitz vector field on a closed $5$-manifold such that all orbits are periodic but they amazingly have unbounded lengths (!). An addendum by Nicholas Kuiper shows that the flow can be chosen smooth, and another by William Thurston gives another construction which is real analytic.
answered 9 hours ago
BS.BS.
8,1992 gold badges30 silver badges45 bronze badges
$endgroup$
2
$begingroup$
It is not clear to me how this answers the question. Why can't this manifold be covered by a manifold with a circle action?
$endgroup$
– Nick L
9 hours ago
1
$begingroup$
@Nick L : I assumed (perhaps wrongly) that the OP asked for the circle action to have orbits the leaves of the lifted foliation.
$endgroup$
– BS.
8 hours ago
add a comment |
$begingroup$
Dennis Sullivan constructed a non-vanishing Lispschitz vector field on a closed $5$-manifold such that all orbits are periodic but they amazingly have unbounded lengths (!). An addendum by Nicholas Kuiper shows that the flow can be chosen smooth, and another by William Thurston gives another construction which is real analytic.
answered 9 hours ago
BS.BS.
8,1992 gold badges30 silver badges45 bronze badges
$endgroup$
2
$begingroup$
It is not clear to me how this answers the question. Why can't this manifold be covered by a manifold with a circle action?
$endgroup$
– Nick L
9 hours ago
1
$begingroup$
@Nick L : I assumed (perhaps wrongly) that the OP asked for the circle action to have orbits the leaves of the lifted foliation.
$endgroup$
– BS.
8 hours ago
add a comment |
$begingroup$
Dennis Sullivan constructed a non-vanishing Lispschitz vector field on a closed $5$-manifold such that all orbits are periodic but they amazingly have unbounded lengths (!). An addendum by Nicholas Kuiper shows that the flow can be chosen smooth, and another by William Thurston gives another construction which is real analytic.
answered 9 hours ago
BS.BS.
8,1992 gold badges30 silver badges45 bronze badges
$endgroup$
Dennis Sullivan constructed a non-vanishing Lispschitz vector field on a closed $5$-manifold such that all orbits are periodic but they amazingly have unbounded lengths (!). An addendum by Nicholas Kuiper shows that the flow can be chosen smooth, and another by William Thurston gives another construction which is real analytic.
answered 9 hours ago
BS.BS.
8,1992 gold badges30 silver badges45 bronze badges
answered 9 hours ago
BS.BS.
8,1992 gold badges30 silver badges45 bronze badges
answered 9 hours ago
BS.BS.
8,1992 gold badges30 silver badges45 bronze badges
answered 9 hours ago
BS.BS.
8,1992 gold badges30 silver badges45 bronze badges
8,1992 gold badges30 silver badges45 bronze badges
2
$begingroup$
It is not clear to me how this answers the question. Why can't this manifold be covered by a manifold with a circle action?
$endgroup$
– Nick L
9 hours ago
1
$begingroup$
@Nick L : I assumed (perhaps wrongly) that the OP asked for the circle action to have orbits the leaves of the lifted foliation.
$endgroup$
– BS.
8 hours ago
add a comment |
2
$begingroup$
It is not clear to me how this answers the question. Why can't this manifold be covered by a manifold with a circle action?
$endgroup$
– Nick L
9 hours ago
1
$begingroup$
@Nick L : I assumed (perhaps wrongly) that the OP asked for the circle action to have orbits the leaves of the lifted foliation.
$endgroup$
– BS.
8 hours ago
2
2
$begingroup$
It is not clear to me how this answers the question. Why can't this manifold be covered by a manifold with a circle action?
$endgroup$
– Nick L
9 hours ago
$begingroup$
It is not clear to me how this answers the question. Why can't this manifold be covered by a manifold with a circle action?
$endgroup$
– Nick L
9 hours ago
1
1
$begingroup$
@Nick L : I assumed (perhaps wrongly) that the OP asked for the circle action to have orbits the leaves of the lifted foliation.
$endgroup$
– BS.
8 hours ago
$begingroup$
@Nick L : I assumed (perhaps wrongly) that the OP asked for the circle action to have orbits the leaves of the lifted foliation.
$endgroup$
– BS.
8 hours ago
add a comment |
$begingroup$
For circle foliations of compact $3$-manifolds, this is essentially answered by a theorem of Epstein: every such foliation is a Seifert fibration. Most Seifert fibrations are finitely covered by a product (surface)x(circle), and in these cases one of course has a free circle action.
One of the exceptional cases of Seifert fibrations not covered by a product is the fibration of $S^3$ obtained as follows. Take the linear foliation of slope p/q on a solid torus (completed by the core of the solid torus as another leaf), the linear foliation of slope q/p on another solid torus (again with the core as another leaf), and glue the two solid tori along their boundaries by sending one meridian to the longitude to the other and one longitude to the meridian of the other. The result is $S^3$ with a circle foliation, which does not seem to come from a free circle action if p and q are bigger than 1.
answered 10 hours ago
ThiKuThiKu
6,7931 gold badge24 silver badges42 bronze badges
$endgroup$
add a comment |
$begingroup$
For circle foliations of compact $3$-manifolds, this is essentially answered by a theorem of Epstein: every such foliation is a Seifert fibration. Most Seifert fibrations are finitely covered by a product (surface)x(circle), and in these cases one of course has a free circle action.
One of the exceptional cases of Seifert fibrations not covered by a product is the fibration of $S^3$ obtained as follows. Take the linear foliation of slope p/q on a solid torus (completed by the core of the solid torus as another leaf), the linear foliation of slope q/p on another solid torus (again with the core as another leaf), and glue the two solid tori along their boundaries by sending one meridian to the longitude to the other and one longitude to the meridian of the other. The result is $S^3$ with a circle foliation, which does not seem to come from a free circle action if p and q are bigger than 1.
answered 10 hours ago
ThiKuThiKu
6,7931 gold badge24 silver badges42 bronze badges
$endgroup$
add a comment |
$begingroup$
For circle foliations of compact $3$-manifolds, this is essentially answered by a theorem of Epstein: every such foliation is a Seifert fibration. Most Seifert fibrations are finitely covered by a product (surface)x(circle), and in these cases one of course has a free circle action.
One of the exceptional cases of Seifert fibrations not covered by a product is the fibration of $S^3$ obtained as follows. Take the linear foliation of slope p/q on a solid torus (completed by the core of the solid torus as another leaf), the linear foliation of slope q/p on another solid torus (again with the core as another leaf), and glue the two solid tori along their boundaries by sending one meridian to the longitude to the other and one longitude to the meridian of the other. The result is $S^3$ with a circle foliation, which does not seem to come from a free circle action if p and q are bigger than 1.
answered 10 hours ago
ThiKuThiKu
6,7931 gold badge24 silver badges42 bronze badges
$endgroup$
For circle foliations of compact $3$-manifolds, this is essentially answered by a theorem of Epstein: every such foliation is a Seifert fibration. Most Seifert fibrations are finitely covered by a product (surface)x(circle), and in these cases one of course has a free circle action.
One of the exceptional cases of Seifert fibrations not covered by a product is the fibration of $S^3$ obtained as follows. Take the linear foliation of slope p/q on a solid torus (completed by the core of the solid torus as another leaf), the linear foliation of slope q/p on another solid torus (again with the core as another leaf), and glue the two solid tori along their boundaries by sending one meridian to the longitude to the other and one longitude to the meridian of the other. The result is $S^3$ with a circle foliation, which does not seem to come from a free circle action if p and q are bigger than 1.
answered 10 hours ago
ThiKuThiKu
6,7931 gold badge24 silver badges42 bronze badges
edited 7 hours ago
answered 10 hours ago
ThiKuThiKu
6,7931 gold badge24 silver badges42 bronze badges
answered 10 hours ago
ThiKuThiKu
6,7931 gold badge24 silver badges42 bronze badges
answered 10 hours ago
ThiKuThiKu
6,7931 gold badge24 silver badges42 bronze badges
6,7931 gold badge24 silver badges42 bronze badges
add a comment |
add a comment |
Caterina C. is a new contributor. Be nice, and check out our Code of Conduct.
Caterina C. is a new contributor. Be nice, and check out our Code of Conduct.
Caterina C. is a new contributor. Be nice, and check out our Code of Conduct.
Caterina C. is a new contributor. Be nice, and check out our Code of Conduct.
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3
$begingroup$
Do you want the action to be free?
$endgroup$
– Thomas Rot
10 hours ago
1
$begingroup$
Maybe look at simply connected $4$-manifolds, by a result of Fintushel if there is circle action their topology is very restricted (just connect sums of $ mathbbCP^2$, $ barmathbbCP^2$,$S^2 times S^2$ and $S^4$). It could be that circle foliations aren't as retricted- I don't know.
$endgroup$
– Nick L
10 hours ago
1
$begingroup$
Do you want the leaves of the circle action to project to the leaves of the initial foliation by circles ?
$endgroup$
– BS.
8 hours ago
1
$begingroup$
The action does not necessarily need to be free. Also the orbits of the circle action do not need to be the lifted leaves of the original foliation. Thanks a lot for your answers till now!
$endgroup$
– Caterina C.
4 hours ago