Example of compact Riemannian manifold with only one closed geodesic. Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Why are we interested in closed geodesics?Existence of geodesic on a compact Riemannian manifoldAre closed geodesics the prime numbers of Riemannian manifolds?Completeness of a Riemannian manifold with boundaryTotally geodesic hypersurface in compact hyperbolic manifoldExistence of closed, non self-intersecting geodesics on compact manifoldsExample for conjugate points with only one connecting geodesicExample for infinitely many points with more than one minimizing geodesic to a point?Compact totally geodesic submanifolds in manifold with positive sectional curvatureClosed geodesic on a non-simply connected Riemannian manifold
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Example of compact Riemannian manifold with only one closed geodesic.
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Why are we interested in closed geodesics?Existence of geodesic on a compact Riemannian manifoldAre closed geodesics the prime numbers of Riemannian manifolds?Completeness of a Riemannian manifold with boundaryTotally geodesic hypersurface in compact hyperbolic manifoldExistence of closed, non self-intersecting geodesics on compact manifoldsExample for conjugate points with only one connecting geodesicExample for infinitely many points with more than one minimizing geodesic to a point?Compact totally geodesic submanifolds in manifold with positive sectional curvatureClosed geodesic on a non-simply connected Riemannian manifold
$begingroup$
The Lyusternik-Fet theorem states that every compact Riemannian manifold has at least one closed geodesic.
Are there any easy-to-construct1 examples of compact Riemannian manifolds for which it is easy to see they only have one closed geodesic?2
If there aren't any such examples, are there any easy-to-construct examples that only have one closed geodesic but where proving this might be difficult?
And if there aren't any examples of this, are there any examples at all of compact manifolds with only one closed geodesic?
1 Of course, the $1$-sphere $S^1$ contains just one closed geodesic, but I'm interested in examples besides this one.
2 By the theorem of the three geodesics, this example cannot be a topological sphere.
differential-geometry examples-counterexamples geodesic
asked 2 days ago
Peter KageyPeter Kagey
1,60072053
$endgroup$
add a comment |
$begingroup$
The Lyusternik-Fet theorem states that every compact Riemannian manifold has at least one closed geodesic.
Are there any easy-to-construct1 examples of compact Riemannian manifolds for which it is easy to see they only have one closed geodesic?2
If there aren't any such examples, are there any easy-to-construct examples that only have one closed geodesic but where proving this might be difficult?
And if there aren't any examples of this, are there any examples at all of compact manifolds with only one closed geodesic?
1 Of course, the $1$-sphere $S^1$ contains just one closed geodesic, but I'm interested in examples besides this one.
2 By the theorem of the three geodesics, this example cannot be a topological sphere.
differential-geometry examples-counterexamples geodesic
asked 2 days ago
Peter KageyPeter Kagey
1,60072053
$endgroup$
add a comment |
$begingroup$
The Lyusternik-Fet theorem states that every compact Riemannian manifold has at least one closed geodesic.
Are there any easy-to-construct1 examples of compact Riemannian manifolds for which it is easy to see they only have one closed geodesic?2
If there aren't any such examples, are there any easy-to-construct examples that only have one closed geodesic but where proving this might be difficult?
And if there aren't any examples of this, are there any examples at all of compact manifolds with only one closed geodesic?
1 Of course, the $1$-sphere $S^1$ contains just one closed geodesic, but I'm interested in examples besides this one.
2 By the theorem of the three geodesics, this example cannot be a topological sphere.
differential-geometry examples-counterexamples geodesic
asked 2 days ago
Peter KageyPeter Kagey
1,60072053
$endgroup$
The Lyusternik-Fet theorem states that every compact Riemannian manifold has at least one closed geodesic.
Are there any easy-to-construct1 examples of compact Riemannian manifolds for which it is easy to see they only have one closed geodesic?2
If there aren't any such examples, are there any easy-to-construct examples that only have one closed geodesic but where proving this might be difficult?
And if there aren't any examples of this, are there any examples at all of compact manifolds with only one closed geodesic?
1 Of course, the $1$-sphere $S^1$ contains just one closed geodesic, but I'm interested in examples besides this one.
2 By the theorem of the three geodesics, this example cannot be a topological sphere.
differential-geometry examples-counterexamples geodesic
differential-geometry examples-counterexamples geodesic
asked 2 days ago
Peter KageyPeter Kagey
1,60072053
asked 2 days ago
Peter KageyPeter Kagey
1,60072053
edited 8 hours ago
Peter Kagey
asked 2 days ago
Peter KageyPeter Kagey
1,60072053
asked 2 days ago
Peter KageyPeter Kagey
1,60072053
asked 2 days ago
Peter KageyPeter Kagey
1,60072053
1,60072053
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
First of all, you have to exclude constant maps $S^1to M$ from consideration: They are all closed geodesics. Secondly, you have to talk about geometrically distinct closed geodesics: Geodesics which have the same image are regarded as "the same". Then, it is a notorious conjecture/open problem:
Conjecture. Every compact Riemannian manifold of dimension $n >1$ contains infinitely many geometrically distinct nonconstant geodesics.
See for instance this survey article by Burns and Matveev.
This is known for surfaces (with the only hard case when the surface is diffeomorphic to $S^2$ in which case the result is due to Bangert and Franks) and for many higher-dimensional manifolds. However, the problem is open already when $M$ is diffeomorphic to the sphere $S^n$, $nge 3$.
answered 2 days ago
Moishe KohanMoishe Kohan
48.8k344111
$endgroup$
add a comment |
$begingroup$
If you analyze the geodesics using Clairaut's relation, you'll find that the only closed geodesic on a hyperboloid of one sheet is the central circle. Indeed, the same holds for a concave surface of revolution of the same "shape" as the hyperboloid of one sheet.
EDIT: Apologies for missing the crucial compactness hypothesis.
answered 2 days ago
Ted ShifrinTed Shifrin
65.1k44792
$endgroup$
$begingroup$
Lovely example, but a hyperboloid isn't compact, right?
$endgroup$
– Peter Kagey
2 days ago
$begingroup$
Oops. Sloppy reading. I'll delete.
$endgroup$
– Ted Shifrin
2 days ago
1
$begingroup$
It's a nice example; you should leave it.
$endgroup$
– Peter Kagey
2 days ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
First of all, you have to exclude constant maps $S^1to M$ from consideration: They are all closed geodesics. Secondly, you have to talk about geometrically distinct closed geodesics: Geodesics which have the same image are regarded as "the same". Then, it is a notorious conjecture/open problem:
Conjecture. Every compact Riemannian manifold of dimension $n >1$ contains infinitely many geometrically distinct nonconstant geodesics.
See for instance this survey article by Burns and Matveev.
This is known for surfaces (with the only hard case when the surface is diffeomorphic to $S^2$ in which case the result is due to Bangert and Franks) and for many higher-dimensional manifolds. However, the problem is open already when $M$ is diffeomorphic to the sphere $S^n$, $nge 3$.
answered 2 days ago
Moishe KohanMoishe Kohan
48.8k344111
$endgroup$
add a comment |
$begingroup$
First of all, you have to exclude constant maps $S^1to M$ from consideration: They are all closed geodesics. Secondly, you have to talk about geometrically distinct closed geodesics: Geodesics which have the same image are regarded as "the same". Then, it is a notorious conjecture/open problem:
Conjecture. Every compact Riemannian manifold of dimension $n >1$ contains infinitely many geometrically distinct nonconstant geodesics.
See for instance this survey article by Burns and Matveev.
This is known for surfaces (with the only hard case when the surface is diffeomorphic to $S^2$ in which case the result is due to Bangert and Franks) and for many higher-dimensional manifolds. However, the problem is open already when $M$ is diffeomorphic to the sphere $S^n$, $nge 3$.
answered 2 days ago
Moishe KohanMoishe Kohan
48.8k344111
$endgroup$
add a comment |
$begingroup$
First of all, you have to exclude constant maps $S^1to M$ from consideration: They are all closed geodesics. Secondly, you have to talk about geometrically distinct closed geodesics: Geodesics which have the same image are regarded as "the same". Then, it is a notorious conjecture/open problem:
Conjecture. Every compact Riemannian manifold of dimension $n >1$ contains infinitely many geometrically distinct nonconstant geodesics.
See for instance this survey article by Burns and Matveev.
This is known for surfaces (with the only hard case when the surface is diffeomorphic to $S^2$ in which case the result is due to Bangert and Franks) and for many higher-dimensional manifolds. However, the problem is open already when $M$ is diffeomorphic to the sphere $S^n$, $nge 3$.
answered 2 days ago
Moishe KohanMoishe Kohan
48.8k344111
$endgroup$
First of all, you have to exclude constant maps $S^1to M$ from consideration: They are all closed geodesics. Secondly, you have to talk about geometrically distinct closed geodesics: Geodesics which have the same image are regarded as "the same". Then, it is a notorious conjecture/open problem:
Conjecture. Every compact Riemannian manifold of dimension $n >1$ contains infinitely many geometrically distinct nonconstant geodesics.
See for instance this survey article by Burns and Matveev.
This is known for surfaces (with the only hard case when the surface is diffeomorphic to $S^2$ in which case the result is due to Bangert and Franks) and for many higher-dimensional manifolds. However, the problem is open already when $M$ is diffeomorphic to the sphere $S^n$, $nge 3$.
answered 2 days ago
Moishe KohanMoishe Kohan
48.8k344111
edited yesterday
answered 2 days ago
Moishe KohanMoishe Kohan
48.8k344111
answered 2 days ago
Moishe KohanMoishe Kohan
48.8k344111
answered 2 days ago
Moishe KohanMoishe Kohan
48.8k344111
48.8k344111
add a comment |
add a comment |
$begingroup$
If you analyze the geodesics using Clairaut's relation, you'll find that the only closed geodesic on a hyperboloid of one sheet is the central circle. Indeed, the same holds for a concave surface of revolution of the same "shape" as the hyperboloid of one sheet.
EDIT: Apologies for missing the crucial compactness hypothesis.
answered 2 days ago
Ted ShifrinTed Shifrin
65.1k44792
$endgroup$
$begingroup$
Lovely example, but a hyperboloid isn't compact, right?
$endgroup$
– Peter Kagey
2 days ago
$begingroup$
Oops. Sloppy reading. I'll delete.
$endgroup$
– Ted Shifrin
2 days ago
1
$begingroup$
It's a nice example; you should leave it.
$endgroup$
– Peter Kagey
2 days ago
add a comment |
$begingroup$
If you analyze the geodesics using Clairaut's relation, you'll find that the only closed geodesic on a hyperboloid of one sheet is the central circle. Indeed, the same holds for a concave surface of revolution of the same "shape" as the hyperboloid of one sheet.
EDIT: Apologies for missing the crucial compactness hypothesis.
answered 2 days ago
Ted ShifrinTed Shifrin
65.1k44792
$endgroup$
$begingroup$
Lovely example, but a hyperboloid isn't compact, right?
$endgroup$
– Peter Kagey
2 days ago
$begingroup$
Oops. Sloppy reading. I'll delete.
$endgroup$
– Ted Shifrin
2 days ago
1
$begingroup$
It's a nice example; you should leave it.
$endgroup$
– Peter Kagey
2 days ago
add a comment |
$begingroup$
If you analyze the geodesics using Clairaut's relation, you'll find that the only closed geodesic on a hyperboloid of one sheet is the central circle. Indeed, the same holds for a concave surface of revolution of the same "shape" as the hyperboloid of one sheet.
EDIT: Apologies for missing the crucial compactness hypothesis.
answered 2 days ago
Ted ShifrinTed Shifrin
65.1k44792
$endgroup$
If you analyze the geodesics using Clairaut's relation, you'll find that the only closed geodesic on a hyperboloid of one sheet is the central circle. Indeed, the same holds for a concave surface of revolution of the same "shape" as the hyperboloid of one sheet.
EDIT: Apologies for missing the crucial compactness hypothesis.
answered 2 days ago
Ted ShifrinTed Shifrin
65.1k44792
edited 2 days ago
answered 2 days ago
Ted ShifrinTed Shifrin
65.1k44792
answered 2 days ago
Ted ShifrinTed Shifrin
65.1k44792
answered 2 days ago
Ted ShifrinTed Shifrin
65.1k44792
65.1k44792
$begingroup$
Lovely example, but a hyperboloid isn't compact, right?
$endgroup$
– Peter Kagey
2 days ago
$begingroup$
Oops. Sloppy reading. I'll delete.
$endgroup$
– Ted Shifrin
2 days ago
1
$begingroup$
It's a nice example; you should leave it.
$endgroup$
– Peter Kagey
2 days ago
add a comment |
$begingroup$
Lovely example, but a hyperboloid isn't compact, right?
$endgroup$
– Peter Kagey
2 days ago
$begingroup$
Oops. Sloppy reading. I'll delete.
$endgroup$
– Ted Shifrin
2 days ago
1
$begingroup$
It's a nice example; you should leave it.
$endgroup$
– Peter Kagey
2 days ago
$begingroup$
Lovely example, but a hyperboloid isn't compact, right?
$endgroup$
– Peter Kagey
2 days ago
$begingroup$
Lovely example, but a hyperboloid isn't compact, right?
$endgroup$
– Peter Kagey
2 days ago
$begingroup$
Oops. Sloppy reading. I'll delete.
$endgroup$
– Ted Shifrin
2 days ago
$begingroup$
Oops. Sloppy reading. I'll delete.
$endgroup$
– Ted Shifrin
2 days ago
1
1
$begingroup$
It's a nice example; you should leave it.
$endgroup$
– Peter Kagey
2 days ago
$begingroup$
It's a nice example; you should leave it.
$endgroup$
– Peter Kagey
2 days ago
add a comment |
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