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Example of compact Riemannian manifold with only one geodesic.

2

$begingroup$

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

share|cite|improve this question

edited 1 hour ago

Peter Kagey

asked 1 hour ago

Peter KageyPeter Kagey

1,57072053

$endgroup$

add a comment | 

2

$begingroup$

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

share|cite|improve this question

edited 1 hour ago

Peter Kagey

asked 1 hour ago

Peter KageyPeter Kagey

1,57072053

$endgroup$

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

share|cite|improve this question

edited 1 hour ago

Peter Kagey

asked 1 hour ago

Peter KageyPeter Kagey

1,57072053

$endgroup$

share|cite|improve this question

edited 1 hour ago

Peter Kagey

asked 1 hour ago

Peter KageyPeter Kagey

1,57072053

share|cite|improve this question

edited 1 hour ago

Peter Kagey

asked 1 hour ago

Peter KageyPeter Kagey

1,57072053

asked 1 hour ago

Peter KageyPeter Kagey

1,57072053

asked 1 hour ago

Peter KageyPeter Kagey

1,57072053

2 Answers
2

active

oldest

votes

4

$begingroup$

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 even unknown if $M$ is diffeomorphic to the sphere $S^n$, $nge 3$.

share|cite|improve this answer

edited 35 mins ago

answered 54 mins ago

Moishe KohanMoishe Kohan

48.6k344110

$endgroup$

add a comment | 

2

$begingroup$

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.

share|cite|improve this answer

edited 57 mins ago

answered 1 hour ago

Ted ShifrinTed Shifrin

65k44792

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Lovely example, but a hyperboloid isn't compact, right?
  $endgroup$
  – Peter Kagey
  59 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oops. Sloppy reading. I'll delete.
  $endgroup$
  – Ted Shifrin
  58 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's a nice example; you should leave it.
  $endgroup$
  – Peter Kagey
  58 mins ago
  

  
  
  
  

add a comment | 

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4

$begingroup$

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 even unknown if $M$ is diffeomorphic to the sphere $S^n$, $nge 3$.

share|cite|improve this answer

edited 35 mins ago

answered 54 mins ago

Moishe KohanMoishe Kohan

48.6k344110

$endgroup$

add a comment | 

4

$begingroup$

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 even unknown if $M$ is diffeomorphic to the sphere $S^n$, $nge 3$.

share|cite|improve this answer

edited 35 mins ago

answered 54 mins ago

Moishe KohanMoishe Kohan

48.6k344110

$endgroup$

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 even unknown if $M$ is diffeomorphic to the sphere $S^n$, $nge 3$.

share|cite|improve this answer

edited 35 mins ago

answered 54 mins ago

Moishe KohanMoishe Kohan

48.6k344110

$endgroup$

share|cite|improve this answer

edited 35 mins ago

answered 54 mins ago

Moishe KohanMoishe Kohan

48.6k344110

answered 54 mins ago

Moishe KohanMoishe Kohan

48.6k344110

answered 54 mins ago

Moishe KohanMoishe Kohan

48.6k344110

2

$begingroup$

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.

share|cite|improve this answer

edited 57 mins ago

answered 1 hour ago

Ted ShifrinTed Shifrin

65k44792

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Lovely example, but a hyperboloid isn't compact, right?
  $endgroup$
  – Peter Kagey
  59 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oops. Sloppy reading. I'll delete.
  $endgroup$
  – Ted Shifrin
  58 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's a nice example; you should leave it.
  $endgroup$
  – Peter Kagey
  58 mins ago
  

  
  
  
  

add a comment | 

2

$begingroup$

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.

share|cite|improve this answer

edited 57 mins ago

answered 1 hour ago

Ted ShifrinTed Shifrin

65k44792

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Lovely example, but a hyperboloid isn't compact, right?
  $endgroup$
  – Peter Kagey
  59 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oops. Sloppy reading. I'll delete.
  $endgroup$
  – Ted Shifrin
  58 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's a nice example; you should leave it.
  $endgroup$
  – Peter Kagey
  58 mins 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.

share|cite|improve this answer

edited 57 mins ago

answered 1 hour ago

Ted ShifrinTed Shifrin

65k44792

$endgroup$

share|cite|improve this answer

edited 57 mins ago

answered 1 hour ago

Ted ShifrinTed Shifrin

65k44792

answered 1 hour ago

Ted ShifrinTed Shifrin

65k44792

answered 1 hour ago

Ted ShifrinTed Shifrin

65k44792