Sfrgttk

Is there a writing software that you can sort scenes like slides in PowerPoint?

Deal with toxic manager when you can't quit

Homework question about an engine pulling a train

Why can't wing-mounted spoilers be used to steepen approaches?

Does Parliament hold absolute power in the UK?

Can a flute soloist sit?

how can a perfect fourth interval be considered either consonant or dissonant?

Why doesn't a hydraulic lever violate conservation of energy?

The following signatures were invalid: EXPKEYSIG 1397BC53640DB551

Example of compact Riemannian manifold with only one geodesic.

Nested ellipses in tikzpicture: Chomsky hierarchy

Is an up-to-date browser secure on an out-of-date OS?

Is every episode of "Where are my Pants?" identical?

Identify 80s or 90s comics with ripped creatures (not dwarves)

Circular reasoning in L'Hopital's rule

Do I have Disadvantage attacking with an off-hand weapon?

My body leaves; my core can stay

What information about me do stores get via my credit card?

What can I do if neighbor is blocking my solar panels intentionally?

What does "spokes" mean in this context?

Can withdrawing asylum be illegal?

Do working physicists consider Newtonian mechanics to be "falsified"?

What other Star Trek series did the main TNG cast show up in?

Match Roman Numerals

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 51 mins 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 51 mins 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 51 mins ago

Peter Kagey

asked 1 hour ago

Peter KageyPeter Kagey

1,57072053

$endgroup$

share|cite|improve this question

edited 51 mins ago

Peter Kagey

asked 1 hour ago

Peter KageyPeter Kagey

1,57072053

share|cite|improve this question

edited 51 mins 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 25 mins ago

answered 43 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 47 mins ago

answered 50 mins ago

Ted ShifrinTed Shifrin

65k44792

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

Your Answer

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
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
});


}
});

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3185649%2fexample-of-compact-riemannian-manifold-with-only-one-geodesic%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

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 25 mins ago

answered 43 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 25 mins ago

answered 43 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 25 mins ago

answered 43 mins ago

Moishe KohanMoishe Kohan

48.6k344110

$endgroup$

share|cite|improve this answer

edited 25 mins ago

answered 43 mins ago

Moishe KohanMoishe Kohan

48.6k344110

answered 43 mins ago

Moishe KohanMoishe Kohan

48.6k344110

answered 43 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 47 mins ago

answered 50 mins ago

Ted ShifrinTed Shifrin

65k44792

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's a nice example; you should leave it.
  $endgroup$
  – Peter Kagey
  47 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 47 mins ago

answered 50 mins ago

Ted ShifrinTed Shifrin

65k44792

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's a nice example; you should leave it.
  $endgroup$
  – Peter Kagey
  47 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 47 mins ago

answered 50 mins ago

Ted ShifrinTed Shifrin

65k44792

$endgroup$

share|cite|improve this answer

edited 47 mins ago

answered 50 mins ago

Ted ShifrinTed Shifrin

65k44792

answered 50 mins ago

Ted ShifrinTed Shifrin

65k44792

answered 50 mins ago

Ted ShifrinTed Shifrin

65k44792