Example of compact Riemannian manifold with only one geodesic.












6












$begingroup$


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











$endgroup$

















    6












    $begingroup$


    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











    $endgroup$















      6












      6








      6





      $begingroup$


      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











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 7 hours ago







      Peter Kagey

















      asked 7 hours ago









      Peter KageyPeter Kagey

      1,59072053




      1,59072053






















          2 Answers
          2






          active

          oldest

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          10












          $begingroup$

          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











          $endgroup$





















            5












            $begingroup$

            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











            $endgroup$













            • $begingroup$
              Lovely example, but a hyperboloid isn't compact, right?
              $endgroup$
              – Peter Kagey
              7 hours ago










            • $begingroup$
              Oops. Sloppy reading. I'll delete.
              $endgroup$
              – Ted Shifrin
              7 hours ago










            • $begingroup$
              It's a nice example; you should leave it.
              $endgroup$
              – Peter Kagey
              7 hours ago












            Your Answer








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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            10












            $begingroup$

            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











            $endgroup$


















              10












              $begingroup$

              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











              $endgroup$
















                10












                10








                10





                $begingroup$

                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











                $endgroup$



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







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 7 hours ago

























                answered 7 hours ago









                Moishe KohanMoishe Kohan

                48.7k344111




                48.7k344111























                    5












                    $begingroup$

                    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











                    $endgroup$













                    • $begingroup$
                      Lovely example, but a hyperboloid isn't compact, right?
                      $endgroup$
                      – Peter Kagey
                      7 hours ago










                    • $begingroup$
                      Oops. Sloppy reading. I'll delete.
                      $endgroup$
                      – Ted Shifrin
                      7 hours ago










                    • $begingroup$
                      It's a nice example; you should leave it.
                      $endgroup$
                      – Peter Kagey
                      7 hours ago
















                    5












                    $begingroup$

                    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











                    $endgroup$













                    • $begingroup$
                      Lovely example, but a hyperboloid isn't compact, right?
                      $endgroup$
                      – Peter Kagey
                      7 hours ago










                    • $begingroup$
                      Oops. Sloppy reading. I'll delete.
                      $endgroup$
                      – Ted Shifrin
                      7 hours ago










                    • $begingroup$
                      It's a nice example; you should leave it.
                      $endgroup$
                      – Peter Kagey
                      7 hours ago














                    5












                    5








                    5





                    $begingroup$

                    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











                    $endgroup$



                    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.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 7 hours ago

























                    answered 7 hours ago









                    Ted ShifrinTed Shifrin

                    65k44792




                    65k44792












                    • $begingroup$
                      Lovely example, but a hyperboloid isn't compact, right?
                      $endgroup$
                      – Peter Kagey
                      7 hours ago










                    • $begingroup$
                      Oops. Sloppy reading. I'll delete.
                      $endgroup$
                      – Ted Shifrin
                      7 hours ago










                    • $begingroup$
                      It's a nice example; you should leave it.
                      $endgroup$
                      – Peter Kagey
                      7 hours ago


















                    • $begingroup$
                      Lovely example, but a hyperboloid isn't compact, right?
                      $endgroup$
                      – Peter Kagey
                      7 hours ago










                    • $begingroup$
                      Oops. Sloppy reading. I'll delete.
                      $endgroup$
                      – Ted Shifrin
                      7 hours ago










                    • $begingroup$
                      It's a nice example; you should leave it.
                      $endgroup$
                      – Peter Kagey
                      7 hours ago
















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




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












                    $begingroup$
                    Oops. Sloppy reading. I'll delete.
                    $endgroup$
                    – Ted Shifrin
                    7 hours ago




                    $begingroup$
                    Oops. Sloppy reading. I'll delete.
                    $endgroup$
                    – Ted Shifrin
                    7 hours ago












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




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


















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