Does every manifold admit a Lagrangian Riemannian metric?

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Let $(M,g)$ be a Riemannian manifold. The $LC$ connection associated to the metric gives an $n$ dimensional distribution $D$ for $TM$. Let $omega$ be the symplectic structure of $TM$ which is obtained by pulling back of the standard structure of the cotangent bundle via the isomorphism between the tangent and cotangent bundle.



We say a Riemannian metric is a Lagrangian metric if the distribution $D$ of $TM$ is a Lagrangian distribution.




Does every manifold admit a Lagrangian metric?











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  • How is $D$ defined from $nabla^LC$?
    – Qfwfq
    Aug 28 at 19:55






  • 2




    @Qfwfq Horizontal curves in TM are parallel vector fields on M.
    – alvarezpaiva
    Aug 28 at 21:41










  • n. manifold is very ambiguous: cdn.nexternal.com/vacmotors/images/VAC-BPSM-M.jpg It would appear my edit was wrong but this still needs fixing.
    – Joshua
    Aug 30 at 18:58














up vote
8
down vote

favorite
1












Let $(M,g)$ be a Riemannian manifold. The $LC$ connection associated to the metric gives an $n$ dimensional distribution $D$ for $TM$. Let $omega$ be the symplectic structure of $TM$ which is obtained by pulling back of the standard structure of the cotangent bundle via the isomorphism between the tangent and cotangent bundle.



We say a Riemannian metric is a Lagrangian metric if the distribution $D$ of $TM$ is a Lagrangian distribution.




Does every manifold admit a Lagrangian metric?











share|cite|improve this question























  • How is $D$ defined from $nabla^LC$?
    – Qfwfq
    Aug 28 at 19:55






  • 2




    @Qfwfq Horizontal curves in TM are parallel vector fields on M.
    – alvarezpaiva
    Aug 28 at 21:41










  • n. manifold is very ambiguous: cdn.nexternal.com/vacmotors/images/VAC-BPSM-M.jpg It would appear my edit was wrong but this still needs fixing.
    – Joshua
    Aug 30 at 18:58












up vote
8
down vote

favorite
1









up vote
8
down vote

favorite
1






1





Let $(M,g)$ be a Riemannian manifold. The $LC$ connection associated to the metric gives an $n$ dimensional distribution $D$ for $TM$. Let $omega$ be the symplectic structure of $TM$ which is obtained by pulling back of the standard structure of the cotangent bundle via the isomorphism between the tangent and cotangent bundle.



We say a Riemannian metric is a Lagrangian metric if the distribution $D$ of $TM$ is a Lagrangian distribution.




Does every manifold admit a Lagrangian metric?











share|cite|improve this question















Let $(M,g)$ be a Riemannian manifold. The $LC$ connection associated to the metric gives an $n$ dimensional distribution $D$ for $TM$. Let $omega$ be the symplectic structure of $TM$ which is obtained by pulling back of the standard structure of the cotangent bundle via the isomorphism between the tangent and cotangent bundle.



We say a Riemannian metric is a Lagrangian metric if the distribution $D$ of $TM$ is a Lagrangian distribution.




Does every manifold admit a Lagrangian metric?








dg.differential-geometry riemannian-geometry sg.symplectic-geometry






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edited Aug 29 at 12:05

























asked Aug 28 at 17:42









Ali Taghavi

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39451882











  • How is $D$ defined from $nabla^LC$?
    – Qfwfq
    Aug 28 at 19:55






  • 2




    @Qfwfq Horizontal curves in TM are parallel vector fields on M.
    – alvarezpaiva
    Aug 28 at 21:41










  • n. manifold is very ambiguous: cdn.nexternal.com/vacmotors/images/VAC-BPSM-M.jpg It would appear my edit was wrong but this still needs fixing.
    – Joshua
    Aug 30 at 18:58
















  • How is $D$ defined from $nabla^LC$?
    – Qfwfq
    Aug 28 at 19:55






  • 2




    @Qfwfq Horizontal curves in TM are parallel vector fields on M.
    – alvarezpaiva
    Aug 28 at 21:41










  • n. manifold is very ambiguous: cdn.nexternal.com/vacmotors/images/VAC-BPSM-M.jpg It would appear my edit was wrong but this still needs fixing.
    – Joshua
    Aug 30 at 18:58















How is $D$ defined from $nabla^LC$?
– Qfwfq
Aug 28 at 19:55




How is $D$ defined from $nabla^LC$?
– Qfwfq
Aug 28 at 19:55




2




2




@Qfwfq Horizontal curves in TM are parallel vector fields on M.
– alvarezpaiva
Aug 28 at 21:41




@Qfwfq Horizontal curves in TM are parallel vector fields on M.
– alvarezpaiva
Aug 28 at 21:41












n. manifold is very ambiguous: cdn.nexternal.com/vacmotors/images/VAC-BPSM-M.jpg It would appear my edit was wrong but this still needs fixing.
– Joshua
Aug 30 at 18:58




n. manifold is very ambiguous: cdn.nexternal.com/vacmotors/images/VAC-BPSM-M.jpg It would appear my edit was wrong but this still needs fixing.
– Joshua
Aug 30 at 18:58










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The distribution of horizontal subspaces is always Lagrangian not only for Riemannian metrics, but also for the Ehresmann connection associated to Finsler metrics and for the Levi-Civita connection of pseudo-Riemannian metrics. It is not always Lagrangian for general sprays though.



In the Riemannian case this is classical (see Klingenberg's book on Riemannian Geometry or Paternain's book on geodesic flows), but you can also see my answer to this question: Intuition for Levi-Civita connection via Hamiltonian flows



There is a much more elementary and geometric version of the construction of the connection given in my answer to When is a flow geodesic and how to construct the connection from it . You can complete it by proving (it's really easy!) that the harmonic conjugate---as defined in the answer---of three Lagrangian subspaces is again a Lagrangian subspace.






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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    11
    down vote



    accepted










    The distribution of horizontal subspaces is always Lagrangian not only for Riemannian metrics, but also for the Ehresmann connection associated to Finsler metrics and for the Levi-Civita connection of pseudo-Riemannian metrics. It is not always Lagrangian for general sprays though.



    In the Riemannian case this is classical (see Klingenberg's book on Riemannian Geometry or Paternain's book on geodesic flows), but you can also see my answer to this question: Intuition for Levi-Civita connection via Hamiltonian flows



    There is a much more elementary and geometric version of the construction of the connection given in my answer to When is a flow geodesic and how to construct the connection from it . You can complete it by proving (it's really easy!) that the harmonic conjugate---as defined in the answer---of three Lagrangian subspaces is again a Lagrangian subspace.






    share|cite|improve this answer


























      up vote
      11
      down vote



      accepted










      The distribution of horizontal subspaces is always Lagrangian not only for Riemannian metrics, but also for the Ehresmann connection associated to Finsler metrics and for the Levi-Civita connection of pseudo-Riemannian metrics. It is not always Lagrangian for general sprays though.



      In the Riemannian case this is classical (see Klingenberg's book on Riemannian Geometry or Paternain's book on geodesic flows), but you can also see my answer to this question: Intuition for Levi-Civita connection via Hamiltonian flows



      There is a much more elementary and geometric version of the construction of the connection given in my answer to When is a flow geodesic and how to construct the connection from it . You can complete it by proving (it's really easy!) that the harmonic conjugate---as defined in the answer---of three Lagrangian subspaces is again a Lagrangian subspace.






      share|cite|improve this answer
























        up vote
        11
        down vote



        accepted







        up vote
        11
        down vote



        accepted






        The distribution of horizontal subspaces is always Lagrangian not only for Riemannian metrics, but also for the Ehresmann connection associated to Finsler metrics and for the Levi-Civita connection of pseudo-Riemannian metrics. It is not always Lagrangian for general sprays though.



        In the Riemannian case this is classical (see Klingenberg's book on Riemannian Geometry or Paternain's book on geodesic flows), but you can also see my answer to this question: Intuition for Levi-Civita connection via Hamiltonian flows



        There is a much more elementary and geometric version of the construction of the connection given in my answer to When is a flow geodesic and how to construct the connection from it . You can complete it by proving (it's really easy!) that the harmonic conjugate---as defined in the answer---of three Lagrangian subspaces is again a Lagrangian subspace.






        share|cite|improve this answer














        The distribution of horizontal subspaces is always Lagrangian not only for Riemannian metrics, but also for the Ehresmann connection associated to Finsler metrics and for the Levi-Civita connection of pseudo-Riemannian metrics. It is not always Lagrangian for general sprays though.



        In the Riemannian case this is classical (see Klingenberg's book on Riemannian Geometry or Paternain's book on geodesic flows), but you can also see my answer to this question: Intuition for Levi-Civita connection via Hamiltonian flows



        There is a much more elementary and geometric version of the construction of the connection given in my answer to When is a flow geodesic and how to construct the connection from it . You can complete it by proving (it's really easy!) that the harmonic conjugate---as defined in the answer---of three Lagrangian subspaces is again a Lagrangian subspace.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 28 at 21:39

























        answered Aug 28 at 19:16









        alvarezpaiva

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