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?

  • $\begingroup$ How is $D$ defined from $\nabla^{LC}$? $\endgroup$
    – Qfwfq
    Aug 28, 2018 at 19:55
  • 2
    $\begingroup$ @Qfwfq Horizontal curves in TM are parallel vector fields on M. $\endgroup$ Aug 28, 2018 at 21:41
  • $\begingroup$ 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. $\endgroup$
    – Joshua
    Aug 30, 2018 at 18:58

1 Answer 1


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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