# Does every manifold admit a Lagrangian Riemannian metric?

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?

• How is $D$ defined from $\nabla^{LC}$? – Qfwfq Aug 28 '18 at 19:55
• @Qfwfq Horizontal curves in TM are parallel vector fields on M. – alvarezpaiva Aug 28 '18 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 '18 at 18:58