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Given a Pseudo-riemaniann metric on ${\cal{M}}$, is it possible to find a Riemaniann metric on ${\cal{M}}$ with the same Levi-Civita connection?

If in general this is not possible, what sufficient or necessary conditions are needed for the Pseudo-Riemaniann metric and the holonomy ?

If I understood correcly from this post. The question can be rephrase:

What conditions an indefinite symmetric quadratic form that is left invariant by the holonomy must satisfy to allow a positive definite quadratic form to be invariant under the same holonomy?

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It is impossible in general. The reason is that LC Riemannian connection on a compact manifold is automatically complete. This is not the case in the Lorentzian setting.

Edit: Another way to think about this problem is to look at the holonomy group of the connection. For a LC connection holonomy on an $n$-manifold $M$ is always relatively compact in $GL(n,R)$. See here for the complete list of closures of holonomy groups for LC connections on simply-connected manifolds. On the other hand, in the Lorentzian setting, holonomy is contained (up to conjugation) in $O(n-1,1)$ (I am assuming that your pseudo-Riemannian metric is of signature $(n-1,1)$). Thus, the necessary condition is for the holonomy group to be conjugate into a compact subgroup of $O(n-1,1)$.

To be more specific: Suppose that $n=2$ and $M$ is simply-connected. Then the holonomy has to be contained in a connected compact subgroup of $O(1,1)$, which means it is trivial. Hence, your pseudo-Riemannian metric is locally flat. I am pretty sure that this condition is also sufficient for existence of a Riemannian metric with the same LC connection (in the simply-connected case).

Edit: One also has the following. Suppose that $M$ is a simply-connected manifold with (torsion-free) compatible connection $\nabla$, $g$ is a flat pseudo-Riemannian metric on $M$ (signature and dimension do not matter). Then $M$ admits a flat Riemannian metric with LC connection $\nabla$. The same is also true if we replace flatness assumption with the assumption that the holonomy of $\nabla$ is relatively compact (and drop simple connectivity).

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  • $\begingroup$ Thank you for your answer. If I understand correctly the necessary condition is that the holonomy ${\cal{O}}(n-1,1)$ must reduced to one of the possible holonomies for the Riemaniann metrics. When you mention this is also sufficient is this for any dimension or just for $n=2$. $\endgroup$ – yess May 13 '14 at 1:53
  • $\begingroup$ Also, can you say something about a neighbourhood of a point instead of the whole manifold? Rephrasing: Is it possible to have a Levi-Civita connection compatible with both types of quadratic forms locally even if you can't extend to the whole manifold? This is motivated by the comments of Bill Thurston in his own answer to this post mathoverflow.net/questions/54434/… or am I missing something? $\endgroup$ – yess May 13 '14 at 1:54
  • $\begingroup$ @yess: the holonomy discussion is purely local. $\endgroup$ – Moishe Kohan May 13 '14 at 2:02

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