Does the group of Diffeomorphisms act transitively on the space of Riemannian metrics?

Let $M$ be a smooth manifold (maybe compact, if that helps). Denote by $\operatorname{Diff}(M)$ the group of diffeomorphisms $M\to M$ and by $R(M)$ the space of Riemannian metrics on $M$. We obtain a canonical group action $$R(M) \times \operatorname{Diff}(M) \to R(M), (g,F) \mapsto F^*g,$$ where $F^*g$ denotes the pullback of $g$ along $F$. Is this action transitive? In other words, is it possible for any two Riemannian metrics $g,h$ on $M$ to find a diffeomorphism $F$ such that $F^*g=h$? Do you know any references for this type of questions?

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The quotient space $R(M)/\operatorname{Diff}{(M)}$ is sometimes called the space of Riemannian structures on $M$. See e.g. Berger's Panoramic view, p.501ff for some of its uses. –  t.b. Apr 6 '12 at 10:28

This map will not be transitive in general. For example, if $g$ is a metric and $\phi \in Diff(M)$ then the curvature of $\phi^* g$ is going to be the pullback of the curvature of $g$. So there's no way for a metric with zero curvature to be diffeomorphic to a manifold with non-zero curvature. Or for example, if $g$ is einstein $(g = \lambda Ric)$ then so is $\phi^* g$. So there are many diffeomorphism invariants of a metric.

Indeed, this should make sense because you can think of a diffeomorphism as passive, i.e. as just a change of coordinates. Then all of the natural things about the Riemannian geometry of a manifold should be coordinate ($\Leftrightarrow$ diffeomorphism) invariant.

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Good argument. :) –  Meneldur Apr 6 '12 at 15:55

What about the action of the group of C^1 diffeomorpshims ? There are no curvature constraints, so that we seem to be close to the framework of Nash's theorem...

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I think it would be better to make this a separate question than asking it in an answer. –  Eric O. Korman Mar 21 '13 at 20:24

The group of C^1 diffeomorphism does not act transitively on the space of Riemannian metrics on a compact manifold. For example, two circles of different radius have different diameters and by pullback we can "copy" the metric of one of them on the other. Doing so we get two metrics on a circle with different diameters, hence they can not be C^1 related.

Thus, the point is that "the diameter" is a C^1-invariant hence an homotety of the metric change the diameter and the "new metric" have different diameter.

For non compact manifolds, by using a theorem of Nomizu, there exist both complete and non-complete riemannian metrics. See http://www.oberlin.edu/faculty/jcalcut/Nomizu_Ozeki_1961.pdf

Since "completeness" is a C^1 invariant we get that the group of C^1 diffeomorphism does not act transitively on the space of riemannian metrics.

h.

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