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Is the space of Riemannian metrics, over a compact manifold, complete when endowed with the $C^k$-topology of metrics?. Is there a good reference for this?

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Completeness usually refers to a metric space. For example, a compact manifold, you can put a $C^k$-metric on the space of Riemann metrics. That's a complete metric space provided $k \geq 1$ is finite. Is that what you're asking about? –  Ryan Budney Mar 14 '13 at 2:29
    
Ryan, I'm not sure what you mean by your comment and how to put a complete metric on the space of $C^k$ Riemannian metrics on a compact manifold. Could you elaborate? –  Deane Yang Mar 14 '13 at 2:48
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To get a complete metric on the space of Riemannian metrics, you first need to put a complete metric on the space of positive definite symmetric matrices. Once you do that, the rest is straightforward. Otherwise, the space is incomplete, because the metric can degenerate into a semi-Riemannian metric. –  Deane Yang Mar 14 '13 at 2:50
    
I think we are referring to the same thing. But just in case let me be more specific. Fix a background metric $g$ and define the distance on the space of Riemannian metrics by $$d_g(g_1,g_2)= \max_{1 \leq \ell \leq k\}\| \nabla^\ell_g (g_1-g_2) \|_\infty.$$ This distance is independent of the background metric and defines the $C^k$ topology I was referring to. Is the space of Riemannian metrics complete with this topology (on a compact manifold)?. Where can I find a reference? –  Cecilia Mar 14 '13 at 2:53
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Another comment, in some sense these kind of convergence of metrics are often not the "best" ones from a geometric point of view. The problem is that they are not invariant under diffeomorphisms. You might be interested in what is called "Cheeger-Gromov convergence", which is "$C^k$ convergence of metrics up to diffeomorphisms". –  Thomas Richard Mar 14 '13 at 6:16
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No it is not complete. It is an open convex cone of the Banach space (Frechet space if $k=\infty$) $\Gamma_{C^k} S^2T^*M$ of $C^k$-sections of the vector bundle $S^2T^*M$. 0 is always in the closure of this cone, and many more things. The norm on this Banach space depends on many choices (charts, metric, etc.), but all these norms are equivalent.

You can also put several natural Riemannian metrics on the space of all Riemannian metrics, but none of them is geodesically complete. Natural means: invariant under the diffeomorphism group. See:

Martin Bauer, Philipp Harms, Peter W. Michor: Sobolev metrics on the manifold of all Riemannian metrics. 20 pages. To appear in: Journal of Differential Geometry.(pdf)

Check also the references there.

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Thank you! Just to make sure: Are you saying that both the space of $C^\infty$ Riemannian metrics and $C^k$ Riemannian metrics over a compact smooth manifold are incomplete with the $C^k$ topology? –  Cecilia Mar 14 '13 at 17:33
    
Yes. Both are incomplete. Symmetric tensor fields which are $\ge 0$ but not $>0$ somewhere are always in the closure. $C^\infty$ metrics are incomplete also since their completion contains $C^k$-metrics. –  Peter Michor Mar 14 '13 at 18:25
    
Thank you so much! –  Cecilia Mar 14 '13 at 20:58
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